Chapter 3 · Chemical Kinetics

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Question

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

Rate of a chemical reaction: The change in concentration of reactant or product in unit time is called rate of a chemical reaction.

It can be expressed in terms of:

(i) The rate in decrease in concentration of any one of the reactants.

(ii) The rate in increase in concentration of any one of the products.

Consider a hypothetical reaction, assuming that the volume of the system remains constant.

R → P

One mole of the reactant R produces one mole of the product P.

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,

∆t = t2 – t1

∆[R] = [R]2 – [R]1

∆[P] = [P]2 – [P]1

(Square bracket expressing molar concentration.)

Rate of disappearance of

R = =

Rate of appearance of

P = = +

Above given equations represent the average rate of a reaction (rav).

Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur.

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.

So, to express the rate at a particular moment of time we determine instantaneous rate.

Instantaneous rate is obtained when we consider the average rate at the smallest time interval say dt (when ∆t approaches zero).

Hence, mathematically for an infinitesimally small dt instantaneous rate is given by:

rav = –

As ∆t → 0 or rinst = –

Unit of rate of a reaction: Units of rate are concentration time–1.

If concentration is in mol L–1 and time is in seconds then the units will be mol L–1sec–1.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Rate of a chemical reaction: The change in concentration of reactant or product in unit time is called rate of a chemical reaction.

It can be expressed in terms of:

(i) The rate in decrease in concentration of any one of the reactants.

(ii) The rate in increase in concentration of any one of the products.

Consider a hypothetical reaction, assuming that the volume of the system remains constant.

R → P

One mole of the reactant R produces one mole of the product P.

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,

∆t = t2 – t1

∆[R] = [R]2 – [R]1

∆[P] = [P]2 – [P]1

(Square bracket expressing molar concentration.)

Rate of disappearance of

R = =

Rate of appearance of

P = = +

Above given equations represent the average rate of a reaction (rav).

Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur.

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.

So, to express the rate at a particular moment of time we determine instantaneous rate.

Instantaneous rate is obtained when we consider the average rate at the smallest time interval say dt (when ∆t approaches zero).

Hence, mathematically for an infinitesimally small dt instantaneous rate is given by:

rav = –

As ∆t → 0 or rinst = –

Unit of rate of a reaction: Units of rate are concentration time–1.

If concentration is in mol L–1 and time is in seconds then the units will be mol L–1sec–1.

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.

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#2 SUB 0M 🖼 1 ✎ Edit ⚠ Log Error

Question

Explain the rate expression for a reaction in which stoichiometric coefficients of reactants and products are not same or unit.

Answer

For reaction, n1A + n2B → n3C + n4D

Rate of reaction:

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.

For reaction, Hg(l) + Cl2(g) → HgCl2(s)

where stoichiometric coefficients of the reactants and products are same, then rate of the reaction is given as:

Rate of reaction =

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:

Rate of reaction =

Similarly for the reaction:

5Br–(aq) + BrO3–(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l)

Rate =

=

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain the rate expression for a reaction in which stoichiometric coefficients of reactants and products are not same or unit.

Answer

For reaction, n1A + n2B → n3C + n4D

Rate of reaction:

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.

For reaction, Hg(l) + Cl2(g) → HgCl2(s)

where stoichiometric coefficients of the reactants and products are same, then rate of the reaction is given as:

Rate of reaction =

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:

Rate of reaction =

Similarly for the reaction:

5Br–(aq) + BrO3–(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l)

Rate =

=

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#3 SUB 3M 🖼 3 ✎ Edit ⚠ Log Error

Question

Explain rate law and rate constant with an example.

Answer

Rate of a reaction depends upon the concentration of reactants.

Consider a general reaction

aA + bB → cC + dD

where a, b, c and d are the stoichiometric coefficients of reactants and products.

The rate expression for this reaction is

Rate α[A]x[B]y

where exponents x and y may or may not be equal to the stoichiometric coefficients (a and b) of the reactants.

Above equation can also be written as

Rate = k[A]x[B]y

= k[A]x[B]y

This form of equation is known as differential rate equation, where k is proportionality constant called rate constant.

The equation which relates the rate of reaction to concentration of reactants is called rate law or rate expression.

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.

For example: 2NO(g) + O2(g) → 2NO2(g)

The rate equation for this reaction will be:

Rate = k[NO]2[O2]

This differential form of this rate expression is given as:

= k[NO]2[O2]

Some other examples are given below:

1. CHCl3 + Cl2 → CCl4 + HCl

2. CH3COOC2H5 + H2O → CH3COOH + C2H5OH

Experimental rate expression:

Rate = k[CHCl3] ·

Rate = k[CH3COOC2H5][H2O]0

In these reactions, the exponents of the concentration terms are not the same as their stoichiometric coefficients. Thus, we can say that:

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain rate law and rate constant with an example.

