{ "chapter_id": 3, "chapter_title": "Chemical Kinetics", "uploaded_filename": "Chapter3.zip", "uploaded_sha256": "109a177c7d9597af2bd69c3b0dc4317f37aa5e0fcf6843a5760aae2908e56ac1", "status": "confirmed", "sections": [ { "section_id": "3.2", "section_title": "Topic-wise Q & A", "heading_text": "Topic-wise Questions and Answers", "detected_type": "SUB", "match_confidence": 0.609, "questions": [ { "sequence_no": 1, "question_text": "
What is rate of a chemical reaction? Explain average and instantaneous rate of reaction with help of graph and also give unit of rate of reaction.
", "answer_text": "
Rate of a chemical reaction: The change in concentration of reactant or product in unit time is called rate of a chemical reaction.
\n
It can be expressed in terms of:

\t(i)\tThe rate in decrease in concentration of any one of the reactants.

\t(ii)\tThe rate in increase in concentration of any one of the products.

\n
Consider a hypothetical reaction, assuming that the volume of the system remains constant.
\n
\tR → P
\n
One mole of the reactant R produces one mole of the product P.
\n
If [R]1 and [P]1 are the concentrations of R and P respectively at time t1 and [R]2 and [P]2 are their concentrations at time t2 then,
\n
∆t = t2 – t1
\n
∆[R] = [R]2 – [R]1
\n
∆[P] = [P]2 – [P]1
\n
(Square bracket expressing molar concentration.)
\n
Rate of disappearance of
\n
\tR = = \"\"
\n
\tRate of appearance of
\n
\tP = = +\"\"
\n
Above given equations represent the average rate of a reaction (rav).
\n
Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur.
\n
\"\"
\n
\"\"
\n
Instantaneous rate of reaction: Average rate cannot be used to predict the rate of a reaction at a particular instant because the rate is generally not constant throughout the reaction.
\n
So, to express the rate at a particular moment of time we determine instantaneous rate.
\n
Instantaneous rate is obtained when we consider the average rate at the smallest time interval say dt (when ∆t approaches zero).
\n
Hence, mathematically for an infinitesimally small dt instantaneous rate is given by:
\n
\trav = –\"\"
\n
As ∆t → 0 or rinst = – \"\"
\n
Unit of rate of a reaction: Units of rate are concentration time–1.
\n
If concentration is in mol L–1 and time is in seconds then the units will be mol L–1sec–1.
\n
However, in gaseous reactions, when the concentration of gases is expressed in terms of their partial pressure, then the units of the rate equation will be atm s–1.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-215853.png", "position": "answer", "context": "R = =" }, { "id": "img-1", "path": "Chapter3/EQ-0124-215910.png", "position": "answer", "context": "P = = +" }, { "id": "img-2", "path": "Chapter3/3.png", "position": "answer", "context": "" }, { "id": "img-3", "path": "Chapter3/4.png", "position": "answer", "context": "" }, { "id": "img-4", "path": "Chapter3/EQ-1221-143044.png", "position": "answer", "context": "r av = –" }, { "id": "img-5", "path": "Chapter3/EQ-1221-143133.png", "position": "answer", "context": "As ∆ t → 0 or r inst = –" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "marks": 4, "source_number": "1", "original_raw_text": " What is rate of a chemical reaction? Explain average and instantaneous rate of reaction with help of graph and also give unit of rate of reaction. ", "deduction_level": 0, "source_html": "Chapter3", "boundary_log": [ "leading-marker-stripped" ], "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 2, "question_text": "
Explain the rate expression for a reaction in which stoichiometric coefficients of reactants and products are not same or unit.
", "answer_text": "
For reaction, n1A + n2B → n3C + n4D
\n
Rate of reaction:
\n
For expressing the rate of such a reaction where stoichiometric coefficients of reactants or product are not equal to one, rate of disappearance of any of the reactants or the rate of appearance of the products is divided by their respective stoichiometric coefficients.
\n
For reaction, Hg(l) + Cl2(g) → HgCl2(s)
\n
\twhere stoichiometric coefficients of the reactants and products are same, then rate of the reaction is given as:
\n
Rate of reaction =
\n
For reaction, 2Hl(g) → H2(g) + I2(g) the rate of decomposition of HI is twice the rate of formation of H2 or I2, to make them equal, the term ∆[HI] is divided by 2. The rate of this reaction is given by:
\n
Rate of reaction =
\n
Similarly for the reaction:
\n
5Br–(aq) + BrO3–(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l)
\n
Rate\t=
\n
\t\t= \"\"
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-012.png", "position": "answer", "context": "=" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "source_number": "2", "original_raw_text": " Explain the rate expression for a reaction in which stoichiometric coefficients of reactants and products are not same or unit. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 3, "question_text": "
Explain rate law and rate constant with an example.
", "answer_text": "
Rate of a reaction depends upon the concentration of reactants.
\n
Consider a general reaction
\n
aA + bB → cC + dD
\n
\twhere a, b, c and d are the stoichiometric coefficients of reactants and products.
\n
The rate expression for this reaction is
\n
\tRate α[A]x[B]y
\n
\twhere exponents x and y may or may not be equal to the stoichiometric coefficients (a and b) of the reactants.
\n
\tAbove equation can also be written as
\n
\tRate = k[A]x[B]y
\n
\"\" = k[A]x[B]y
\n
This form of equation is known as differential rate equation, where k is proportionality constant called rate constant.
\n
The equation which relates the rate of reaction to concentration of reactants is called rate law or rate expression.
\n
Thus, rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as stoichiometric coefficient of the reacting species in a balanced equation.
\n
For example: 2NO(g) + O2(g) → 2NO2(g)
\n
\tThe rate equation for this reaction will be:
\n
\tRate = k[NO]2[O2]
\n
This differential form of this rate expression is given as:
\n
\"\" = k[NO]2[O2]
\n
Some other examples are given below:

\t1.\tCHCl3 + Cl2 → CCl4 + HCl

\t2. CH3COOC2H5 + H2O → \tCH3COOH + C2H5OH

\n
Experimental rate expression:
\n
\tRate = k[CHCl3] ·\"\"
\n
\tRate = k[CH3COOC2H5][H2O]0
\n
In these reactions, the exponents of the concentration terms are not the same as their stoichiometric coefficients. Thus, we can say that:
\n
Rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation. i.e. theoretically but must be determined experimentally.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-220915.png", "position": "answer", "context": "= k [A] x [B] y" }, { "id": "img-1", "path": "Chapter3/EQ-0124-220958.png", "position": "answer", "context": "= k [NO] 2 [O 2 ]" }, { "id": "img-2", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-027.png", "position": "answer", "context": "Rate = k[CHCl 3 ] ·" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "marks": 3, "source_number": "3", "original_raw_text": "Explain rate law and rate constant with an example. ", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 4, "question_text": "
What is order of reaction ? Discuss elementary and complex reactions, Explain units of rate constant in detail.
", "answer_text": "
The sum of power of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.
\n
For reaction: aA + bB → product
\n
Rate = k[A]x[B]y x and y indicate how sensitive the rate is to change in concentration of A and B.
\n
Sum of these exponents, i.e. x + y gives the overall order of a reaction whereas x and y represent the order with respect to the reactants A and B respectively.
\n
Order of reaction can be 0, 1, 2, 3 and even a fraction.
\n
A zero order reaction means that the rate of reaction is independent of the concentration of reactants.
\n
The reactions taking place in one step are called elementary reactions.
\n
When a sequence of elementary reactions (called mechanism) gives us the products, that reactions are called complex reactions.
\n
Units of rate constant:
\n
\tFor a general reaction
\n
\taA + bB → cC + dD
\n
\tRate = k[A]x[B]y
\n
\twhere x + y = n = order of the reaction
\n
k = \"\" =
\n
\t(where [A] = [B])
\n
Taking SI unit of concentration, mol L–1 and time, s, the unit of k for different reaction order listed in table.

Reaction

Order

Units of rate constant

Zero order reaction

0

\"\"

= mol L–1 s–1

First order reaction

1

\"\"

= s–1

Second order reaction

2

\"\"

= mol–1 L s–1

", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-022.png", "position": "answer", "context": "k = =" }, { "id": "img-1", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-024.png", "position": "answer", "context": "Taking SI unit of concentration, mol L –1 and time, s, the u…" }, { "id": "img-2", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-025.png", "position": "answer", "context": "Taking SI unit of concentration, mol L –1 and time, s, the u…" }, { "id": "img-3", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-026.png", "position": "answer", "context": "Taking SI unit of concentration, mol L –1 and time, s, the u…" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "marks": 3, "source_number": "4", "original_raw_text": " What is order of reaction ? Discuss elementary and complex reactions, Explain units of rate constant in detail. ", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 5, "question_text": "
Write a note on molecularity of a reaction and its different types.
", "answer_text": "
Molecularity of a reaction: The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction.
\n
For elementary reaction, order of a reaction and molecularity of a reaction are same.
\n
The reaction can be unimolecular when one reacting species is involved. For example, decomposition of ammonium nitrite.
\n
\tNH4NO2 → N2 + 2H2O\t
\n
Bimolecular reactions involve simultaneous collision between two species, for example, dissociation of hydrogen iodide. 2HI → H2 + I2
\n
Trimolecular or tetramolecular reactions involve simultaneous collision between three reacting species. For example, 2NO + O2 → 2NO2
\n
The probability that more than three molecules can collide and react simultaneously is very small. Hence, the reaction with the molecularity three are very rare and slow to proceed.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "marks": 3, "source_number": "5", "original_raw_text": " Write a note on molecularity of a reaction and its different types. ", "deduction_level": 0, "source_html": "Chapter3", "boundary_log": [ "leading-marker-stripped" ], "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 6, "question_text": "
Explain how the molecularity of a complex reaction determined? Give the points showing difference between molecularity of a reaction and order of a reaction.
", "answer_text": "
Term molecularity is only applicable to elementary reactions, even though to determine molecularity of a complex reaction it is necessary to know the mechanism of that reaction.
\n
Through mechanism of a reaction we can have information about slowest step of reaction.
\n
The overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step.
\n
Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in an alkaline medium,
\n
The rate equation for this reaction is found to be: Rate = – \"\"
\n
This reaction is first order with respect to both \"\" and \"\".
\n
Mechanism involve two steps:
\n
\t(1)
\n
\t(2)
\n
The first step, being slow, is the rate determining step.
\n
Thus, the rate of formation of intermediate will determine the rate of this reaction.
\n
Thus, it can be said that, for complex reaction, order is given by the slowest step and molecularity of the slowest step is same as the order of the overall reaction.
\n
Difference between molecularity and order of a reaction:
\n
\t(i) Order of a reaction is an experimental quantity. It can be zero and even a fraction but molecularity cannot be zero or a non integer.
\n
\t(ii) Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions. For complex reaction molecularity has no meaning.
\n
\t(iii) For complex reaction, order is given by the slowest step and generally molecularity of the slowest step is same as the order of the overall reaction.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1221-144240.png", "position": "answer", "context": "The rate equation for this reaction is found to be: Rate = –" }, { "id": "img-1", "path": "Chapter3/EQ-0311-213753.png", "position": "answer", "context": "This reaction is first order with respect to both and ." }, { "id": "img-2", "path": "Chapter3/EQ-0311-213834.png", "position": "answer", "context": "This reaction is first order with respect to both and ." } ], "tables": [], "metadata": { "source_class": "Quest", "source_number": "6", "original_raw_text": " Explain how the molecularity of a complex reaction determined? Give the points showing difference between molecularity of a reaction and order of a reaction. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 7, "question_text": "
Derive integrated rate equation for zero order reaction and also explain how the rate constant can be determine with help of graph.
", "answer_text": "
Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants.
\n
Consider the reaction.
\n
\tR → P
\n
Rate of reaction for this reaction can be expressed as
\n
\tRate = \"\" = k[R]0
\n
As any quantity raised to power zero is units.
\n
\tRate = \"\" = k × 1
\n
Thus, the rate of zero order reaction is independent from concentration of reactants.
\n
\td[R] = –k dt
\n
Integrating both sides
\n
\t[R] = –kt + I\t \t... ... Eq. (1)
\n
\tWhere, I is the constant of integration
\n
At t = 0, the concentration of the reactant
R = [R]0, where [R]0 is initial concentration of the reactant.
\n
Substituting in equation (1)
\n
\t[R]0 = –kx 0 + I
\n
\t[R]0 = I
\n
Substituting the value of I in the equation (1)
\n
\t[R] = –kt + [R]0\t\t... ... Eq. (2)
\n
Further simplifying equation (2)
\n
k = \"\"\t\t... ... Eq. (3)
\n
Comparing equation (2) with equation of straight line, y = mx + c, if we plot [R] against t, we get a straight line with slope = –k and intercept equal to [R]0
\n
\"\"
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-223338.png", "position": "answer", "context": "Rate = = k [R] 0" }, { "id": "img-1", "path": "Chapter3/EQ-0124-223338.png", "position": "answer", "context": "Rate = = k × 1" }, { "id": "img-2", "path": "Chapter3/EQ-0125-153346.png", "position": "answer", "context": "k = ... ... Eq. (3)" }, { "id": "img-3", "path": "Chapter3/20.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "marks": 3, "source_number": "7", "original_raw_text": " Derive integrated rate equation for zero order reaction and also explain how the rate constant can be determine with help of graph. ", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 8, "question_text": "
Give examples of zero order reaction.
", "answer_text": "
Zero order reaction are relatively uncommon but they occur under special condition.
\n
Some enzyme catalysed reaction and reaction which occur on metal surface are a few examples of zero order reactions.
\n
The decomposition of gaseous ammonia on hot platinum surface is a zero order reaction at high pressure.
\n
\t2NH3(g) \"\" N2(g) + 3H2(g)
\n
\tRate = k[NH3]0 = k
\n
In this reaction, platinum metal acts as a catalyst.
\n
At high pressure, the metal surface gets saturated with gas molecules.
\n
So, a further change in reaction condition is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration.
\n
The thermal decomposition of HI on gold surface is another example of zero order reaction.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-113134.png", "position": "answer", "context": "2NH 3(g) N 2(g) + 3H 2(g)" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 2, "source_number": "8", "original_raw_text": " Give examples of zero order reaction. ", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 9, "question_text": "
Derive integrated rate equation for first order reaction and also explain how the rate constant can be determine with help of graph.
\n
OR
\n
\tDerive the formula for Rate constant (k) and half life period \"\" for first order reaction.
", "answer_text": "
First order reaction means that the rate of reaction is proportional to the first power of the concentration of the reactant R.
\n
For example, consider the following reaction
\n
\tR → P
\n
Rate of reaction for this reaction can be expressed as
\n
\tRate = \"\" = k[R]
\n
\tOr \"\" = –kdt
\n
Integrating this equation, we get
\n
\tln [R] = –kt + I \t\t... ... Eq. (1)
\n
\twhere, I is the constant of integration and its value can be determined easily.
\n
When t = 0, R = [R]0, where [R]0 is the initial concentration of the reactant.
\n
Therefore, equation (1) can be written as
\n
\tln [R]0 = –k × 0 + I
\n
\tln [R]0 = I
\n
Substituting the value of I in equation (1)
\n
\tln [R] = –kt + ln [R]0\t... ... Eq. (2)
\n
Rearranging this equation
\n
\tln \"\" = –kt
\n
\tor k = \"\" ln \"\" \t... ... Eq. (3)
\n
At time t1 from equation (1)
\n
\tln [R]1 = – kt1 + ln [R]0 \t... ... Eq. (4)
\n
\tAt time t2
\n
\tln [R]2 = – kt2 + ln [R]0 \t... ... Eq. (5)
\n
\twhere, [R]1 and [R]2 are the concentration of the reactants at time t1 and t2 respectively.
\n
Subtracting Eq. (5) from (4)
\n
\tln [R]1 – ln [R]2 = –kt1 – (–kt2)
\n
\tln \"\" = k(t2 – t1)
\n
k = \"\"ln \"\"\t... ... Eq. (6)
\n
So, equation (2) can also be written as
\n
\tln \"\" = –kt
\n
Taking antilog of both the sides
\n
\t[R] = [R]0 e–kt ... ... Eq. (7)
\n
Comparing equation (2) with y = mx + c, if we plot ln [R] against t, we get a straight line with slope = – k and intercept equal to ln [R]0
\n
\"\"
\n
The first order rate equation (3) can also be written in the form
\n
k = \"\" log \"\"\t... ... Eq. (8)
\n
\tor log \"\"\"\"
\n
If we plot a graph between log \"\" vs t,
\n
\tthe slope = \"\"
\n
\"\"
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0119-170409.png", "position": "stem", "context": "Derive the formula for Rate constant ( k ) and half life per…" }, { "id": "img-1", "path": "Chapter3/EQ-0124-223338.png", "position": "answer", "context": "Rate = = k [R]" }, { "id": "img-2", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-058.png", "position": "answer", "context": "Or = – k dt" }, { "id": "img-3", "path": "Chapter3/EQ-0125-161037.png", "position": "answer", "context": "ln = – kt" }, { "id": "img-4", "path": "Chapter3/EQ-0124-223707.png", "position": "answer", "context": "or k = ln ... ... Eq. (3)" }, { "id": "img-5", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-060.png", "position": "answer", "context": "or k = ln ... ... Eq. (3)" }, { "id": "img-6", "path": "Chapter3/EQ-0125-161212.png", "position": "answer", "context": "ln = k ( t 2 – t 1 )" }, { "id": "img-7", "path": "Chapter3/EQ-0124-223808.png", "position": "answer", "context": "k = ln ... ... Eq. (6)" }, { "id": "img-8", "path": "Chapter3/EQ-0125-161212.png", "position": "answer", "context": "k = ln ... ... Eq. (6)" }, { "id": "img-9", "path": "Chapter3/EQ-0125-161037.png", "position": "answer", "context": "ln = – kt" }, { "id": "img-10", "path": "Chapter3/24.png", "position": "answer", "context": "" }, { "id": "img-11", "path": "Chapter3/EQ-0124-223910.png", "position": "answer", "context": "k = log ... ... Eq. (8)" }, { "id": "img-12", "path": "Chapter3/EQ-0125-161301.png", "position": "answer", "context": "k = log ... ... Eq. (8)" }, { "id": "img-13", "path": "Chapter3/EQ-0125-161301.png", "position": "answer", "context": "or log" }, { "id": "img-14", "path": "Chapter3/EQ-0125-161350.png", "position": "answer", "context": "or log" }, { "id": "img-15", "path": "Chapter3/EQ-0125-161301.png", "position": "answer", "context": "If we plot a graph between log vs t," }, { "id": "img-16", "path": "Chapter3/EQ-0124-224029.png", "position": "answer", "context": "the slope =" }, { "id": "img-17", "path": "Chapter3/25.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 4, "source_number": "9", "boundary_log": [ "or_alt→stem" ], "original_raw_text": " Derive integrated rate equation for first order reaction and also explain how the rate constant can be determine with help of graph.\nOR\n\tDerive the formula for Rate constant (k) and half life period for first order reaction. ", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 10, "question_text": "
Give examples of first order reaction.
", "answer_text": "
Hydrogenation of ethene is an example of first order reaction.
\n
\tC2H4(g) + H2(g) → C2H6(g)
\n
\tRate = k[C2H4]
\n
All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.
\n
\"\" → \"\" + \"\"
\n
Rate = k[Ra]
\n
Decomposition of N2O5 and N2O are some more examples of first order reactions.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-114028.png", "position": "answer", "context": "→ +" }, { "id": "img-1", "path": "Chapter3/EQ-1122-114045.png", "position": "answer", "context": "→ +" }, { "id": "img-2", "path": "Chapter3/EQ-1122-114104.png", "position": "answer", "context": "→ +" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-33", "marks": 2, "source_number": "10", "original_raw_text": " Give examples of first order reaction.", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 11, "question_text": "
Derive integrated rate equation for first order reaction containing gaseous components.
", "answer_text": "
Let us consider a typical first order gas phase reaction.
\n
A(g) → B(g) + C(g)
\n
Let pi be the initial pressure of A and pt the total pressure at time ‘t’. Integrated rate equation for such a reaction can be derived as
\n
Total pressure pt = pA + pB + pC (pressure units)
\n
pA, pB and pC are partial pressures of A, B and C respectively. If x atm can be the decrease in pressure of A at time t and one mole each of B and C is being formed. The increase in pressure of B and C will also be X atm each
\n
\t\t\tA(g) →\tB(g)\t+\tC(g)
\n
\tAt t = 0\tpi atm\t\t0 atm\t\t0 atm
\n
\tAt time t\t(pi– x) atm\t\tx atm\t\tx atm
\n
\twhere, pi is the initial pressure at time t = 0.
\n
pt = (pi – x) + x + x = pi + x
\n
\tx = (pt – pi)
\n
\twhere, pA = pi – x = pi – (pt – pi) = 2pi – pt
\n
k\t= \"\"
\n
\t\t= \"\" log
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-224516.png", "position": "answer", "context": "k =" }, { "id": "img-1", "path": "Chapter3/EQ-0124-223910.png", "position": "answer", "context": "= log" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "source_number": "11", "original_raw_text": " Derive integrated rate equation for first order reaction containing gaseous components. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 12, "question_text": "
What is half-life of a reaction? Derive formula for half-life of zero and first order reaction.
\n
OR
\n
\tDerive equation of rate constant and half life of reaction for the zeroth order reaction.
", "answer_text": "
Half-life of a reaction: The time in which the concentration of a reactant is reduced to one half of its initial concentration is called half-life \"\" of a reaction.
\n
Half-life for zero order reaction:
\n
For a zero order reaction, rate constant is given by following equation
\n
k = \"\"
\n
\tAt t = \"\"\"\", [R] = \"\"[R]0
\n
\tThe rate constant at \"\" becomes
\n
k =
\n
\"\" = \"\"
\n
It is clear that \"\"for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant.
\n
Half-life for first order reaction:
\n
For the first order reaction,
\n
k = \"\" log\"\"
\n
\tat t = \"\" [R] = \"\"
\n
So, the above equation becomes
\n
k = \"\" log\"\"
\n
\tor \"\" = \"\" log2
\n
\"\"= \"\" × 0.301
\n
\"\"= \"\"
\n
It can be seen that for a first order reaction, half-life period is constant, i.e. it is independent of initial concentration of the reacting species.
\n
The half-life of a first order equation is readily calculated from the rate constant and vice versa.
\n
\tFor zero order reaction \"\" × [R]0 for first order reaction \"\" is independent of [R]0
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-224605.png", "position": "answer", "context": "Ans. Half-life of a reaction: The time in which the concentr…" }, { "id": "img-1", "path": "Chapter3/EQ-0124-224630.png", "position": "answer", "context": "k =" }, { "id": "img-2", "path": "Chapter3/EQ-0215-161548.png", "position": "answer", "context": "At t = , [R] = [R] 0" }, { "id": "img-3", "path": "Chapter3/EQ-0215-161609.png", "position": "answer", "context": "At t = , [R] = [R] 0" }, { "id": "img-4", "path": "Chapter3/EQ-1122-114653.png", "position": "answer", "context": "At t = , [R] = [R] 0" }, { "id": "img-5", "path": "Chapter3/EQ-0124-224701.png", "position": "answer", "context": "The rate constant at becomes" }, { "id": "img-6", "path": "Chapter3/EQ-0124-224729.png", "position": "answer", "context": "=" }, { "id": "img-7", "path": "Chapter3/EQ-0125-161617.png", "position": "answer", "context": "=" }, { "id": "img-8", "path": "Chapter3/EQ-0124-224758.png", "position": "answer", "context": "It is clear that for a zero order reaction is directly propo…" }, { "id": "img-9", "path": "Chapter3/EQ-0124-223910.png", "position": "answer", "context": "k = log" }, { "id": "img-10", "path": "Chapter3/EQ-0125-161301.png", "position": "answer", "context": "k = log" }, { "id": "img-11", "path": "Chapter3/EQ-0124-224844.png", "position": "answer", "context": "at t = [R] =" }, { "id": "img-12", "path": "Chapter3/EQ-0125-161839.png", "position": "answer", "context": "at t = [R] =" }, { "id": "img-13", "path": "Chapter3/EQ-0124-224906.png", "position": "answer", "context": "k = log" }, { "id": "img-14", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-071.png", "position": "answer", "context": "k = log" }, { "id": "img-15", "path": "Chapter3/EQ-0124-224920.png", "position": "answer", "context": "or = log2" }, { "id": "img-16", "path": "Chapter3/EQ-0124-225005.png", "position": "answer", "context": "or = log2" }, { "id": "img-17", "path": "Chapter3/EQ-0124-224934.png", "position": "answer", "context": "= × 0.301" }, { "id": "img-18", "path": "Chapter3/EQ-0124-225005.png", "position": "answer", "context": "= × 0.301" }, { "id": "img-19", "path": "Chapter3/EQ-0124-224729.png", "position": "answer", "context": "=" }, { "id": "img-20", "path": "Chapter3/EQ-0124-225018.png", "position": "answer", "context": "=" }, { "id": "img-21", "path": "Chapter3/EQ-0124-225051.png", "position": "answer", "context": "For zero order reaction × [R] 0 for first order reaction is …" }, { "id": "img-22", "path": "Chapter3/EQ-0124-225051.png", "position": "answer", "context": "For zero order reaction × [R] 0 for first order reaction is …" } ], "tables": [], "metadata": { "source_class": "Quest", "marks": 3, "source_number": "12", "original_raw_text": " What is half-life of a reaction? Derive formula for half-life of zero and first order reaction. \nOR\n\tDerive equation of rate constant and half life of reaction for the zeroth order reaction. ", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 13, "question_text": "
Explain pseudo first order reaction with example.
", "answer_text": "
The order of a reaction is sometimes altered by conditions.
\n
There are many reactions which obey first order rate law although they are higher order reactions.
\n
Consider the hydrolysis of ethyl acetate which is a chemical reaction between ethyl acetate and water. In reality, it is second order reaction and concentration of both ethyl acetate and water affect the rate of the reaction.
\n
But water is taken in large excess for hydrolysis, therefore, concentration of water is not altered much during the reaction.
\n
Thus, the rate of reaction is affected by concentration of ethyl acetate.
\n
For e.g. 0.01 mol ethyl acetate react with 10 mol of water amounts of the reactants and products at the beginning (t = 0) and completion (t) of the reaction are give as under.
\n
\tCH3COOC2H5 + H2O \"\" CH3COOH + C2H5OH
\n
t = 0 0.01 mol\t 10 mol\t0 mol\t0 mol
\n
t = t\t 0 mol\t 9.99 mol\t 0.01 mol\t 0.01 mol
\n
The concentration of water does not get altered much during the course of the reaction. So, the reaction behaves as first order reaction. Such reactions are called pseudo first order reactions.
\n
\tInversion of cane sugar is another pseudo first order reaction.
\n
\tC12H22O11 + H2O \"\" C6H12O6 + C6H12O6
\n
\tCane sugar\t Glucose\t Fructose
\n
\tRate = k [C12H22O11]
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-115444.png", "position": "answer", "context": "CH 3 COOC 2 H 5 + H 2 O CH 3 COOH + C 2 H 5 OH" }, { "id": "img-1", "path": "Chapter3/EQ-1122-115444.png", "position": "answer", "context": "C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "source_number": "13", "original_raw_text": " Explain pseudo first order reaction with example. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 14, "question_text": "
What is activation energy? Explain with the help of graph by suitable example.
", "answer_text": "
Activation energy can be understood clearly using the formation reaction of hydrogen iodide.
\n
\tH2(g) + I2(g) → 2HI(g)
\n
According to Arrhenius, this reaction can take place only when a molecule hydrogen and molecule of iodine collide to form an unstable intermediate. It exists for a very short time and then breaks up to form two molecules of hydrogen iodide.
\n
\"\"
\n
Intermediate
\n
The energy required to form this intermediate, called activated complex (C), is known as activation energy (Ea).
\n
Below figure is obtained by plotting potential energy vs reaction coordinate. Reaction coordinate represents the profile of energy change when reactant change into products.
\n
\"\"
\n
Some energy is released when the complex decomposes to form products. So, the final enthalpy of the reaction depends upon the nature of reactants and products.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/39.png", "position": "answer", "context": "" }, { "id": "img-1", "path": "Chapter3/40.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "marks": 3, "source_number": "14", "original_raw_text": " What is activation energy? Explain with the help of graph by suitable example. ", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 15, "question_text": "
Derive the formula for determining activation energy from Arrhenius equation using rate constant at different temperatures and also explain how activation energy can be determined using graph.
", "answer_text": "
Activation energy using graph:
\n
Taking natural logarithm on both the side of Arrhenius equation k = Ae\"\"
\n
ln k = \"\" + ln A \t... ... Eq. (1)
\n
The plot of ln k vs 1/T gives a straight line as shown in figure.
\n
\"\"
\n
In figure, slope = \"\" and intercept = lnA. So, we can calculate activation energy (Ea) and Arrhenius constant A using these values.
\n
Formula of activation energy: It has been found from Arrhenius equation that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in rate constant.
\n
Thus, at temperature T1, equation (1) is
\n
\tln k1 = \"\" + lnA\t... ... Eq. (2)
\n
at temperature T2, equation (1) is
\n
\tln k2 = \"\" + lnA \t... ... Eq. (3)
\n
\t(since A is constant for given reaction)
\n
k1 and k2 are rate constant at temperatures T1 and T2 respectively.
\n
Subtracting equation (2) from (3), we obtain
\n
\tlnk2 – lnk1 = \"\"
\n
\tln\"\"
\n
\tlog\"\"
\n
\tlog\"\"
\n
From above formula activation energy can be calculate using measured values of rate constants at different temperatures.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0215-160301.png", "position": "answer", "context": "Taking natural logarithm on both the side of Arrhenius equat…" }, { "id": "img-1", "path": "Chapter3/EQ-1122-144403.png", "position": "answer", "context": "ln k = + ln A \t... ... Eq. (1)" }, { "id": "img-2", "path": "Chapter3/42.png", "position": "answer", "context": "" }, { "id": "img-3", "path": "Chapter3/EQ-1122-144433.png", "position": "answer", "context": "In figure, slope = and intercept = lnA. So, we can calculate…" }, { "id": "img-4", "path": "Chapter3/EQ-1122-144458.png", "position": "answer", "context": "ln k 1 = + lnA\t... ... Eq. (2)" }, { "id": "img-5", "path": "Chapter3/EQ-1122-144515.png", "position": "answer", "context": "ln k 2 = + lnA \t... ... Eq. (3)" }, { "id": "img-6", "path": "Chapter3/EQ-0314-134245.png", "position": "answer", "context": "ln k 2 – ln k 1 =" }, { "id": "img-7", "path": "Chapter3/EQ-0124-232320.png", "position": "answer", "context": "ln" }, { "id": "img-8", "path": "Chapter3/EQ-0124-232335.png", "position": "answer", "context": "log" }, { "id": "img-9", "path": "Chapter3/EQ-0124-232401.png", "position": "answer", "context": "log" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-33", "marks": 4, "source_number": "15", "original_raw_text": " Derive the formula for determining activation energy from Arrhenius equation using rate constant at different temperatures and also explain how activation energy can be determined using graph.", "deduction_level": 0, "source_html": "Chapter3", "boundary_log": [ "leading-marker-stripped" ], "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 16, "question_text": "
Explain the effect of catalyst on rate of reaction and also give its characteristics.
\n
Or
", "answer_text": "
Answer the following questions on catalyst:
\n
\t(i)\tHow it increases the rate of reaction.
\n
\t(ii)\tWrite any 4 characteristics of catalyst.
\n
Ans.\tA catalyst is a substance which increase the rate of a reaction without itself undergoing any permanent chemical change.
\n
For example, MnO2 catalyses the following reaction so as to increase its rate considerably.
\n
\t2KClO3 \"\"2KCl + 3O2
\n
The word catalyst should not be used when the added substance reduces the rate of reaction. The substance is then called inhibitor.
\n
The action of catalyst can be explained by intermediate complex theory.
\n
According to this theory, a catalyst participates in a chemical reaction by forming temporary bond with the reactants resulting in an intermediate complex.
\n
It is believed that the catalyst provides an alternate path or reaction mechanism by reducing the activation energy between reactants and products hence lowering the potential energy barrier as shown in figure.
\n
It is clear from Arrhenius equation that lower the value of activation energy faster will be the rate of reaction.
\n
This has transitory existence and decomposes to yield products and the catalyst.
\n
\"\"
\n
Characteristics of catalyst:
\n
A small amount of catalyst can catalyses a large amount of reactants.
\n
A catalyst does not alter Gibbs energy of a reaction.
\n
It catalysis the spontaneous reaction but does not catalyze the non-spontaneous reactions.
\n
It is also found that a catalyst does not change the equilibrium constant of a reaction rather, it helps in attaining the equilibrium faster, that is, it catalyses the forward as well as backward reaction to the same extent so that the equilibrium state remains same but is reached earlier.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-145030.png", "position": "answer", "context": "2KClO 3 2KCl + 3O 2" }, { "id": "img-1", "path": "Chapter3/48.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 2, "source_number": "16", "original_raw_text": " Explain the effect of catalyst on rate of reaction and also give its characteristics.\nOr", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2", "needs_review": true, "needs_or_review": true, "or_review_reason": "question ends with a dangling standalone અથવા/OR" } }, { "sequence_no": 17, "question_text": "
Write Arrhenius equation. Explain the terms involved in it.
", "answer_text": "
The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation.
\n
k = A\"\"
\n
Where, A is the Arrhenius factor or frequency factor. It is also called per-exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol–1).
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-142011.png", "position": "answer", "context": "k = A" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "marks": 2, "source_number": "17", "original_raw_text": " Write Arrhenius equation. Explain the terms involved in it. ", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } }, { "sequence_no": 18, "question_text": "
Write a note on collision theory of chemical reaction.
\n
Or
\n
What is meant by collision theory? Explain.
", "answer_text": "
Collision theory, which was developed by Max Trautz and William Lewis in 1916-18, provides greater insight into the energetic and mechanistic aspects of reaction.
\n
It is based on kinetic theory of gases.
\n
According to this theory, the reactant molecules are assumed to be hard spheres and reaction is postulated to occur when molecules collide with each other.
\n
The number of collision per second per unit volume of the reaction mixture is known as collision frequency (Z).
\n
Another factor which affects the rate of chemical reaction is activation energy.
\n
For a bimolecular elementary reaction;
\n
\tA + B → Products
\n
Rate of reaction can be expressed as:
\n
Rate = ZAB\"\"\t\t... ... Eq. (1)
\n
where ZAB represents the collision frequency of reactants A and B, and \"\" represents the fraction of molecules with energies equal to or greater than Ea.
\n
Comparing equation (1) with Arrhenius equation, we can say that A is related to collision frequency.
\n
Equation (1) predicts the value of rate constant fairly accurately for the reaction that involve atomic species or simple molecule but for complex molecules significant deviations are observed.
\n
The reason could be that all collision do not lead to the formation of products.
\n
The collision in which molecules collide with sufficient kinetic energy (called threshold energy) and proper orientation, so as to facilitate breaking of bonds between reacting species and formation of new bonds to form products are called as effective collision.
\n
For example, formation of methanol from bromomethane CH3Br + OH – → CH3OH + Br– depends upon the orientation of reactant molecule as shown in figure.
\n
\"\"
\n
The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no product are formed.
\n
To account for effective collision, another factor P, called the probability factor or steric factor is introduced.
\n
\tIt takes into account the fact that in a collision, molecules be properly oriented i.e.
\n
Rate = PZAB\"\"
\n
Thus, in collision theory activation energy and proper orientation of the molecules together determine the criteria for an effective collision and hence the rate of chemical reaction.
\n
Class 12 Chemistry (Part 1) 012
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-142011.png", "position": "answer", "context": "Rate = Z AB ... ... Eq. (1)" }, { "id": "img-1", "path": "Chapter3/EQ-1122-142011.png", "position": "answer", "context": "where Z AB represents the collision frequency of reactants A…" }, { "id": "img-2", "path": "Chapter3/55.png", "position": "answer", "context": "" }, { "id": "img-3", "path": "Chapter3/EQ-1122-142011.png", "position": "answer", "context": "Rate = PZ AB" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-33", "source_number": "18", "marks": 4, "boundary_log": [ "or_alt→stem" ], "original_raw_text": " Write a note on collision theory of chemical reaction.\nOr\n What is meant by collision theory? Explain.", "deduction_level": 0, "source_html": "Chapter3", "section": "Topic-wise Q & A", "section_id": "3.2" } } ] }, { "section_id": "", "section_title": "NCERT Intext Questions and Answers", "heading_text": "NCERT Intext Questions and Answers", "detected_type": "SUB", "match_confidence": 0.542, "questions": [ { "sequence_no": 19, "question_text": "
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
", "answer_text": "
Average rate =
\n
\t\t\t = \"\"
\n
\t\t\t = \"\"
\n
\t\t\t = \"\"
\n
\t\t\t\t= 4 × 10–4 M min–1
\n
\tRate of reaction in terms of second
\n
\t\t= \"\"
\n
\t\t= 6.67 ×10–6 M s–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-215853.png", "position": "answer", "context": "=" }, { "id": "img-1", "path": "Chapter3/EQ-0123-194248.png", "position": "answer", "context": "=" }, { "id": "img-2", "path": "Chapter3/EQ-1119-155410.png", "position": "answer", "context": "=" }, { "id": "img-3", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-017.png", "position": "answer", "context": "=" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "marks": 2, "source_number": "1", "original_raw_text": " For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 20, "question_text": "
In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L–1 to
0.4 mol L–1 in 10 minutes. Calculate the rate during this interval ?
", "answer_text": "
For reaction 2A → Product,
\n
∆t = 10 min
\n
\tRate\t= \"\"
\n
\t\t= \"\"
\n
\t\t= \"\" mol L–1 min–1
\n
\t\t= 5.0 ×10–3 M min–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-220340.png", "position": "answer", "context": "Rate\t=" }, { "id": "img-1", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-019.png", "position": "answer", "context": "=" }, { "id": "img-2", "path": "Chapter3/EQ-1119-160018.png", "position": "answer", "context": "= mol L –1 min –1" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "marks": 2, "source_number": "2", "original_raw_text": " In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L–1 to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval ? ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 21, "question_text": "
For reaction A + B → Product; the rate law is given by, r = k[A]\"\"[B]2 What is the order of the reaction?
", "answer_text": "
Overall order of reaction is equal to the sum of powers of the concentration of reactants
\n
\tHere, order of the reaction = \"\" + 2 = 2.5
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0129-134432.png", "position": "stem", "context": "Q.3. For reaction A + B → Product; the rate law is given by,…" }, { "id": "img-1", "path": "Chapter3/EQ-1122-114653.png", "position": "answer", "context": "Here, order of the reaction = + 2 = 2.5" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "marks": 1, "source_number": "3", "original_raw_text": " For reaction A + B → Product; the rate law is given by, r = k[A][B]2 What is the order of the reaction? ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 22, "question_text": "
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times How will it affect that rate of formation of Y?
", "answer_text": "
X → Y is a second order reaction,
\n
\tTherefore, rate law can be expressed as
\n
\tV1 = k[X]2
\n
If the concentration of X is increased three time, then the new concentration will be 3X and the new rate will be
\n
\tV2 = k[3X]2 = 9k[X]2
\n
Taking ratio of V2 and V1, we get
\n
\"\"
\n
\tV2 = 9V1
\n
Hence the rate of formation of Y will increase by 9 times.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-223304.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 2, "source_number": "4", "original_raw_text": " The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times How will it affect that rate of formation of Y?", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 23, "question_text": "
A first order reaction has a rate constant
1.15 × 10–3 s–1. How long will 5 g of this reactant take to reduce to 3 g ?
", "answer_text": "
Integrated rate equation for first order reaction,
\n
k = \"\" log\"\"
\n
[R]0 = 5 g and [R] = 3 g, k = 1.15 × 10–3 s–1
\n
Substituting above values in equation
\n
t = \"\" log\"\"
\n
t = log\"\"
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-223910.png", "position": "answer", "context": "k = log" }, { "id": "img-1", "path": "Chapter3/EQ-0125-161301.png", "position": "answer", "context": "k = log" }, { "id": "img-2", "path": "Chapter3/EQ-0124-225005.png", "position": "answer", "context": "t = log" }, { "id": "img-3", "path": "Chapter3/EQ-0125-161301.png", "position": "answer", "context": "t = log" }, { "id": "img-4", "path": "Chapter3/EQ-1122-122024.png", "position": "answer", "context": "t = log" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-53", "marks": 3, "source_number": "5", "original_raw_text": " A first order reaction has a rate constant 1.15 × 10–3 s–1. How long will 5 g of this reactant take to reduce to 3 g ? ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 24, "question_text": "
Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, then calculate the rate constant of the reaction.
", "answer_text": "
Time required to decompose half of the reagent is 60 minutes, i.e. \"\" = 60 min.
\n
For first order reaction, relation between half-life \"\" and rate constant k is
\n
k = \"\" = 0.01155 min–1 OR
\n
k = \"\" = 1.925 ×10– 4 s–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-232128.png", "position": "answer", "context": "Ans. Time required to decompose half of the reagent is 60 mi…" }, { "id": "img-1", "path": "Chapter3/EQ-0124-232141.png", "position": "answer", "context": "For first order reaction, relation between half-life and rat…" }, { "id": "img-2", "path": "Chapter3/EQ-1221-155226.png", "position": "answer", "context": "k = = 0.01155 min –1 OR" }, { "id": "img-3", "path": "Chapter3/EQ-1122-122201.png", "position": "answer", "context": "k = = 1.925 ×10 – 4 s –1" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-33", "marks": 2, "source_number": "6", "original_raw_text": " Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, then calculate the rate constant of the reaction.", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 25, "question_text": "
Explain effect of temperature on rate constant.
\n
Or
\n
\tWhat will be the effect of temperature on rate constant?
", "answer_text": "
Most of the chemical reactions are accelerated by increase in temperature.
\n
For example, in decomposition of N2O5, the time taken for half of the original amount of material to decompose is 12 min at 50°C, 5 h at 25°C and 10 days at 0°C.
\n
In a mixture of potassium permanganate and oxalic acid, potassium permanganate gets decolourised faster at a higher temperature than that at a lower temperature.
\n
It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "source_number": "7", "marks": 2, "boundary_log": [ "or_alt→stem" ], "original_raw_text": " Explain effect of temperature on rate constant. \nOr\n\tWhat will be the effect of temperature on rate constant? ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 26, "question_text": "
The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K, calculate Ea.
", "answer_text": "
log\"\" = \"\"
\n
Here, T1 = 298K, T2 = 298 + 10K = 308K
\n
R = 8.314 J K–1 mol–1
\n
The rate of chemical reaction doubles for an increase of 10K in absolute temperature, therefore \"\" = 2.
\n
Substituting the values in above equation, we get
\n
\tlog 2.0 =
\n
\tEa =
\n
\tEa =
\n
\tEa = 52897.77 J mol–1 = 52.89777 kJ mol–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-233314.png", "position": "answer", "context": "Ans. log =" }, { "id": "img-1", "path": "Chapter3/EQ-1122-150108.png", "position": "answer", "context": "Ans. log =" }, { "id": "img-2", "path": "Chapter3/EQ-1122-150134.png", "position": "answer", "context": "The rate of chemical reaction doubles for an increase of 10K…" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-15", "marks": 3, "source_number": "8", "original_raw_text": " The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K, calculate Ea. ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 27, "question_text": "
The activation energy for the reaction 2HI(g) → H2 + I2(g) is 209.5 kJ mol–1 at
581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
", "answer_text": "
The fraction of molecules of reactants having energy equal to or greater than activation energy x = \"\"
\n
Taking 10 base log on both the sides,
\n
\tlog x = \"\"
\n
∴ log x = \"\"
\n
∴ log x = –18.8323
\n
∴ x = antilog (–18.8323) = \"\".1677
\n
∴ The fraction of molecules of reactants having energy equal to or greater than activation energy
x = 1.471 × 10–19
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-142011.png", "position": "answer", "context": "Ans. The fraction of molecules of reactants having energy eq…" }, { "id": "img-1", "path": "Chapter3/EQ-1122-150520.png", "position": "answer", "context": "log x =" }, { "id": "img-2", "path": "Chapter3/EQ-1122-150549.png", "position": "answer", "context": "∴ log x =" }, { "id": "img-3", "path": "Chapter3/EQ-1122-150628.png", "position": "answer", "context": "∴ x = antilog (–18.8323) = .1677" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 3, "source_number": "9", "original_raw_text": " The activation energy for the reaction 2HI(g) → H2 + I2(g) is 209.5 kJ mol–1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Intext Questions and Answers", "section_id": "" } } ] }, { "section_id": "3.5", "section_title": "NCERT Textbook Exercise Q & A", "heading_text": "NCERT Textbook Exercise Questions And Answers", "detected_type": "SUB", "match_confidence": 0.75, "questions": [ { "sequence_no": 28, "question_text": "
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
\n
\t(i) 3NO(g) → N2O(g) + NO2(g); Rate = k[NO]2
\n
\t(ii) H2O2(aq) + 3l–(aq) + 2H+ → 2H2O(l) + I3–;
\n
Rate = k[H2O2][I –]
\n
\t(iii) CH3CHO(g) → CH4(g) + CO(g);
\n
Rate = k[CH3CHO]\"\"
\n
\t(iv) C2H5Cl(g) → C2H4(g) + HCl(g);
\n
Rate = k[C2H5Cl]
", "answer_text": "
(i) 3NO(g) → N2O(g) + NO2(g)
\n
\t\tRate = k[NO]2
\n
∴ Order of reaction = 2
\n
\t\tDimension of the rate constant:
\n
\t\tRate = k[NO]2
\n
k = \"\" = = mol–1 L s–1
\n
(ii) H2O2(aq) + 3I –(aq) + 2H+ → 2H2O(l) + I3–
\n
\t\tRate = k[H2O2][I –]
\n
\t\tOrder of reaction with respect to H2O2 = 1
\n
\t\torder of reaction with respect to l–1 = 1
\n
∴ overall order of reaction = 2
\n
\t\tDimension of the rate constant.
\n
\tRate = k[H2O2][I–]
\n
k\t= \"\"
\n
\t\t\t\t= \"\"
\n
\t\t\t\t= mol–1 L s–1
\n
(iii) CH3CHO(g) → CH4(g) + CO(g)
\n
\t\tRate = k[CH3CHO]\"\"
\n
∴ Order of reaction = \"\"
\n
\t\t\tDimension of the rate constant
\n
\t\t\tRate = k[CH3CHO]\"\"
\n
k\t= \"\"
\n
\t\t\t\t\t=
\n
\t\t\t\t\t= \"\"
\n
\t(iv) C2H5Cl(g) → C2H4(g) + HCl(g)
\n
\t\tRate = k[C2H5Cl]
\n
∴ Order of reaction = 1
\n
\t\tDimension of the rate constant.
\n
\t\tRate = k[C2H5Cl]
\n
k\t= \"\"
\n
\t\t\t= = s–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0206-161444.png", "position": "stem", "context": "Rate = k [CH 3 CHO]" }, { "id": "img-1", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-029.png", "position": "answer", "context": "k = = = mol –1 L s –1" }, { "id": "img-2", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-031.png", "position": "answer", "context": "k =" }, { "id": "img-3", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-032.png", "position": "answer", "context": "=" }, { "id": "img-4", "path": "Chapter3/EQ-1122-103732.png", "position": "answer", "context": "Rate = k [CH 3 CHO]" }, { "id": "img-5", "path": "Chapter3/EQ-1122-103945.png", "position": "answer", "context": "∴ Order of reaction =" }, { "id": "img-6", "path": "Chapter3/EQ-1122-104007.png", "position": "answer", "context": "Rate = k [CH 3 CHO]" }, { "id": "img-7", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-033.png", "position": "answer", "context": "k =" }, { "id": "img-8", "path": "Chapter3/EQ-0311-214738.png", "position": "answer", "context": "=" }, { "id": "img-9", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-035.png", "position": "answer", "context": "k =" } ], "tables": [], "metadata": { "source_class": "Quest", "source_number": "1", "marks": 2, "original_raw_text": " From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.\n\t(i) 3NO(g) → N2O(g) + NO2(g); Rate = k[NO]2\n\t(ii) H2O2(aq) + 3l–(aq) + 2H+ → 2H2O(l) + I3–; \nRate = k[H2O2][I –]\n\t(iii) CH3CHO(g) → CH4(g) + CO(g);\nRate = k[CH3CHO]\n\t(iv) C2H5Cl(g) → C2H4(g) + HCl(g);\nRate = k[C2H5Cl]", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 29, "question_text": "
For the reaction: 2A + B → A2B
\n
\tthe rate = k[A][B]2 with k = 2.0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1,
[B] = 0.2 mol L–1, Calculate the rate of reaction after [A] is reduce to 0.06 mol L–1
", "answer_text": "
Initial rate = k[A][B]2
\n
\t\t\t = 2.0 × 10–6 mol–2 L2 s–1 × 0.1 mol L–1
\n
× (0.2 mol L–1)2
\n
\t\t\t = 8.0 × 10–9 mol L–1 s–1
\n
Concentration of A is reduce to 0.06 mol L–1
\n
Concentration of [A] reacted\t= 0.10 – 0.06
\n
\t\t\t\t= 0.04 mol L–1
\n
As per the reaction, 2A + B → A2B
\n
Half of the concentration of B is reduce then A
\n
∴ Concentration of [B] reacted = \"\" X 0.04
\n
\t\t\t\t= 0.02 mol L–1
\n
[B] = 0.2 – 0.02 = 0.18 mol L–1
\n
\tRate\t= k [A][B]2
\n
\t\t= 2.0 × 10–6 mol–2 L2 s–1 × 0.06 mol L–1
\n
× (0.18 mol L–1)2
\n
\t \t= 3.88 × 10–9 mol L–1 s–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-114653.png", "position": "answer", "context": "∴ Concentration of [B] reacted = X 0.04" } ], "tables": [], "metadata": { "source_class": "Quest", "source_number": "2", "marks": 3, "original_raw_text": " For the reaction: 2A + B → A2B\n\tthe rate = k[A][B]2 with k = 2.0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1, [B] = 0.2 mol L–1, Calculate the rate of reaction after [A] is reduce to 0.06 mol L–1 ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 30, "question_text": "
The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10– 4
mol L–1 s–1 ?
", "answer_text": "
The decomposition reaction of ammonia is
\n
\t2NH3(g) → N2(g) + 3H2(g),
\n
This is zero order reaction,
\n
Rate = \"\"\t= \"\"
\n
\t\t\t\t= \"\" = k[NH3]0
\n
Rate of appearance of N2 = \"\"
\n
\t\t\t\t\t= k[NH3]0 = k
\n
\t\t\t\t\t= 2.5 × 10–4 mol L–1 s–1
\n
Rate of appearance of H2 = \"\"
\n
\t\t\t\t = k[NH3]0 = k
\n
∴ \"\" = 3k = 3 × 2.5 × 10– 4 mol L–1 s–1
\n
\t\t\t = 7.5 × 10–4 mol L–1 s–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-222058.png", "position": "answer", "context": "Rate = =" }, { "id": "img-1", "path": "Chapter3/EQ-0124-222112.png", "position": "answer", "context": "Rate = =" }, { "id": "img-2", "path": "Chapter3/EQ-0124-222127.png", "position": "answer", "context": "= = k[NH 3 ] 0" }, { "id": "img-3", "path": "Chapter3/EQ-0124-222144.png", "position": "answer", "context": "Rate of appearance of N 2 =" }, { "id": "img-4", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-043.png", "position": "answer", "context": "Rate of appearance of H 2 =" }, { "id": "img-5", "path": "Chapter3/EQ-0124-222354.png", "position": "answer", "context": "∴ = 3 k = 3 × 2.5 × 10 – 4 mol L –1 s –1" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 2, "source_number": "3", "original_raw_text": " The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10– 4 mol L–1 s–1 ?", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 31, "question_text": "
The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k[CH3OCH3]\"\"
\n
\tThe rate of reaction is followed by increase in pressure in a closed vessel. So the rate can also be expressed in terms of the partial pressure of dimethyl ether i.e. Rate = \"\"
\n
\tIf the pressure is measured in bar and time in minutes, then what are the units of rate and rate costants ?
", "answer_text": "
The pressure is measured in bar and time in minutes,
\n
\tSo, the unit of rate = Bar min–1
\n
\tUnit of K\t= \"\"
\n
\t\t\t\t= \"\"
\n
\t\t\t\t= \"\" min–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0119-162258.png", "position": "stem", "context": "Q.4. The decomposition of dimethyl ether leads to the format…" }, { "id": "img-1", "path": "Chapter3/EQ-0119-162136.png", "position": "stem", "context": "The rate of reaction is followed by increase in pressure in …" }, { "id": "img-2", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-037.png", "position": "answer", "context": "Unit of K\t=" }, { "id": "img-3", "path": "Chapter3/EQ-1221-145021.png", "position": "answer", "context": "=" }, { "id": "img-4", "path": "Chapter3/EQ-1221-145030.png", "position": "answer", "context": "= min –1" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "source_number": "4", "original_raw_text": " The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k[CH3OCH3]\n\tThe rate of reaction is followed by increase in pressure in a closed vessel. So the rate can also be expressed in terms of the partial pressure of dimethyl ether i.e. Rate = \n\tIf the pressure is measured in bar and time in minutes, then what are the units of rate and rate costants ? ", "marks": 0, "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 32, "question_text": "
Mention the factors that affect the rate of a chemical reaction.
", "answer_text": "
Factors affecting the rate of a chemical reaction:
\n
(i)\tNature of reactant: If intermolecular attraction force between reactant molecules is more or bond enthalpy of reactant molecule is more than more energy of activation is required, so that the rate of reaction becomes slow.
\n
(ii)\tConcentration of reactant: More is the concentration of reactant, more will be the rate of reaction.
\n
(iii)\tPressure: On increasing pressure, the collision between reactant molecule increases, which results in to increase in the rate of reaction.
\n
(iv)\tTemperature: In most of the cases, on increasing temperature the rate of reaction and rate constant both increase.
\n
For endothermic reaction the rate of reaction increases on increasing temperature, whereas for exothermic reaction the rate of reaction decreases on increasing temperature.
\n
(v)\tCatalyst: The energy of activation decreases on using catalyst. Thus the value of energy barrier decreases for reaction, hence the rate of reaction increases.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "source_number": "5", "original_raw_text": " Mention the factors that affect the rate of a chemical reaction. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 33, "question_text": "
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled (ii) reduced to half ?
", "answer_text": "
Reaction → Products
\n
Rate r1 = k[A]2
\n
(i)\tWhen concentration of reactant is doubled;
\n
\t\tRate r2 = k[2A]2 = 4k[A]2 = 4r1
\n
∴\tWhen concentration of the reactant is doubled, the rate of reaction will become 4 times of the initial rate.
\n
(ii)\tWhen concentration of reactant is reduced to half;
\n
\t\tRate r2 = \"\" = \"\"k[A]2 = \"\"
\n
∴\twhen concentration of the reactant is reduced by half, the rate of reaction will be reduce to \"\" of the initial rate.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-222426.png", "position": "answer", "context": "Rate r 2 = = k [A] 2 =" }, { "id": "img-1", "path": "Chapter3/EQ-1122-110043.png", "position": "answer", "context": "Rate r 2 = = k [A] 2 =" }, { "id": "img-2", "path": "Chapter3/EQ-1122-110050.png", "position": "answer", "context": "Rate r 2 = = k [A] 2 =" }, { "id": "img-3", "path": "Chapter3/EQ-1122-110043.png", "position": "answer", "context": "∴ when concentration of the reactant is reduced by half, the…" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 2, "source_number": "6", "original_raw_text": " A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ? ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 34, "question_text": "
What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?
", "answer_text": "
Most of the chemical reactions are accelerated by increase in temperature.
\n
Increasing the temperature will result in an exponential increase in the rate constant.
\n
It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.
\n
\tRate = (2)\"\"
\n
According to Swedish chemist Arrhenius, the quantitative effect of temperature on the rate constant can be express by following equation.
\n
k = A\"\"
\n
\tWhere, A = Arrhenius factor or frequency factor.
\n
\t\t R = gas constant
\n
Ea = activation energy in terms of
joules/mole (J mol–1).
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-107.png", "position": "answer", "context": "Rate = (2)" }, { "id": "img-1", "path": "Chapter3/EQ-1122-142011.png", "position": "answer", "context": "k = A" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-33", "marks": 2, "source_number": "7", "original_raw_text": " What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively? ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 35, "question_text": "
In a pseudo first order reaction in water, the following results were obtained:
\n
\tcalculate the average rate of reaction between the time interval 30 to 60 second.
", "answer_text": "
Hydrolysis of ester is pseudo first order reaction.
\n
RCOOR' + H2O \"\" RCOOH + R'OH
\n
\tEster Water\tAcid\tAlcohol
\n
Average rate of reaction rav = \"\"
\n
rav = –
\n
[R]2 = concentration at 60 s = 0.17 M
\n
[R]1 = concentration at 30 s = 0.31 M
\n
rav = \"\"
\n
rav = \"\" = 4.67 × 10–3 mol L–1 s–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-122337.png", "position": "answer", "context": "RCOOR' + H 2 O RCOOH + R'OH" }, { "id": "img-1", "path": "Chapter3/EQ-0212-165455.png", "position": "answer", "context": "Average rate of reaction r av =" }, { "id": "img-2", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-075.png", "position": "answer", "context": "r av =" }, { "id": "img-3", "path": "Chapter3/EQ-0311-220449.png", "position": "answer", "context": "r av = = 4.67 × 10 –3 mol L –1 s –1" } ], "tables": [ { "id": "tbl-1", "rows": 2, "columns": 5, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Time/s