Answer

Rate of a reaction depends upon the concentration of reactants.

Consider a general reaction

aA + bB → cC + dD

where a, b, c and d are the stoichiometric coefficients of reactants and products.

The rate expression for this reaction is

Rate α[A]x[B]y

where exponents x and y may or may not be equal to the stoichiometric coefficients (a and b) of the reactants.

Above equation can also be written as

Rate = k[A]x[B]y

= k[A]x[B]y

This form of equation is known as differential rate equation, where k is proportionality constant called rate constant.

The equation which relates the rate of reaction to concentration of reactants is called rate law or rate expression.

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.

For example: 2NO(g) + O2(g) → 2NO2(g)

The rate equation for this reaction will be:

Rate = k[NO]2[O2]

This differential form of this rate expression is given as:

= k[NO]2[O2]

Some other examples are given below:

1. CHCl3 + Cl2 → CCl4 + HCl

2. CH3COOC2H5 + H2O → CH3COOH + C2H5OH

Experimental rate expression:

Rate = k[CHCl3] ·

Rate = k[CH3COOC2H5][H2O]0

In these reactions, the exponents of the concentration terms are not the same as their stoichiometric coefficients. Thus, we can say that:

Rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation. i.e. theoretically but must be determined experimentally.

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Question

What is order of reaction ? Discuss elementary and complex reactions, Explain units of rate constant in detail.

Answer

The sum of power of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.

For reaction: aA + bB → product

Rate = k[A]x[B]y x and y indicate how sensitive the rate is to change in concentration of A and B.

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.

Order of reaction can be 0, 1, 2, 3 and even a fraction.

A zero order reaction means that the rate of reaction is independent of the concentration of reactants.

The reactions taking place in one step are called elementary reactions.

When a sequence of elementary reactions (called mechanism) gives us the products, that reactions are called complex reactions.

Units of rate constant:

For a general reaction

aA + bB → cC + dD

Rate = k[A]x[B]y

where x + y = n = order of the reaction

k = =

(where [A] = [B])

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

What is order of reaction ? Discuss elementary and complex reactions, Explain units of rate constant in detail.

Answer

The sum of power of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.

For reaction: aA + bB → product

Rate = k[A]x[B]y x and y indicate how sensitive the rate is to change in concentration of A and B.

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.

Order of reaction can be 0, 1, 2, 3 and even a fraction.

A zero order reaction means that the rate of reaction is independent of the concentration of reactants.

The reactions taking place in one step are called elementary reactions.

When a sequence of elementary reactions (called mechanism) gives us the products, that reactions are called complex reactions.

Units of rate constant:

For a general reaction

aA + bB → cC + dD

Rate = k[A]x[B]y

where x + y = n = order of the reaction

k = =

(where [A] = [B])

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

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#5 SUB 3M ✎ Edit ⚠ Log Error

Question

Write a note on molecularity of a reaction and its different types.

Answer

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.

For elementary reaction, order of a reaction and molecularity of a reaction are same.

The reaction can be unimolecular when one reacting species is involved. For example, decomposition of ammonium nitrite.

NH4NO2 → N2 + 2H2O

Bimolecular reactions involve simultaneous collision between two species, for example, dissociation of hydrogen iodide. 2HI → H2 + I2

Trimolecular or tetramolecular reactions involve simultaneous collision between three reacting species. For example, 2NO + O2 → 2NO2

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write a note on molecularity of a reaction and its different types.

Answer

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.

For elementary reaction, order of a reaction and molecularity of a reaction are same.

The reaction can be unimolecular when one reacting species is involved. For example, decomposition of ammonium nitrite.

NH4NO2 → N2 + 2H2O

Bimolecular reactions involve simultaneous collision between two species, for example, dissociation of hydrogen iodide. 2HI → H2 + I2

Trimolecular or tetramolecular reactions involve simultaneous collision between three reacting species. For example, 2NO + O2 → 2NO2

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.

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#6 SUB 0M 🖼 3 ✎ Edit ⚠ Log Error

Question

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

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.

Through mechanism of a reaction we can have information about slowest step of reaction.

The overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step.

Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in an alkaline medium,

The rate equation for this reaction is found to be: Rate = –

This reaction is first order with respect to both and .

Mechanism involve two steps:

(1)

(2)

The first step, being slow, is the rate determining step.

Thus, the rate of formation of intermediate will determine the rate of this reaction.

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.

Difference between molecularity and order of a reaction:

(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.

(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.

(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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

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.

Through mechanism of a reaction we can have information about slowest step of reaction.

The overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step.

Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in an alkaline medium,

The rate equation for this reaction is found to be: Rate = –

This reaction is first order with respect to both and .

Mechanism involve two steps:

(1)

(2)

The first step, being slow, is the rate determining step.

Thus, the rate of formation of intermediate will determine the rate of this reaction.

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.

Difference between molecularity and order of a reaction:

(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.

(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.

(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.

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#7 SUB 3M 🖼 4 ✎ Edit ⚠ Log Error

Question

Derive integrated rate equation for zero order reaction and also explain how the rate constant can be determine with help of graph.

Answer

Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants.

Consider the reaction.

R → P

Rate of reaction for this reaction can be expressed as

Rate = = k[R]0

As any quantity raised to power zero is units.

Rate = = k × 1

Thus, the rate of zero order reaction is independent from concentration of reactants.

d[R] = –k dt

Integrating both sides

[R] = –kt + I ... ... Eq. (1)

Where, I is the constant of integration

At t = 0, the concentration of the reactant
R = [R]0, where [R]0 is initial concentration of the reactant.

Substituting in equation (1)

[R]0 = –kx 0 + I

[R]0 = I

Substituting the value of I in the equation (1)

[R] = –kt + [R]0 ... ... Eq. (2)

Further simplifying equation (2)

k = ... ... Eq. (3)

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

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Derive integrated rate equation for zero order reaction and also explain how the rate constant can be determine with help of graph.

Answer

Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants.

Consider the reaction.

R → P

Rate of reaction for this reaction can be expressed as

Rate = = k[R]0

As any quantity raised to power zero is units.

Rate = = k × 1

Thus, the rate of zero order reaction is independent from concentration of reactants.

d[R] = –k dt

Integrating both sides

[R] = –kt + I ... ... Eq. (1)

Where, I is the constant of integration

At t = 0, the concentration of the reactant
R = [R]0, where [R]0 is initial concentration of the reactant.

Substituting in equation (1)

[R]0 = –kx 0 + I

[R]0 = I

Substituting the value of I in the equation (1)

[R] = –kt + [R]0 ... ... Eq. (2)

Further simplifying equation (2)

k = ... ... Eq. (3)

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

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#8 SUB 2M 🖼 1 ✎ Edit ⚠ Log Error

Question

Give examples of zero order reaction.

Answer

Zero order reaction are relatively uncommon but they occur under special condition.

Some enzyme catalysed reaction and reaction which occur on metal surface are a few examples of zero order reactions.

The decomposition of gaseous ammonia on hot platinum surface is a zero order reaction at high pressure.

2NH3(g) N2(g) + 3H2(g)

Rate = k[NH3]0 = k

In this reaction, platinum metal acts as a catalyst.

At high pressure, the metal surface gets saturated with gas molecules.

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.

The thermal decomposition of HI on gold surface is another example of zero order reaction.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Give examples of zero order reaction.

Answer

Zero order reaction are relatively uncommon but they occur under special condition.

Some enzyme catalysed reaction and reaction which occur on metal surface are a few examples of zero order reactions.

The decomposition of gaseous ammonia on hot platinum surface is a zero order reaction at high pressure.

2NH3(g) N2(g) + 3H2(g)

Rate = k[NH3]0 = k

In this reaction, platinum metal acts as a catalyst.

At high pressure, the metal surface gets saturated with gas molecules.

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.

The thermal decomposition of HI on gold surface is another example of zero order reaction.

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#9 SUB 4M 🖼 18 ✎ Edit ⚠ Log Error

Question

Derive integrated rate equation for first order reaction and also explain how the rate constant can be determine with help of graph.

OR

Derive the formula for Rate constant (k) and half life period for first order reaction.

Answer

First order reaction means that the rate of reaction is proportional to the first power of the concentration of the reactant R.

For example, consider the following reaction

R → P

Rate of reaction for this reaction can be expressed as

Rate = = k[R]

Or = –kdt

Integrating this equation, we get

ln [R] = –kt + I ... ... Eq. (1)

where, I is the constant of integration and its value can be determined easily.

When t = 0, R = [R]0, where [R]0 is the initial concentration of the reactant.

Therefore, equation (1) can be written as

ln [R]0 = –k × 0 + I

ln [R]0 = I

Substituting the value of I in equation (1)

ln [R] = –kt + ln [R]0 ... ... Eq. (2)

Rearranging this equation

ln = –kt

or k = ln ... ... Eq. (3)

At time t1 from equation (1)

ln [R]1 = – kt1 + ln [R]0 ... ... Eq. (4)

At time t2

ln [R]2 = – kt2 + ln [R]0 ... ... Eq. (5)

where, [R]1 and [R]2 are the concentration of the reactants at time t1 and t2 respectively.

Subtracting Eq. (5) from (4)

ln [R]1 – ln [R]2 = –kt1 – (–kt2)

ln = k(t2 – t1)

k = ln ... ... Eq. (6)

So, equation (2) can also be written as

ln = –kt

Taking antilog of both the sides

[R] = [R]0 e–kt ... ... Eq. (7)

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

The first order rate equation (3) can also be written in the form

k = log ... ... Eq. (8)

or log

If we plot a graph between log vs t,

the slope =

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Derive integrated rate equation for first order reaction and also explain how the rate constant can be determine with help of graph.