\n		\n

0

\n		\n

30

\n		\n

60

\n		\n

90

\n
\n

[A]/mol L–1

\n		\n

0.55

\n		\n

0.31

\n		\n

0.17

\n		\n

0.085

\n
", "position": "stem", "context": "Time/s 0 30 60 90 [A]/mol L –1 0.55 0.31 0.17 0.085" } ], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "source_number": "8", "marks": 2, "original_raw_text": " In a pseudo first order reaction in water, the following results were obtained:\n\tcalculate the average rate of reaction between the time interval 30 to 60 second. ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 36, "question_text": "
A reaction is first order in A and second
order in B.
\n
\t(i)\tWrite the differential rate equation. [June 2025]
\n
\t(ii)\tHow is the rate affected on increasing the concentration of B three times ?
\n
(iii)\tHow is the rate affected when the concentration of both A and B are doubled ?
\n
[Topic 3.2] [June 2025] [2 Marks]
", "answer_text": "
The given reaction is first order in A and second order in B.
\n
(i) \tRate = k[A][B]2
\n
(ii)\tInitial rate r1 = k[A][B]2
\n
When concentration of B is increased to three times,
\n
Rate r2 = k[A][3B]2 = 9k [A][B]2 = 9r1

∴ Rate becomes nine times.

\n
(iii)\tInitial rate r1 = k[A][B]2
\n
When concentration of both A and B is increased to two times,
\n
Rate r2 = k[2A][2B]2 = 8k [A][B]2 = 8r1
\n
∴ Rate becomes eight times.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "source_number": "9", "marks": 2, "original_raw_text": " A reaction is first order in A and secondorder in B. \n\t(i)\tWrite the differential rate equation. [June 2025]\n\t(ii)\tHow is the rate affected on increasing the concentration of B three times ?\n (iii)\tHow is the rate affected when the concentration of both A and B are doubled ?\n[Topic 3.2] [June 2025] [2 Marks]", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 37, "question_text": "
In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
", "answer_text": "
Suppose,
\n
\tOrder of reaction with respect to A = x
\n
\tOrder of reaction with respect to B = y
\n
Then,
\n
\tRate = k[A]x[B]y
\n
∴ r1 = k[0.20]x[0.30]y = 5.07 ×10–5 ... ... (1)
\n
\t\t\tr2= k[0.20]x[0.10]y = 5.07 ×10–5 ... ... (2)
\n
\t\t\tr3 = k[0.40]x [0.05]y = 1.43 ×10– 4 ... ... (3)
\n
\tDividing (1) and (2) we get,
\n
\t∴ \"\" = [3]y = 1
\n
\t∴ y = 0, because [3]0 = 1 = [3]y
\n
\tDividing (3) and (2) we get,
\n
= 2.82
\n
∴ \"\" = 2.82
\n
∴ [2]x = 2.82
\n
\tTaking log of both the side we get,
\n
\tlog[2]x = log2.82
\n
∴ x log2 = log2.82
\n
x × 0.3010 = 0.4503
\n
∴ x = \"\" = 1.5
\n
∴ order of reaction with respect to A = 1.5
\n
\torder of reaction with respect to B = 0
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-110718.png", "position": "answer", "context": "∴ = [3] y = 1" }, { "id": "img-1", "path": "Chapter3/EQ-0124-222750.png", "position": "answer", "context": "∴ = 2.82" }, { "id": "img-2", "path": "Chapter3/EQ-1122-111017.png", "position": "answer", "context": "∴ x = = 1.5" } ], "tables": [ { "id": "tbl-1", "rows": 3, "columns": 4, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

A/mol L–1

\n		\n

0.20

\n		\n

0.20

\n		\n

0.40

\n
\n

B/mol L–1

\n		\n

0.30

\n		\n

0.10

\n		\n

0.05

\n
\n

r0/mol L–1 s–1

\n		\n

5.07 × 10–5

\n		\n

5.07 × 10–5

\n		\n

1.43 × 10–4

\n
", "position": "stem", "context": "A/mol L –1 0.20 0.20 0.40 B/mol L –1 0.30 0.10 0.05 r 0 /mol…" } ], "metadata": { "source_class": "Quest", "source_number": "10", "marks": 4, "original_raw_text": " In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 38, "question_text": "
The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D
\n
\tDetermine the rate law and the rate constant for the reaction.
", "answer_text": "
Suppose,
\n
\tOrder of reaction with respect to A = x
\n
\tOrder of reaction with respect to B = y
\n
∴ Rate = k[A]x[B]y
\n
∴ r1 = k[0.1]x[0.1]y = 6.0 × 10–3\t... ... (1)
\n
\t r2 = k[0.3]x[0.2]y = 7.2 × 10–2\t... ... (2)
\n
\t r3 = k[0.3]x[0.4]y = 2.88 × 10–1... ... (3)
\n
\t r4 = k[0.4]x[0.1]y = 2.40 × 10–2... ... (4)
\n
\tDividing (4) and (1) we get,
\n
∴ \"\" = [4]x = [4]1
\n
Order of reaction with respect to A = 1
\n
\tDividing (3) and (2) we get,
\n
∴ \"\" = [2]y = 4 = [2]2
\n
∴ y = 2
\n
∴ order of reaction with respect to A = 1
\n
order of reaction with respect to B = 2
\n
overall order of reaction = 3
\n
∴ rate equation,
\n
\tRate = k[A]1[B]2
\n
Calculation of rate constant k
\n
\tFrom equation (1)
\n
r1 = k[0.1]1[0.1]2 = 6.0 × 10–3
\n
\tk\t=
\n
\t \t=
\n
\t \t= 6.0 mol–2 L–2 min–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-111548.png", "position": "answer", "context": "∴ = [4] x = [4] 1" }, { "id": "img-1", "path": "Chapter3/EQ-1122-111731.png", "position": "answer", "context": "∴ = [2] y = 4 = [2] 2" } ], "tables": [ { "id": "tbl-1", "rows": 6, "columns": 4, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Experiment

\n		\n

[A]

\n		\n

[B]

\n		\n

Initial rate of formation of
D/mol L–1 min–1

\n
\n

mol L–1

\n		\n

mol L–1

\n
\n

I

\n		\n

0.1

\n		\n

0.1

\n		\n

6.0 × 10–3

\n
\n

II

\n		\n

0.3

\n		\n

0.2

\n		\n

7.2 × 10–2

\n
\n

III

\n		\n

0.3

\n		\n

0.4

\n		\n

2.88 × 10–1

\n
\n

IV

\n		\n

0.4

\n		\n

0.1

\n		\n

2.40 × 10–2

\n
", "position": "stem", "context": "Experiment [A] [B] Initial rate of formation of D/mol L –1 m…" } ], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "source_number": "11", "marks": 4, "original_raw_text": " The following results have been obtained during the kinetic studies of the reaction:2A + B → C + D\n\tDetermine the rate law and the rate constant for the reaction.", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 39, "question_text": "
The reaction between A and B first order with respect to A and zero order with respect to B. Fill the blanks in the following table:
", "answer_text": "
Reaction A → B
\n
Rate = k[A]1[B]0
\n
Tate constant in Experiment-I:
\n
\tRate = k[A]
\n
∴ k = \"\"
\n
∴ k = 2.0 × 10–1 = 0.2 min–1
\n
[A] in Experiment-II:
\n
\tRate = k[A]
\n
∴ [A] = \"\"
\n
∴ [A] = 0.2 mol L–1
\n
Initial rate in Experiment-III:
\n
\tRate = k[A]
\n
∴ rate = 0.2 × 0.4 = 0.08 mol L–1 min–1
\n
[A] in Experiment-IV:
\n
\tRate = k[A]
\n
∴ [A] = \"\"
\n
∴ [A] = 0.1 mol L–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-053.png", "position": "answer", "context": "∴ k =" }, { "id": "img-1", "path": "Chapter3/EQ-0124-223124.png", "position": "answer", "context": "∴ [A] =" }, { "id": "img-2", "path": "Chapter3/EQ-0124-223147.png", "position": "answer", "context": "∴ [A] =" } ], "tables": [ { "id": "tbl-1", "rows": 6, "columns": 4, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Experiment

\n		\n

[A]

\n		\n

[B]