OR

Derive the formula for Rate constant (k) and half life period for first order reaction.

Answer

First order reaction means that the rate of reaction is proportional to the first power of the concentration of the reactant R.

For example, consider the following reaction

R → P

Rate of reaction for this reaction can be expressed as

Rate = = k[R]

Or = –kdt

Integrating this equation, we get

ln [R] = –kt + I ... ... Eq. (1)

where, I is the constant of integration and its value can be determined easily.

When t = 0, R = [R]0, where [R]0 is the initial concentration of the reactant.

Therefore, equation (1) can be written as

ln [R]0 = –k × 0 + I

ln [R]0 = I

Substituting the value of I in equation (1)

ln [R] = –kt + ln [R]0 ... ... Eq. (2)

Rearranging this equation

ln = –kt

or k = ln ... ... Eq. (3)

At time t1 from equation (1)

ln [R]1 = – kt1 + ln [R]0 ... ... Eq. (4)

At time t2

ln [R]2 = – kt2 + ln [R]0 ... ... Eq. (5)

where, [R]1 and [R]2 are the concentration of the reactants at time t1 and t2 respectively.

Subtracting Eq. (5) from (4)

ln [R]1 – ln [R]2 = –kt1 – (–kt2)

ln = k(t2 – t1)

k = ln ... ... Eq. (6)

So, equation (2) can also be written as

ln = –kt

Taking antilog of both the sides

[R] = [R]0 e–kt ... ... Eq. (7)

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

The first order rate equation (3) can also be written in the form

k = log ... ... Eq. (8)

or log

If we plot a graph between log vs t,

the slope =

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#10 SUB 2M 🖼 3 ✎ Edit ⚠ Log Error

Question

Give examples of first order reaction.

Answer

Hydrogenation of ethene is an example of first order reaction.

C2H4(g) + H2(g) → C2H6(g)

Rate = k[C2H4]

All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.

→ +

Rate = k[Ra]

Decomposition of N2O5 and N2O are some more examples of first order reactions.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Give examples of first order reaction.

Answer

Hydrogenation of ethene is an example of first order reaction.

C2H4(g) + H2(g) → C2H6(g)

Rate = k[C2H4]

All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.

→ +

Rate = k[Ra]

Decomposition of N2O5 and N2O are some more examples of first order reactions.

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#11 SUB 0M 🖼 2 ✎ Edit ⚠ Log Error

Question

Derive integrated rate equation for first order reaction containing gaseous components.

Answer

Let us consider a typical first order gas phase reaction.

A(g) → B(g) + C(g)

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

Total pressure pt = pA + pB + pC (pressure units)

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

A(g) → B(g) + C(g)

At t = 0 pi atm 0 atm 0 atm

At time t (pi– x) atm x atm x atm

where, pi is the initial pressure at time t = 0.

pt = (pi – x) + x + x = pi + x

x = (pt – pi)

where, pA = pi – x = pi – (pt – pi) = 2pi – pt

k =

= log

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Derive integrated rate equation for first order reaction containing gaseous components.

Answer

Let us consider a typical first order gas phase reaction.

A(g) → B(g) + C(g)

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

Total pressure pt = pA + pB + pC (pressure units)

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

A(g) → B(g) + C(g)

At t = 0 pi atm 0 atm 0 atm

At time t (pi– x) atm x atm x atm

where, pi is the initial pressure at time t = 0.

pt = (pi – x) + x + x = pi + x

x = (pt – pi)

where, pA = pi – x = pi – (pt – pi) = 2pi – pt

k =

= log

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#12 SUB 3M 🖼 23 ✎ Edit ⚠ Log Error

Question

What is half-life of a reaction? Derive formula for half-life of zero and first order reaction.

OR

Derive equation of rate constant and half life of reaction for the zeroth order reaction.

Answer

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.

Half-life for zero order reaction:

For a zero order reaction, rate constant is given by following equation

k =

At t = , [R] = [R]0

The rate constant at becomes

k =

=

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.

Half-life for first order reaction:

For the first order reaction,

k = log

at t = [R] =

So, the above equation becomes

k = log

or = log2

= × 0.301

=

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.

The half-life of a first order equation is readily calculated from the rate constant and vice versa.

For zero order reaction × [R]0 for first order reaction is independent of [R]0

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

What is half-life of a reaction? Derive formula for half-life of zero and first order reaction.

OR

Derive equation of rate constant and half life of reaction for the zeroth order reaction.

Answer

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.

Half-life for zero order reaction:

For a zero order reaction, rate constant is given by following equation

k =

At t = , [R] = [R]0

The rate constant at becomes

k =

=

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.

Half-life for first order reaction:

For the first order reaction,

k = log

at t = [R] =

So, the above equation becomes

k = log

or = log2

= × 0.301

=

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.

The half-life of a first order equation is readily calculated from the rate constant and vice versa.

For zero order reaction × [R]0 for first order reaction is independent of [R]0

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#13 SUB 0M 🖼 2 ✎ Edit ⚠ Log Error

Question

Explain pseudo first order reaction with example.

Answer

The order of a reaction is sometimes altered by conditions.

There are many reactions which obey first order rate law although they are higher order reactions.

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.

But water is taken in large excess for hydrolysis, therefore, concentration of water is not altered much during the reaction.

Thus, the rate of reaction is affected by concentration of ethyl acetate.

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.

CH3COOC2H5 + H2O CH3COOH + C2H5OH

t = 0 0.01 mol 10 mol 0 mol 0 mol

t = t 0 mol 9.99 mol 0.01 mol 0.01 mol

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.

Inversion of cane sugar is another pseudo first order reaction.

C12H22O11 + H2O C6H12O6 + C6H12O6

Cane sugar Glucose Fructose

Rate = k [C12H22O11]

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain pseudo first order reaction with example.

Answer

The order of a reaction is sometimes altered by conditions.

There are many reactions which obey first order rate law although they are higher order reactions.

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.

But water is taken in large excess for hydrolysis, therefore, concentration of water is not altered much during the reaction.

Thus, the rate of reaction is affected by concentration of ethyl acetate.

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.

CH3COOC2H5 + H2O CH3COOH + C2H5OH

t = 0 0.01 mol 10 mol 0 mol 0 mol

t = t 0 mol 9.99 mol 0.01 mol 0.01 mol

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.

Inversion of cane sugar is another pseudo first order reaction.

C12H22O11 + H2O C6H12O6 + C6H12O6

Cane sugar Glucose Fructose

Rate = k [C12H22O11]

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Question

What is activation energy? Explain with the help of graph by suitable example.

Answer

Activation energy can be understood clearly using the formation reaction of hydrogen iodide.

H2(g) + I2(g) → 2HI(g)

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.

Intermediate

The energy required to form this intermediate, called activated complex (C), is known as activation energy (Ea).

Below figure is obtained by plotting potential energy vs reaction coordinate. Reaction coordinate represents the profile of energy change when reactant change into products.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

What is activation energy? Explain with the help of graph by suitable example.

Answer

Activation energy can be understood clearly using the formation reaction of hydrogen iodide.

H2(g) + I2(g) → 2HI(g)

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.

Intermediate

The energy required to form this intermediate, called activated complex (C), is known as activation energy (Ea).

Below figure is obtained by plotting potential energy vs reaction coordinate. Reaction coordinate represents the profile of energy change when reactant change into products.

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.

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#15 SUB 4M 🖼 10 ✎ Edit ⚠ Log Error

Question

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

Activation energy using graph:

Taking natural logarithm on both the side of Arrhenius equation k = Ae

ln k = + ln A ... ... Eq. (1)

The plot of ln k vs 1/T gives a straight line as shown in figure.

In figure, slope = and intercept = lnA. So, we can calculate activation energy (Ea) and Arrhenius constant A using these values.

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.

Thus, at temperature T1, equation (1) is

ln k1 = + lnA ... ... Eq. (2)

at temperature T2, equation (1) is

ln k2 = + lnA ... ... Eq. (3)

(since A is constant for given reaction)

k1 and k2 are rate constant at temperatures T1 and T2 respectively.

Subtracting equation (2) from (3), we obtain

lnk2 – lnk1 =

ln

log

log

From above formula activation energy can be calculate using measured values of rate constants at different temperatures.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Activation energy using graph:

Taking natural logarithm on both the side of Arrhenius equation k = Ae

ln k = + ln A ... ... Eq. (1)

The plot of ln k vs 1/T gives a straight line as shown in figure.

In figure, slope = and intercept = lnA. So, we can calculate activation energy (Ea) and Arrhenius constant A using these values.

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.

Thus, at temperature T1, equation (1) is

ln k1 = + lnA ... ... Eq. (2)

at temperature T2, equation (1) is

ln k2 = + lnA ... ... Eq. (3)

(since A is constant for given reaction)

k1 and k2 are rate constant at temperatures T1 and T2 respectively.

Subtracting equation (2) from (3), we obtain

lnk2 – lnk1 =

ln

log

log

From above formula activation energy can be calculate using measured values of rate constants at different temperatures.

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#16 SUB 2M 🖼 2 ✎ Edit ⚠ Log Error

Question

Explain the effect of catalyst on rate of reaction and also give its characteristics.

Or

Answer

Answer the following questions on catalyst:

(i) How it increases the rate of reaction.

(ii) Write any 4 characteristics of catalyst.

Ans. A catalyst is a substance which increase the rate of a reaction without itself undergoing any permanent chemical change.