\n		\n

Initial rate/

\n

mol L–1 min–1

\n
\n

mol L–1

\n		\n

mol L–1

\n
\n

I

\n		\n

0.1

\n		\n

0.1

\n		\n

2.0 × 10–2

\n
\n

II

\n		\n

–

\n		\n

0.2

\n		\n

4.0 × 10–2

\n
\n

III

\n		\n

0.4

\n		\n

0.4

\n		\n

–

\n
\n

IV

\n		\n

–

\n		\n

0.2

\n		\n

2.0 × 10–2

\n
", "position": "stem", "context": "Experiment [A] [B] Initial rate/ mol L –1 min –1 mol L –1 mo…" } ], "metadata": { "source_class": "Quest", "source_number": "12", "marks": 3, "original_raw_text": " The reaction between A and B first order with respect to A and zero order with respect to B. Fill the blanks in the following table:", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 40, "question_text": "
Calculate the half-life of a first order reaction from their rate constants given below:
\n
\t(1) 200 s–1 (ii) 2 min–1 (iii) 4 years–1
", "answer_text": "
For first order reaction \"\" = \"\"
\n
\t(i) \"\"\"\" = 3.465 × 10–3 s
\n
\t(ii) \"\"\"\" = 0.3465 min
\n
\t(iii) \"\"\"\" = 0.17325 year
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-225051.png", "position": "answer", "context": "Ans. For first order reaction =" }, { "id": "img-1", "path": "Chapter3/EQ-1122-123316.png", "position": "answer", "context": "Ans. For first order reaction =" }, { "id": "img-2", "path": "Chapter3/EQ-0124-225051.png", "position": "answer", "context": "(i) = 3.465 × 10 –3 s" }, { "id": "img-3", "path": "Chapter3/EQ-1221-151644.png", "position": "answer", "context": "(i) = 3.465 × 10 –3 s" }, { "id": "img-4", "path": "Chapter3/EQ-0124-225051.png", "position": "answer", "context": "(ii) = 0.3465 min" }, { "id": "img-5", "path": "Chapter3/EQ-1221-151658.png", "position": "answer", "context": "(ii) = 0.3465 min" }, { "id": "img-6", "path": "Chapter3/EQ-0124-225051.png", "position": "answer", "context": "(iii) = 0.17325 year" }, { "id": "img-7", "path": "Chapter3/EQ-1221-151740.png", "position": "answer", "context": "(iii) = 0.17325 year" } ], "tables": [], "metadata": { "source_class": "Quest", "source_number": "13", "marks": 1, "original_raw_text": " Calculate the half-life of a first order reaction from their rate constants given below:\n\t(1) 200 s–1 (ii) 2 min–1 (iii) 4 years–1", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 41, "question_text": "
The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
", "answer_text": "
Half-life of 14C = 5730 years.
\n
All radioactive decay follows first order kinetics.
\n
\"\"
\n
k = \"\"
\n
k = \"\" year–1
\n
For first order reaction.
\n
t = \"\" log\"\"
\n
t = \"\" log\"\"
\n
t = \"\" log1.25
\n
t = \"\" × 0.0969
\n
t = 1845.2 years.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-225247.png", "position": "answer", "context": "" }, { "id": "img-1", "path": "Chapter3/EQ-1221-151847.png", "position": "answer", "context": "k =" }, { "id": "img-2", "path": "Chapter3/EQ-1122-123540.png", "position": "answer", "context": "k = year –1" }, { "id": "img-3", "path": "Chapter3/EQ-0124-225005.png", "position": "answer", "context": "t = log" }, { "id": "img-4", "path": "Chapter3/EQ-0125-161301.png", "position": "answer", "context": "t = log" }, { "id": "img-5", "path": "Chapter3/EQ-1122-123633.png", "position": "answer", "context": "t = log" }, { "id": "img-6", "path": "Chapter3/EQ-1122-123639.png", "position": "answer", "context": "t = log" }, { "id": "img-7", "path": "Chapter3/EQ-1122-123633.png", "position": "answer", "context": "t = log1.25" }, { "id": "img-8", "path": "Chapter3/EQ-1122-123633.png", "position": "answer", "context": "t = × 0.0969" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "source_number": "14", "marks": 4, "original_raw_text": " The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 42, "question_text": "
The experimental data for decomposition of N2O5; [2N2O5 → 4NO2 + O2]
\n
\tin gas phase at 318K are given below:
\n
\t(i)\tPlot [N2O5] against t.
\n
\t(ii)\tFind the half-life period for the reaction.
\n
\t(iii)\tDraw a graph between log [N2O5] and t
\n
\t(iv)\tWhat is the rate law?\t
\n
\t(v)\tCalculate the rate constant.
\n
\t(vi)\tCalculate the half-life period from k and compare it with (ii).
", "answer_text": "
(i)
\n
\"\"
\n
(ii) Initial concentration of N2O5
\n
\t\t\t\t\t= 1.63 × 10–2 M
\n
\t\tHalf of this concentration = 0.815 × 10–2 M
\n
\t\tFrom graph, time corresponding to this concentration is 1400 s.
\n
\t\tHence, \"\" = 1400 s.
\n
(iii) First we will find values of log [N2O5],

Time/(sec)

[N2O5] × 10+2

mol L–1

log[N2O5]

0

1.63

– 1.79

400

1.36

– 1.87

800

1.14

– 1.94

1200

0.93

– 2.03

1600

0.78

– 2.11

2000

0.64

– 2.19

2400

0.53

– 2.28

2800

0.43

– 2.37

3200

0.35

– 2.46

\n
\"\"
\n
(iv) As plot of log [N2O5] → t is straight line. Hence, it is first order reaction.
\n
\t\t\t∴ Rate = k [N2O5]
\n
(v) From plot of log [N2O5] → t
\n
\t\t\tSlope \t= \"\"
\n
\t\t\t∴ k\t= – slope × 2.303
\n
\t\t\t\t= \"\" × 2.303
\n
\t\t\t\t= \"\" × 2.303
\n
\t\t\t\t= 4.82 × 10–4 s–1
\n
OR
\n
From formula of integrated rate law for first order reaction,
\n
k = \"\" log\"\"
\n
\t\t\t∴ k = \"\" log
\n
\t\t\t∴ k = \"\" log4.6571
\n
\t\t\t∴ k = \"\" × 0.6681
\n
\t\t\t∴ k = 4.81 × 10– 4 s–1
\n
\t(vi)\t \"\" = \"\" =
\n
\t\t = 0.1437 × 104 s = 1437 s
\n
\t\tHence, half-life from both (ii) and (vi) are nearly same.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/80.png", "position": "answer", "context": "" }, { "id": "img-1", "path": "Chapter3/EQ-0124-224729.png", "position": "answer", "context": "Hence, = 1400 s." }, { "id": "img-2", "path": "Chapter3/81.png", "position": "answer", "context": "" }, { "id": "img-3", "path": "Chapter3/EQ-0124-225516.png", "position": "answer", "context": "Slope \t=" }, { "id": "img-4", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-077.png", "position": "answer", "context": "= × 2.303" }, { "id": "img-5", "path": "Chapter3/EQ-1122-124909.png", "position": "answer", "context": "= × 2.303" }, { "id": "img-6", "path": "Chapter3/EQ-0124-225555.png", "position": "answer", "context": "k = log" }, { "id": "img-7", "path": "Chapter3/EQ-0125-161301.png", "position": "answer", "context": "k = log" }, { "id": "img-8", "path": "Chapter3/EQ-1122-125104.png", "position": "answer", "context": "∴ k = log" }, { "id": "img-9", "path": "Chapter3/EQ-1122-125131.png", "position": "answer", "context": "∴ k = log4.6571" }, { "id": "img-10", "path": "Chapter3/EQ-1122-125131.png", "position": "answer", "context": "∴ k = × 0.6681" }, { "id": "img-11", "path": "Chapter3/EQ-0124-224729.png", "position": "answer", "context": "(vi) = =" }, { "id": "img-12", "path": "Chapter3/EQ-0124-225018.png", "position": "answer", "context": "(vi) = =" } ], "tables": [ { "id": "tbl-1", "rows": 2, "columns": 10, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Time/s

\n		\n

0

\n		\n

400

\n		\n

800

\n		\n

1200

\n		\n

1600

\n		\n

2000

\n		\n

2400

\n		\n

2800

\n		\n

3200

\n
\n

102× [N2O5]

\n

mol L–1

\n		\n

1.63

\n		\n

1.36

\n		\n

1.14

\n		\n

0.93

\n		\n

0.78

\n		\n

0.64

\n		\n

0.53

\n		\n

0.43

\n		\n

0.35

\n
", "position": "stem", "context": "Time/s 0 400 800 1200 1600 2000 2400 2800 3200 10 2 × [N 2 O…" } ], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-33", "source_number": "15", "marks": 4, "original_raw_text": " The experimental data for decomposition of N2O5; [2N2O5 → 4NO2 + O2]\n\tin gas phase at 318K are given below:\n\t(i)\tPlot [N2O5] against t.\n\t(ii)\tFind the half-life period for the reaction.\n\t(iii)\tDraw a graph between log [N2O5] and t\n\t(iv)\tWhat is the rate law?\t\n\t(v)\tCalculate the rate constant.\n\t(vi)\tCalculate the half-life period from k and compare it with (ii). ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 43, "question_text": "
The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value ?
", "answer_text": "
Rate constant k = 60 s–1
\n
If, initial concentration = [R]0 M
\n
Then, concentration at time t = \"\" M
\n
\tFor first order reaction,
\n
t = \"\" log\"\"
\n
\t\t\t= \"\" log \"\"
\n
\t\t\t= \"\" log16
\n
\t\t\t= \"\" × 1.2044
\n
\t\t\t= 4.62 × 10-2 Sec.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-079.png", "position": "answer", "context": "Then, concentration at time t = M" }, { "id": "img-1", "path": "Chapter3/EQ-0124-225005.png", "position": "answer", "context": "t = log" }, { "id": "img-2", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-183.png", "position": "answer", "context": "t = log" }, { "id": "img-3", "path": "Chapter3/EQ-0124-225005.png", "position": "answer", "context": "= log" }, { "id": "img-4", "path": "Chapter3/EQ-0125-162359.png", "position": "answer", "context": "= log" }, { "id": "img-5", "path": "Chapter3/EQ-0124-225005.png", "position": "answer", "context": "= log16" }, { "id": "img-6", "path": "Chapter3/EQ-0311-220851.png", "position": "answer", "context": "= × 1.2044" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 3, "source_number": "16", "original_raw_text": " The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value ? ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 44, "question_text": "
During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1 mg of 90Sr was absorbed in the bones of a newly born baby instead of calcium. How much of it will remain after 10 years and 60 years if it is not lost metabolically.
", "answer_text": "
Nuclear explosion is first order reaction.
\n
\"\" = 28.1 years and [R]0 = 1 µg.
\n
For first order reaction,
\n
k = \"\" = \"\" = 2.466 × 10–2 year–1
\n
Calculation for amount of 90Sr remain after 10 years:
\n
t = \"\" log\"\"
\n
\t∴ log\"\" = \"\"
\n
\t∴ log\"\" = \"\"
\n
\t∴ log\"\" = 0.1070
\n
\t∴ \"\" = anti log0.1070 = 1.2794
\n
∴ [R]t = \"\" = 0.7816 µg 90Sr is left after 10 years.
\n
Calculation for amount of 90Sr remain after
60 years:
\n
t = \"\" log\"\"
\n
\t∴ log\"\" = \"\"
\n
\t∴ log\"\" = \"\"
\n
\t∴ log\"\" = 0.6425
\n
\t∴ \"\" = antilog 0.6425 = 4.39
\n
\t∴ [R]t = \"\" = 0.228 µg 90Sr is left after 60 years.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-224729.png", "position": "answer", "context": "= 28.1 years and [R] 0 = 1 µg." }, { "id": "img-1", "path": "Chapter3/EQ-1221-153518.png", "position": "answer", "context": "k = = = 2.466 × 10 – 2 year –1" }, { "id": "img-2", "path": "Chapter3/EQ-1122-133327.png", "position": "answer", "context": "k = = = 2.466 × 10 – 2 year –1" }, { "id": "img-3", "path": "Chapter3/EQ-1122-133412.png", "position": "answer", "context": "t = log" }, { "id": "img-4", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-084.png", "position": "answer", "context": "t = log" }, { "id": "img-5", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-084.png", "position": "answer", "context": "∴ log =" }, { "id": "img-6", "path": "Chapter3/EQ-0124-225806.png", "position": "answer", "context": "∴ log =" }, { "id": "img-7", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-084.png", "position": "answer", "context": "∴ log =" }, { "id": "img-8", "path": "Chapter3/EQ-1122-133522.png", "position": "answer", "context": "∴ log =" }, { "id": "img-9", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-087.png", "position": "answer", "context": "∴ log = 0.1070" }, { "id": "img-10", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-087.png", "position": "answer", "context": "∴ = anti log0.1070 = 1.2794" }, { "id": "img-11", "path": "Chapter3/EQ-1122-133607.png", "position": "answer", "context": "∴ [R] t = = 0.7816 µg 90 Sr is left after 10 years." }, { "id": "img-12", "path": "Chapter3/EQ-1122-133412.png", "position": "answer", "context": "t = log" }, { "id": "img-13", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-183.png", "position": "answer", "context": "t = log" }, { "id": "img-14", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-084.png", "position": "answer", "context": "∴ log =" }, { "id": "img-15", "path": "Chapter3/EQ-1122-133705.png", "position": "answer", "context": "∴ log =" }, { "id": "img-16", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-084.png", "position": "answer", "context": "∴ log =" }, { "id": "img-17", "path": "Chapter3/EQ-1122-133732.png", "position": "answer", "context": "∴ log =" }, { "id": "img-18", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-087.png", "position": "answer", "context": "∴ log = 0.6425" }, { "id": "img-19", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-087.png", "position": "answer", "context": "∴ = antilog 0.6425 = 4.39" }, { "id": "img-20", "path": "Chapter3/EQ-1122-133807.png", "position": "answer", "context": "∴ [R] t = = 0.228 µg 90 Sr is left after 60 years." } ], "tables": [], "metadata": { "source_class": "Quest", "marks": 3, "source_number": "17", "original_raw_text": " During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1 mg of 90Sr was absorbed in the bones of a newly born baby instead of calcium. How much of it will remain after 10 years and 60 years if it is not lost metabolically. ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 45, "question_text": "
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
", "answer_text": "
For first order reaction,
\n
t99% = \"\" log\"\"\t\t... ... (1)
\n
t90% = \"\" log\"\" \t\t... ... (2)
\n
Taking ratio of equation (1) and (2)
\n
\"\" = \"\"
\n
\t∴ \"\" = \"\"
\n
\t∴ \"\" = \"\"
\n
\t∴ \"\" = \"\"
\n
\t∴ \"\" = 2
\n
\t∴ t99% = 2 × t90%
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-225005.png", "position": "answer", "context": "t 99% = log ... ... (1)" }, { "id": "img-1", "path": "Chapter3/EQ-1122-133855.png", "position": "answer", "context": "t 99% = log ... ... (1)" }, { "id": "img-2", "path": "Chapter3/EQ-0124-225005.png", "position": "answer", "context": "t 90% = log ... ... (2)" }, { "id": "img-3", "path": "Chapter3/EQ-1122-133943.png", "position": "answer", "context": "t 90% = log ... ... (2)" }, { "id": "img-4", "path": "Chapter3/EQ-0124-225919.png", "position": "answer", "context": "=" }, { "id": "img-5", "path": "Chapter3/EQ-0124-225937.png", "position": "answer", "context": "=" }, { "id": "img-6", "path": "Chapter3/EQ-0124-225919.png", "position": "answer", "context": "∴ =" }, { "id": "img-7", "path": "Chapter3/EQ-1122-134145.png", "position": "answer", "context": "∴ =" }, { "id": "img-8", "path": "Chapter3/EQ-0124-225919.png", "position": "answer", "context": "∴ =" }, { "id": "img-9", "path": "Chapter3/EQ-1122-134233.png", "position": "answer", "context": "∴ =" }, { "id": "img-10", "path": "Chapter3/EQ-1122-134006.png", "position": "answer", "context": "∴ =" }, { "id": "img-11", "path": "Chapter3/EQ-1122-134252.png", "position": "answer", "context": "∴ =" }, { "id": "img-12", "path": "Chapter3/EQ-0124-225919.png", "position": "answer", "context": "∴ = 2" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 3, "source_number": "18", "original_raw_text": " For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 46, "question_text": "
A first order reaction takes 40 min for 30% decomposition. Calculate the half-life period for the reaction \"\".
", "answer_text": "
uppose initial concentration [R]0 = 100
\n
∴ Concentration at time t [R]t = 70, (30% decomposition).
\n
t = 40 min.
\n
Comparing k = \"\" log\"\" and
\n
k = \"\" we get,
\n
\"\" = \"\" log\"\"
\n
∴ \"\" = \"\" log\"\" = \"\" × 0.1549
\n
∴ \"\" = 0.0089185
\n
∴ \"\" = \"\" = 77.7 min
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0119-180531.png", "position": "stem", "context": "Q.19. A first order reaction takes 40 min for 30% decomposit…" }, { "id": "img-1", "path": "Chapter3/EQ-0124-223910.png", "position": "answer", "context": "Comparing k = log and" }, { "id": "img-2", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-183.png", "position": "answer", "context": "Comparing k = log and" }, { "id": "img-3", "path": "Chapter3/EQ-0124-230123.png", "position": "answer", "context": "k = we get," }, { "id": "img-4", "path": "Chapter3/EQ-1221-153826.png", "position": "answer", "context": "= log" }, { "id": "img-5", "path": "Chapter3/EQ-0124-223910.png", "position": "answer", "context": "= log" }, { "id": "img-6", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-183.png", "position": "answer", "context": "= log" }, { "id": "img-7", "path": "Chapter3/EQ-1221-153826.png", "position": "answer", "context": "∴ = log = × 0.1549" }, { "id": "img-8", "path": "Chapter3/EQ-1122-134654.png", "position": "answer", "context": "∴ = log = × 0.1549" }, { "id": "img-9", "path": "Chapter3/EQ-1122-134702.png", "position": "answer", "context": "∴ = log = × 0.1549" }, { "id": "img-10", "path": "Chapter3/EQ-1122-134654.png", "position": "answer", "context": "∴ = log = × 0.1549" }, { "id": "img-11", "path": "Chapter3/EQ-1221-153826.png", "position": "answer", "context": "∴ = 0.0089185" }, { "id": "img-12", "path": "Chapter3/EQ-1221-153922.png", "position": "answer", "context": "∴ = = 77.7 min" }, { "id": "img-13", "path": "Chapter3/EQ-0211-172742.png", "position": "answer", "context": "∴ = = 77.7 min" } ], "tables": [], "metadata": { "source_class": "Quest", "marks": 2, "source_number": "19", "original_raw_text": " A first order reaction takes 40 min for 30% decomposition. Calculate the half-life period for the reaction . ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 47, "question_text": "
For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.
\n
\tCalculate the rate constant.
", "answer_text": "
Decomposition reaction of azoisopropane,
\n
\t(CH3)2CHN = NCH(CH3)2 → C6H14(g) + N2(g)
\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Pressure at

\n

time t = 0

\n		\n

35 mm Hg

\n		\n

0

\n		\n

0

\n
\n

Change in

\n

pressure

\n		\n

– x

\n		\n

+x

\n		\n

+x

\n
\n

Pressure after

\n

t = 360 s

\n		\n

35 – x

\n		\n

+x

\n		\n

+x

\n
\n
\t∴ Total pressure after 360 s = 35 – x + x + x = 54 mm Hg
\n
\t∴ Pressure after t = 360 s;\t35 – 19 mm Hg\t\t\t= 16 mm Hg
\n
Rate constant at 360 s:
\n
Initial pressure [R]0 = pi = 35 mm Hg
\n
Pressure at time t [R]t = pt = 16 mm Hg
\n
For first order reaction,
\n
k\t= \"\" log\"\"
\n
\t\t= \"\" log\"\"
\n
\t\t= \"\" log2.1875
\n
\t\t= \"\" × 0.3399
\n
∴ k = 2.1747 × 10–3 sec–1
\n
k at t = 720 s:
\n
Total pressure at t = 720 s,
\n
Total pressure at equilibrium = 35 – x + x + x
\n
\t\t\t\t = 63 mm Hg
\n
Initial pressure at time t = 0 s, [R]0 = pi = 35 mm Hg
\n
Pressure at time t = 720 s, [R]t = pt = 35 – 28
\n
\t\t\t\t = 7 mm Hg
\n
For first order reaction,
\n
k\t= \"\" log\"\"
\n
\t\t= \"\" log\"\"
\n
\t\t= \"\" log5
\n
\t\t= \"\" × 0.6990
\n
\t∴ k = 2.21 × 10–3 sec–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-135806.png", "position": "answer", "context": "k = log" }, { "id": "img-1", "path": "Chapter3/EQ-0124-230428.png", "position": "answer", "context": "k = log" }, { "id": "img-2", "path": "Chapter3/EQ-1122-135827.png", "position": "answer", "context": "= log" }, { "id": "img-3", "path": "Chapter3/EQ-1122-135833.png", "position": "answer", "context": "= log" }, { "id": "img-4", "path": "Chapter3/EQ-1122-135827.png", "position": "answer", "context": "= log2.1875" }, { "id": "img-5", "path": "Chapter3/EQ-1122-135827.png", "position": "answer", "context": "= × 0.3399" }, { "id": "img-6", "path": "Chapter3/EQ-1122-135806.png", "position": "answer", "context": "k = log" }, { "id": "img-7", "path": "Chapter3/EQ-0124-230428.png", "position": "answer", "context": "k = log" }, { "id": "img-8", "path": "Chapter3/EQ-1122-140102.png", "position": "answer", "context": "= log" }, { "id": "img-9", "path": "Chapter3/EQ-1122-140108.png", "position": "answer", "context": "= log" }, { "id": "img-10", "path": "Chapter3/EQ-1122-140102.png", "position": "answer", "context": "= log5" }, { "id": "img-11", "path": "Chapter3/EQ-1122-140102.png", "position": "answer", "context": "= × 0.6990" } ], "tables": [ { "id": "tbl-1", "rows": 2, "columns": 4, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Time (sec)

\n		\n

0

\n		\n

360

\n		\n

720

\n
\n

P(mm Hg)

\n		\n

35.0

\n		\n

54.0

\n		\n

63.0

\n
", "position": "stem", "context": "Time (sec) 0 360 720 P(mm Hg) 35.0 54.0 63.0" } ], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "source_number": "20", "marks": 4, "original_raw_text": " For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.\n\tCalculate the rate constant. ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 48, "question_text": "
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.
\n
\tSO2Cl2(g) → SO2(g) + Cl2(g)
\n
Calculate the rate of reaction when total pressure is 0.65 atm.
", "answer_text": "
Decomposition reaction of SO2Cl2,
\n
\t\t\tSO2Cl2(g) →\tSO2(g)\t+\tCl2(g)
\n
Pressure at time t = 0 s\tpi atm\t0 atm\t0 atm
\n
Pressure at time t pi – x atm\tx atm\tx atm
\n
pi is initial pressure at time t = 0 s
\n
\t∴ Total pressure at time t,
\n
pt = pi – x + x + x = pi + x
\n
\t∴ x = pt – pi
\n
\t∴ p(SO2Cl2 ) = pi – x = pi – (pt – pi) = 2pi – pt atm.
\n
For experiment - 1
\n
At time t = 0 s, Initial pressure pi = 0.5atm
\n
\tAt time t = 100 s, p(SO2Cl2 ) = 2pi – pt
\n
\t\t\t\t= 2 × 0.5 – 0.6
\n
\t\t\t\t= 0.4 atm
\n
For first order reaction,
\n
\t∴ k = \"\" log \"\"
\n
\t\t\t= \"\" log \"\"
\n
\t\t\t= \"\" × 0.0969
\n
\t∴ k = 2.2318 × 10–3 sec–1
\n
For experiment-2
\n
At time t = 0 s, Initial pressure pi = 0.5 atm
\n
pSO2Cl2 = 2pi – pt = 2 × 0.5 – 0.65 = 0.35 atm
\n
\t∴ rate = k ∙ pSO2Cl2
\n
\t\t\t= 2.2318 × 0.35 × 10–3 sec–1
\n
\t\t\t= 0.78113 × 10–3
\n
\t\t\t= 7.8113 × 10–4 atm sec–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-223910.png", "position": "answer", "context": "∴ k = log" }, { "id": "img-1", "path": "Chapter3/EQ-0124-231158.png", "position": "answer", "context": "∴ k = log" }, { "id": "img-2", "path": "Chapter3/EQ-1122-141008.png", "position": "answer", "context": "= log" }, { "id": "img-3", "path": "Chapter3/EQ-1122-141015.png", "position": "answer", "context": "= log" }, { "id": "img-4", "path": "Chapter3/EQ-1122-141008.png", "position": "answer", "context": "= × 0.0969" } ], "tables": [ { "id": "tbl-1", "rows": 3, "columns": 3, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Experiment

\n		\n

Time/s–1

\n		\n

Total pressure/atm

\n
\n

1

\n		\n

0

\n		\n

0.5

\n
\n

2

\n		\n

100

\n		\n

0.6

\n
", "position": "stem", "context": "Experiment Time/s –1 Total pressure/atm 1 0 0.5 2 100 0.6" } ], "metadata": { "source_class": "Quest", "source_number": "21", "marks": 4, "original_raw_text": " The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.\n\tSO2Cl2(g) → SO2(g) + Cl2(g)\n Calculate the rate of reaction when total pressure is 0.65 atm.", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 49, "question_text": "
The rate constant for the decomposition of N2O5 at various temperatures is given below:
\n
Draw a graph between lnk and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.
", "answer_text": "
\"\"
\n
Ea from the graph:
\n
\tSlope = \"\"
\n
\t∴ Ea = – slope × R
\n
\t∴ Ea = × 8.314
\n
\t∴ Ea = 94523.47 J mol-1 = 94.52347 kJ mol–1
\n
Arrhenius constant from graph:
\n
Intercept of graph shows value of ln A.
\n
\tIntercept = 21.5
\n
\t∴ ln A = 21.5
\n
\t∴ 2.303 log A = 21.5
\n
\t∴ log A = \"\" = 9.3356
\n
\t∴ A = antilog 9.3356
\n
\t∴ A = 2.1657 × 109
\n
The rate constant at 30°C and 50°C temperature:
\n
\tT = 30°C = 303K
\n
\t∴ \"\" = \"\" = 3.3 × 10–3K–1
\n
From graph, at \"\" = 3.3 × 10–3K–1, ln k = 13.8
\n
\t∴ 2.303 log k = 13.8
\n
\t∴ log k = \"\" = 6
\n
\t∴ k = antilog 6
\n
∴ k = 1 × 106 s–1
\n
\tT = 50°C = 323K
\n
\t∴ \"\" = \"\" = 3.09 × 10–3K–1
\n
From graph, at \"\" = 3.09 × 10–3K–1, ln k = 16
\n
\t∴ 2.303 log k = 16
\n
\t∴ log k = \"\" = 6.9474
\n
\t∴ k = antilog 6.9474
\n
∴ k = 8.86 × 106 s-1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/90.png", "position": "answer", "context": "" }, { "id": "img-1", "path": "Chapter3/EQ-1122-144433.png", "position": "answer", "context": "Slope =" }, { "id": "img-2", "path": "Chapter3/EQ-1122-151604.png", "position": "answer", "context": "∴ log A = = 9.3356" }, { "id": "img-3", "path": "Chapter3/EQ-1122-151626.png", "position": "answer", "context": "∴ = = 3.3 × 10 –3 K –1" }, { "id": "img-4", "path": "Chapter3/EQ-1122-151637.png", "position": "answer", "context": "∴ = = 3.3 × 10 –3 K –1" }, { "id": "img-5", "path": "Chapter3/EQ-1122-151626.png", "position": "answer", "context": "From graph, at = 3.3 × 10 – 3 K –1 , ln k = 13.8" }, { "id": "img-6", "path": "Chapter3/EQ-1122-151710.png", "position": "answer", "context": "∴ log k = = 6" }, { "id": "img-7", "path": "Chapter3/EQ-1122-151626.png", "position": "answer", "context": "∴ = = 3.09 × 10 –3 K –1" }, { "id": "img-8", "path": "Chapter3/EQ-1122-151744.png", "position": "answer", "context": "∴ = = 3.09 × 10 –3 K –1" }, { "id": "img-9", "path": "Chapter3/EQ-1122-151626.png", "position": "answer", "context": "From graph, at = 3.09 × 10 –3 K –1 , ln k = 16" }, { "id": "img-10", "path": "Chapter3/EQ-1122-151810.png", "position": "answer", "context": "∴ log k = = 6.9474" } ], "tables": [ { "id": "tbl-1", "rows": 2, "columns": 6, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

T/°C

\n		\n

0

\n		\n

20

\n		\n

40

\n		\n

60

\n		\n

80

\n
\n

105 × k/s–1

\n		\n

0.0787

\n		\n

1.70

\n		\n

25.7

\n		\n

178

\n		\n

2140

\n
", "position": "stem", "context": "T/°C 0 20 40 60 80 10 5 × k/s –1 0.0787 1.70 25.7 178 2140" }, { "id": "tbl-2", "rows": 6, "columns": 5, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