For example, MnO2 catalyses the following reaction so as to increase its rate considerably.

2KClO3 2KCl + 3O2

The word catalyst should not be used when the added substance reduces the rate of reaction. The substance is then called inhibitor.

The action of catalyst can be explained by intermediate complex theory.

According to this theory, a catalyst participates in a chemical reaction by forming temporary bond with the reactants resulting in an intermediate complex.

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.

It is clear from Arrhenius equation that lower the value of activation energy faster will be the rate of reaction.

This has transitory existence and decomposes to yield products and the catalyst.

Characteristics of catalyst:

A small amount of catalyst can catalyses a large amount of reactants.

A catalyst does not alter Gibbs energy of a reaction.

It catalysis the spontaneous reaction but does not catalyze the non-spontaneous reactions.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain the effect of catalyst on rate of reaction and also give its characteristics.

Or

Answer

Answer the following questions on catalyst:

(i) How it increases the rate of reaction.

(ii) Write any 4 characteristics of catalyst.

Ans. A catalyst is a substance which increase the rate of a reaction without itself undergoing any permanent chemical change.

For example, MnO2 catalyses the following reaction so as to increase its rate considerably.

2KClO3 2KCl + 3O2

The word catalyst should not be used when the added substance reduces the rate of reaction. The substance is then called inhibitor.

The action of catalyst can be explained by intermediate complex theory.

According to this theory, a catalyst participates in a chemical reaction by forming temporary bond with the reactants resulting in an intermediate complex.

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.

It is clear from Arrhenius equation that lower the value of activation energy faster will be the rate of reaction.

This has transitory existence and decomposes to yield products and the catalyst.

Characteristics of catalyst:

A small amount of catalyst can catalyses a large amount of reactants.

A catalyst does not alter Gibbs energy of a reaction.

It catalysis the spontaneous reaction but does not catalyze the non-spontaneous reactions.

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.

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#17 SUB 2M 🖼 1 ✎ Edit ⚠ Log Error

Question

Write Arrhenius equation. Explain the terms involved in it.

Answer

The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation.

k = A

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write Arrhenius equation. Explain the terms involved in it.

Answer

The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation.

k = A

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).

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#18 SUB 4M 🖼 4 ✎ Edit ⚠ Log Error

Question

Write a note on collision theory of chemical reaction.

Or

What is meant by collision theory? Explain.

Answer

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.

It is based on kinetic theory of gases.

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.

The number of collision per second per unit volume of the reaction mixture is known as collision frequency (Z).

Another factor which affects the rate of chemical reaction is activation energy.

For a bimolecular elementary reaction;

A + B → Products

Rate of reaction can be expressed as:

Rate = ZAB ... ... Eq. (1)

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.

Comparing equation (1) with Arrhenius equation, we can say that A is related to collision frequency.

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.

The reason could be that all collision do not lead to the formation of products.

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.

For example, formation of methanol from bromomethane CH3Br + OH – → CH3OH + Br– depends upon the orientation of reactant molecule as shown in figure.

The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no product are formed.

To account for effective collision, another factor P, called the probability factor or steric factor is introduced.

It takes into account the fact that in a collision, molecules be properly oriented i.e.

Rate = PZAB

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.

Class 12 Chemistry (Part 1) 012

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write a note on collision theory of chemical reaction.

Or

What is meant by collision theory? Explain.

Answer

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.

It is based on kinetic theory of gases.

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.

The number of collision per second per unit volume of the reaction mixture is known as collision frequency (Z).

Another factor which affects the rate of chemical reaction is activation energy.

For a bimolecular elementary reaction;

A + B → Products

Rate of reaction can be expressed as:

Rate = ZAB ... ... Eq. (1)

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.

Comparing equation (1) with Arrhenius equation, we can say that A is related to collision frequency.

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.

The reason could be that all collision do not lead to the formation of products.

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.

For example, formation of methanol from bromomethane CH3Br + OH – → CH3OH + Br– depends upon the orientation of reactant molecule as shown in figure.

The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no product are formed.

To account for effective collision, another factor P, called the probability factor or steric factor is introduced.

It takes into account the fact that in a collision, molecules be properly oriented i.e.

Rate = PZAB

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.

Class 12 Chemistry (Part 1) 012

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#19 SUB 2M 🖼 4 ✎ Edit ⚠ Log Error

Question

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

Average rate =

=

=

=

= 4 × 10–4 M min–1

Rate of reaction in terms of second

=

= 6.67 ×10–6 M s–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Average rate =

=

=

=

= 4 × 10–4 M min–1

Rate of reaction in terms of second

=

= 6.67 ×10–6 M s–1

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#20 SUB 2M 🖼 3 ✎ Edit ⚠ Log Error

Question

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

For reaction 2A → Product,

∆t = 10 min

Rate =

=

= mol L–1 min–1

= 5.0 ×10–3 M min–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

For reaction 2A → Product,

∆t = 10 min

Rate =

=

= mol L–1 min–1

= 5.0 ×10–3 M min–1

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#21 SUB 1M 🖼 2 ✎ Edit ⚠ Log Error

Question

For reaction A + B → Product; the rate law is given by, r = k[A][B]2 What is the order of the reaction?