T/°C

\n		\n

T/K

\n		\n

k/s–1

\n		\n

lnk

\n		\n

1/T

\n
\n

0

\n		\n

273

\n		\n

0.0787 × 105

\n

= 78700

\n		\n

11.27

\n		\n

3.663 × 10–3

\n
\n

20

\n		\n

293

\n		\n

1.70 × 105

\n

= 170000

\n		\n

12.04

\n		\n

3.412 × 10–3

\n
\n

40

\n		\n

313

\n		\n

25.7 × 105

\n

= 2570000

\n		\n

14.76

\n		\n

3.195 × 10–3

\n
\n

60

\n		\n

333

\n		\n

178 × 105

\n

= 17800000

\n		\n

16.69

\n		\n

3.003 × 10–3

\n
\n

80

\n		\n

353

\n		\n

2140 × 105

\n

= 214000000

\n		\n

19.18

\n		\n

2.832 × 10–3

\n
", "position": "stem", "context": "T/°C T/K k/s –1 lnk 1/T 0 273 0.0787 × 10 5 = 78700 11.27 3.…" } ], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-3", "source_number": "22", "marks": 4, "original_raw_text": " The rate constant for the decomposition of N2O5 at various temperatures is given below:\n Draw a graph between lnk and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 50, "question_text": "
The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5 s–1 546K. If the energy of activation is 179.9 kJ/mol. What will be the value of pre-exponential factor.
", "answer_text": "
Arrhenius equation,
\n
logk = logA – \"\"
\n
∴ logA = logk + \"\"
\n
\tHere,\tT = 546K, k = 2.418× 10–5 s–1
\n
\t\t\tEa = 179.9 kJ/mol = 179900 J/mol
\n
\t\t\tA = ?
\n
∴ logA = log2.418 × 10–5 + \"\"
\n
∴ logA = log2.418 + log 10–5 + 17.2082
\n
∴ logA = 0.3834 – 5.0 + 17.2082
\n
∴ logA = 12.5917
\n
∴ A = antilog12.5917 = 3.9057 × 1012 sec–1.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-151837.png", "position": "answer", "context": "log k = logA –" }, { "id": "img-1", "path": "Chapter3/EQ-1122-151837.png", "position": "answer", "context": "∴ logA = log k +" }, { "id": "img-2", "path": "Chapter3/EQ-1122-151916.png", "position": "answer", "context": "∴ logA = log2.418 × 10 –5 +" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 3, "source_number": "23", "original_raw_text": " The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5 s–1 546K. If the energy of activation is 179.9 kJ/mol. What will be the value of pre-exponential factor. ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 51, "question_text": "
Consider a certain reaction A → Products with
k = 2.0 × 10–2 s–1. Calculate the concentration of A remaining after 100 sec if the initial concentration of A is 1.0 mol L–1.
", "answer_text": "
k = 2.0 × 10–2 sec–1
\n
\t[R]0 = 1 mol L–1, t = 100 sec
\n
\t[R]t = ?
\n
k = \"\" log\"\"
\n
\t∴ log\"\" = \"\"
\n
\t∴ log\"\" = \"\"
\n
\t∴ log\"\" = 0.8684
\n
\t∴ \"\" = anti log0.8684 = 7.3863
\n
\t∴ [R]t = \"\" = \"\" = 0.1354 mol L–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-135806.png", "position": "answer", "context": "k = log" }, { "id": "img-1", "path": "Chapter3/EQ-0124-230859.png", "position": "answer", "context": "k = log" }, { "id": "img-2", "path": "Chapter3/EQ-0124-230859.png", "position": "answer", "context": "∴ log =" }, { "id": "img-3", "path": "Chapter3/EQ-0124-225806.png", "position": "answer", "context": "∴ log =" }, { "id": "img-4", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-084.png", "position": "answer", "context": "∴ log =" }, { "id": "img-5", "path": "Chapter3/EQ-1122-141257.png", "position": "answer", "context": "∴ log =" }, { "id": "img-6", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-084.png", "position": "answer", "context": "∴ log = 0.8684" }, { "id": "img-7", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-084.png", "position": "answer", "context": "∴ = anti log0.8684 = 7.3863" }, { "id": "img-8", "path": "Chapter3/EQ-0212-165812.png", "position": "answer", "context": "∴ [R] t = = = 0.1354 mol L –1" }, { "id": "img-9", "path": "Chapter3/EQ-1122-141403.png", "position": "answer", "context": "∴ [R] t = = = 0.1354 mol L –1" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "source_number": "24", "marks": 3, "original_raw_text": " Consider a certain reaction A → Products with k = 2.0 × 10–2 s–1. Calculate the concentration of A remaining after 100 sec if the initial concentration of A is 1.0 mol L–1.", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 52, "question_text": "
Sucrose decompose in acid solution into glucose and fructose according to the first order rate law, with \"\" = 3.00 hours. What reaction of sample of sucrose remains after
8 hours?
", "answer_text": "
For first order reaction,
\n
k = \"\" = \"\" = 0.231 hour–1
\n
Fraction of sample of sucrose remains after 8 hours:
\n
Taking initial concentration [R]0 = 1 M,
\n
Concentration after 8 hours [R]t = ?
\n
k = \"\" log\"\"
\n
\t∴ log\"\" = \"\"
\n
\t∴ log\"\" = \"\"
\n
\t∴ log\"\" = 0.8024
\n
\t∴ \"\" = antilog 0.8024 = 6.3445
\n
\t∴ [R]t = \"\" = \"\" = 0.1576 M
\n
\t∴ The fraction of sample of sucrose that remains after 8 hours is 0.1576 M.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0119-180531.png", "position": "stem", "context": "Q.25. Sucrose decompose in acid solution into glucose and fr…" }, { "id": "img-1", "path": "Chapter3/EQ-1221-154659.png", "position": "answer", "context": "k = = = 0.231 hour –1" }, { "id": "img-2", "path": "Chapter3/EQ-1122-141609.png", "position": "answer", "context": "k = = = 0.231 hour –1" }, { "id": "img-3", "path": "Chapter3/EQ-1122-135806.png", "position": "answer", "context": "k = log" }, { "id": "img-4", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-183.png", "position": "answer", "context": "k = log" }, { "id": "img-5", "path": "Chapter3/EQ-0124-230859.png", "position": "answer", "context": "∴ log =" }, { "id": "img-6", "path": "Chapter3/EQ-0124-225806.png", "position": "answer", "context": "∴ log =" }, { "id": "img-7", "path": "Chapter3/EQ-0124-230859.png", "position": "answer", "context": "∴ log =" }, { "id": "img-8", "path": "Chapter3/EQ-1122-141807.png", "position": "answer", "context": "∴ log =" }, { "id": "img-9", "path": "Chapter3/EQ-0124-230859.png", "position": "answer", "context": "∴ log = 0.8024" }, { "id": "img-10", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-183.png", "position": "answer", "context": "∴ = antilog 0.8024 = 6.3445" }, { "id": "img-11", "path": "Chapter3/3_2_Std._12_Sci_EM_CHEMISTRY_Chapter_3_Que_&_Ans_Anis-0203-101.png", "position": "answer", "context": "∴ [R] t = = = 0.1576 M" }, { "id": "img-12", "path": "Chapter3/EQ-1122-141906.png", "position": "answer", "context": "∴ [R] t = = = 0.1576 M" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 3, "source_number": "25", "original_raw_text": " Sucrose decompose in acid solution into glucose and fructose according to the first order rate law, with = 3.00 hours. What reaction of sample of sucrose remains after 8 hours? ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 53, "question_text": "
The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s–1) \"\". Calculate
activation energy Ea.
", "answer_text": "
Arrhenius equation,
\n
k = A\"\"\t\t\t... ... (I)
\n
k = 4.5 × 1011 s–1 \"\"\t... ... (II)
\n
Taking ratio of Eq. (I) and Eq. (II) we get,
\n
\"\"
\n
\tEa = 28000K × R
\n
\tEa = 28000K × 8.314 J K–1 mol–1
\n
\tEa = 232792 J mol-1 = 232.792 kJ mol–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0119-190823.png", "position": "stem", "context": "Q.26. The decomposition of hydrocarbon follows the equation …" }, { "id": "img-1", "path": "Chapter3/EQ-1122-142011.png", "position": "answer", "context": "k = A ... ... (I)" }, { "id": "img-2", "path": "Chapter3/EQ-1221-161522.png", "position": "answer", "context": "k = 4.5 × 10 11 s – 1 ... ... (II)" }, { "id": "img-3", "path": "Chapter3/EQ-1221-161531.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 3, "source_number": "26", "original_raw_text": " The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s–1) . Calculateactivation energy Ea. ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 54, "question_text": "
The rate constant for the first order decomposition of H2O2 is given by the following equation:
\n
\tlog k = 14.34 – 1.25 × 104 K/T
\n
\tCalculate Ea for this reaction and at what temperature will its half-period be 256 minutes ?
", "answer_text": "
Ea is the activation energy.
\n
According to Arrhenius equation,
\n
k = A\"\"
\n
\tlog k = logA – \"\" \t... ... (I)
\n
Now, log k = 14.34 – ... ... (II)
\n
Comparing Eq. (I) and Eq. (II) we get,
\n
\"\"
\n
\tEa = 1.25 × 104K × 2.303 × R
\n
\tEa = 1.25 × 104K × 2.303 × 8.314 J K–1 mol–1
\n
\tEa = 23.93 × 104 J mol–1 = 239.3 kJ mol–1
\n
For first order reaction
\n
k = \"\"
\n
k = \"\" = 4.51 × 10–5 s–1
\n
\tFor given equation,
\n
\tlog k = 14.34 –
\n
\tlog 4.51 × 10-5 = 14.34 –
\n
\t– 4.35 = 14.34 –
\n
= 14.34 + 4.35 = 18.69
\n
\tT = = 669K
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1122-152448.png", "position": "answer", "context": "k = A" }, { "id": "img-1", "path": "Chapter3/EQ-1122-151837.png", "position": "answer", "context": "log k = logA – ... ... (I)" }, { "id": "img-2", "path": "Chapter3/EQ-1221-161029.png", "position": "answer", "context": "" }, { "id": "img-3", "path": "Chapter3/EQ-1122-152742.png", "position": "answer", "context": "k =" }, { "id": "img-4", "path": "Chapter3/EQ-1122-152815.png", "position": "answer", "context": "k = = 4.51 × 10 –5 s –1" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-33", "source_number": "27", "marks": 4, "original_raw_text": " The rate constant for the first order decomposition of H2O2 is given by the following equation: \n\tlog k = 14.34 – 1.25 × 104 K/T\n\tCalculate Ea for this reaction and at what temperature will its half-period be 256 minutes ? ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 55, "question_text": "
The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104 s–1 ?
", "answer_text": "
A → product
\n
\tHere, T1 = 10°C = 283K
\n
k1 = 4.5 × 103 s–1
\n
\tEa = 60 kJ mol-1= 60000 J mol–1
\n
k2 = 1.5 × 104 s–1
\n
\tT2 = ?
\n
\tR = 8.314 J K–1 mol–1
\n
log\"\"
\n
\tlog
\n
\t0.5228 = \"\"
\n
= 0.0001668
\n
– 0.0001668
\n
\"\" = 0.003534 – 0.0001668 = 0.003367
\n
\tT2 = \"\" = 297.00 K
\n
\tT2 = 297 – 273 = 24°C
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-232335.png", "position": "answer", "context": "log" }, { "id": "img-1", "path": "Chapter3/EQ-1122-153316.png", "position": "answer", "context": "0.5228 =" }, { "id": "img-2", "path": "Chapter3/EQ-1122-153455.png", "position": "answer", "context": "= 0.003534 – 0.0001668 = 0.003367" }, { "id": "img-3", "path": "Chapter3/EQ-1122-154439.png", "position": "answer", "context": "T 2 = = 297.00 K" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 3, "source_number": "28", "original_raw_text": " The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104 s–1 ? ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 56, "question_text": "
The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010 s–1. Calculate k and activation energy (Ea)at 318K.
", "answer_text": "
(A) Calculation of Ea:
\n
(A)\tFor first order reaction, k = \"\" log \"\"
\n
At 298K temperature, k1 = \"\" log \"\"
... ... (I)
\n
At 308K temperature, k2 = \"\" log \"\"
\n
... ... (II)
\n
Dividing Eq. (II) by Eq.(I) we get,
\n
\"\" = 2.73
\n
According to Arrhenius equation,
\n
\tlog\"\"
\n
\tlog2.73 =
\n
\tEa =
\n
\tEa = 76640 J/mol = 76.640 kJ/mol
\n
(B)\tAccording to Arrhenius equation,
\n
\tlog k = logA – \"\"
\n
\tlog k = log4 × 1010 – \"\"
\n
\tlog k = 10.6021 – 12.5870 = –1.9849 = \"\".0151
\n
k = antilog\"\".0151
\n
k = 1.035 × 10–2 s–1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-223910.png", "position": "answer", "context": "(A)\tFor first order reaction, k = log" }, { "id": "img-1", "path": "Chapter3/EQ-0125-161301.png", "position": "answer", "context": "(A)\tFor first order reaction, k = log" }, { "id": "img-2", "path": "Chapter3/EQ-0124-223910.png", "position": "answer", "context": "At 298K temperature, k 1 = log ... ... (I)" }, { "id": "img-3", "path": "Chapter3/EQ-1122-154558.png", "position": "answer", "context": "At 298K temperature, k 1 = log ... ... (I)" }, { "id": "img-4", "path": "Chapter3/EQ-0124-223910.png", "position": "answer", "context": "At 308K temperature, k 2 = log" }, { "id": "img-5", "path": "Chapter3/EQ-1122-154627.png", "position": "answer", "context": "At 308K temperature, k 2 = log" }, { "id": "img-6", "path": "Chapter3/EQ-0124-233029.png", "position": "answer", "context": "= 2.73" }, { "id": "img-7", "path": "Chapter3/EQ-0124-233049.png", "position": "answer", "context": "log" }, { "id": "img-8", "path": "Chapter3/EQ-1122-151837.png", "position": "answer", "context": "log k = logA –" }, { "id": "img-9", "path": "Chapter3/EQ-1122-155152.png", "position": "answer", "context": "log k = log4 × 10 10 –" }, { "id": "img-10", "path": "Chapter3/EQ-1122-155211.png", "position": "answer", "context": "log k = 10.6021 – 12.5870 = –1.9849 = .0151" }, { "id": "img-11", "path": "Chapter3/EQ-1122-155211.png", "position": "answer", "context": "k = antilog .0151" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-31", "marks": 4, "source_number": "29", "original_raw_text": " The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010 s–1. Calculate k and activation energy (Ea)at 318K.", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } }, { "sequence_no": 57, "question_text": "
The rate of a reaction quadruples when the temperature changes from 293K to 313K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
", "answer_text": "
At T1 = 293K, rate constant = k1
\n
\tAt T2 = 313K, rate constant = k2 = 4k1
\n
\tR = 8.314 J K–1 mol–1
\n
According to Arrhenius equation,
\n
\tlog\"\"
\n
\tlog
\n
\tEa =
\n
\tEa =
\n
\tEa = 52863.33 J/mol = 52.863 kJ/mol
\n
Class 12 Chemistry (Part 1) 013
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0124-233049.png", "position": "answer", "context": "log" } ], "tables": [], "metadata": { "source_class": "Quest idf7b8abf61-ParaOverride-29", "marks": 3, "source_number": "30", "original_raw_text": " The rate of a reaction quadruples when the temperature changes from 293K to 313K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. ", "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Exercise Q & A", "section_id": "3.5" } } ] }, { "section_id": "3.7", "section_title": "Multiple Choice Questions (MCQs)", "heading_text": "Multiple Choice Questions (MCQs)", "detected_type": "MCQ", "match_confidence": 1.0, "questions": [ { "sequence_no": 58, "question_text": "
(i) \"\"
\n
\t(ii) \"\"
\n
\t(iii) \"\"
\n
1.\tDuring the decomposition of gas on the surface of a solid catalyst, at constant temperature the pressure of the gas at different time was observed to be as follows:
\n
Then, what would be an order of reaction ?
\n
2.\tIf concentration of reactant changes from 0.03 M to 0.02 M in 25 seconds, then average rate of that reaction will be ______.
\n
3.\tIn the reaction 2A + 3B → C + 4D, the rate of formation of product D is ______ times than that the rate of decrease in concentration of reactant B.
\n
4.\tThe units of rate of reaction are:
\n
5.\t2N2O5 4NO2 + O2
\n
For the above reaction which of the following is not correct about rate of reaction?
\n
6.\tOn addition of AgNO3 to NaCl, white ppt. occurs:
\n
7.\tObserve the following reaction,
\n
2A + B → C. The rate of formation of C is 2.2 × 10–3 mol L–1 min–1. What is the value of –\"\" (mol L–1 min–1) ?
\n
8.\tFor reaction 2A + B → 3C + D
\n
Which of the following does not express the reaction of rate?
\n
9.\tThe differential rate expression for the reaction H2 + I2 → 2HI is:
\n
10.\tIn the reaction
\n
\"\"(aq) + 5Br–(aq) + 6H+ → 3Br2(l) + 3H2O(l)
\n
The rate of appearance of bromine (Br2) is related to rate of disappearance of bromide ions as following:
\n
11.\tFor a reaction, A+ 2B → C, rate is given by +\"\" = k[A][B], hence, the order of the reaction is:
\n
12.\tThe rate of chemical reaction:
\n
13.\tFor a reaction \"\" A → 2B, rate of disappearance of 'A' is related to the rate of appearance of B by the expression:
\n
14.\tFor a chemical reaction 2X + Y → Z, the rate of appearance of Z is 0.05 mol L–1. The rate of disappearance of X will be:
\n
15.\tFor the reaction
\n
N2 + 3H2 2NH3
\n
The rate of change of concentration for hydrogen is 0.3 × 10–4 M s–1. The rates of change of concentration of ammonia is:
\n
16.\tThe reaction; N2O5 in 2NO2 + \"\" O2(g) is of first order for N2O5 with rate constant
6.2 × 10–4 s–1. What is the value of rate of reaction when [N2O5] = 1.25 mole L–1 ?
\n
17.\tFor the reaction N2 + 3H2 → 2NH3, the rate
\n
= 2 × 10–4 M s–1. Therefore, the rate
\n
– is given as:
\n
18.\tThe rate of a reaction is expressed in different ways as follows:
\n
+ = \"\" = +\"\" = \"\"
\n
The reaction is:
\n
19.\tFor the reaction, Cl2 + 2I– → I2 + 2Cl–, the initial concentration of I– was 0.20 mol L–1 and the concentration after 20 min was 0.18 mol L–1. Then the rate of formation of I2 in mol L–1 would be:
\n
20.\tFor the reaction, N2O5 → 2NO2 + \"\"O2
\n
Given, –\"\" = k1[N2O5]
\n
\"\" = k2[N2O5]
\n
and \"\" = k3[N2O5].
\n
The relation in between k1, k2 and k3 is:
\n
21.\tThe rate constant for the reaction,
\n
2N2O5 → 4NO2 + O2 is 3.0 × 10–5 s–1. If the rate is 2.40 × 10–5 then the concentration of N2O5 (in mol/L) is:
\n
22. Among the following reactions, the fastest
one is:
\n
23.\tFor the reaction A + 2B → C, the rate of reaction at a given instant can be given by:
\n
24.\tThe ionic reaction are usually very fast because:
\n
25.\tIn a reaction 2A → Products; the concentration of A decreases from 0.5 mol litre–1 to
0.4 mol litre–1 in 10 minute. The rate of reaction during this interval is:\t
\n
26.\tIn the synthesis of ammonia by Haber process, if 60 moles of ammonia is obtained in one hour, then the rate of disappearance of nitrogen is:
\n
27.\tThe rate constant of a first order reaction is
3 × 10–6 per second and initial concentration is 0.10 M. Then the initial rate of reaction is:
\n
28.\tFor the reaction N2(g) + 3H2(g) → 2NH3(g) under certain condition of temperature and partial pressure of the reactants, the rate of formation of NH3 is 10–3 kg hr–1. The rate of conversion of H2 under same condition is:
\n
29.\tWith increase in temperature, rate of reaction:
\n
30.\t\tIn a gaseous phase reaction:
\n
A2(g) → B(g) + (1/2)C(g), the increase in pressure from 100 mm to 120 mm is noticed in 5 minutes. The rate of disappearance of A2 in mm
min–1 is:
\n
31.\t\tThe branch of chemistry which deals with the reaction rates and reaction mechanism is called:
\n
32.\tFor the reaction A → B, if rate increase by 100 times on increasing concentration of reactant A from 0.1 M to 1 M. then what would be an order of reaction ?\t
\n
33.\tThe rate law for a reaction A(g) + B(g) → Product (takes place in a close vessel) is
rate = k[A]2[B]. If the volume of vessel becomes half then what effect will be observed in the rate of reaction?\t
\n
34. The value of rate constant for reaction
A + B → Product is 4.5 × 10-2 lit. mole-1 sec-1. If the concentration of A increase by two times the rate of reaction becomes half, then what will be increase in rate of reaction if concentration of B is doubled ?
\n
35.\tFor which type of reaction the order of reaction and molecularity are same?
\n
36.\tThe value of rate constant for a reaction is 2.75 sec–1, then the order of reaction is:
\n
37.\tFor reaction 2SO2(g) + O2(g) \"\" Products,
if pressure of gaseous components is increased by three times then reaction rate:
\n
38.\tIn the reaction A → B if the concentration of A is doubled then the reaction rate increases by 1.59 times then what will be the order of reaction?
\n
39. The value of rate constant for the reaction
A + B → Product is 1.25 × 10-2 \"\" sec-1. if the concentration of A is doubled the reaction rate increases by two times, then what will be increase in rate of reaction if the concentration of B is increased by four times?
\n
40.\tUnits of rate constant of first and zero order reactions in terms of molarity M unit are respectively:
\n
41.\tThe reaction 2A + B + C → D + E is found to be first order in A, second order in B and zero order in C. What is the effect on the rate of increasing concentration of A, B and C two times?
\n
42.\tThe rate of the reaction
\n
CCl3CHO + NO → CHCl3 + NO + CO is equal to rate k[CCl3CHO][NO]. If concentration is expressed in mol/L. The unit of k is:
\n
43.\tThe unit of rate constant of a third order chemical reaction is:
\n
44. The rate law for a reaction between the substances A and B is given by rate = k[A]n [B]m. On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as:
\n
45.\tFor the reaction H2(g) + Br2(g) → 2HBr(g). The experimental data suggest rate = k[H2][Br2]1/2. The molecularity and order of the reaction are respectively:
\n
46.\tThe following mechanism has been proposed for the reaction of NO with Br2 to form NOBr2 (NO(g) + Br2(g) NOBr2(g))
\n
\tNOBr2(g) + NO(g) → 2NOBr(g)
\n
If the second step is the rate determining step, the order of the reaction with respect to NO(g) is:
\n
47.\tA reaction involving two different reactants can never be:\t
\n
48.\tThe burning of coal represented by the equation; C(s) + O2(g) → CO2(g). The rate of this reaction is increased by:
\n
49.\tWhich of the following statement is incorrect about the molecularity of a reaction?
\n
50.\tFor a reaction A + B → Products, the rate of the reaction was doubled. When the concentration of A and B were doubled, the rate was again doubled, the order of the reaction with respect to A and B are:
\n
51.\tFor the reaction A → B, the rate expression is r = k[A]n. When the concentration of A is doubled, the rate of reaction is quadrupled. The value of n is:
\n
52.\tFor a chemical reaction, ______ can never be a fractional.
\n
53.\tIf the rate of reaction A → B doubles on increasing the concentration of A by 4 times, the order of the reaction is:
\n
54.\tThe rate constant for a chemical reaction has units L mol–1s–1, order of the reaction will be:
\n
55.\tWhich statement about molecularity of a reaction is wrong?
\n
56.\tIf the volume of the vessel in which the reaction 2NO + O2 → 2NO2 is occurring is diminished to 1/3rd of its initial volume. The rate of the reaction will be increased by:
\n
57.\tFor a reaction A + B → C + D, if the concentration of A is doubled without altering the concentration of B, the rate gets doubled. If the concentration of B is increased by nine times without altering the concentration of A, the rate gets tripled. The order of the reaction is:
\n
58.\tFor the reaction, 2N2O5(g) → 4NO2(g) + O2(g)
\n
If the concentration of NO2 increase by
5.2 × 10–3 M in 100 s then the rate of the reactions:
\n
59.\tThe rate of the reaction A → product, at the initial concentration of 3.24 × 10–2 M is nine times its rate at another initial concentration of 1.2 × 10–3 M. The order of the reaction is:
\n
60.\tConsider the reaction 2A + B → product
\n
When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is:
\n
61.\tThe rate of reaction between two reactants A and B decreases by a factor 4, if the concentration of reactant B is doubled, than the order of this reaction with respect
to B i:
\n
62.\tRate of reaction can be expressed by following rate expression, rate = k[A]2[B].
If concentration of A is increased by 3 times and concentration of B is increased by 2 times, how many times rate of reaction increases?
\n
63.\tThe data for the reaction, A + B → C;
\n
The rate law corresponds to the above data is:
\n
64.\tIn a reaction, when the concentration of reactant is increased two times, the increase in rate of reaction was four times. Order of reaction is:
\n
65.\tFor a reaction A + 2B → C, rate is given by
\n
r = k[A][B]2. The order of reaction is:
\n
66.\tDuring the kinetic study of the reaction
2A + B → C + D following results were obtained.
\n
\tOn the basis of above data which one is correct:
\n
67.\tFor the reaction A → B, when concentration of A is made 1.5 times, the rate of reaction becomes 1.837 times. The order of reaction is
\n
68. A reaction involving A, B and C as reactants is found to obey the rate law, rate = k[A]x[B]y[C]z. When the concentration of A, B and C are doubled separately, the rate is also found to increase two, zero and four times respectively. The overall order of the reaction is
\n
69.\tIn a chemical reaction two reactants take part. The rate of reaction is directly proportional to the concentration of one of them and inversely proportional to the concentration of the other. The order of reaction is:
\n
70.\t_________ of a reaction cannot be determined experimentally.
\n
71.\tWhat is the order of a reaction which has an expression rate = k[A]3/2[B]–1?
\n
72.\tThe rate of elementary reaction A → B increases by 100 times when the concentration of A is increased ten folds. The order of the reaction with respect to A is:
\n
73.\tThe order of a gaseous phase reaction for which rate becomes half if volume of container having same amount of reactant is doubled is:
\n
74.\tConsider a reaction; aG + bH → Products
\n
When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is:
\n
75.\tIn a reaction, the rate expression is,
\n
rate = K[A][B]2/3[C]0, the order of reaction is:
\n
76.\tWhich one of the following statements for the order of a reaction is incorrect?
\n
77.\tIf the rate of reaction between A and B is given by, rate = K[A][B]n, then the reaction is:
\n
78.\tIn a reaction, A + B → Product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of
8 when the concentrations of both the reactants
(A and B) are doubled, rate law for the reaction can be written as:
\n
79.\tFor a reaction between gaseous compounds,
\n
\t2A + B → C + D
\n
\tThe reaction rate = k[A][B]. If the volume of the container is made \"\" of the initial, then what will be the rate of reaction as compared to the initial rate?
\n
80.\tIn a reaction A + B → C, the rate expression is R = k[A][B]2. If the concentration of both the reaction is doubled at constant volume then the rate of the reaction will be:
\n
81.\tFor the reaction, 2A + B → products, the active mass of B is kept constant, and that of A is doubled. The rate of reaction will be then;
\n
82.\tFor the fourth order reaction, what is the unit of k?
\n
83.\tFor the reaction; 2N2O5 → 4NO2 + O2, rate and rate constant are 1.02 × 10–4 M sec–1 and
3.4 × 10–5 sec–1 respectively, then concentration of N2O5, at that time will be:
\n
84.\tThe rate of the elementary reaction,
2NO + O2 → 2NO2, when the volume of the reaction vessel is doubled:
\n
85.\tIf the concentration of reactants is increased by 'x' then rate constant k becomes.
\n
86.\tThe reaction X → product, follows first order kinetics. In 40 minutes, the concentration of X changes from 0.1 M to 0.025 M, then the rate of reaction when concentration of X is
0.01 M is:
\n
87.\tFor first order reaction, \"\" = _____?
\n
88.\tMention the value of slope in the graph of rate of reaction versus time for zero order reactions.
\n
89.\tBy which formula the value of \"\"can be found out for the first order reaction, where k = Rate Constant.
\n
90.\tIn the first order reaction R → P if the initial concentration of reactant is [R]0, then what would be the concentration of the reactant left after time \"\" ?\t
\n
91.\tFor the first order reaction what would be the value of Y, if t96% ÷ ty% = 2?
\n
92.\tWhat is the formula to find value of t1/2 for a zero order reaction?
\n
93.\tFor a first order reaction the graph log
[A] versus t is given below:
\n
\"\"
\n
x is equal to
\n
94.\tThe rate constant of a first order reaction is 4 × 10–3sec–1. At a reactant concentration of 0.02 M, the rate of reaction would be:
\n
95.\tThe rate of first order reaction, A → Products, is 7.5 × 10–4mol litre–1sec–1. If the concentration of A is 0.5 mol litre–1. The rate constant is:
\n
96.\tHalf-life period of a first order reaction is 1386 seconds. The specific rate constant of the reaction is:
\n
97.\tWhich is correct about zero order reaction?
\n
98.\tConsider the following statements, the rate law for the acid catalysed hydrolysis of an ester being given as: Rate = k[H+][ester] = k' [ester].
\n
\tIf the acid concentration is doubled at constant ester concentration:
\n
\tI.\tThe second order rate constant, k is doubled.
\n
II.\tThe pseudo first order rate constant, k is double.
\n
\tIII.\tThe rate of the reaction is doubled. Which of the above statements are correct?
\n
99.\tHalf-life of two samples is 0.1 and 0.8 s. Their respective concentration is 400 and 50 respectively. The order of reaction is:
\n
100.\tA reaction proceeds by first order, 75% of this reaction was completed in 32 min. The time required for 50% completion is:
\n
101.\tThe half-life period of a first order reaction is 1 min 40 s. Calculate its rate constant.
\n
102.\tThe halftime of a second order reaction is:
\n
103.\tAcid hydrolysis of sucrose is:
\n
104.\tMathematical expression for t1/4 i.e., when (1/4)th reaction is over following first order kinetics can be given by:
\n
105.\tThe rate of first order reaction is
1.5 × 10–2 mol L–1 min–1 at 0.5 M concentration of the reactant. The half-life of reaction is:
\n
106.\tWhich one is not correct?
\n
107.\tIn a first order reaction, the concentration of the reactant is decreased from 1.0 M to 0.25 M in 20 minute. The rate constant of the reaction would be:
\n
108.\t\tIn the reaction,
\n
\t2N2O5 → 4NO2 + O2 initial pressure is
500 atm and rate constant k is\t3.38 × 10–5s–1 after 10 min the final pressure of N2O5 is:
\n
109.\tHalf-life of a reaction is found to be inversely proportional to the cube of initial concentration. The order of reaction is:
\n
110.\tFor a zero order reaction:
\n
111.\tThe rate constant for the first order reaction is 60 s–1. How much time will it take to reduce the concentration of the reaction to
1/16 M value?
\n
112.\tThe unit and value of rate constant and that of rate of reaction are same for.
\n
113.\tThe time taken for the completion of 3/4 of a first order reaction is:
\n
114.\tA zero order reaction is one:
\n
115.\tFor zero order reaction, the integrated rate equation is:
\n
116.\tThe half-life period of a first order reaction is 69.3 s. What is the rate constant?
\n
117.\tA reactiotn has a rate constant of 0.5 mol–1 dm³ min–1. If initial concentration of the reactant is 0.2 mol dm–3, half-life of the reaction is:
\n
118. A first order reaction is 20% complete in 10 min. What is the rate constant of the reaction?
\n
119.\tThe rate constant of a zero order reaction is 0.2 mol dm–3 h–1. If the concentration of the reactant after 30 min is 0.05 mol dm–3. Then its initial concentration would be:
\n
120.\tFor a reaction, x(g) → y(g) + z(g) the half-life period is 10 min. in what period of time would the concentration of X be reduce to 10% of original concentration?
\n
121.\tFor the first order reaction with the rate constant k, which expression gives the rate half-life period? (Initial conc. = a)
\n
122.\tFor a given reaction of first order, it takes
15 minute for the concentration to drop from 0.8 M litre–1 to 0.4 M litre–1. The time required for the concentration to drop from 0.1 M litre–1 to 0.025 M litre–1 will be:
\n
123.\tAt 500K, the half-life period of a gaseous reaction at an initial pressure of 80 kPa is
350 s. When the pressure is 40 kPa, the
half-life period is 175 s. The order of the reaction is:
\n
124.\tThe half-life period for zero order reaction
A → product, is 100 min. How long will it take in 80% completion?
\n
125.\tIn a first order reaction the concentration of reactant decreases form 800 mol/dm6 to
50 mol/dm6 in 2 × 104 s. The rate constant of reaction in s–1 is:
\n
126.\tThe rate constant of a first order reaction whose half-life is 480 s is:
\n
127.\t2A → B + C; it would be a zero order
reaction when:
\n
128.\tThe plot between concentration versus time for zero order reaction is represented by:
\n
129.\tIn the first order reaction, the concentration of the reactants is reduced to 25% in one hour. The half-life period of the reaction is:
\n
130.\tFor a first order reaction, the initial concentration of a reactant is 0.05 M. After 45 min it is decreased by 0.015 M. Calculate half reaction time (\"\").
\n
131.\tIn the reaction A + B → products, if B is taken in excess, then it is an example of.
\n
132.\tThe ratio of the times for 99.9% of the reaction to complete and half of the reaction to complete is:
\n
133.\tAfter how many second will the concentration of the reactant in a first order reaction be halved if the rate constant is 1.155 × 10–3 s–1?
\n
134.\tThe inversion of cane sugar into glucose and fructose is:
\n
135.\tWhat is the two third life of a first order reaction having k = 5.48 × 10–14 s–1?
\n
136.\tOrder of radioactive disintegration reaction is:
\n
137.\tThe rate constant of a first order reaction is
6.9 × 10–3 s–1. How much time will it take
to reduce the initial concentration to its
1/8th value?
\n
138.\tThe \"\" of the first order reaction is.
\n
139.\t\tThe time required for 100% completion of a zero order reaction is.
\n
140.\tThe thermal decomposition of a compound is of first order. If a sample of the compound decompose 50% in 120 min, then, what time will it take to undergo 90% decomposition?
\n
141.\tFor a first order reaction, A → products, the rate of reaction at [A] = 0.2 M is 1.0 × 10–2 mol L–1 min–1. The half-life period for the reaction is:
\n
142.\tFor a first order reaction k = 100 s–1. The time for completion of 50% reaction is:
\n
143.\tThe rate constant for a first order reaction whose half-life is 480 s is:
\n
144.\tFor a reaction, the rate constant is 2.34 s–1. The half-life period for reaction is:
\n
145.\tIn a first order reaction A → B, if k is the rate constant initial concentration of the reactant is 0.5 M, then half-life is:
\n
146.\tDDT on exposure to water decomposes.
Half-life is 10 year. How much time it will take for its decomposition to 99%?
\n
147.\tA first order reaction is 20% complete in 10 min. Calculate the time for 75% completion of the reaction.
\n
148.\tThe half-life period of a first order chemical reaction is 6.93 min. The time required for the completion of 99% of the chemical reaction will be: (log2 = 0.302)
\n
149.\tA first order reaction has a rate constant 1.15 × 10–3 s–1. How long will 5g of this reactant take to reduce to 3 g?
\n
150.\tFor the reaction,
\n
\tA + 2B → Product.
\n
\tThe rate law is given by \"\" = K[A]2⋅[B].
If A is taken in large excess, the order of the reaction will be:
\n
151.\tIf rate of reaction doubles on increasing temperature by 10°C, then how many times the rate of reaction will be increased
if temperature of reaction increases to 50°C?
\n
152.\tWhich of the following is the correct relation for endothermic reaction?\t
\n
153.\tActivation energy of a reaction is:
\n
154.\tThe activation energy for a reaction is
9.0 kcal/mol. The increase in the rate constant when its temperature is increased from
298K to 308K is:
\n
155.\tIn Arrhenius plot intercept is equal to.
\n
156.\tThe activation energy of a reaction is zero. The rate constant for the reaction.
\n
157.\tAccording to the Arrhenius equation a straight line is to be obtained by plotting the logarithm of the rate constant of a chemical reaction (logk) against:
\n
158.\tIn respect of the equation k = Ae–Ea/RT in chemical kinetics, which one of the statement is correct?
\n
159.\tEffect of temperature on reaction rate is
given by:
\n
160.\tAn endothermic reaction A → B has an activation energy of 15 kcal/mol and the energy of reaction is 5 kcal/mol. The activation energy for the reaction B → A is:
\n
161.\tThe Arrhenius equation expressing the effect of temperature on the rate constant of reaction is:
\n
162.\tThe rate constant of a reaction at temperature 200K is 10 times less than the rate constant at 400K. What is the activation energy (Ea) of the reaction?
\n
163.\tActivation energy of a chemical reaction can be determined by:
\n
164.\tThe rate constants k1 and k2 for two different reactions are 1016 e–2000/T and 1015 e–1000/T, respectively. The temperature at which k1 = k2 is:
\n
165.\tThe activation energy of exothermic reaction A → B 80 kJ mol–1. The heat of reaction is 200 kJmol–1. The activation energy for the reaction B → A(in kJ mol–1) will be:
\n
166.\tTemperature coefficient of a reaction is 2. When temperature is increased from 30°C to 100°C, rate of the reaction increases by:
\n
167.\tThe slope in Arrhenius plot, is equal to.
\n
168.\tA first order reaction is 50% complete in 30 min at 27°C and in 10 min at 47°C. The energy of activation of the reaction is.
\n
169.\tWhat is the activation energy for the decomposition of N2O5 as,
\n
\tN2O5 2NO2 + \"\" O2,
\n
\tif the values of rate constant = 3.45 × 10–5 at 27°C and rate constant = 6.9 × 10–3 at 67°C?
\n
170.\tThe rate constant of a reaction increases by 5% when its temperature is raised from
27°C to 28°C. The activation energy of the reaction is
\n
171.\tTemperature dependent equation can be written as:
\n
172.\tWhen a graph is plotted between lnk and 1/T for a first order reaction, a straight line is obtained. The slope of the line is equal to:
\n
173.\tFor the two gaseous reactions, following data are given;
\n
\tA → B; k1 = 1010 e–20,000/T,
\n
\tC → D; k2 = 1012 e–24,606/T.
\n
The temperature at which k1 becomes equal to k2 is:
\n
174.\tFor a first order reaction A → P, the temperature (T) dependent rate constant (k) was found to follow the equation.
\n
logk = –(2000)/T + 6.0
\n
\tThe pre-exponential factor A and the activation energy Ea, respectively, are:
\n
175.\tIn Arrhenius equation k = Ae–Ea/RT, the quantity – Ea/RT is referred as:
\n
176.\tThe rate of a chemical reaction doubled for every 10°C rise in temperature. If the temperature is increased by 80°C the rate of reaction increases by:
\n
177.\tThe rate constant of a reaction is given by
k = 2.1 × 1010 exp(–2700/RT). It means that,
\n
178.\tThe activation energy of a reaction at a given temperature is found to be 2.303 RT J mol–1. The ratio of rate constant to the Arrhenius factor is:
\n
179.\tChemical reactions with very high Ea values are generally:
\n
180.\tThe rate constant is doubled when temperature increases from 27°C to 37°C. Activation energy in kJ is
\n
181.\tEffective collisions are those in which molecules must:
\n
182.\tIn a reaction, the threshold energy is equal to:
\n
183.\tCollision theory is applicable to:
\n
184.\tFor producing the effective collisions the colliding molecules must have:
\n
185.\tAccording to collision theory of reaction rates:
\n
186.\tThe minimum energy required for the reacting molecules to undergo reaction is:
\n
187.\tIncrease in the concentration of the reactants leads to the change in:
\n
188.\tWhich of the following theory is not related to chemical kinetics?
\n
189.\tThe minimum energy required for a molecule to take part in a reaction is called
\n
190.\tAssertion\t:\tThe rate of chemical reaction increases in the presence of catalyst.
\n
Reason :\tCatalyst lowers the activation energy of the reactants.
\n
191.\tAssertion\t:\tThe hydrolysis of ethyl acetate with aqueous HCI is a pseudo first-order process.
\n
Reason :\tHCl acts as a catalyst in the hydrolysis reaction.
\n
192. Assertion\t:\tMixing an aqueous solution of silver nitrate and sodium chloride immediately precipitates silver chloride.
\n
Reason :\tIonic reaction are very fast.
\n
193. Assertion\t:\tFor a first order reaction, \"\"is dependent on the rate constant.
\n
Reason :\tFor a first order process, \"\" ∝ [R]0.
\n
194. Assertion\t:\tRate of a reaction is the change in concentration
of reactant or product per unit time.
\n
Reason :\tRate of reaction remains constant throughout the reaction.
\n
195. Assertion\t:\tAll collisions of molecules lead to the formation of products.
\n
Reason :\tReactant molecules do not undergo chemical change regardless of every collision.
\n
196. Assertion\t:\tThe rate of a reaction never depends on the concentration.
\n
Reason :\tLower the activation energy, faster the reaction.
\n
197. Assertion\t:\tNot more than three molecularities are found.
\n
Reason :\tThe overall molecularity of a complex process is equal to the molecularity of the slowest step.
\n
198. Assertion\t:\tFor reaction CHCl3 + Cl2 → CCl4 + HCl, Rate = k[CHCl3] [Cl2]\"\".
\n
Reason :\tThe rate of a reaction is always equal to the sum of the stoichiometric coefficients of the reacting species in a balanced chemical equation.
\n
199. Assertion\t:\tFollowing reaction is first order reaction.
\n
C12H22O11 + H2O \"\" C6H12O6 + C6H12O6
\n
Sucrose Glucose Fructose
\n
Reason :\tChange in concentration of H2O is negated.
\n
200. Assertion\t:\tA catalyst increases the rate of a reaction without causing any permanent chemical change.
\n
Reason :\tA catalyst changes the free energy of a reaction and the equilibrium constant of the reaction.
\n
201. Assertion\t:\tA complex reaction takes place in different steps and the slowest step determines the rate of the reaction .
\n
Reason :\tOrder of reaction and molecularity are always the same .
\n
202. Assertion\t:\tWith respect to any reaction, the order of reaction can be zero, positive, or fractional.
\n
Reason :\tIncrease in concentration of reactants or products does not decrease the rate of the reaction.
\n
203. Assertion\t:\t10°C in temperature increases For a chemical reaction rate constant become almost doubles.
\n
Reason :\tt + 10°C, the fraction of molecules with energies equal to or greater than the activation energy doubles.
\n
204. Assertion\t:\tThe decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure.
\n
Reason :\tFor a zero order reaction , the rate of the reaction is independent of the initial concentration.
\n
205. Assertion\t:\tIn a reversible endothermic reaction the Ea of the forward reaction is greater than that of the backward reaction.
\n
Reason :\tIncreasing the temperature of a substance increases the fraction of molecules, which collide with energy greater than Ea.
\n
206. Assertion\t:\tIn a zero order reaction, if concentration of the reactant is doubled, half life period is also doubled.
\n
Reason :\tRate of reaction of zero order reaction is independent from concentration of the reactants.
\n
207.\tUse the appropriate symbols T (True) and
F (False) for the following statements regarding chemical equilibrium.
\n
(i) \tConcentration of reactant and product are constant.
\n
(ii)\tRate of forward and backward reaction are same.
\n
(iii) The reaction is moving fast.
\n
(iv) Factors like pressures and temperature are constant.
\n
208. Select the correct option using T (True) or F (False) symbol for the following statements regarding Elementary reaction:
\n
(i)\tIt is a multi-step reaction.
\n
(ii)\tIts rate constant is very small.
\n
(iii)\tIt has the same molecularity and order of reaction.
\n
(iv)\tIt is always endothermic . So it is a high temperature reaction .
\n
209. Select the correct option using T for (True)statements and F for (False) statements given below.
\n
(i)\tMolecularity means the sum of
mole-numbers of reactants in a balanced process.
\n
(ii)\tMolecularity means number of molecules associated with slow step of the reaction.
\n
(iii)\tMolecularity means order of reaction.
\n
(iv)\tMolecularity can be defined only for elementary reaction.
\n
210.\tChoose the correct option using T (True) and F (False) notation for the following statements:
\n
(i)\tRate = = k [A] [B]\"\" [C]\"\" Is rate equation for initial reaction.
\n
(ii)\tIonic reactions are instantaneous.
\n
(iii)\tThe unit of rate of reaction varies with the order of the reaction.
\n
(iv)\tEvery first order process is unimolecular.
\n
211. Choose T (True) or F (False) for the following statements.
\n
(i)\tMolecularity can be fractional.
\n
(ii)\tMolecularity and order can be 0, 1,2 and 3.
\n
(iii)\tFor a gaseous reaction the rate of reaction depends on the pressure of reactant or product.
\n
(iv)\tIntegral rate equation is more practical than differential rate equation for determining rate of reaction.
\n
212.\tChoose T (True) or F (False) for the following statements.
\n
(i)\tMolecularity is a theoretical derivation, while order of reaction is practical derivation.
\n
(ii)\tThe value of molecularity is a positive integer, while the value of order of reaction can be positive, negative or zero.
\n
(iii)\tThe molecularity explains the methodology of the reaction , whereas the order does not provide any information about the methodology.
\n
(iv)\tOrder of reaction and molecularity are same for elementary reaction.
\n
213.\tChoose T (True) or F (False) for the following statements depending on the order of reaction.
\n
(i)\tThe molecularity of a second order reaction can be two.
\n
(ii)\tThe value of the order of reaction can be positive, zero, negative or fractional and is determined experimentally.
\n
(iii)\tReaction of higher order are rare.
\n
(iv)\tThe order of reaction increases with increasing temperature.
\n
214.\tChoose T (True) or F (False) for the following statements related to the order of reaction:
\n
(i)\tOrder of reaction increases with increasing temperature.
\n
(ii)\tHigher order of reaction are rare.
\n
(iii)\tThe order of reaction depends on the concentration of the product.
\n
(iv)\tThe overall reaction order of
H2 + Br2 → 2HBr is 2.
\n
215.\tFor the reaction A2(g) + 3B2(g) → 2AB3(g) use notation T or F depending on the statements given below. (Molecular weight of A = 14, Molecular weight of B = 1)
\n
(i)\tWhen 7 g of A is used 17 g of AB3 is produced.
\n
(ii)\twhen 28 g of A is used 34 g of AB3 is produced.
\n
(iii)\t–\"\" \"\" = + \"\" \"\"
\n
(iv)\t– 2 \"\" = + \"\"
\n
216.\tChoose the correct option using T (True) and F (False) notation for the following statements.
\n
(i)\tThe cooking time of rice in a closed pressure cooker on a mountain top or at sea level is the same at both places.
\n
(ii)\tBornvita powder dissolves faster in milk than frozen Bornvita pieces.
\n
(iii)\tCatalysts reduce the amount of energy that has to be supplied externally to the reaction.
\n
(iv)\tThe rate of reaction is same whether the reactant is in liquid state or gas state.
\n
217.\tDetermine the statements True (T) and
False (F), for the first order reaction, \t
2N2O5 → 4NO2 + O2.
\n
(i)\tThe concentration of the reactant decreases exponentially with time.
\n
(ii)\tBoth the half-time of the reaction decreases with increasing temperature.
\n
(iii)\tThe half-time of the reaction depends on the initial concentration of the reactant.
\n
(iv)\tThe reaction proceeds to 99.6% completion in eight half-life duration.
\n
218.\tDetermine the True (T) and False (F) statements from the relationship between the rate of consumption of A and the rate of production of B for the reaction \"\"A → 2B.
\n
\t(i) – \"\" = \"\" (ii) – \"\" = \"\"
\n
(iii) – \"\" = \"\" \"\" (iv)–\"\"= \"\" \"\"
\n
219.\tWhen can nitrogen and oxygen combine? Determine the True (T) and False (F) statements for it.
\n
(i)\tA collision between N2 and O2 must occur.
\n
(ii)\tThe molecules do not need to have maximum total activation energy.
\n
(iii)\tThe molecules must have a minimum amount of kinetic energy.
\n
(iv)\tThe molecules must have the proper orientation.
\n
220.\tThe rate equation for the reaction 2A + B → C is found to be: rate = k [A] [B] determine the statements True (T) and False (F).
\n
(i)\tThe value of k is independent of the concentrations of A and B.
\n
(ii)\tUnit time of k is sec–1.
\n
(iii)\tThe half-life time of the reaction is constant.
\n
(iv)\tThe rate of production of B is twice the rate of consumption of A.
\n
Class 12 Chemistry (Part 1) 014
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": { "A": "TFFF", "B": "TFTF", "C": "TTFF", "D": "FTTT" }, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/142.png", "position": "stem", "context": "(i)" }, { "id": "img-1", "path": "Chapter3/143.png", "position": "stem", "context": "(ii)" }, { "id": "img-2", "path": "Chapter3/144.png", "position": "stem", "context": "(iii)" }, { "id": "img-3", "path": "Chapter3/EQ-1122-164529.png", "position": "option:A", "context": "(A) 6.67 ×10 -6 (B) 2.4 ×10 –3" }, { "id": "img-4", "path": "Chapter3/EQ-1122-164543.png", "position": "option:A", "context": "(A) 6.67 ×10 -6 (B) 2.4 ×10 –3" }, { "id": "img-5", "path": "Chapter3/EQ-1122-164529.png", "position": "option:C", "context": "(C) 2.4 ×10 –2 (D) 2.4 ×10 –3" }, { "id": "img-6", "path": "Chapter3/EQ-1122-164529.png", "position": "option:C", "context": "(C) 2.4 ×10 –2 (D) 2.4 ×10 –3" }, { "id": "img-7", "path": "Chapter3/EQ-1122-163231.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-8", "path": "Chapter3/EQ-1122-163238.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-9", "path": "Chapter3/EQ-1122-110043.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-10", "path": "Chapter3/EQ-1122-103945.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-11", "path": "Chapter3/EQ-0125-093918.png", "position": "option:A", "context": "(A) =" }, { "id": "img-12", "path": "Chapter3/EQ-0125-093953.png", "position": "option:B", "context": "(B) =" }, { "id": "img-13", "path": "Chapter3/EQ-0125-093953.png", "position": "option:C", "context": "(C) =" }, { "id": "img-14", "path": "Chapter3/EQ-0125-094043.png", "position": "option:C", "context": "(C) =" }, { "id": "img-15", "path": "Chapter3/EQ-0125-094134.png", "position": "option:D", "context": "(D) = =" }, { "id": "img-16", "path": "Chapter3/EQ-0125-094205.png", "position": "stem", "context": "2A + B → C. The rate of formation of C is 2.2 × 10 –3 mol L …" }, { "id": "img-17", "path": "Chapter3/EQ-0125-094235.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-18", "path": "Chapter3/EQ-0125-094317.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-19", "path": "Chapter3/EQ-0125-094330.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-20", "path": "Chapter3/EQ-0125-094343.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-21", "path": "Chapter3/EQ-0125-094357.png", "position": "option:A", "context": "(A) = =" }, { "id": "img-22", "path": "Chapter3/EQ-0125-094414.png", "position": "option:A", "context": "(A) = =" }, { "id": "img-23", "path": "Chapter3/EQ-0125-094427.png", "position": "option:A", "context": "(A) = =" }, { "id": "img-24", "path": "Chapter3/EQ-0125-094439.png", "position": "option:B", "context": "(B) = =" }, { "id": "img-25", "path": "Chapter3/EQ-0125-094452.png", "position": "option:B", "context": "(B) = =" }, { "id": "img-26", "path": "Chapter3/EQ-0125-094506.png", "position": "option:B", "context": "(B) = =" }, { "id": "img-27", "path": "Chapter3/EQ-0125-094520.png", "position": "option:C", "context": "(C) = =" }, { "id": "img-28", "path": "Chapter3/EQ-0125-094533.png", "position": "option:C", "context": "(C) = =" }, { "id": "img-29", "path": "Chapter3/EQ-0125-094506.png", "position": "option:C", "context": "(C) = =" }, { "id": "img-30", "path": "Chapter3/EQ-0125-094439.png", "position": "option:D", "context": "(D) –2 = –2 =" }, { "id": "img-31", "path": "Chapter3/EQ-0125-094452.png", "position": "option:D", "context": "(D) –2 = –2 =" }, { "id": "img-32", "path": "Chapter3/EQ-0125-094506.png", "position": "option:D", "context": "(D) –2 = –2 =" }, { "id": "img-33", "path": "Chapter3/EQ-1029-100559.png", "position": "stem", "context": "(aq) + 5Br – (aq) + 6H + → 3Br 2( l ) + 3H 2 O ( l )" }, { "id": "img-34", "path": "Chapter3/EQ-0125-094721.png", "position": "option:A", "context": "(A) =" }, { "id": "img-35", "path": "Chapter3/EQ-0125-094733.png", "position": "option:A", "context": "(A) =" }, { "id": "img-36", "path": "Chapter3/EQ-0125-094721.png", "position": "option:B", "context": "(B) =" }, { "id": "img-37", "path": "Chapter3/EQ-0125-094721.png", "position": "option:C", "context": "(C) =" }, { "id": "img-38", "path": "Chapter3/EQ-0125-094721.png", "position": "option:D", "context": "(D) =" }, { "id": "img-39", "path": "Chapter3/EQ-0125-094855.png", "position": "option:D", "context": "(D) =" }, { "id": "img-40", "path": "Chapter3/EQ-0125-094912.png", "position": "stem", "context": "11. For a reaction, A+ 2B → C, rate is given by + = k[A][B],…" }, { "id": "img-41", "path": "Chapter3/EQ-1029-100644.png", "position": "stem", "context": "13. For a reaction A → 2B, rate of disappearance of 'A' is r…" }, { "id": "img-42", "path": "Chapter3/EQ-0304-110127.png", "position": "option:A", "context": "(A) – = (B) – =" }, { "id": "img-43", "path": "Chapter3/EQ-0212-172120.png", "position": "option:A", "context": "(A) – = (B) – =" }, { "id": "img-44", "path": "Chapter3/EQ-0304-110155.png", "position": "option:A", "context": "(A) – = (B) – =" }, { "id": "img-45", "path": "Chapter3/EQ-0304-110127.png", "position": "option:A", "context": "(A) – = (B) – =" }, { "id": "img-46", "path": "Chapter3/EQ-0304-110259.png", "position": "option:A", "context": "(A) – = (B) – =" }, { "id": "img-47", "path": "Chapter3/EQ-0304-110155.png", "position": "option:A", "context": "(A) – = (B) – =" }, { "id": "img-48", "path": "Chapter3/EQ-0304-110127.png", "position": "option:C", "context": "(C) – = (D) – = 4" }, { "id": "img-49", "path": "Chapter3/EQ-0304-110155.png", "position": "option:C", "context": "(C) – = (D) – = 4" }, { "id": "img-50", "path": "Chapter3/EQ-0304-110127.png", "position": "option:C", "context": "(C) – = (D) – = 4" }, { "id": "img-51", "path": "Chapter3/EQ-0304-110155.png", "position": "option:C", "context": "(C) – = (D) – = 4" }, { "id": "img-52", "path": "Chapter3/EQ-0110-101419.png", "position": "stem", "context": "16. The reaction; N 2 O 5 in 2NO 2 + O 2 (g) is of first ord…" }, { "id": "img-53", "path": "Chapter3/EQ-1029-101848.png", "position": "stem", "context": "+ = = + =" }, { "id": "img-54", "path": "Chapter3/EQ-1029-101902.png", "position": "stem", "context": "+ = = + =" }, { "id": "img-55", "path": "Chapter3/EQ-1029-101916.png", "position": "stem", "context": "+ = = + =" }, { "id": "img-56", "path": "Chapter3/EQ-0212-172120.png", "position": "option:D", "context": "(D) B + D → 4A + 2C" }, { "id": "img-57", "path": "Chapter3/EQ-1029-100644.png", "position": "stem", "context": "20. For the reaction, N 2 O 5 → 2NO 2 + O 2" }, { "id": "img-58", "path": "Chapter3/EQ-1029-121018.png", "position": "stem", "context": "Given, – = k 1 [N 2 O 5 ]" }, { "id": "img-59", "path": "Chapter3/EQ-1029-121031.png", "position": "stem", "context": "= k 2 [N 2 O 5 ]" }, { "id": "img-60", "path": "Chapter3/EQ-1029-121041.png", "position": "stem", "context": "and = k 3 [N 2 O 5 ]." }, { "id": "img-61", "path": "Chapter3/EQ-0304-110127.png", "position": "option:A", "context": "(A) + = + = +" }, { "id": "img-62", "path": "Chapter3/EQ-0304-113829.png", "position": "option:A", "context": "(A) + = + = +" }, { "id": "img-63", "path": "Chapter3/EQ-0304-113843.png", "position": "option:A", "context": "(A) + = + = +" }, { "id": "img-64", "path": "Chapter3/EQ-0304-110127.png", "position": "option:B", "context": "(B) + = – = –" }, { "id": "img-65", "path": "Chapter3/EQ-0304-113829.png", "position": "option:B", "context": "(B) + = – = –" }, { "id": "img-66", "path": "Chapter3/EQ-0304-113843.png", "position": "option:B", "context": "(B) + = – = –" }, { "id": "img-67", "path": "Chapter3/EQ-0304-110127.png", "position": "option:C", "context": "(C) – = – = +" }, { "id": "img-68", "path": "Chapter3/EQ-0304-113829.png", "position": "option:C", "context": "(C) – = – = +" }, { "id": "img-69", "path": "Chapter3/EQ-0304-113843.png", "position": "option:C", "context": "(C) – = – = +" }, { "id": "img-70", "path": "Chapter3/EQ-0304-110127.png", "position": "option:D", "context": "(D) + = + = +" }, { "id": "img-71", "path": "Chapter3/EQ-0304-113829.png", "position": "option:D", "context": "(D) + = + = +" }, { "id": "img-72", "path": "Chapter3/EQ-0304-113843.png", "position": "option:D", "context": "(D) + = + = +" }, { "id": "img-73", "path": "Chapter3/EQ-1122-110043.png", "position": "option:A", "context": "(A) Becomes times\t\t(B) Becomes times" }, { "id": "img-74", "path": "Chapter3/EQ-1122-164227.png", "position": "option:A", "context": "(A) Becomes times\t\t(B) Becomes times" }, { "id": "img-75", "path": "Chapter3/EQ-1122-114653.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-76", "path": "Chapter3/EQ-1122-164017.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-77", "path": "Chapter3/EQ-1122-103945.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-78", "path": "Chapter3/EQ-1122-164029.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-79", "path": "Chapter3/EQ-1122-122337.png", "position": "stem", "context": "37. For reaction 2SO 2(g) + O 2 (g) Products, if pressure of…" }, { "id": "img-80", "path": "Chapter3/EQ-1122-163821.png", "position": "option:A", "context": "(A) (B) (C) (1.59) 2 (D) 1.59" }, { "id": "img-81", "path": "Chapter3/EQ-1122-103945.png", "position": "option:A", "context": "(A) (B) (C) (1.59) 2 (D) 1.59" }, { "id": "img-82", "path": "Chapter3/EQ-1029-121348.png", "position": "stem", "context": "39. The value of rate constant for the reaction A + B → Prod…" }, { "id": "img-83", "path": "Chapter3/EQ-0227-161728.png", "position": "option:A", "context": "(A) (B) ( m + n )" }, { "id": "img-84", "path": "Chapter3/EQ-0227-162252.png", "position": "option:A", "context": "(A) (B) 1, 1\t(C) (D) 2," }, { "id": "img-85", "path": "Chapter3/EQ-0227-162326.png", "position": "option:A", "context": "(A) (B) 1, 1\t(C) (D) 2," }, { "id": "img-86", "path": "Chapter3/EQ-0227-162347.png", "position": "option:A", "context": "(A) (B) 1, 1\t(C) (D) 2," }, { "id": "img-87", "path": "Chapter3/EQ-0212-172120.png", "position": "option:A", "context": "(A) 2\t(B) 1 (C) (D) 4" }, { "id": "img-88", "path": "Chapter3/EQ-0227-162347.png", "position": "option:A", "context": "(A) 2\t(B) 1 (C) (D)" }, { "id": "img-89", "path": "Chapter3/EQ-0304-121208.png", "position": "option:A", "context": "(A) 2\t(B) 1 (C) (D)" }, { "id": "img-90", "path": "Chapter3/EQ-0212-172120.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-91", "path": "Chapter3/EQ-0304-122058.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-92", "path": "Chapter3/EQ-0227-162347.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-93", "path": "Chapter3/EQ-0304-122117.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-94", "path": "Chapter3/148.png", "position": "option:C", "context": "(C) rate = [A][B] 3 (D) rate = k [A] 2 [B] 2" }, { "id": "img-95", "path": "Chapter3/EQ-0227-162347.png", "position": "option:A", "context": "(A) (B) Zero" }, { "id": "img-96", "path": "Chapter3/EQ-0212-172120.png", "position": "option:C", "context": "(C) (D) None of these" }, { "id": "img-97", "path": "Chapter3/EQ-0212-172120.png", "position": "option:A", "context": "(A) 1 (B) (C) 2\t(D)" }, { "id": "img-98", "path": "Chapter3/EQ-0304-125117.png", "position": "option:A", "context": "(A) 1 (B) (C) 2\t(D)" }, { "id": "img-99", "path": "Chapter3/EQ-0304-125341.png", "position": "option:A", "context": "(A) 1\t(B) 2 (C) (D) Zero" }, { "id": "img-100", "path": "Chapter3/EQ-1029-135612.png", "position": "stem", "context": "The reaction rate = k [A][B]. If the volume of the container…" }, { "id": "img-101", "path": "Chapter3/EQ-0304-133738.png", "position": "option:C", "context": "(C) times\t\t(D) times" }, { "id": "img-102", "path": "Chapter3/EQ-0304-133754.png", "position": "option:C", "context": "(C) times\t\t(D) times" }, { "id": "img-103", "path": "Chapter3/EQ-0110-104256.png", "position": "option:A", "context": "(A) s –1 (B) s –1" }, { "id": "img-104", "path": "Chapter3/EQ-0110-104308.png", "position": "option:A", "context": "(A) s –1 (B) s –1" }, { "id": "img-105", "path": "Chapter3/EQ-0110-104256.png", "position": "option:C", "context": "(C) s\t\t(D)" }, { "id": "img-106", "path": "Chapter3/EQ-0110-104256.png", "position": "option:C", "context": "(C) s\t\t(D)" }, { "id": "img-107", "path": "Chapter3/EQ-0304-134655.png", "position": "option:A", "context": "(A) In (B) (C) k + x (D) k" }, { "id": "img-108", "path": "Chapter3/EQ-0304-134655.png", "position": "option:A", "context": "(A) In (B) (C) k + x (D) k" }, { "id": "img-109", "path": "Chapter3/EQ-1122-164705.png", "position": "option:A", "context": "(A) 1.73 × 10 –4 (B) 3.47 × 10 –5" }, { "id": "img-110", "path": "Chapter3/EQ-1122-164705.png", "position": "option:A", "context": "(A) 1.73 × 10 –4 (B) 3.47 × 10 –5" }, { "id": "img-111", "path": "Chapter3/EQ-1122-164705.png", "position": "option:C", "context": "(C) 3.47 × 10 –4 (D) 1.73 × 10 –5" }, { "id": "img-112", "path": "Chapter3/EQ-1122-164705.png", "position": "option:C", "context": "(C) 3.47 × 10 –4 (D) 1.73 × 10 –5" }, { "id": "img-113", "path": "Chapter3/EQ-0125-095758.png", "position": "stem", "context": "87. For first order reaction, = _____?" }, { "id": "img-114", "path": "Chapter3/EQ-0110-104537.png", "position": "option:A", "context": "(A) k\t(B) (C) –2.303k (D) 0" }, { "id": "img-115", "path": "Chapter3/EQ-0129-164410.png", "position": "stem", "context": "89. By which formula the value of can be found out for the f…" }, { "id": "img-116", "path": "Chapter3/EQ-1221-170158.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-117", "path": "Chapter3/EQ-1221-170208.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-118", "path": "Chapter3/EQ-1221-170244.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-119", "path": "Chapter3/EQ-1221-170256.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-120", "path": "Chapter3/EQ-0129-164510.png", "position": "stem", "context": "90. In the first order reaction R → P if the initial concent…" }, { "id": "img-121", "path": "Chapter3/EQ-0220-110636.png", "position": "option:A", "context": "(A) (B)" }, { "id": "img-122", "path": "Chapter3/EQ-0220-110724.png", "position": "option:A", "context": "(A) (B)" }, { "id": "img-123", "path": "Chapter3/EQ-0220-110736.png", "position": "option:C", "context": "(C) (D)" }, { "id": "img-124", "path": "Chapter3/EQ-0227-165406.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-125", "path": "Chapter3/EQ-0227-165427.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-126", "path": "Chapter3/EQ-0227-165451.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-127", "path": "Chapter3/EQ-0227-165509.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-128", "path": "Chapter3/150.png", "position": "stem", "context": "" }, { "id": "img-129", "path": "Chapter3/EQ-0227-165509.png", "position": "option:A", "context": "(A) (B) (C) – (D) log [A] 0" }, { "id": "img-130", "path": "Chapter3/EQ-0227-165945.png", "position": "option:A", "context": "(A) (B) (C) – (D) log [A] 0" }, { "id": "img-131", "path": "Chapter3/EQ-0227-165945.png", "position": "option:A", "context": "(A) (B) (C) – (D) log [A] 0" }, { "id": "img-132", "path": "Chapter3/EQ-0228-094431.png", "position": "option:A", "context": "(A) t 1/4 = log4\t\t(B) t 1/4 = log2" }, { "id": "img-133", "path": "Chapter3/EQ-0228-094431.png", "position": "option:A", "context": "(A) t 1/4 = log4\t\t(B) t 1/4 = log2" }, { "id": "img-134", "path": "Chapter3/EQ-0228-094431.png", "position": "option:C", "context": "(C) t 1/4 = log (D) t 1/4 = log" }, { "id": "img-135", "path": "Chapter3/EQ-0304-121208.png", "position": "option:C", "context": "(C) t 1/4 = log (D) t 1/4 = log" }, { "id": "img-136", "path": "Chapter3/EQ-0228-094431.png", "position": "option:C", "context": "(C) t 1/4 = log (D) t 1/4 = log" }, { "id": "img-137", "path": "Chapter3/EQ-0304-122058.png", "position": "option:C", "context": "(C) t 1/4 = log (D) t 1/4 = log" }, { "id": "img-138", "path": "Chapter3/EQ-0228-102954.png", "position": "option:C", "context": "(C) t 1/2 α (D) t 1/2 α 1/" }, { "id": "img-139", "path": "Chapter3/EQ-0228-102954.png", "position": "option:C", "context": "(C) t 1/2 α (D) t 1/2 α 1/" }, { "id": "img-140", "path": "Chapter3/EQ-0228-110203.png", "position": "option:A", "context": "(A) kt = (B) kt = [A] – [A] 0" }, { "id": "img-141", "path": "Chapter3/EQ-0227-162347.png", "position": "option:A", "context": "(A) ka 2 (B) (C) (D)" }, { "id": "img-142", "path": "Chapter3/EQ-0304-140301.png", "position": "option:A", "context": "(A) ka 2 (B) (C) (D)" }, { "id": "img-143", "path": "Chapter3/EQ-0304-140313.png", "position": "option:A", "context": "(A) ka 2 (B) (C) (D)" }, { "id": "img-144", "path": "Chapter3/EQ-0227-165509.png", "position": "option:A", "context": "(A) ka 2 (B) (C) (D)" }, { "id": "img-145", "path": "Chapter3/151.png", "position": "option:A", "context": "(A) (B)" }, { "id": "img-146", "path": "Chapter3/152.png", "position": "option:A", "context": "(A) (B)" }, { "id": "img-147", "path": "Chapter3/153.png", "position": "option:C", "context": "(C) (D)" }, { "id": "img-148", "path": "Chapter3/154.png", "position": "option:C", "context": "(C) (D)" }, { "id": "img-149", "path": "Chapter3/EQ-0212-172120.png", "position": "option:A", "context": "(A) 2 h\t(B) 4 h (C) h (D) h" }, { "id": "img-150", "path": "Chapter3/EQ-0304-110259.png", "position": "option:A", "context": "(A) 2 h\t(B) 4 h (C) h (D) h" }, { "id": "img-151", "path": "Chapter3/EQ-0129-164734.png", "position": "stem", "context": "130. For a first order reaction, the initial concentration o…" }, { "id": "img-152", "path": "Chapter3/EQ-0129-164734.png", "position": "stem", "context": "138. The of the first order reaction is." }, { "id": "img-153", "path": "Chapter3/EQ-0304-151255.png", "position": "option:A", "context": "(A) ak (B) (C) (D)" }, { "id": "img-154", "path": "Chapter3/EQ-0304-151307.png", "position": "option:A", "context": "(A) ak (B) (C) (D)" }, { "id": "img-155", "path": "Chapter3/EQ-0304-151315.png", "position": "option:A", "context": "(A) ak (B) (C) (D)" }, { "id": "img-156", "path": "Chapter3/EQ-0110-113105.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-157", "path": "Chapter3/EQ-0110-113117.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-158", "path": "Chapter3/EQ-0110-113134.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-159", "path": "Chapter3/EQ-0110-113154.png", "position": "option:A", "context": "(A) (B) (C) (D)" }, { "id": "img-160", "path": "Chapter3/EQ-1029-140621.png", "position": "stem", "context": "The rate law is given by = K[A] 2 ⋅ [B]. If A is taken in la…" }, { "id": "img-161", "path": "Chapter3/EQ-0228-121505.png", "position": "option:A", "context": "(A) T\t(B) log T (C) (D) log" }, { "id": "img-162", "path": "Chapter3/EQ-0228-121505.png", "position": "option:A", "context": "(A) T\t(B) log T (C) (D) log" }, { "id": "img-163", "path": "Chapter3/EQ-0228-122716.png", "position": "option:A", "context": "(A) k = (B) k = Ae –E a /RT" }, { "id": "img-164", "path": "Chapter3/EQ-0228-122716.png", "position": "option:C", "context": "(C) k = log e (D) k = e –Ea/RT" }, { "id": "img-165", "path": "Chapter3/EQ-0304-162417.png", "position": "option:A", "context": "(A) K (B) 2000K" }, { "id": "img-166", "path": "Chapter3/EQ-0304-162436.png", "position": "option:C", "context": "(C) K (D) 1000K" }, { "id": "img-167", "path": "Chapter3/EQ-0304-162056.png", "position": "option:A", "context": "(A) – (B)" }, { "id": "img-168", "path": "Chapter3/EQ-0304-162106.png", "position": "option:A", "context": "(A) – (B)" }, { "id": "img-169", "path": "Chapter3/EQ-0304-162124.png", "position": "option:C", "context": "(C) – (D) None of the above" }, { "id": "img-170", "path": "Chapter3/EQ-1029-100644.png", "position": "stem", "context": "N 2 O 5  2NO 2 + O 2 ," }, { "id": "img-171", "path": "Chapter3/EQ-0304-165735.png", "position": "option:A", "context": "(A) – (B) –" }, { "id": "img-172", "path": "Chapter3/EQ-0304-162056.png", "position": "option:A", "context": "(A) – (B) –" }, { "id": "img-173", "path": "Chapter3/EQ-0304-165821.png", "position": "option:C", "context": "(C) – (D) –" }, { "id": "img-174", "path": "Chapter3/EQ-0304-162106.png", "position": "option:C", "context": "(C) – (D) –" }, { "id": "img-175", "path": "Chapter3/EQ-0228-121505.png", "position": "option:A", "context": "(A) logk vs will be a curved line with slope = ." }, { "id": "img-176", "path": "Chapter3/EQ-0304-171636.png", "position": "option:A", "context": "(A) logk vs will be a curved line with slope = ." }, { "id": "img-177", "path": "Chapter3/EQ-0228-121505.png", "position": "option:B", "context": "(B)\tlog k vs will be a straight line with intercept on log k…" }, { "id": "img-178", "path": "Chapter3/EQ-0110-123454.png", "position": "stem", "context": "193. Assertion\t:\tFor a first order reaction, is dependent on…" }, { "id": "img-179", "path": "Chapter3/EQ-0110-123454.png", "position": "stem", "context": "Reason : For a fir st o rder process, ∝ [R] 0 ." }, { "id": "img-180", "path": "Chapter3/EQ-1113-190133.png", "position": "stem", "context": "198. Assertion\t:\tFor reaction CHCl 3 + Cl 2 → CCl 4 + HCl , …" }, { "id": "img-181", "path": "Chapter3/EQ-1113-190449.png", "position": "stem", "context": "C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6" }, { "id": "img-182", "path": "Chapter3/EQ-1113-183648.png", "position": "stem", "context": "(i)\tRate = = k [A] [B] [C] Is rate equation for initial reac…" }, { "id": "img-183", "path": "Chapter3/EQ-1113-201357.png", "position": "stem", "context": "(i)\tRate = = k [A] [B] [C] Is rate equation for initial reac…" }, { "id": "img-184", "path": "Chapter3/EQ-1113-201931.png", "position": "stem", "context": "(iii) – = +" }, { "id": "img-185", "path": "Chapter3/EQ-1113-201957.png", "position": "stem", "context": "(iii) – = +" }, { "id": "img-186", "path": "Chapter3/EQ-0212-172120.png", "position": "stem", "context": "(iii) – = +" }, { "id": "img-187", "path": "Chapter3/EQ-1113-202036.png", "position": "stem", "context": "(iii) – = +" }, { "id": "img-188", "path": "Chapter3/EQ-1113-202138.png", "position": "stem", "context": "(iv) – 2 = +" }, { "id": "img-189", "path": "Chapter3/EQ-1113-202036.png", "position": "stem", "context": "(iv) – 2 = +" }, { "id": "img-190", "path": "Chapter3/EQ-0110-101419.png", "position": "stem", "context": "218. Determine the True (T) and False (F) statements from th…" }, { "id": "img-191", "path": "Chapter3/EQ-1113-214121.png", "position": "stem", "context": "(i) – = (ii) – =" }, { "id": "img-192", "path": "Chapter3/EQ-1113-214137.png", "position": "stem", "context": "(i) – = (ii) – =" }, { "id": "img-193", "path": "Chapter3/EQ-1113-214121.png", "position": "stem", "context": "(i) – = (ii) – =" }, { "id": "img-194", "path": "Chapter3/EQ-1113-214227.png", "position": "stem", "context": "(i) – = (ii) – =" }, { "id": "img-195", "path": "Chapter3/EQ-1113-214121.png", "position": "stem", "context": "(iii) – = (iv ) – =" }, { "id": "img-196", "path": "Chapter3/EQ-1029-100644.png", "position": "stem", "context": "(iii) – = (iv ) – =" }, { "id": "img-197", "path": "Chapter3/EQ-1113-214137.png", "position": "stem", "context": "(iii) – = (iv ) – =" }, { "id": "img-198", "path": "Chapter3/EQ-1113-214121.png", "position": "stem", "context": "(iii) – = (iv ) – =" }, { "id": "img-199", "path": "Chapter3/EQ-1029-135612.png", "position": "stem", "context": "(iii) – = (iv ) – =" }, { "id": "img-200", "path": "Chapter3/EQ-1113-214137.png", "position": "stem", "context": "(iii) – = (iv ) – =" } ], "tables": [ { "id": "tbl-1", "rows": 2, "columns": 5, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Time (sec)