Answer

Overall order of reaction is equal to the sum of powers of the concentration of reactants

Here, order of the reaction = + 2 = 2.5

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

For reaction A + B → Product; the rate law is given by, r = k[A][B]2 What is the order of the reaction?

Answer

Overall order of reaction is equal to the sum of powers of the concentration of reactants

Here, order of the reaction = + 2 = 2.5

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#22 SUB 2M 🖼 1 ✎ Edit ⚠ Log Error

Question

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

X → Y is a second order reaction,

Therefore, rate law can be expressed as

V1 = k[X]2

If the concentration of X is increased three time, then the new concentration will be 3X and the new rate will be

V2 = k[3X]2 = 9k[X]2

Taking ratio of V2 and V1, we get

V2 = 9V1

Hence the rate of formation of Y will increase by 9 times.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

X → Y is a second order reaction,

Therefore, rate law can be expressed as

V1 = k[X]2

If the concentration of X is increased three time, then the new concentration will be 3X and the new rate will be

V2 = k[3X]2 = 9k[X]2

Taking ratio of V2 and V1, we get

V2 = 9V1

Hence the rate of formation of Y will increase by 9 times.

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#23 SUB 3M 🖼 5 ✎ Edit ⚠ Log Error

Question

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

Integrated rate equation for first order reaction,

k = log

[R]0 = 5 g and [R] = 3 g, k = 1.15 × 10–3 s–1

Substituting above values in equation

t = log

t = log

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Integrated rate equation for first order reaction,

k = log

[R]0 = 5 g and [R] = 3 g, k = 1.15 × 10–3 s–1

Substituting above values in equation

t = log

t = log

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#24 SUB 2M 🖼 4 ✎ Edit ⚠ Log Error

Question

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

Time required to decompose half of the reagent is 60 minutes, i.e. = 60 min.

For first order reaction, relation between half-life and rate constant k is

k = = 0.01155 min–1 OR

k = = 1.925 ×10– 4 s–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Time required to decompose half of the reagent is 60 minutes, i.e. = 60 min.

For first order reaction, relation between half-life and rate constant k is

k = = 0.01155 min–1 OR

k = = 1.925 ×10– 4 s–1

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#25 SUB 2M ✎ Edit ⚠ Log Error

Question

Explain effect of temperature on rate constant.

Or

What will be the effect of temperature on rate constant?

Answer

Most of the chemical reactions are accelerated by increase in temperature.

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.

In a mixture of potassium permanganate and oxalic acid, potassium permanganate gets decolourised faster at a higher temperature than that at a lower temperature.

It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain effect of temperature on rate constant.

Or

What will be the effect of temperature on rate constant?

Answer

Most of the chemical reactions are accelerated by increase in temperature.

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.

In a mixture of potassium permanganate and oxalic acid, potassium permanganate gets decolourised faster at a higher temperature than that at a lower temperature.

It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.

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#26 SUB 3M 🖼 3 ✎ Edit ⚠ Log Error

Question

The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K, calculate Ea.

Answer

log =

Here, T1 = 298K, T2 = 298 + 10K = 308K

R = 8.314 J K–1 mol–1

The rate of chemical reaction doubles for an increase of 10K in absolute temperature, therefore = 2.

Substituting the values in above equation, we get

log 2.0 =

Ea =

Ea =

Ea = 52897.77 J mol–1 = 52.89777 kJ mol–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K, calculate Ea.

Answer

log =

Here, T1 = 298K, T2 = 298 + 10K = 308K

R = 8.314 J K–1 mol–1

The rate of chemical reaction doubles for an increase of 10K in absolute temperature, therefore = 2.

Substituting the values in above equation, we get

log 2.0 =

Ea =

Ea =

Ea = 52897.77 J mol–1 = 52.89777 kJ mol–1

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#27 SUB 3M 🖼 4 ✎ Edit ⚠ Log Error

Question

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

The fraction of molecules of reactants having energy equal to or greater than activation energy x =

Taking 10 base log on both the sides,

log x =

∴ log x =

∴ log x = –18.8323

∴ x = antilog (–18.8323) = .1677

∴ The fraction of molecules of reactants having energy equal to or greater than activation energy
x = 1.471 × 10–19

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

The fraction of molecules of reactants having energy equal to or greater than activation energy x =

Taking 10 base log on both the sides,

log x =

∴ log x =

∴ log x = –18.8323

∴ x = antilog (–18.8323) = .1677

∴ The fraction of molecules of reactants having energy equal to or greater than activation energy
x = 1.471 × 10–19