\n		\n

0

\n		\n

100

\n		\n

200

\n		\n

300

\n
\n

Pressure (Pa)

\n		\n

4 × 103

\n		\n

3.5 × 103

\n		\n

3 × 103

\n		\n

2.5 × 103

\n
", "position": "stem", "context": "Time (sec) 0 100 200 300 Pressure (P a ) 4 × 10 3 3.5 × 10 3…" }, { "id": "tbl-2", "rows": 5, "columns": 4, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Exp.

\n		\n

[A]0

\n		\n

[B]0

\n		\n

Initial rate

\n
\n

1

\n		\n

0.012

\n		\n

0.035

\n		\n

0.10

\n
\n

2

\n		\n

0.024

\n		\n

0.070

\n		\n

0.80

\n
\n

3

\n		\n

0.024

\n		\n

0.035

\n		\n

0.10

\n
\n

4

\n		\n

0.012

\n		\n

0.070

\n		\n

0.80

\n
", "position": "stem", "context": "Exp. [A] 0 [B] 0 Initial rate 1 0.012 0.035 0.10 2 0.024 0.0…" }, { "id": "tbl-3", "rows": 5, "columns": 4, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Exp.

\n		\n

[A] in

\n

M

\n		\n

[B] in

\n

M

\n		\n

Initial rate of

\n

formation

\n

D in M s–1

\n
\n

I

\n		\n

0.1

\n		\n

0.1

\n		\n

6.0 × 10–3

\n
\n

II

\n		\n

0.3

\n		\n

0.2

\n		\n

7.2 × 10–2

\n
\n

III

\n		\n

0.3

\n		\n

0.4

\n		\n

2.88 × 10–1

\n
\n

IV

\n		\n

0.4

\n		\n

0.1

\n		\n

2.40 × 10–2

\n
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(B) 152. (A) 153. (C) 154. (D) 155. (B) 156. (D) 157. (C) 158. (C) 159. (B) 160. (C)
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(B) 162. (B) 163. (C) 164. (C) 165. (D) 166. (A) 167. (A) 168. (D) 169. (A) 170. (A)
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(B) 192. (A) 193. (C) 194. (C) 195. (D) 196. (D) 197. (B) 198. (C) 199. (A) 200. (C)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idf7b8abf61-ParaOverride-22", "source_number": "191", "original_raw_text": "(B) 192. (A) 193. (C) 194. (C) 195. (D) 196. (D) 197. (B) 198. (C) 199. (A) 200. (C)", "marks": 1, "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Based MCQs", "section_id": "3.9", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } }, { "sequence_no": 81, "question_text": "
(C) 202. (C) 203. (A) 204. (B) 205. (B) 206. (B) 207. (B) 208.\t(B) 209.\t(B) 210.\t(D)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idf7b8abf61-ParaOverride-22", "source_number": "201", "original_raw_text": "(C) 202. (C) 203. (A) 204. (B) 205. (B) 206. (B) 207. (B) 208.\t(B) 209.\t(B) 210.\t(D)", "marks": 1, "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Based MCQs", "section_id": "3.9", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } }, { "sequence_no": 82, "question_text": "
(B) 212.\t(A) 213.\t(A) 214.\t(A) 215.\t(A) 216.\t(C) 217.\t(A) 218.\t(B) 219.\t(A) 220.\t(A)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idf7b8abf61-ParaOverride-22", "source_number": "211", "original_raw_text": "(B) 212.\t(A) 213.\t(A) 214.\t(A) 215.\t(A) 216.\t(C) 217.\t(A) 218.\t(B) 219.\t(A) 220.\t(A)", "marks": 1, "deduction_level": 0, "source_html": "Chapter3", "section": "NCERT Textbook Based MCQs", "section_id": "3.9", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } } ] }, { "section_id": "3.12", "section_title": "Case Study Based Q & A", "heading_text": "Case Study Based Questions and Answers", "detected_type": "CS", "match_confidence": 0.69, "questions": [ { "sequence_no": 83, "question_text": "
Rate of reaction relates to the concentration of reactants and products changing per unit time. “Rate of a reaction is the change in concentration of reactants or products per unit time,”
\n
Rate = \"\"
\n
Average Rate: Average rate is taken as the ratio of change in total concentration to total time taken.
\n
Average Rate = \"\"
\n
Instantaneous Rate: The instantaneous rate of a reaction is the rate at a specific time. The instantaneous rate of a reaction is the ratio of the change in concentration over a number of small time periods.
\n
Instantaneous Rate = \"\"
\n
(a)\tFor the following reaction rate of appearance of I2 is 1.14 × 10–2. Then find out rate of disappearance of MnO–4.
\n
2MnO–4 + 10I– + 16H+ → 2Mn2+ + 5I2 + 8H2O
\n
(b)\tIn kinetic study of a chemical reaction, slopes are drawn at different times in the plot of concentration of reactants versus time. The magnitude of slopes with increase of time.
\n
(c)\t2NO(g) + O2(g) → 2NO2(g); If \"\" = 0.052 Ms–1 then, \"\" = _________.
\n
(d)\tIn the reaction 2A + B → A2 reactant B will disappear _________.
", "answer_text": "
(a)\t4.56 × 10–3 Ms–1
\n
\t2MnO–4 + 10I– + 16H+ → 2Mn2+ + 5I2 + 8H2O
\n
\t–\"\"\t= +\"\"
\n
\"\"\t= \"\"× (1.14 × 10–2)
\n
\"\"\t= 4.56 × 10–3 m/sec
\n
(b)\tDecrease
\n
\tAs the reaction progresses, the concentration of reactants decrease. So, the rate of reaction slow down over time.
\n
\tTherefore, the magnitude of slops decrease with the increase of time.
\n
(c)\t0.026 Ms–1
\n
–\"\"\t= –\"\" = +\"\"
\n
\"\"\t= \"\" × 0.052 ⇒ 0.026 m/sec
\n
(d)\tAt half-life the disappearance of rate of A
\n
2A + B → A2
\n
\t–\"\" = –\"\"
\n
\tReactant B will disappear half as fast as reactant A.
", "type": "CS", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/167.png", "position": "stem", "context": "Rate =" }, { "id": "img-1", "path": "Chapter3/168.png", "position": "stem", "context": "Average Rate =" }, { "id": "img-2", "path": "Chapter3/169.png", "position": "stem", "context": "Instantaneous Rate =" }, { "id": "img-3", "path": "Chapter3/EQ-0119-203324.png", "position": "stem", "context": "(c) 2NO ( g ) + O 2( g ) → 2NO 2( g ) ; If = 0.052 Ms –1 the…" }, { "id": "img-4", "path": "Chapter3/EQ-0119-203339.png", "position": "stem", "context": "(c) 2NO ( g ) + O 2( g ) → 2NO 2( g ) ; If = 0.052 Ms –1 the…" }, { "id": "img-5", "path": "Chapter3/EQ-0206-180516.png", "position": "answer", "context": "– = +" }, { "id": "img-6", "path": "Chapter3/EQ-0206-180457.png", "position": "answer", "context": "– = +" }, { "id": "img-7", "path": "Chapter3/EQ-0206-180612.png", "position": "answer", "context": "= × (1.14 × 10 –2 )" }, { "id": "img-8", "path": "Chapter3/EQ-0206-180632.png", "position": "answer", "context": "= × (1.14 × 10 –2 )" }, { "id": "img-9", "path": "Chapter3/EQ-0206-180612.png", "position": "answer", "context": "= 4.56 × 10 –3 m/sec" }, { "id": "img-10", "path": "Chapter3/EQ-0206-181501.png", "position": "answer", "context": "– = – = +" }, { "id": "img-11", "path": "Chapter3/EQ-0206-181533.png", "position": "answer", "context": "– = – = +" }, { "id": "img-12", "path": "Chapter3/EQ-0206-181605.png", "position": "answer", "context": "– = – = +" }, { "id": "img-13", "path": "Chapter3/EQ-0206-181533.png", "position": "answer", "context": "= × 0.052 ⇒ 0.026 m/sec" }, { "id": "img-14", "path": "Chapter3/EQ-0112-124037.png", "position": "answer", "context": "= × 0.052 ⇒ 0.026 m/sec" }, { "id": "img-15", "path": "Chapter3/EQ-0206-181958.png", "position": "answer", "context": "– = –" }, { "id": "img-16", "path": "Chapter3/EQ-0206-182028.png", "position": "answer", "context": "– = –" } ], "tables": [], "metadata": { "source_class": "Question_S2-to-S4_Que_Casestudy_01", "marks": 0, "deduction_level": 0, "original_raw_text": " Rate of reaction relates to the concentration of reactants and products changing per unit time. “Rate of a reaction is the change in concentration of reactants or products per unit time,”\nRate = \nAverage Rate: Average rate is taken as the ratio of change in total concentration to total time taken.\nAverage Rate = \nInstantaneous Rate: The instantaneous rate of a reaction is the rate at a specific time. The instantaneous rate of a reaction is the ratio of the change in concentration over a number of small time periods.\nInstantaneous Rate = \n(a)\tFor the following reaction rate of appearance of I2", "source_html": "Chapter3", "sub_questions": [ { "label": "(a)", "question": "For the following reaction rate of appearance of I 2 is 1.14 × 10 –2 . Then find out rate of disappearance of MnO – 4 .\n2MnO – 4 + 10I – + 16H + → 2Mn 2+ + 5I 2 + 8H 2 O", "answer": "4.56 × 10 –3 Ms –1\n2MnO – 4 + 10I – + 16H + → 2Mn 2+ + 5I 2 + 8H 2 O\n– = +\n= × (1.14 × 10 –2 )\n= 4.56 × 10 –3 m/sec" }, { "label": "(b)", "question": "In kinetic study of a chemical reaction, slopes are drawn at different times in the plot of concentration of reactants versus time. The magnitude of slopes with increase of time.", "answer": "Decrease\nAs the reaction progresses, the concentration of reactants decrease. So, the rate of reaction slow down over time.\nTherefore, the magnitude of slops decrease with the increase of time." }, { "label": "(c)", "question": "2NO ( g ) + O 2( g ) → 2NO 2( g ) ; If = 0.052 Ms –1 then, = _________ .", "answer": "0.026 Ms –1\n– = – = +\n= × 0.052 ⇒ 0.026 m/sec" }, { "label": "(d)", "question": "In the reaction 2A + B → A 2 reactant B will disappear _________.", "answer": "At half-life the disappearance of rate of A\n2A + B → A 2\n– = –\nReactant B will disappear half as fast as reactant A." } ], "section": "Case Study Based Q & A", "section_id": "3.12", "needs_review": true, "needs_optsplit_review": true, "optsplit_review_reason": "answer body is an option list (A/B/C/D) — options mis-routed into the answer; they belong to the question" } }, { "sequence_no": 84, "question_text": "
In Arrhenius equation K = A\"\", Ea is activation energy.
\n
According to the collision theory, a reaction occurs only when reactant molecules come close and collide with each other at the same time. During the collisions rearrangement of atoms takes place leading to the formation of products. The rearrangement of an atom involves the breaking of a bond and the formation of a new bond. Molecules rearrangements can only occur when the colliding molecules have energy equal to or greater than the minimum energy required for rearrangement. If the atom/molecule do not have this minimum energy, no rearrangement of molecules will occur and no product will be formed. The difference between the minimum energy required to rearrange molecule and the average energy of the reaction is called the activation energy (Ea). That is, the additional energy of a reactant molecule over and above the energy it already possesses to bring about a chemical change is called activation energy.
\n
(a)\tCollision frequency is _________ .
\n
(b)\tFor the process R → P the potential energy versus reaction axis is given. The enthalpy change of the reaction is ________ to energy corresponding.
\n
\"\"
\n
(c)\tActivation energy required for a reaction can be reduced by what?
\n
(d)\tWhich of the following statements is not true?
\n
(A)\tFor an endothermic reaction, the heat of reaction is less than the activation energy.
\n
(B) \tFor an exothermic reaction, the heat of reaction is greater than the activation energy.
\n
(C)\tFor an exothermic reaction, the activation energy of the forward reaction is lower than that of the backward reaction.
\n
(D) \tFor an endothermic process, the activation energy of the forward reaction is higher than that of the backward reaction.
", "answer_text": "
(a)\tCollision frequency (Z) is defined as the total number of collisions that occur per unit volume per unit time in a reaction mixture.
\n
(b)\tb – a
\n
Since the product energy (b) is lower than the reactant energy (a) in this specific graph, the reaction is exothermic, and ΔH would be negative.
\n
\tThe enthalpy change corresponds to the value (b−a).
\n
(c)\tAdding of catalyst
\n
\tThe activation energy required for a reaction can be reduced by adding a catalyst.
\n
(d)\t(B) For an exothermic reaction, the heat of reaction is greater than the activation energy.
\n
\tFor an exothermic reaction, the heat of reaction (ΔH) is the difference between reactants and products. It is not inherently greater than the activation energy.
", "type": "CS", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1113-185835.png", "position": "stem", "context": "In Arrhenius equation K = A , E a is activation energy." }, { "id": "img-1", "path": "Chapter3/171.png", "position": "stem", "context": "" } ], "tables": [], "metadata": { "source_class": "Question_S2-to-S4_Que_Casestudy_01", "marks": 0, "deduction_level": 0, "original_raw_text": "In Arrhenius equation K = A, Ea is activation energy.\nAccording to the collision theory, a reaction occurs only when reactant molecules come close and collide with each other at the same time. During the collisions rearrangement of atoms takes place leading to the formation of products. The rearrangement of an atom involves the breaking of a bond and the formation of a new bond. Molecules rearrangements can only occur when the colliding molecules have energy equal to or greater than the minimum energy required for rearrangement. If the atom/molecule do not have this minimum energy, no rearrang", "source_html": "Chapter3", "sub_questions": [ { "label": "(a)", "question": "Collision frequency is _________ .", "answer": "Collision frequency (Z) is defined as the total number of collisions that occur per unit volume per unit time in a reaction mixture." }, { "label": "(b)", "question": "For the process R → P the potential energy versus reaction axis is given. The enthalpy change of the reaction is ________ to energy corresponding.", "answer": "b – a\nSince the product energy (b) is lower than the reactant energy (a) in this specific graph, the reaction is exothermic, and Δ H would be negative.\nThe enthalpy change corresponds to the value (b−a)." }, { "label": "(c)", "question": "Activation energy required for a reaction can be reduced by what?", "answer": "Adding of catalyst\nThe activation energy required for a reaction can be reduced by adding a catalyst." }, { "label": "(d)", "question": "Which of the following statements is not true?", "answer": "(B) For an exothermic reaction, the heat of reaction is greater than the activation energy.\nFor an exothermic reaction, the heat of reaction ( Δ H) is the difference between reactants and products. It is not inherently greater than the activation energy." }, { "label": "(A)", "question": "For an endothermic reaction, the heat of reaction is less than the activation energy.", "answer": "" }, { "label": "(B)", "question": "For an exothermic reaction, the heat of reaction is greater than the activation energy.", "answer": "" }, { "label": "(C)", "question": "For an exothermic reaction, the activation energy of the forward reaction is lower than that of the backward reaction.", "answer": "" }, { "label": "(D)", "question": "For an endothermic process, the activation energy of the forward reaction is higher than that of the backward reaction.", "answer": "" } ], "section": "Case Study Based Q & A", "section_id": "3.12", "needs_review": true, "needs_optsplit_review": true, "optsplit_review_reason": "answer body is an option list (A/B/C/D) — options mis-routed into the answer; they belong to the question" } }, { "sequence_no": 85, "question_text": "
A pharmaceutical company is studying how the rate of drug decomposition changes with temperature. The drug breaks down following first order kinetics. At different temperatures, the following rate constants were measured:
\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Temperature (0°C)

\n		\n

Rate Constant k(s–1)

\n
\n

25°C

\n		\n

3.0 × 10–4

\n
\n

35°C

\n		\n

8.1 × 10–4

\n
\n
\tThe drug must remain stable during storage, so scientist want to predict:
\n
The activation energy (Ea) using the Arrhenius equation.
\n
The half-life of the drug at 25°C.
\n
Whether the drug decomposes significantly in 6 hours.
\n
How increasing temperature affects shelf life.
\n
\t\t(R = 8.314 J mol–1 K–1).
\n
(a) Using the Arrhenius equation, calculate the activation energy (Ea).
\n
(b) Calculate the half-life (t1/2) of the drug at 25°C.
\n
(c) What fractions of the drug will remain after 6 hours at 25°C?
\n
(d) Explain how an increase from 25°C to 35°C affects the drug's shelf life.
", "answer_text": "
(a) Arrhenius equation
\n
ln = \"\" =
\n
k1 = 3.0 × 10–4\tT1 = 298 K
\n
k2 = 8.1 × 10–4\tT2 = 308 K
\n
∴\tln (2.7)\t= \"\"
\n
∴ 0.993\t= \"\" (0.000109)
\n
∴ Ea\t= \"\"
\n
\t\t= 75.8 kJ / mol
\n
(b) t1/2\t= \"\" = \"\"
\n
\t= 2310 second
\n
\t= 38.5 minutes
\n
(c) Fraction remaining after 6 hours
\n
First - order: N\t\t= Noe–kt (t = 21600 sec)
\n
\t\tN\t= e–(3 × 10–4) (21600)
\n
\t\tN\t= e–6.48
\n
= 0.015
\n
\t\t\t= 0.15%
\n
(d) Rate increases from 3 × 10–4 to 8.1 × 10–4 (Almost 2.7 times).
", "type": "CS", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-1213-201907.png", "position": "answer", "context": "l n = =" }, { "id": "img-1", "path": "Chapter3/EQ-1213-202247.png", "position": "answer", "context": "∴ ln (2.7)\t=" }, { "id": "img-2", "path": "Chapter3/EQ-1213-211715.png", "position": "answer", "context": "∴ 0.993\t= (0.000109)" }, { "id": "img-3", "path": "Chapter3/EQ-1213-211909.png", "position": "answer", "context": "∴ E a =" }, { "id": "img-4", "path": "Chapter3/EQ-0227-165509.png", "position": "answer", "context": "(b) t 1/2 = =" }, { "id": "img-5", "path": "Chapter3/EQ-1213-212051.png", "position": "answer", "context": "(b) t 1/2 = =" } ], "tables": [], "metadata": { "source_class": "Question_S2-to-S4_Que_Casestudy_01", "marks": 0, "deduction_level": 0, "original_raw_text": " A pharmaceutical company is studying how the rate of drug decomposition changes with temperature. The drug breaks down following first order kinetics. At different temperatures, the following rate constants were measured:\n\n\n\n\n\n\n\n\nTemperature (0°C)\n\n\nRate Constant k(s–1)\n\n\n\n\n25°C\n\n\n3.0 × 10–4\n\n\n\n\n35°C\n\n\n8.1 × 10–4\n\n\n\n\n\tThe drug must remain stable during storage, so scientist want to predict:\nThe activation energy (Ea) using the Arrhenius equation.\nThe half-life of the drug at 25°C.\nWhether the drug decomposes significantly in 6 hours.\nHow increasing temperature affects shelf life.\n\t\t(R = 8.314", "source_html": "Chapter3", "sub_questions": [ { "label": "(a)", "question": "Using the Arrhenius equation, calculate the activation energy (E a ).", "answer": "Arrhenius equation\nl n = =\nk 1 = 3.0 × 10 –4 T 1 = 298 K\nk 2 = 8.1 × 10 –4 T 2 = 308 K\n∴ ln (2.7) =\n∴ 0.993 = (0.000109)\n∴ E a =\n= 75.8 kJ / mol" }, { "label": "(b)", "question": "Calculate the half-life (t 1/2 ) of the drug at 25°C.", "answer": "t 1/2 = =\n= 2310 second\n= 38.5 minutes" }, { "label": "(c)", "question": "What fractions of the drug will remain after 6 hours at 25°C?", "answer": "Fraction remaining after 6 hours\nFirst - order: N = Noe –kt (t = 21600 sec)\nN = e –(3 × 10 –4 ) (21600)\nN = e –6.48\n= 0.015\n= 0.15%" }, { "label": "(d)", "question": "Explain how an increase from 25°C to 35°C affects the drug's shelf life.", "answer": "Rate increases from 3 × 10 –4 to 8.1 × 10 –4 (Almost 2.7 times)." } ], "section": "Case Study Based Q & A", "section_id": "3.12", "needs_review": true, "needs_optsplit_review": true, "optsplit_review_reason": "answer body is an option list (A/B/C/D) — options mis-routed into the answer; they belong to the question" } }, { "sequence_no": 86, "question_text": "
\tHigher temperature accelerates decomposition, reducing shelf life drastically.
\n
A chemical industry studies the decomposition of N2O5, which follows first order kinetics:
\n
\t2N2O5 → 4NO2 + O2
\n
\tAt 25°C, the rate constant is 6.5 × 10–3s–1.
\n
\tSamples show the following data:
\n
Initial concentration: 0.50 M
\n
After t seconds, conc. reduces to 0.10 M
\n
\tThe industry aims to determine:
\n
\t1.\tTime required to reach 0.10 M.
\n
\t2.\tTotal gas evolved.
\n
\t3.\tWhether the reaction is feasible at low temperatures.
\n
\t4.\tImportance of first order kinetics in industrial reactors.
\n
(a) Calculate the time required for concentration to fall from 0.50 M to 0.10 M.
\n
(b) How many moles of O2 are formed from 0.50 mol of N2O5?
\n
(c) If temperature decrease, what happens to the rate constant?
\n
(d) State one industrial advantage of first order reactions.
", "answer_text": "
(a) t\t= \"\" ln\"\"
\n
\t= \"\" ln\"\"
\n
\t= 153.8 (1.609)
\n
t\t= 247.5 seconds.
\n
(b) 2 mole N2O5 → 1 mol O2
\n
∴ 0.50 mole N2O5 → 0.25 mole O2
\n
(c) Temp ↓ → Rate constant decreases exponentially (Arrhenius eq.)
\n
Reaction slow down drastically.
\n
(d) The half-life of first order reaction is independent of the initial reactant concentration. So, it is easy to control and predictable. \"\"
\n
Reaction Rate: Change in concentration of reactants or products per unit time.
\n
Average Rate: Rate calculated over a long-time interval from overall concentration change.
\n
Instantaneous Rate: Rate at a specific moment, obtained from the slope of tangent on a concentration–time curve.
\n
Rate Law: Experimentally derived expression relating rate to reactant concentrations.
\n
Rate Constant (k): Proportionality constant in the rate law; depends on temperature and catalyst.
\n
Order of Reaction: Sum of powers of concentration terms appearing in the rate law.
\n
Molecularity: Number of species colliding in a single elementary step to form products.
\n
Zero-Order Reaction: Reaction whose rate is independent of reactant concentration.
\n
First-Order Reaction: Reaction whose rate is proportional to the first power of one reactant.
\n
Half-Life (t½): Time required for the concentration of a reactant to become half of its initial value.
\n
Pseudo First-Order Reaction: Higher-order reaction that behaves like first order because one reactant is in excess.
\n
Activation Energy (Ea): Minimum energy required for reactants to form the activated complex.
\n
Activated Complex / Transition State: High-energy, unstable intermediate formed during reaction progression.
\n
Arrhenius Equation: Equation relating rate constant (k) with temperature and activation energy.
\n
Collision Theory: Reaction occurs only when molecules collide with sufficient energy and correct orientation.
\n
Steric Factor (P): Fraction of collisions with the proper orientation to form products.
\n
Reaction Coordinate: Path showing energy changes as reactants convert into products.
\n
Rate-Determining Step: Slowest step in a multi-step mechanism controlling total reaction rate.
\n
Exothermic Reaction: Reaction releasing heat where products have lower energy than reactants (ΔH < 0).
\n
Endothermic Reaction: Reaction absorbing heat where products have higher energy than reactants (ΔH > 0).
", "type": "CS", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter3/EQ-0304-140313.png", "position": "answer", "context": "(a) t = ln" }, { "id": "img-1", "path": "Chapter3/EQ-1213-213919.png", "position": "answer", "context": "(a) t = ln" }, { "id": "img-2", "path": "Chapter3/EQ-1213-214023.png", "position": "answer", "context": "= ln" }, { "id": "img-3", "path": "Chapter3/EQ-1213-214000.png", "position": "answer", "context": "= ln" }, { "id": "img-4", "path": "Chapter3/EQ-0206-183648.png", "position": "answer", "context": "(d) The half-life of f irst order reaction is independent of…" } ], "tables": [], "metadata": { "source_class": "Question_S2-to-S4_Que_Casestudy_01", "marks": 0, "deduction_level": 0, "original_raw_text": "\tHigher temperature accelerates decomposition, reducing shelf life drastically.\n A chemical industry studies the decomposition of N2O5, which follows first order kinetics:\n\t2N2O5 → 4NO2 + O2\n\tAt 25°C, the rate constant is 6.5 × 10–3s–1.\n\tSamples show the following data:\nInitial concentration: 0.50 M\nAfter t seconds, conc. reduces to 0.10 M\n\tThe industry aims to determine:\n\t1.\tTime required to reach 0.10 M.\n\t2.\tTotal gas evolved.\n\t3.\tWhether the reaction is feasible at low temperatures.\n\t4.\tImportance of first order kinetics in industrial reactors.\n(a) Calculate the time required for concentrat", "source_html": "Chapter3", "sub_questions": [ { "label": "(a)", "question": "Calculate the time required for concentration to fall from 0.50 M to 0.10 M.", "answer": "t = ln\n= ln\n= 153.8 (1.609)\nt = 247.5 seconds." }, { "label": "(b)", "question": "How many moles of O 2 are formed from 0.50 mol of N 2 O 5 ?", "answer": "2 mole N 2 O 5 → 1 mol O 2\n∴ 0.50 mole N 2 O 5 → 0.25 mole O 2" }, { "label": "(c)", "question": "If temperature decrease, what happens to the rate constant?", "answer": "Temp ↓ → Rate constant decreases exponentially (Arrhenius eq.)\nReaction slow down drastically." }, { "label": "(d)", "question": "State one industrial advantage of first order reactions.", "answer": "The half-life of f irst order reaction is independent of the initial reactant concentration. So, it is easy to control and predictable.\nReaction Rate: Change in concentration of reactants or products per unit time.\nAverage Rate: Rate calculated over a long-time interval from overall concentration change.\nInstantaneous Rate: Rate at a specific moment, obtained from the slope of tangent on a concentration–time curve.\nRate Law: Experimentally derived expression relating rate to reactant concentrations.\nRate Constant ( k ): Proportionality constant in the rate law; depends on temperature and catalyst.\nOrder of Reaction: Sum of powers of concentration terms appearing in the rate law.\nMolecularity: Number of species colliding in a single elementary step to form products.\nZero-Order Reaction: Reaction whose rate is independent of reactant concentration.\nFirst-Order Reaction: Reaction whose rate is proportional to the first power of one reactant.\nHalf-Life (t½): Time required for the concentration of a reactant to become half of its initial value.\nPseudo First-Order Reaction: Higher-order reaction that behaves like first order because one reactant is in excess.\nActivation Energy (E a ): Minimum energy required for reactants to form the activated complex.\nActivated Complex / Transition State: High-energy, unstable intermediate formed during reaction progression.\nArrhenius Equation: Equation relating rate constant ( k ) with temperature and activation energy.\nCollision Theory: Reaction occurs only when molecules collide with sufficient energy and correct orientation.\nSteric Factor (P): Fraction of collisions with the proper orientation to form products.\nReaction Coordinate: Path showing energy changes as reactants convert into products.\nRate-Determining Step: Slowest step in a multi-step mechanism controlling total reaction rate.\nExothermic Reaction: Reaction releasing heat where products have lower energy than reactants (ΔH Endothermic Reaction: Reaction absorbing heat where products have higher energy than reactants (ΔH > 0)." } ], 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{\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.33;\n\tmargin-bottom:0;\n\tmargin-left:48px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-48px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Normal-copy {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.3;\n\tmargin-bottom:0;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-28px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Basic-Paragraph {\n\tcolor:#000000;\n\tfont-family:\"Minion Pro\", serif;\n\tfont-size:12px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.2;\n\tmargin-bottom:0;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Quest {\n\t-epub-hyphens:none;\n\tcolor:#008bbf;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\tline-height:1.5;\n\tmargin-bottom:3px;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:9px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-28px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Quest_Black {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\tline-height:1.31;\n\tmargin-bottom:3px;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-28px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Question_S1-to-S4_Questions_S1-to-S4 {\n\t-epub-hyphens:none;\n\tcolor:#ef4d89;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:23px;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-23px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Question_S2-to-S4_Que_Casestudy_01 {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:23px;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-23px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Question_S2-to-S4_Questions_S2-to-S4 {\n\t-epub-hyphens:none;\n\tcolor:#00aeef;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:23px;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-23px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Spacing {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.5;\n\tmargin-bottom:0;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:4px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Summary-Style_Normal {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Summary-Style_spacing {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"B Bharati TitleLightTwo\", sans-serif;\n\tfont-size:12px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:6px;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\nli.body-text-copy {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.5;\n\tlist-style-position:outside;\n\tmargin-bottom:0;\n\tmargin-left:20px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\nspan.Biology-Character-Style-1_CaseStudy-Number {\n\tcolor:#ec008c;\n\tfont-family:\"Myriad Pro\", sans-serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Biology-Character-Style-1_English {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.Biology-Character-Style-1_Enlighs-Bold {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Biology-Character-Style-1_Ques-Number {\n\tcolor:#11aea7;\n\tfont-family:\"Myriad Pro\", sans-serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Character-Style-12 {\n\tcolor:#006cb7;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Character-Style-13-copy {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:italic;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-transform:none;\n}\nspan.Character-Style-2 {\n\tcolor:#00aeef;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:8px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-transform:none;\n}\nspan.Character-Style-5 {\n\tcolor:#939598;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-transform:none;\n}\nspan.Chemistry-Character-Style-1_English {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.Chemistry-Character-Style-1_Enlighs-Bold {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Chemistry-Character-Style-1_Ques-Number {\n\tcolor:#008bbf;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Chemistry-Character-Style-1_Ques-Number-2 {\n\tcolor:#ec008c;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.English {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.English-bold {\n\tfont-family:Calibri, sans-serif;\n\tfont-size:12px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Marks-Magenta {\n\tcolor:#ec008c;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:8px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Mind-Map-English-Bold {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Myriad-Pro {\n\tfont-family:\"Myriad Pro\", sans-serif;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Normal-English {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-transform:none;\n}\nspan.Normal-English-Bold {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\ttext-transform:none;\n}\nspan.Physics-Character-Style-1_Title-Two {\n\tcolor:#000000;\n\tfont-family:\"B Bharati TitleTwo\", sans-serif;\n\tfont-size:12px;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.Symbol {\n\tfont-family:Symbol, sans-serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-transform:none;\n}\ntable.TableOverride-1 {\n\tborder-collapse:collapse;\n}\ntable.TableOverride-2 {\n\tborder-collapse:collapse;\n\tmargin-left:28px;\n}\ntable.TableOverride-3 {\n\tborder-collapse:collapse;\n\tmargin-left:auto;\n}\ntable.TableOverride-4 {\n\tborder-collapse:collapse;\n\tmargin-left:auto;\n\tmargin-right:auto;\n}\ntable.TableOverride-5 {\n\tborder-collapse:collapse;\n\tborder-color:#ec008c;\n\tborder-width:1px;\n}\ntable.TableOverride-6 {\n\tborder-collapse:collapse;\n\tmargin-left:23px;\n}\ntable.TableOverride-7 {\n\tborder-collapse:collapse;\n\tborder-width:1px;\n\tmargin-bottom:-4px;\n\tmargin-top:4px;\n}\ntd.CellOverride-1 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n}\ntd.CellOverride-2 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tvertical-align:middle;\n}\ntd.CellOverride-3 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n}\ntd.CellOverride-4 {\n\tborder-bottom-color:#62b9dc;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#62b9dc;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#62b9dc;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#62b9dc;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n}\ntd.CellOverride-5 {\n\tborder-bottom-color:#62b9dc;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#62b9dc;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#62b9dc;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#62b9dc;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-6 {\n\tborder-bottom-color:#62b9dc;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#62b9dc;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#62b9dc;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#62b9dc;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n\tvertical-align:middle;\n}\ntd.CellOverride-7 {\n\tborder-bottom-color:#62b9dc;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#62b9dc;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#62b9dc;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#62b9dc;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tvertical-align:middle;\n}\ntd.CellOverride-8 {\n\tborder-bottom-color:#62b9dc;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#62b9dc;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#62b9dc;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#62b9dc;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-left:3px;\n\tpadding-right:0px;\n\tpadding-top:6px;\n}\ntd.CellOverride-9 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#76aa80;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#76aa80;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#76aa80;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#76aa80;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:4px;\n\tpadding-top:4px;\n\tvertical-align:middle;\n}\ntd.CellOverride-10 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#76aa80;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#76aa80;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#76aa80;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#76aa80;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:4px;\n\tpadding-top:4px;\n}\ntd.CellOverride-11 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#76aa80;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#76aa80;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#76aa80;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#76aa80;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:4px;\n\tpadding-top:4px;\n}\ntd.CellOverride-12 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#76aa80;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#76aa80;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#76aa80;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#76aa80;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:3px;\n\tpadding-top:3px;\n}\ntd.CellOverride-13 {\n\tborder-bottom-color:#62b9dc;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#62b9dc;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#62b9dc;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#62b9dc;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:1px;\n\tpadding-top:1px;\n\tvertical-align:middle;\n}\ntd.CellOverride-14 {\n\tborder-bottom-width:1px;\n\tborder-left-width:1px;\n\tborder-right-width:1px;\n\tborder-top-width:1px;\n}\ntd.CellOverride-15 {\n\tborder-bottom-width:1px;\n\tborder-left-width:1px;\n\tborder-right-width:1px;\n\tborder-top-width:1px;\n\tvertical-align:middle;\n}\ntd.CellOverride-16 {\n\tborder-bottom-color:#62b9dc;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#62b9dc;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#62b9dc;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#62b9dc;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-top:6px;\n}\ntd.CellOverride-17 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-18 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{\n\twidth:44px;\n}\ntr._idGenTableRowColumn-58 {\n\tmin-height:24px;\n}\ntr._idGenTableRowColumn-59 {\n\tmin-height:49px;\n}\ntr._idGenTableRowColumn-60 {\n\tmin-height:113px;\n}\ntr._idGenTableRowColumn-61 {\n\tmin-height:66px;\n}\ntr._idGenTableRowColumn-62 {\n\tmin-height:71px;\n}\ntr._idGenTableRowColumn-63 {\n\tmin-height:38px;\n}\ntr._idGenTableRowColumn-64 {\n\tmin-height:22px;\n}\ntr._idGenTableRowColumn-65 {\n\tmin-height:35px;\n}\ncol._idGenTableRowColumn-66 {\n\twidth:35px;\n}\ncol._idGenTableRowColumn-67 {\n\twidth:106px;\n}\ntr._idGenTableRowColumn-68 {\n\tmin-height:112px;\n}\ntr._idGenTableRowColumn-69 {\n\tmin-height:62px;\n}\ntr._idGenTableRowColumn-70 {\n\tmin-height:21px;\n}\ncol._idGenTableRowColumn-71 {\n\twidth:94px;\n}\ncol._idGenTableRowColumn-72 {\n\twidth:71px;\n}\ncol._idGenTableRowColumn-73 {\n\twidth:84px;\n}\ncol._idGenTableRowColumn-74 {\n\twidth:140px;\n}\ntr._idGenTableRowColumn-75 {\n\tmin-height:154px;\n}\ntr._idGenTableRowColumn-76 {\n\tmin-height:44px;\n}\ncol._idGenTableRowColumn-77 {\n\twidth:66px;\n}\ncol._idGenTableRowColumn-78 {\n\twidth:57px;\n}\ncol._idGenTableRowColumn-79 {\n\twidth:46px;\n}\ntr._idGenTableRowColumn-80 {\n\tmin-height:48px;\n}\ncol._idGenTableRowColumn-81 {\n\twidth:228px;\n}\ntr._idGenTableRowColumn-82 {\n\tmin-height:96px;\n}\ncol._idGenTableRowColumn-83 {\n\twidth:114px;\n}\ncol._idGenTableRowColumn-84 {\n\twidth:26px;\n}\ncol._idGenTableRowColumn-85 {\n\twidth:82px;\n}\ntr._idGenTableRowColumn-86 {\n\tmin-height:26px;\n}\ntr._idGenTableRowColumn-87 {\n\tmin-height:46px;\n}\ntr._idGenTableRowColumn-88 {\n\tmin-height:42px;\n}\ncol._idGenTableRowColumn-89 {\n\twidth:28px;\n}\ncol._idGenTableRowColumn-90 {\n\twidth:142px;\n}\ntr._idGenTableRowColumn-91 {\n\tmin-height:10px;\n}\ntr._idGenTableRowColumn-92 {\n\tmin-height:29px;\n}\ntr._idGenTableRowColumn-93 {\n\tmin-height:39px;\n}\ncol._idGenTableRowColumn-94 {\n\twidth:113px;\n}\ncol._idGenTableRowColumn-95 {\n\twidth:85px;\n}\ncol._idGenTableRowColumn-96 {\n\twidth:518px;\n}\n#_idContainer000, #_idContainer157, #_idContainer208, #_idContainer508, #_idContainer1280 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:37.98px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:518.74px;\n\tz-index:0;\n}\n#_idContainer001, #_idContainer158, #_idContainer209, #_idContainer509, #_idContainer1281 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:26.68px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:26.68px;\n\tz-index:0;\n}\n#_idContainer002, #_idContainer159, #_idContainer210, #_idContainer510, #_idContainer1282 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.46px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:17.58px;\n\tz-index:0;\n}\n#_idContainer003, #_idContainer160, #_idContainer211, #_idContainer511, #_idContainer1283 {\n\t-ms-transform:translate(3.244px,10.458px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(3.244px,10.458px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:9.41px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(3.244px,10.458px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(3.244px,10.458px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:11.09px;\n\tz-index:1;\n}\n#_idContainer004, #_idContainer161, #_idContainer212, #_idContainer512, #_idContainer1284 {\n\t-ms-transform:translate(6.062px,3.121px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(6.062px,3.121px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.46px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(6.062px,3.121px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(6.062px,3.121px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:17.58px;\n\tz-index:1;\n}\n#_idContainer005, #_idContainer162, #_idContainer213, #_idContainer513, #_idContainer1285 {\n\t-ms-transform:translate(0.000px,-0.001px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,-0.001px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,-0.001px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,-0.001px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:29.70px;\n\tz-index:1;\n}\n#_idContainer006 {\n\t-ms-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:267.58px;\n\tz-index:0;\n}\n#_idContainer007 {\n\t-ms-transform:translate(122.054px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(122.054px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(122.054px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(122.054px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:274.63px;\n\tz-index:1;\n}\n#_idContainer008, #_idContainer165, #_idContainer216, #_idContainer516, #_idContainer1288 {\n\tdisplay:inline-block;\n\theight:38px;\n\tposition:relative;\n\twidth:519px;\n}\n#_idContainer163 {\n\t-ms-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:284.30px;\n\tz-index:0;\n}\n#_idContainer164 {\n\t-ms-transform:translate(113.692px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(113.692px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(113.692px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(113.692px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:291.35px;\n\tz-index:1;\n}\n#_idContainer214 {\n\t-ms-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:366.13px;\n\tz-index:0;\n}\n#_idContainer215 {\n\t-ms-transform:translate(72.779px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(72.779px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(72.779px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(72.779px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:373.18px;\n\tz-index:1;\n}\n#_idContainer514 {\n\t-ms-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:264.10px;\n\tz-index:0;\n}\n#_idContainer515 {\n\t-ms-transform:translate(123.792px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(123.792px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(123.792px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(123.792px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:271.16px;\n\tz-index:1;\n}\n#_idContainer537 {\n\t-ms-transform:translate(0.000px,11.943px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,11.943px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:5.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,11.943px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,11.943px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.90px;\n\tz-index:11;\n}\n#_idContainer538 {\n\t-ms-transform:translate(0.000px,22.734px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,22.734px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:5.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,22.734px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,22.734px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.90px;\n\tz-index:12;\n}\n#_idContainer539 {\n\t-ms-transform:translate(0.000px,40.865px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,40.865px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:5.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,40.865px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,40.865px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.90px;\n\tz-index:13;\n}\n#_idContainer540 {\n\t-ms-transform:translate(0.000px,62.737px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,62.737px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:5.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,62.737px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,62.737px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.90px;\n\tz-index:14;\n}\n#_idContainer541 {\n\t-ms-transform:translate(0.000px,72.809px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,72.809px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:5.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,72.809px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,72.809px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.90px;\n\tz-index:15;\n}\n#_idContainer542 {\n\t-ms-transform:translate(3.675px,111.372px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(3.675px,111.372px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:6.20px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(3.675px,111.372px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(3.675px,111.372px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:7.69px;\n\tz-index:16;\n}\n#_idContainer543 {\n\t-ms-transform:translate(11.209px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(11.209px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 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{\n\t-ms-transform:translate(11.321px,43.638px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(11.321px,43.638px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:1.00px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(11.321px,43.638px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(11.321px,43.638px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:42.57px;\n\tz-index:3;\n}\n#_idContainer546 {\n\t-ms-transform:translate(11.468px,17.588px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(11.468px,17.588px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 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{\n\t-ms-transform:translate(11.468px,65.687px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(11.468px,65.687px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:1.00px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(11.468px,65.687px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(11.468px,65.687px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:66.13px;\n\tz-index:5;\n}\n#_idContainer549 {\n\t-ms-transform:translate(11.762px,74.896px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(11.762px,74.896px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:1.00px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(11.762px,74.896px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(11.762px,74.896px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:41.46px;\n\tz-index:6;\n}\n#_idContainer550 {\n\t-ms-transform:translate(15.072px,8.893px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(15.072px,8.893px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:75.72px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(15.072px,8.893px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(15.072px,8.893px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:71.13px;\n\tz-index:10;\n}\n#_idContainer551 {\n\t-ms-transform:translate(17.350px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(17.350px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:5.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(17.350px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(17.350px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.90px;\n\tz-index:17;\n}\n#_idContainer552 {\n\t-ms-transform:translate(23.898px,74.824px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(23.898px,74.824px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:35.40px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(23.898px,74.824px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(23.898px,74.824px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.43px;\n\tz-index:7;\n}\n#_idContainer553 {\n\t-ms-transform:translate(30.583px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(30.583px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:5.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(30.583px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(30.583px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.90px;\n\tz-index:18;\n}\n#_idContainer554 {\n\t-ms-transform:translate(33.014px,66.190px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(33.014px,66.190px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:44.03px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(33.014px,66.190px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(33.014px,66.190px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.43px;\n\tz-index:8;\n}\n#_idContainer555 {\n\t-ms-transform:translate(49.256px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(49.256px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:5.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(49.256px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(49.256px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.90px;\n\tz-index:19;\n}\n#_idContainer556 {\n\t-ms-transform:translate(53.305px,43.455px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(53.305px,43.455px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:66.77px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(53.305px,43.455px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(53.305px,43.455px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.43px;\n\tz-index:4;\n}\n#_idContainer557 {\n\t-ms-transform:translate(72.782px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(72.782px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:5.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(72.782px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(72.782px,111.660px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.90px;\n\tz-index:20;\n}\n#_idContainer558 {\n\tdisplay:inline-block;\n\theight:118px;\n\tposition:relative;\n\twidth:103px;\n}\n#_idContainer799 {\n\t-ms-transform:translate(0.000px,9.609px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,9.609px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:452.56px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,9.609px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,9.609px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:512.74px;\n\tz-index:0;\n}\n#_idContainer800 {\n\t-ms-transform:translate(7.370px,31.181px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.370px,31.181px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:433.31px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.370px,31.181px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.370px,31.181px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:504.00px;\n\tz-index:2;\n}\n#_idContainer801 {\n\t-ms-transform:translate(229.731px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(229.731px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:20.30px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(229.731px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(229.731px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:59.53px;\n\tz-index:1;\n}\n#_idContainer802 {\n\tdisplay:inline-block;\n\theight:468px;\n\tposition:relative;\n\twidth:519px;\n}\n#_idContainer803 {\n\t-ms-transform:translate(0.000px,11.175px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,11.175px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:0.00px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,11.175px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,11.175px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:517.74px;\n\tz-index:0;\n}\n#_idContainer804 {\n\t-ms-transform:translate(202.677px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(202.677px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:22.85px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(202.677px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(202.677px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:113.39px;\n\tz-index:1;\n}\n#_idContainer805 {\n\tdisplay:inline-block;\n\theight:23px;\n\tposition:relative;\n\twidth:519px;\n}\n#_idContainer1243 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:674.50px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:0;\n}\n#_idContainer1244 {\n\t-ms-transform:translate(177.982px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(177.982px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:674.50px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(177.982px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(177.982px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:1;\n}\n#_idContainer1245 {\n\tdisplay:inline-block;\n\theight:675px;\n\tposition:relative;\n\twidth:178px;\n}\n#_idContainer1246, #_idContainer1249, #_idContainer1252, #_idContainer1255, #_idContainer1258, #_idContainer1261, #_idContainer1265, #_idContainer1268, #_idContainer1271, #_idContainer1274, #_idContainer1277 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:707.66px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:0;\n}\n#_idContainer1247, #_idContainer1250, #_idContainer1253, #_idContainer1256, #_idContainer1259, #_idContainer1262, #_idContainer1266, #_idContainer1269, #_idContainer1272, #_idContainer1275, #_idContainer1278 {\n\t-ms-transform:translate(177.982px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(177.982px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:707.66px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(177.982px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(177.982px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:1;\n}\n#_idContainer1248, #_idContainer1251, #_idContainer1254, #_idContainer1257, #_idContainer1260, #_idContainer1263, #_idContainer1267, #_idContainer1270, #_idContainer1273, #_idContainer1276, #_idContainer1279 {\n\tdisplay:inline-block;\n\theight:709px;\n\tposition:relative;\n\twidth:178px;\n}\n#_idContainer1286 {\n\t-ms-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:313.22px;\n\tz-index:0;\n}\n#_idContainer1287 {\n\t-ms-transform:translate(99.236px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(99.236px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(99.236px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(99.236px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:320.27px;\n\tz-index:1;\n}\n#_idContainer1325 {\n\t-ms-transform:translate(0.000px,11.338px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,11.338px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:409.31px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,11.338px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,11.338px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:517.74px;\n\tz-index:0;\n}\n#_idContainer1326 {\n\t-ms-transform:translate(213.068px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(213.068px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(213.068px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(213.068px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:94.02px;\n\tz-index:1;\n}\n#_idContainer1327 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:17.69px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:17.69px;\n\tz-index:0;\n}\n#_idContainer1328 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:15.55px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:11.65px;\n\tz-index:0;\n}\n#_idContainer1329 {\n\t-ms-transform:translate(2.150px,6.932px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(2.150px,6.932px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:6.24px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(2.150px,6.932px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(2.150px,6.932px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:7.35px;\n\tz-index:1;\n}\n#_idContainer1330 {\n\t-ms-transform:translate(4.018px,2.069px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(4.018px,2.069px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:15.55px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(4.018px,2.069px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(4.018px,2.069px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:11.65px;\n\tz-index:1;\n}\n#_idContainer1331 {\n\t-ms-transform:translate(217.723px,2.334px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(217.723px,2.334px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:19.69px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(217.723px,2.334px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(217.723px,2.334px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:19.69px;\n\tz-index:2;\n}\n#_idContainer1332 {\n\t-ms-transform:translate(259.370px,40.113px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(259.370px,40.113px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:367.88px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(259.370px,40.113px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(259.370px,40.113px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:3;\n}\n#_idContainer1333 {\n\tdisplay:inline-block;\n\theight:422px;\n\tposition:relative;\n\twidth:519px;\n}\n#_idContainer1334 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:39.69px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:39.69px;\n\tz-index:0;\n}\n#_idContainer1335 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:27.44px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:27.25px;\n\tz-index:0;\n}\n#_idContainer1336 {\n\t-ms-transform:translate(11.049px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(11.049px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.26px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(11.049px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(11.049px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:10.15px;\n\tz-index:2;\n}\n#_idContainer1337 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.26px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:10.15px;\n\tz-index:0;\n}\n#_idContainer1338 {\n\t-ms-transform:translate(10.693px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(10.693px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:16.30px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(10.693px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(10.693px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.25px;\n\tz-index:3;\n}\n#_idContainer1339 {\n\t-ms-transform:translate(0.753px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.753px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:16.30px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.753px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.753px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.25px;\n\tz-index:1;\n}\n#_idContainer1340 {\n\t-ms-transform:translate(3.023px,2.091px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(3.023px,2.091px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.26px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(3.023px,2.091px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(3.023px,2.091px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:21.20px;\n\tz-index:1;\n}\n#_idContainer1341 {\n\t-ms-transform:translate(6.218px,5.984px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(6.218px,5.984px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:27.44px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(6.218px,5.984px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(6.218px,5.984px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:27.25px;\n\tz-index:1;\n}\n#_idContainer1342 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:39.69px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:39.69px;\n\tz-index:1;\n}\n#_idContainer1343 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:30.05px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:117.82px;\n\tz-index:0;\n}\n#_idContainer1344 {\n\t-ms-transform:translate(20.143px,6.586px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(20.143px,6.586px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:18.66px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(20.143px,6.586px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(20.143px,6.586px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:91.98px;\n\tz-index:1;\n}\n#_idContainer1345 {\n\t-ms-transform:translate(19.842px,9.637px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(19.842px,9.637px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:30.05px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(19.842px,9.637px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(19.842px,9.637px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:117.82px;\n\tz-index:0;\n}\n#_idContainer1346 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:40px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:138px;\n}\n#_idContainer1371 {\n\t-ms-transform:translate(0.000px,2.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,2.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:56.82px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,2.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,2.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:1;\n}\n#_idContainer1372 {\n\t-ms-transform:translate(-0.001px,0.839px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(-0.001px,0.839px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:20.16px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(-0.001px,0.839px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(-0.001px,0.839px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:0;\n}\n#_idContainer1373 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:22px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:58px;\n}\n#_idContainer1374 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:24px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1376 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:84px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer1377 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:9px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer1378 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:326px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer1380 {\n\t-ms-transform:translate(0.000px,12.959px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,12.959px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:17.44px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,12.959px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,12.959px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:1;\n}\n#_idContainer1381 {\n\t-ms-transform:translate(19.208px,12.959px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(19.208px,12.959px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:11.96px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(19.208px,12.959px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(19.208px,12.959px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:0;\n}\n#_idContainer1382 {\n\tdisplay:inline-block;\n\theight:13px;\n\tposition:relative;\n\twidth:19px;\n}\n#_idContainer1383 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:80px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1384 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:199px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer1385 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:209px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer1387 {\n\t-ms-transform:translate(0.000px,51.334px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,51.334px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:4.85px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,51.334px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,51.334px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:51.33px;\n\tz-index:3;\n}\n#_idContainer1388 {\n\t-ms-transform:translate(6.236px,0.316px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(6.236px,0.316px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:50.42px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(6.236px,0.316px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(6.236px,0.316px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:0;\n}\n#_idContainer1389 {\n\t-ms-transform:translate(7.511px,3.118px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.511px,3.118px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:5.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.511px,3.118px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.511px,3.118px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.93px;\n\tz-index:4;\n}\n#_idContainer1390 {\n\t-ms-transform:translate(6.377px,7.888px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(6.377px,7.888px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:36.51px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(6.377px,7.888px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(6.377px,7.888px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:42.67px;\n\tz-index:2;\n}\n#_idContainer1391 {\n\t-ms-transform:translate(43.370px,20.692px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(43.370px,20.692px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:4.85px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(43.370px,20.692px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(43.370px,20.692px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:21.97px;\n\tz-index:6;\n}\n#_idContainer1392 {\n\t-ms-transform:translate(34.135px,24.595px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(34.135px,24.595px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:6.17px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(34.135px,24.595px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(34.135px,24.595px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.37px;\n\tz-index:5;\n}\n#_idContainer1393 {\n\t-ms-transform:translate(6.236px,51.732px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(6.236px,51.732px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:50.42px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(6.236px,51.732px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(6.236px,51.732px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:1;\n}\n#_idContainer1394 {\n\t-ms-transform:translate(20.834px,52.582px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(20.834px,52.582px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:4.85px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(20.834px,52.582px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(20.834px,52.582px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:21.97px;\n\tz-index:7;\n}\n#_idContainer1395 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:57px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:65px;\n}\n#_idContainer1400, #_idContainer1412 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:10px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1402, #_idContainer1422 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:7px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1403 {\n\t-ms-transform:translate(0.000px,0.500px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.500px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:9.46px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.500px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.500px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:1;\n}\n#_idContainer1404 {\n\t-ms-transform:translate(11.298px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(11.298px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:131.31px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(11.298px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(11.298px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:0;\n}\n#_idContainer1405 {\n\tdisplay:inline-block;\n\theight:132px;\n\tposition:relative;\n\twidth:11px;\n}\n#_idContainer1406 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:16px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1423 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:156px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer1424 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:87px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1426 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:6px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1427 {\n\t-ms-transform:translate(0.000px,111.616px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,111.616px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:67.51px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,111.616px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,111.616px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:111.62px;\n\tz-index:0;\n}\n#_idContainer1428 {\n\t-ms-transform:translate(10.948px,101.366px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(10.948px,101.366px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:49.61px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(10.948px,101.366px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(10.948px,101.366px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:88.23px;\n\tz-index:1;\n}\n#_idContainer1429 {\n\tdisplay:inline-block;\n\theight:112px;\n\tposition:relative;\n\twidth:68px;\n}\n#_idContainer1432 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:11px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1433 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:4px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1434, #_idContainer1435 {\n\tdisplay:inline-block;\n\theight:0px;\n\tposition:relative;\n\twidth:24px;\n}\n#_idContainer1436 {\n\tdisplay:inline-block;\n\theight:0px;\n\tposition:relative;\n\twidth:71px;\n}\n#_idContainer1437 {\n\tdisplay:inline-block;\n\theight:0px;\n\tposition:relative;\n\twidth:348px;\n}\n#_idContainer1438 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:165px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1442 {\n\t-ms-transform:translate(-0.001px,51.554px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(-0.001px,51.554px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:49.33px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(-0.001px,51.554px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(-0.001px,51.554px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:0;\n}\n#_idContainer1443 {\n\t-ms-transform:translate(0.000px,52.760px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,52.760px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:51.76px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,52.760px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,52.760px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:1;\n}\n#_idContainer1444 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:54px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:50px;\n}\n#_idContainer1445 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:16px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1454, #_idContainer1455 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:26px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1456 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:63px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1458 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:7px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1460 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:34px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer1461 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:33px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer1466 {\n\tdisplay:inline-block;\n\theight:39px;\n\tposition:relative;\n\twidth:1px;\n}\n#_idContainer1467 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:27.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:517.61px;\n\tz-index:0;\n}\n#_idContainer1468 {\n\t-ms-transform:translate(0.000px,38.534px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,38.534px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:670.13px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,38.534px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,38.534px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:518.74px;\n\tz-index:1;\n}\n#_idContainer1469 {\n\tdisplay:inline-block;\n\theight:709px;\n\tposition:relative;\n\twidth:519px;\n}\nimg._idGenObjectAttribute-1 {\n\theight:100.00%;\n\tmin-width:100%;\n\twidth:100.00%;\n}\nimg._idGenObjectAttribute-2 {\n\theight:24px;\n\twidth:253px;\n}\nimg._idGenObjectAttribute-3 {\n\theight:13px;\n\twidth:35px;\n}\nimg._idGenObjectAttribute-4 {\n\theight:22px;\n\twidth:138px;\n}\nimg._idGenObjectAttribute-5 {\n\theight:23px;\n\twidth:35px;\n}\nimg._idGenObjectAttribute-6 {\n\theight:22px;\n\twidth:133px;\n}\nimg._idGenObjectAttribute-7 {\n\theight:23px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-8 {\n\theight:178px;\n\twidth:234px;\n}\nimg._idGenObjectAttribute-9 {\n\theight:174px;\n\twidth:250px;\n}\nimg._idGenObjectAttribute-10 {\n\theight:23px;\n\twidth:63px;\n}\nimg._idGenObjectAttribute-11 {\n\theight:23px;\n\twidth:59px;\n}\nimg._idGenObjectAttribute-12 {\n\theight:473px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-13 {\n\theight:18px;\n\twidth:40px;\n}\nimg._idGenObjectAttribute-14 {\n\theight:26px;\n\twidth:198px;\n}\nimg._idGenObjectAttribute-15 {\n\theight:27px;\n\twidth:145px;\n}\nimg._idGenObjectAttribute-16 {\n\theight:27px;\n\twidth:123px;\n}\nimg._idGenObjectAttribute-17 {\n\theight:29px;\n\twidth:167px;\n}\nimg._idGenObjectAttribute-18 {\n\theight:27px;\n\twidth:103px;\n}\nimg._idGenObjectAttribute-19 {\n\theight:443px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-20 {\n\theight:24px;\n\twidth:257px;\n}\nimg._idGenObjectAttribute-21 {\n\theight:23px;\n\twidth:34px;\n}\nimg._idGenObjectAttribute-22 {\n\theight:23px;\n\twidth:33px;\n}\nimg._idGenObjectAttribute-23 {\n\theight:21px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-24 {\n\theight:76px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-25 {\n\theight:15px;\n\twidth:35px;\n}\nimg._idGenObjectAttribute-26 {\n\theight:23px;\n\twidth:45px;\n}\nimg._idGenObjectAttribute-27 {\n\theight:23px;\n\twidth:147px;\n}\nimg._idGenObjectAttribute-28 {\n\theight:30px;\n\twidth:96px;\n}\nimg._idGenObjectAttribute-29 {\n\theight:30px;\n\twidth:97px;\n}\nimg._idGenObjectAttribute-30 {\n\theight:171px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-31 {\n\theight:188px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-32 {\n\theight:17px;\n\twidth:35px;\n}\nimg._idGenObjectAttribute-33 {\n\theight:21px;\n\twidth:159px;\n}\nimg._idGenObjectAttribute-34 {\n\theight:27px;\n\twidth:112px;\n}\nimg._idGenObjectAttribute-35 {\n\theight:13px;\n\twidth:28px;\n}\nimg._idGenObjectAttribute-36 {\n\theight:10px;\n\twidth:12px;\n}\nimg._idGenObjectAttribute-37 {\n\theight:20px;\n\twidth:149px;\n}\nimg._idGenObjectAttribute-38 {\n\theight:18px;\n\twidth:175px;\n}\nimg._idGenObjectAttribute-39 {\n\theight:13px;\n\twidth:176px;\n}\nimg._idGenObjectAttribute-40 {\n\theight:60px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-41 {\n\theight:13px;\n\twidth:77px;\n}\nimg._idGenObjectAttribute-42 {\n\theight:22px;\n\twidth:46px;\n}\nimg._idGenObjectAttribute-43 {\n\theight:169px;\n\twidth:208px;\n}\nimg._idGenObjectAttribute-44 {\n\theight:21px;\n\twidth:38px;\n}\nimg._idGenObjectAttribute-45 {\n\theight:13px;\n\twidth:223px;\n}\nimg._idGenObjectAttribute-46 {\n\theight:19px;\n\twidth:20px;\n}\nimg._idGenObjectAttribute-47 {\n\theight:13px;\n\twidth:177px;\n}\nimg._idGenObjectAttribute-48 {\n\theight:25px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-49 {\n\theight:28px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-50 {\n\theight:20px;\n\twidth:11px;\n}\nimg._idGenObjectAttribute-51 {\n\theight:25px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-52 {\n\theight:27px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-53 {\n\theight:28px;\n\twidth:38px;\n}\nimg._idGenObjectAttribute-54 {\n\theight:160px;\n\twidth:195px;\n}\nimg._idGenObjectAttribute-55 {\n\theight:20px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-56 {\n\theight:21px;\n\twidth:39px;\n}\nimg._idGenObjectAttribute-57 {\n\theight:146px;\n\twidth:180px;\n}\nimg._idGenObjectAttribute-58 {\n\theight:15px;\n\twidth:28px;\n}\nimg._idGenObjectAttribute-59 {\n\theight:14px;\n\twidth:21px;\n}\nimg._idGenObjectAttribute-60 {\n\theight:14px;\n\twidth:28px;\n}\nimg._idGenObjectAttribute-61 {\n\theight:300px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-62 {\n\theight:30px;\n\twidth:80px;\n}\nimg._idGenObjectAttribute-63 {\n\theight:27px;\n\twidth:46px;\n}\nimg._idGenObjectAttribute-64 {\n\theight:497px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-65 {\n\theight:14px;\n\twidth:160px;\n}\nimg._idGenObjectAttribute-66 {\n\theight:19px;\n\twidth:21px;\n}\nimg._idGenObjectAttribute-67 {\n\theight:22px;\n\twidth:49px;\n}\nimg._idGenObjectAttribute-68 {\n\theight:8px;\n\twidth:7px;\n}\nimg._idGenObjectAttribute-69 {\n\theight:11px;\n\twidth:7px;\n}\nimg._idGenObjectAttribute-70 {\n\theight:21px;\n\twidth:12px;\n}\nimg._idGenObjectAttribute-71 {\n\theight:16px;\n\twidth:12px;\n}\nimg._idGenObjectAttribute-72 {\n\theight:38px;\n\twidth:59px;\n}\nimg._idGenObjectAttribute-73 {\n\theight:20px;\n\twidth:12px;\n}\nimg._idGenObjectAttribute-74 {\n\theight:24px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-75 {\n\theight:19px;\n\twidth:12px;\n}\nimg._idGenObjectAttribute-76 {\n\theight:13px;\n\twidth:47px;\n}\nimg._idGenObjectAttribute-77 {\n\theight:18px;\n\twidth:12px;\n}\nimg._idGenObjectAttribute-78 {\n\theight:23px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-79 {\n\theight:28px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-80 {\n\theight:30px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-81 {\n\theight:158px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-82 {\n\theight:292px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-83 {\n\theight:31px;\n\twidth:165px;\n}\nimg._idGenObjectAttribute-84 {\n\theight:16px;\n\twidth:21px;\n}\nimg._idGenObjectAttribute-85 {\n\theight:126px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-86 {\n\theight:37px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-87 {\n\theight:36px;\n\twidth:103px;\n}\nimg._idGenObjectAttribute-88 {\n\theight:166px;\n\twidth:223px;\n}\nimg._idGenObjectAttribute-89 {\n\theight:15px;\n\twidth:16px;\n}\nimg._idGenObjectAttribute-90 {\n\theight:151px;\n\twidth:170px;\n}\nimg._idGenObjectAttribute-91 {\n\theight:25px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-92 {\n\theight:29px;\n\twidth:29px;\n}\nimg._idGenObjectAttribute-93 {\n\theight:29px;\n\twidth:30px;\n}\nimg._idGenObjectAttribute-94 {\n\theight:29px;\n\twidth:54px;\n}\nimg._idGenObjectAttribute-95 {\n\theight:30px;\n\twidth:82px;\n}\nimg._idGenObjectAttribute-96 {\n\theight:30px;\n\twidth:103px;\n}\nimg._idGenObjectAttribute-97 {\n\theight:32px;\n\twidth:100px;\n}\nimg._idGenObjectAttribute-98 {\n\theight:279px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-99 {\n\theight:61px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-100 {\n\theight:298px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-101 {\n\theight:14px;\n\twidth:86px;\n}\nimg._idGenObjectAttribute-102 {\n\theight:14px;\n\twidth:79px;\n}\nimg._idGenObjectAttribute-103 {\n\theight:14px;\n\twidth:29px;\n}\nimg._idGenObjectAttribute-104 {\n\theight:128px;\n\twidth:223px;\n}\nimg._idGenObjectAttribute-105 {\n\theight:20px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-106 {\n\theight:131px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-107 {\n\theight:161px;\n\twidth:249px;\n}\nimg._idGenObjectAttribute-108 {\n\theight:14px;\n\twidth:78px;\n}\nimg._idGenObjectAttribute-109 {\n\theight:22px;\n\twidth:142px;\n}\nimg._idGenObjectAttribute-110 {\n\theight:23px;\n\twidth:67px;\n}\nimg._idGenObjectAttribute-111 {\n\theight:21px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-112 {\n\theight:24px;\n\twidth:53px;\n}\nimg._idGenObjectAttribute-113 {\n\theight:24px;\n\twidth:69px;\n}\nimg._idGenObjectAttribute-114 {\n\theight:21px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-115 {\n\theight:15px;\n\twidth:10px;\n}\nimg._idGenObjectAttribute-116 {\n\theight:14px;\n\twidth:76px;\n}\nimg._idGenObjectAttribute-117 {\n\theight:29px;\n\twidth:61px;\n}\nimg._idGenObjectAttribute-118 {\n\theight:30px;\n\twidth:68px;\n}\nimg._idGenObjectAttribute-119 {\n\theight:30px;\n\twidth:15px;\n}\nimg._idGenObjectAttribute-120 {\n\theight:32px;\n\twidth:77px;\n}\nimg._idGenObjectAttribute-121 {\n\theight:30px;\n\twidth:17px;\n}\nimg._idGenObjectAttribute-122 {\n\theight:27px;\n\twidth:117px;\n}\nimg._idGenObjectAttribute-123 {\n\theight:23px;\n\twidth:157px;\n}\nimg._idGenObjectAttribute-124 {\n\theight:21px;\n\twidth:157px;\n}\nimg._idGenObjectAttribute-125 {\n\theight:25px;\n\twidth:51px;\n}\nimg._idGenObjectAttribute-126 {\n\theight:24px;\n\twidth:97px;\n}\nimg._idGenObjectAttribute-127 {\n\theight:11px;\n\twidth:14px;\n}\nimg._idGenObjectAttribute-128 {\n\theight:14px;\n\twidth:75px;\n}\nimg._idGenObjectAttribute-129 {\n\theight:24px;\n\twidth:33px;\n}\nimg._idGenObjectAttribute-130 {\n\theight:30px;\n\twidth:53px;\n}\nimg._idGenObjectAttribute-131 {\n\theight:27px;\n\twidth:53px;\n}\nimg._idGenObjectAttribute-132 {\n\theight:29px;\n\twidth:87px;\n}\nimg._idGenObjectAttribute-133 {\n\theight:15px;\n\twidth:7px;\n}\nimg._idGenObjectAttribute-134 {\n\theight:15px;\n\twidth:6px;\n}\nimg._idGenObjectAttribute-135 {\n\theight:33px;\n\twidth:61px;\n}\nimg._idGenObjectAttribute-136 {\n\theight:34px;\n\twidth:53px;\n}\nimg._idGenObjectAttribute-137 {\n\theight:17px;\n\twidth:65px;\n}\nimg._idGenObjectAttribute-138 {\n\theight:27px;\n\twidth:48px;\n}\nimg._idGenObjectAttribute-139 {\n\theight:27px;\n\twidth:55px;\n}\nimg._idGenObjectAttribute-140 {\n\theight:27px;\n\twidth:38px;\n}\nimg._idGenObjectAttribute-141 {\n\theight:27px;\n\twidth:47px;\n}\nimg._idGenObjectAttribute-142 {\n\theight:27px;\n\twidth:32px;\n}\nimg._idGenObjectAttribute-143 {\n\theight:27px;\n\twidth:40px;\n}\nimg._idGenObjectAttribute-144 {\n\theight:27px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-145 {\n\theight:24px;\n\twidth:67px;\n}\nimg._idGenObjectAttribute-146 {\n\theight:18px;\n\twidth:76px;\n}\nimg._idGenObjectAttribute-147 {\n\theight:36px;\n\twidth:61px;\n}\nimg._idGenObjectAttribute-148 {\n\theight:32px;\n\twidth:49px;\n}\nimg._idGenObjectAttribute-149 {\n\theight:17px;\n\twidth:30px;\n}\nimg._idGenObjectAttribute-150 {\n\theight:18px;\n\twidth:80px;\n}\nimg._idGenObjectAttribute-151 {\n\theight:24px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-152 {\n\theight:23px;\n\twidth:14px;\n}\nimg._idGenObjectAttribute-153 {\n\theight:16px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-154 {\n\theight:23px;\n\twidth:49px;\n}\nimg._idGenObjectAttribute-155 {\n\theight:27px;\n\twidth:59px;\n}\nimg._idGenObjectAttribute-156 {\n\theight:28px;\n\twidth:103px;\n}\nimg._idGenObjectAttribute-157 {\n\theight:21px;\n\twidth:32px;\n}\nimg._idGenObjectAttribute-158 {\n\theight:27px;\n\twidth:155px;\n}\nimg._idGenObjectAttribute-159 {\n\theight:26px;\n\twidth:14px;\n}\nimg._idGenObjectAttribute-160 {\n\theight:27px;\n\twidth:93px;\n}\nimg._idGenObjectAttribute-161 {\n\theight:21px;\n\twidth:36px;\n}\nimg._idGenObjectAttribute-162 {\n\theight:27px;\n\twidth:144px;\n}\nimg._idGenObjectAttribute-163 {\n\theight:14px;\n\twidth:127px;\n}\nimg._idGenObjectAttribute-164 {\n\theight:26px;\n\twidth:79px;\n}\nimg._idGenObjectAttribute-165 {\n\theight:24px;\n\twidth:79px;\n}\nimg._idGenObjectAttribute-166 {\n\theight:14px;\n\twidth:72px;\n}\nimg._idGenObjectAttribute-167 {\n\theight:24px;\n\twidth:83px;\n}\nimg._idGenObjectAttribute-168 {\n\theight:24px;\n\twidth:85px;\n}\nimg._idGenObjectAttribute-169 {\n\theight:26px;\n\twidth:77px;\n}\nimg._idGenObjectAttribute-170 {\n\theight:13px;\n\twidth:146px;\n}\nimg._idGenObjectAttribute-171 {\n\theight:24px;\n\twidth:50px;\n}\nimg._idGenObjectAttribute-172 {\n\theight:31px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-173 {\n\theight:21px;\n\twidth:62px;\n}\nimg._idGenObjectAttribute-174 {\n\theight:21px;\n\twidth:22px;\n}\nimg._idGenObjectAttribute-175 {\n\theight:164px;\n\twidth:170px;\n}\nimg._idGenObjectAttribute-176 {\n\theight:221px;\n\twidth:226px;\n}\nimg._idGenObjectAttribute-177 {\n\theight:21px;\n\twidth:37px;\n}\nimg._idGenObjectAttribute-178 {\n\theight:24px;\n\twidth:76px;\n}\nimg._idGenObjectAttribute-179 {\n\theight:25px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-180 {\n\theight:21px;\n\twidth:44px;\n}\nimg._idGenObjectAttribute-181 {\n\theight:27px;\n\twidth:54px;\n}\nimg._idGenObjectAttribute-182 {\n\theight:24px;\n\twidth:54px;\n}\nimg._idGenObjectAttribute-183 {\n\theight:13px;\n\twidth:75px;\n}\nimg._idGenObjectAttribute-184 {\n\theight:23px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-185 {\n\theight:33px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-186 {\n\theight:14px;\n\twidth:81px;\n}\nimg._idGenObjectAttribute-187 {\n\theight:30px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-188 {\n\theight:24px;\n\twidth:80px;\n}\nimg._idGenObjectAttribute-189 {\n\theight:23px;\n\twidth:23px;\n}\nimg._idGenObjectAttribute-190 {\n\theight:21px;\n\twidth:35px;\n}\nimg._idGenObjectAttribute-191 {\n\theight:24px;\n\twidth:81px;\n}\nimg._idGenObjectAttribute-192 {\n\theight:21px;\n\twidth:26px;\n}\nimg._idGenObjectAttribute-193 {\n\theight:27px;\n\twidth:23px;\n}\nimg._idGenObjectAttribute-194 {\n\theight:34px;\n\twidth:57px;\n}\nimg._idGenObjectAttribute-195 {\n\theight:26px;\n\twidth:38px;\n}\nimg._idGenObjectAttribute-196 {\n\theight:29px;\n\twidth:36px;\n}\nimg._idGenObjectAttribute-197 {\n\theight:26px;\n\twidth:40px;\n}\nimg._idGenObjectAttribute-198 {\n\theight:16px;\n\twidth:13px;\n}\nimg._idGenObjectAttribute-199 {\n\theight:14px;\n\twidth:123px;\n}\nimg._idGenObjectAttribute-200 {\n\theight:19px;\n\twidth:49px;\n}\nimg._idGenObjectAttribute-201 {\n\theight:26px;\n\twidth:16px;\n}\nimg._idGenObjectAttribute-202 {\n\theight:21px;\n\twidth:17px;\n}\nimg._idGenObjectAttribute-203 {\n\theight:13px;\n\twidth:125px;\n}\nimg._idGenObjectAttribute-204 {\n\theight:28px;\n\twidth:43px;\n}\nimg._idGenObjectAttribute-205 {\n\theight:21px;\n\twidth:20px;\n}\nimg._idGenObjectAttribute-206 {\n\theight:198px;\n\twidth:234px;\n}\nimg._idGenObjectAttribute-207 {\n\theight:24px;\n\twidth:59px;\n}\nimg._idGenObjectAttribute-208 {\n\theight:21px;\n\twidth:13px;\n}\nimg._idGenObjectAttribute-209 {\n\theight:21px;\n\twidth:23px;\n}\nimg._idGenObjectAttribute-210 {\n\theight:25px;\n\twidth:44px;\n}\nimg._idGenObjectAttribute-211 {\n\theight:21px;\n\twidth:91px;\n}\nimg._idGenObjectAttribute-212 {\n\theight:24px;\n\twidth:75px;\n}\nimg._idGenObjectAttribute-213 {\n\theight:23px;\n\twidth:36px;\n}\nimg._idGenObjectAttribute-214 {\n\theight:21px;\n\twidth:45px;\n}\nimg._idGenObjectAttribute-215 {\n\theight:17px;\n\twidth:40px;\n}\nimg._idGenObjectAttribute-216 {\n\theight:14px;\n\twidth:74px;\n}\nimg._idGenObjectAttribute-217 {\n\theight:17px;\n\twidth:39px;\n}\nimg._idGenObjectAttribute-218 {\n\theight:25px;\n\twidth:75px;\n}\nimg._idGenObjectAttribute-219 {\n\theight:13px;\n\twidth:119px;\n}\nimg._idGenObjectAttribute-220 {\n\theight:20px;\n\twidth:23px;\n}\nimg._idGenObjectAttribute-221 {\n\theight:24px;\n\twidth:58px;\n}\nimg._idGenObjectAttribute-222 {\n\theight:25px;\n\twidth:109px;\n}\nimg._idGenObjectAttribute-223 {\n\theight:24px;\n\twidth:94px;\n}\nimg._idGenObjectAttribute-224 {\n\theight:27px;\n\twidth:166px;\n}\nimg._idGenObjectAttribute-225 {\n\theight:27px;\n\twidth:114px;\n}\nimg._idGenObjectAttribute-226 {\n\theight:27px;\n\twidth:164px;\n}\nimg._idGenObjectAttribute-227 {\n\theight:25px;\n\twidth:46px;\n}\nimg._idGenObjectAttribute-228 {\n\theight:25px;\n\twidth:16px;\n}\nimg._idGenObjectAttribute-229 {\n\theight:21px;\n\twidth:46px;\n}\nimg._idGenObjectAttribute-230 {\n\theight:14px;\n\twidth:120px;\n}\nimg._idGenObjectAttribute-231 {\n\theight:33px;\n\twidth:101px;\n}\nimg._idGenObjectAttribute-232 {\n\theight:21px;\n\twidth:156px;\n}\nimg._idGenObjectAttribute-233 {\n\theight:11px;\n\twidth:10px;\n}\nimg._idGenObjectAttribute-234 {\n\theight:30px;\n\twidth:144px;\n}\nimg._idGenObjectAttribute-235 {\n\theight:23px;\n\twidth:149px;\n}\nimg._idGenObjectAttribute-236 {\n\theight:17px;\n\twidth:171px;\n}\nimg._idGenObjectAttribute-237 {\n\theight:14px;\n\twidth:124px;\n}\nimg._idGenObjectAttribute-238 {\n\theight:13px;\n\twidth:163px;\n}\nimg._idGenObjectAttribute-239 {\n\theight:14px;\n\twidth:170px;\n}\nimg._idGenObjectAttribute-240 {\n\theight:14px;\n\twidth:77px;\n}\nimg._idGenObjectAttribute-241 {\n\theight:17px;\n\twidth:78px;\n}\nimg._idGenObjectAttribute-242 {\n\theight:17px;\n\twidth:217px;\n}\nimg._idGenObjectAttribute-243 {\n\theight:14px;\n\twidth:87px;\n}\nimg._idGenObjectAttribute-244 {\n\theight:22px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-245 {\n\theight:114px;\n\twidth:124px;\n}\nimg._idGenObjectAttribute-246 {\n\theight:19px;\n\twidth:29px;\n}\nimg._idGenObjectAttribute-247 {\n\theight:26px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-248 {\n\theight:26px;\n\twidth:37px;\n}\nimg._idGenObjectAttribute-249 {\n\theight:20px;\n\twidth:14px;\n}\nimg._idGenObjectAttribute-250 {\n\theight:72px;\n\twidth:73px;\n}\nimg._idGenObjectAttribute-251 {\n\theight:115px;\n\twidth:106px;\n}\nimg._idGenObjectAttribute-252 {\n\theight:24px;\n\twidth:37px;\n}\nimg._idGenObjectAttribute-253 {\n\theight:24px;\n\twidth:17px;\n}\nimg._idGenObjectAttribute-254 {\n\theight:24px;\n\twidth:36px;\n}\nimg._idGenObjectAttribute-255 {\n\theight:24px;\n\twidth:32px;\n}\nimg._idGenObjectAttribute-256 {\n\theight:75px;\n\twidth:84px;\n}\nimg._idGenObjectAttribute-257 {\n\theight:78px;\n\twidth:94px;\n}\nimg._idGenObjectAttribute-258 {\n\theight:87px;\n\twidth:97px;\n}\nimg._idGenObjectAttribute-259 {\n\theight:80px;\n\twidth:89px;\n}\nimg._idGenObjectAttribute-260 {\n\theight:109px;\n\twidth:115px;\n}\nimg._idGenObjectAttribute-261 {\n\theight:101px;\n\twidth:130px;\n}\nimg._idGenObjectAttribute-262 {\n\theight:98px;\n\twidth:116px;\n}\nimg._idGenObjectAttribute-263 {\n\theight:19px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-264 {\n\theight:21px;\n\twidth:33px;\n}\nimg._idGenObjectAttribute-265 {\n\theight:21px;\n\twidth:29px;\n}\nimg._idGenObjectAttribute-266 {\n\theight:26px;\n\twidth:49px;\n}\nimg._idGenObjectAttribute-267 {\n\theight:26px;\n\twidth:36px;\n}\nimg._idGenObjectAttribute-268 {\n\theight:26px;\n\twidth:54px;\n}\nimg._idGenObjectAttribute-269 {\n\theight:26px;\n\twidth:43px;\n}\nimg._idGenObjectAttribute-270 {\n\theight:26px;\n\twidth:30px;\n}\nimg._idGenObjectAttribute-271 {\n\theight:22px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-272 {\n\theight:22px;\n\twidth:26px;\n}\nimg._idGenObjectAttribute-273 {\n\theight:22px;\n\twidth:33px;\n}\nimg._idGenObjectAttribute-274 {\n\theight:22px;\n\twidth:32px;\n}\nimg._idGenObjectAttribute-275 {\n\theight:26px;\n\twidth:33px;\n}\nimg._idGenObjectAttribute-276 {\n\theight:22px;\n\twidth:36px;\n}\nimg._idGenObjectAttribute-277 {\n\theight:26px;\n\twidth:26px;\n}\nimg._idGenObjectAttribute-278 {\n\theight:22px;\n\twidth:29px;\n}\nimg._idGenObjectAttribute-279 {\n\theight:26px;\n\twidth:39px;\n}\nimg._idGenObjectAttribute-280 {\n\theight:26px;\n\twidth:35px;\n}\nimg._idGenObjectAttribute-281 {\n\theight:14px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-282 {\n\theight:26px;\n\twidth:32px;\n}\nimg._idGenObjectAttribute-283 {\n\theight:25px;\n\twidth:42px;\n}\nimg._idGenObjectAttribute-284 {\n\theight:25px;\n\twidth:48px;\n}\nimg._idGenObjectAttribute-285 {\n\theight:22px;\n\twidth:37px;\n}\nimg._idGenObjectAttribute-286 {\n\theight:22px;\n\twidth:45px;\n}\nimg._idGenObjectAttribute-287 {\n\theight:22px;\n\twidth:34px;\n}\nimg._idGenObjectAttribute-288 {\n\theight:22px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-289 {\n\theight:29px;\n\twidth:43px;\n}\nimg._idGenObjectAttribute-290 {\n\theight:21px;\n\twidth:19px;\n}\nimg._idGenObjectAttribute-291 {\n\theight:17px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-292 {\n\theight:20px;\n\twidth:19px;\n}\nimg._idGenObjectAttribute-293 {\n\theight:7px;\n\twidth:5px;\n}\nimg._idGenObjectAttribute-294 {\n\theight:20px;\n\twidth:16px;\n}\nimg._idGenObjectAttribute-295 {\n\theight:23px;\n\twidth:38px;\n}\nimg._idGenObjectAttribute-296 {\n\theight:21px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-297 {\n\theight:28px;\n\twidth:33px;\n}\nimg._idGenObjectAttribute-298 {\n\theight:16px;\n\twidth:17px;\n}\nimg._idGenObjectAttribute-299 {\n\theight:28px;\n\twidth:20px;\n}\nimg._idGenObjectAttribute-300 {\n\theight:27px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-301 {\n\theight:30px;\n\twidth:63px;\n}\nimg._idGenObjectAttribute-302 {\n\theight:27px;\n\twidth:26px;\n}\nimg._idGenObjectAttribute-303 {\n\theight:22px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-304 {\n\theight:20px;\n\twidth:29px;\n}\nimg._idGenObjectAttribute-305 {\n\theight:111px;\n\twidth:116px;\n}\nimg._idGenObjectAttribute-306 {\n\theight:20px;\n\twidth:17px;\n}\nimg._idGenObjectAttribute-307 {\n\theight:63px;\n\twidth:64px;\n}\nimg._idGenObjectAttribute-308 {\n\theight:63px;\n\twidth:77px;\n}\nimg._idGenObjectAttribute-309 {\n\theight:63px;\n\twidth:65px;\n}\nimg._idGenObjectAttribute-310 {\n\theight:16px;\n\twidth:16px;\n}\nimg._idGenObjectAttribute-311 {\n\theight:18px;\n\twidth:17px;\n}\nimg._idGenObjectAttribute-312 {\n\theight:19px;\n\twidth:17px;\n}\nimg._idGenObjectAttribute-313 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{\n\theight:25px;\n\twidth:15px;\n}\nimg._idGenObjectAttribute-328 {\n\theight:28px;\n\twidth:55px;\n}\nimg._idGenObjectAttribute-329 {\n\theight:31px;\n\twidth:38px;\n}\nimg._idGenObjectAttribute-330 {\n\theight:17px;\n\twidth:42px;\n}\nimg._idGenObjectAttribute-331 {\n\theight:26px;\n\twidth:18px;\n}\nimg._idGenObjectAttribute-332 {\n\theight:24px;\n\twidth:23px;\n}\nimg._idGenObjectAttribute-333 {\n\theight:17px;\n\twidth:34px;\n}\nimg._idGenObjectAttribute-334 {\n\theight:33px;\n\twidth:35px;\n}\nimg._idGenObjectAttribute-335 {\n\theight:26px;\n\twidth:34px;\n}\nimg._idGenObjectAttribute-336 {\n\theight:24px;\n\twidth:28px;\n}\nimg._idGenObjectAttribute-337 {\n\theight:22px;\n\twidth:13px;\n}\nimg._idGenObjectAttribute-338 {\n\theight:16px;\n\twidth:34px;\n}\nimg._idGenObjectAttribute-339 {\n\theight:16px;\n\twidth:42px;\n}\nimg._idGenObjectAttribute-340 {\n\theight:18px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-341 {\n\theight:22px;\n\twidth:12px;\n}\nimg._idGenObjectAttribute-342 {\n\theight:27px;\n\twidth:37px;\n}\nimg._idGenObjectAttribute-343 {\n\theight:27px;\n\twidth:43px;\n}\nimg._idGenObjectAttribute-344 {\n\theight:29px;\n\twidth:41px;\n}\nimg._idGenObjectAttribute-345 {\n\theight:27px;\n\twidth:34px;\n}\nimg._idGenObjectAttribute-346 {\n\theight:22px;\n\twidth:59px;\n}\nimg._idGenObjectAttribute-347 {\n\theight:22px;\n\twidth:61px;\n}\nimg._idGenObjectAttribute-348 {\n\theight:9px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-349 {\n\theight:29px;\n\twidth:45px;\n}\nimg._idGenObjectAttribute-350 {\n\theight:28px;\n\twidth:40px;\n}\nimg._idGenObjectAttribute-351 {\n\theight:28px;\n\twidth:32px;\n}\nimg._idGenObjectAttribute-352 {\n\theight:22px;\n\twidth:41px;\n}\nimg._idGenObjectAttribute-353 {\n\theight:22px;\n\twidth:71px;\n}\nimg._idGenObjectAttribute-354 {\n\theight:28px;\n\twidth:29px;\n}\nimg._idGenObjectAttribute-355 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{\n\theight:15px;\n\twidth:32px;\n}\nimg._idGenObjectAttribute-426 {\n\theight:28px;\n\twidth:16px;\n}\nimg._idGenObjectAttribute-427 {\n\theight:31px;\n\twidth:54px;\n}\nimg._idGenObjectAttribute-428 {\n\theight:24px;\n\twidth:93px;\n}\nimg._idGenObjectAttribute-429 {\n\theight:28px;\n\twidth:18px;\n}\nimg._idGenObjectAttribute-430 {\n\theight:18px;\n\twidth:20px;\n}\nimg._idGenObjectAttribute-431 {\n\theight:96px;\n\twidth:150px;\n}\nimg._idGenObjectAttribute-432 {\n\theight:24px;\n\twidth:16px;\n}\nimg._idGenObjectAttribute-433 {\n\theight:28px;\n\twidth:15px;\n}\nimg._idGenObjectAttribute-434 {\n\theight:26px;\n\twidth:44px;\n}\nimg._idGenObjectAttribute-435 {\n\theight:34px;\n\twidth:16px;\n}\nimg._idGenObjectAttribute-436 {\n\theight:22px;\n\twidth:57px;\n}\nimg._idGenObjectAttribute-437 {\n\theight:24px;\n\twidth:90px;\n}\nimg._idGenObjectAttribute-438 {\n\theight:24px;\n\twidth:64px;\n}\nimg._idGenObjectAttribute-439 {\n\theight:20px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-440 {\n\theight:20px;\n\twidth:26px;\n}\nimg._idGenObjectAttribute-441 {\n\theight:31px;\n\twidth:16px;\n}\nimg._idGenObjectAttribute-442 {\n\theight:26px;\n\twidth:41px;\n}\nimg._idGenObjectAttribute-443 {\n\theight:28px;\n\twidth:44px;\n}\nimg._idGenObjectAttribute-444 {\n\theight:113px;\n\twidth:138px;\n}\nimg._idGenObjectAttribute-445 {\n\theight:29px;\n\twidth:26px;\n}\nimg._idGenObjectAttribute-446 {\n\theight:26px;\n\twidth:47px;\n}\nimg._idGenObjectAttribute-447 {\n\theight:28px;\n\twidth:96px;\n}\nimg._idGenObjectAttribute-448 {\n\theight:20px;\n\twidth:96px;\n}\nimg._idGenObjectAttribute-449 {\n\theight:29px;\n\twidth:58px;\n}\nimg._idGenObjectAttribute-450 {\n\theight:24px;\n\twidth:60px;\n}\nimg._idGenObjectAttribute-451 {\n\theight:26px;\n\twidth:69px;\n}\nimg._idGenObjectAttribute-452 {\n\theight:28px;\n\twidth:36px;\n}\nimg._idGenObjectAttribute-453 {\n\theight:24px;\n\twidth:63px;\n}\nimg._idGenObjectAttribute-454 {\n\theight:20px;\n\twidth:49px;\n}\nimg._idGenObjectAttribute-455 {\n\theight:17px;\n\twidth:32px;\n}\nimg._idGenObjectAttribute-456 {\n\theight:15px;\n\twidth:31px;\n}\nimg._idGenObjectAttribute-457 {\n\theight:22px;\n\twidth:42px;\n}\nimg._idGenObjectAttribute-458 {\n\theight:26px;\n\twidth:48px;\n}\nimg._idGenObjectAttribute-459 {\n\theight:22px;\n\twidth:14px;\n}\nimg._idGenObjectAttribute-460 {\n\theight:32px;\n\twidth:98px;\n}\nimg._idGenObjectAttribute-461 {\n\theight:20px;\n\twidth:36px;\n}\nimg._idGenObjectAttribute-462 {\n\theight:17px;\n\twidth:71px;\n}\nimg._idGenObjectAttribute-463 {\n\theight:31px;\n\twidth:109px;\n}\nimg._idGenObjectAttribute-464 {\n\theight:31px;\n\twidth:89px;\n}\nimg._idGenObjectAttribute-465 {\n\theight:26px;\n\twidth:55px;\n}\nimg._idGenObjectAttribute-466 {\n\theight:20px;\n\twidth:44px;\n}\nimg._idGenObjectAttribute-467 {\n\theight:24px;\n\twidth:30px;\n}\nimg._idGenObjectAttribute-468 {\n\theight:24px;\n\twidth:48px;\n}\nimg._idGenObjectAttribute-469 {\n\theight:16px;\n\twidth:22px;\n}\nimg._idGenObjectAttribute-470 {\n\theight:125px;\n\twidth:196px;\n}\nimg._idGenObjectAttribute-471 {\n\theight:30px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-472 {\n\theight:27px;\n\twidth:61px;\n}\nimg._idGenObjectAttribute-473 {\n\theight:25px;\n\twidth:86px;\n}\nimg._idGenObjectAttribute-474 {\n\theight:20px;\n\twidth:62px;\n}\nimg._idGenObjectAttribute-475 {\n\theight:23px;\n\twidth:39px;\n}\nimg._idGenObjectAttribute-476 {\n\theight:101px;\n\twidth:21px;\n}\nimg._idGenObjectAttribute-477 {\n\theight:22px;\n\twidth:21px;\n}\nimg._idGenObjectAttribute-478 {\n\theight:100px;\n\twidth:21px;\n}\nimg._idGenObjectAttribute-479 {\n\theight:10px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-480 {\n\theight:18px;\n\twidth:21px;\n}\nimg._idGenObjectAttribute-481 {\n\theight:18px;\n\twidth:22px;\n}\nimg._idGenObjectAttribute-482 {\n\theight:10px;\n\twidth:22px;\n}\nimg._idGenObjectAttribute-483 {\n\theight:35px;\n\twidth:19px;\n}\nimg._idGenObjectAttribute-484 {\n\theight:33px;\n\twidth:19px;\n}\nimg._idGenObjectAttribute-485 {\n\theight:74px;\n\twidth:30px;\n}\nimg._idGenObjectAttribute-486 {\n\theight:32px;\n\twidth:23px;\n}\nimg._idGenObjectAttribute-487 {\n\theight:28px;\n\twidth:23px;\n}\nimg._idGenObjectAttribute-488 {\n\theight:38px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-489 {\n\theight:29px;\n\twidth:23px;\n}\nimg._idGenObjectAttribute-490 {\n\theight:40px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-491 {\n\theight:42px;\n\twidth:23px;\n}\nimg._idGenObjectAttribute-492 {\n\theight:35px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-493 {\n\theight:33px;\n\twidth:30px;\n}\nimg._idGenObjectAttribute-494 {\n\theight:12px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-495 {\n\theight:34px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-496 {\n\theight:229px;\n\twidth:106px;\n}\ndiv._idGenObjectStyleOverride-1 {\n\tbackground-color:#dae6f5;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-2 {\n\tbackground-color:#5aa0d7;\n\tborder-color:#ffffff;\n\tborder-style:solid;\n\tborder-width:2px;\n}\ndiv._idGenObjectStyleOverride-3 {\n\tbackground-color:#ffffff;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-4 {\n\tbackground-color:#cadbf1;\n\tborder-bottom-left-radius:4px;\n\tborder-bottom-right-radius:4px;\n\tborder-top-left-radius:4px;\n\tborder-top-right-radius:4px;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-5 {\n\tbackground-color:#f3f6fc;\n\tborder-color:#bad2ed;\n\tborder-style:solid;\n\tborder-width:3px;\n}\ndiv._idGenObjectStyleOverride-6 {\n\tbackground-color:#5aa0d7;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n}\ndiv._idGenObjectStyleOverride-7 {\n\tbackground-color:#ffffff;\n\tborder-color:#5aa0d7;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-8 {\n\tbackground-color:#5aa0d7;\n\tborder-bottom-left-radius:4px;\n\tborder-bottom-right-radius:4px;\n\tborder-top-left-radius:4px;\n\tborder-top-right-radius:4px;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-9 {\n\tbackground-color:#000000;\n\tborder-bottom-left-radius:0px;\n\tborder-bottom-right-radius:0px;\n\tborder-style:dotted;\n\tborder-top-left-radius:9px;\n\tborder-top-right-radius:1px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-10 {\n\tbackground-color:#f3f6fc;\n\tborder-color:#5aa0d7;\n\tborder-style:solid;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-11 {\n\tbackground-color:#5aa0d7;\n\tborder-color:#ffffff;\n\tborder-style:solid;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-12 {\n\tborder-style:dotted;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-13 {\n\tbackground-color:#58595b;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-14 {\n\tbackground-color:#ffffff;\n\tborder-bottom-right-radius:2px;\n\tborder-top-right-radius:2px;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-15 {\n\tbackground-color:#4ec8eb;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-16 {\n\tbackground-color:#bd202e;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-17 {\n\tborder-bottom-left-radius:12px;\n\tborder-bottom-right-radius:12px;\n\tborder-color:#ffffff;\n\tborder-style:solid;\n\tborder-top-left-radius:12px;\n\tborder-top-right-radius:12px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-18 {\n\tbackground-color:#f0e4f1;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#b055a0;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-19 {\n\tbackground-color:#eaf4e7;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#6dc067;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-20 {\n\tbackground-color:#ededee;\n\tborder-color:#b055a0;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-21 {\n\tbackground-color:#ededee;\n\tborder-color:#39b54a;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-22 {\n\tbackground-color:#a3238e;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#a3238e;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-23 {\n\tbackground-color:#ededee;\n\tborder-color:#a3238e;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-24 {\n\tbackground-color:#39b54a;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#39b54a;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-25 {\n\tbackground-color:#d8ecd4;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#6dc067;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-26 {\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-27 {\n\tbackground-color:#ccdbf1;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#6dc067;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-28 {\n\tbackground-color:#eff7e9;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#6dc067;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-29 {\n\tbackground-color:#ccdbf1;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#2d7ec2;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-30 {\n\tbackground-color:#ededee;\n\tborder-color:#006cb7;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-31 {\n\tbackground-color:#e4ebf7;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#3f92cf;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-32 {\n\tbackground-color:#0080c6;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#0080c6;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-33 {\n\tbackground-color:#ed1c24;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-34 {\n\tborder-bottom-left-radius:8px;\n\tborder-bottom-right-radius:8px;\n\tborder-top-left-radius:8px;\n\tborder-top-right-radius:8px;\n}\ndiv._idGenObjectStyleOverride-35 {\n\tborder-color:#ec008c;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-36 {\n\tborder-color:#f58232;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-37 {\n\tbackground-color:#fde9f1;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#ef5ba1;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-38 {\n\tbackground-color:#ededee;\n\tborder-color:#ec008c;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-39 {\n\tbackground-color:#ec0085;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#ec0085;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-40 {\n\tbackground-color:#fad5e5;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#ef5ba1;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-41 {\n\tbackground-color:#ffefe3;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#f89958;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-42 {\n\tbackground-color:#f58232;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#f58232;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-43 {\n\tbackground-color:#ededee;\n\tborder-color:#f89958;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyle-Disabled {\n\tbackground-color:transparent;\n\tborder-width:0px;\n}\ndiv._idGenObjectLayout-1 {\n\ttext-align:center;\n}\ndiv._idGenObjectLayout-2 {\n\tmargin:49px -49px 49px -49px;\n}\ndiv._idGenObjectLayout-3 {\n\tmargin:18px -18px 18px -18px;\n}\ndiv._idGenObjectLayout-4 {\n\tmargin:-42px 42px -42px 42px;\n}\ndiv._idGenObjectLayout-5 {\n\tmargin:-163px 163px -163px 163px;\n}\ndiv._idGenObjectLayout-6 {\n\tmargin:-39px 39px -39px 39px;\n}\ndiv._idGenObjectLayout-7 {\n\tmargin:-100px 100px -100px 100px;\n}\ndiv._idGenObjectLayout-8 {\n\tmargin:-105px 105px -105px 105px;\n}\ndiv._idGenObjectLayout-9 {\n\tmargin:4px -4px 4px -4px;\n}\ndiv._idGenObjectLayout-10 {\n\tmargin:-4px 4px -4px 4px;\n}\ndiv._idGenObjectLayout-11 {\n\tmargin:-7px 7px -7px 7px;\n}\ndiv._idGenObjectLayout-12 {\n\tmargin:-78px 78px -78px 78px;\n}\ndiv._idGenObjectLayout-13 {\n\tmargin:-43px 43px -43px 43px;\n}\ndiv._idGenObjectLayout-14 {\n\tmargin:-82px 82px -82px 82px;\n}\ndiv._idGenObjectLayout-15 {\n\tmargin:-2px 2px -2px 2px;\n}\ndiv._idGenObjectLayout-16 {\n\tmargin:-31px 31px -31px 31px;\n}\ndiv._idGenObjectLayout-17 {\n\tmargin:-17px 17px -17px 17px;\n}\n" }