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#28 SUB 2M 🖼 10 ✎ Edit ⚠ Log Error

Question

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3NO(g) → N2O(g) + NO2(g); Rate = k[NO]2

(ii) H2O2(aq) + 3l–(aq) + 2H+ → 2H2O(l) + I3–;

Rate = k[H2O2][I –]

(iii) CH3CHO(g) → CH4(g) + CO(g);

Rate = k[CH3CHO]

(iv) C2H5Cl(g) → C2H4(g) + HCl(g);

Rate = k[C2H5Cl]

Answer

(i) 3NO(g) → N2O(g) + NO2(g)

Rate = k[NO]2

∴ Order of reaction = 2

Dimension of the rate constant:

Rate = k[NO]2

k = = = mol–1 L s–1

(ii) H2O2(aq) + 3I –(aq) + 2H+ → 2H2O(l) + I3–

Rate = k[H2O2][I –]

Order of reaction with respect to H2O2 = 1

order of reaction with respect to l–1 = 1

∴ overall order of reaction = 2

Dimension of the rate constant.

Rate = k[H2O2][I–]

k =

=

= mol–1 L s–1

(iii) CH3CHO(g) → CH4(g) + CO(g)

Rate = k[CH3CHO]

∴ Order of reaction =

Dimension of the rate constant

Rate = k[CH3CHO]

k =

=

=

(iv) C2H5Cl(g) → C2H4(g) + HCl(g)

Rate = k[C2H5Cl]

∴ Order of reaction = 1

Dimension of the rate constant.

Rate = k[C2H5Cl]

k =

= = s–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3NO(g) → N2O(g) + NO2(g); Rate = k[NO]2

(ii) H2O2(aq) + 3l–(aq) + 2H+ → 2H2O(l) + I3–;

Rate = k[H2O2][I –]

(iii) CH3CHO(g) → CH4(g) + CO(g);

Rate = k[CH3CHO]

(iv) C2H5Cl(g) → C2H4(g) + HCl(g);

Rate = k[C2H5Cl]

Answer

(i) 3NO(g) → N2O(g) + NO2(g)

Rate = k[NO]2

∴ Order of reaction = 2

Dimension of the rate constant:

Rate = k[NO]2

k = = = mol–1 L s–1

(ii) H2O2(aq) + 3I –(aq) + 2H+ → 2H2O(l) + I3–

Rate = k[H2O2][I –]

Order of reaction with respect to H2O2 = 1

order of reaction with respect to l–1 = 1

∴ overall order of reaction = 2

Dimension of the rate constant.

Rate = k[H2O2][I–]

k =

=

= mol–1 L s–1

(iii) CH3CHO(g) → CH4(g) + CO(g)

Rate = k[CH3CHO]

∴ Order of reaction =

Dimension of the rate constant

Rate = k[CH3CHO]

k =

=

=

(iv) C2H5Cl(g) → C2H4(g) + HCl(g)

Rate = k[C2H5Cl]

∴ Order of reaction = 1

Dimension of the rate constant.

Rate = k[C2H5Cl]

k =

= = s–1

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#29 SUB 3M 🖼 1 ✎ Edit ⚠ Log Error

Question

For the reaction: 2A + B → A2B

the 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

Initial rate = k[A][B]2

= 2.0 × 10–6 mol–2 L2 s–1 × 0.1 mol L–1

× (0.2 mol L–1)2

= 8.0 × 10–9 mol L–1 s–1

Concentration of A is reduce to 0.06 mol L–1

Concentration of [A] reacted = 0.10 – 0.06

= 0.04 mol L–1

As per the reaction, 2A + B → A2B

Half of the concentration of B is reduce then A

∴ Concentration of [B] reacted = X 0.04

= 0.02 mol L–1

[B] = 0.2 – 0.02 = 0.18 mol L–1

Rate = k [A][B]2

= 2.0 × 10–6 mol–2 L2 s–1 × 0.06 mol L–1

× (0.18 mol L–1)2

= 3.88 × 10–9 mol L–1 s–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

For the reaction: 2A + B → A2B

the 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

Initial rate = k[A][B]2

= 2.0 × 10–6 mol–2 L2 s–1 × 0.1 mol L–1

× (0.2 mol L–1)2

= 8.0 × 10–9 mol L–1 s–1

Concentration of A is reduce to 0.06 mol L–1

Concentration of [A] reacted = 0.10 – 0.06

= 0.04 mol L–1

As per the reaction, 2A + B → A2B

Half of the concentration of B is reduce then A

∴ Concentration of [B] reacted = X 0.04

= 0.02 mol L–1

[B] = 0.2 – 0.02 = 0.18 mol L–1

Rate = k [A][B]2

= 2.0 × 10–6 mol–2 L2 s–1 × 0.06 mol L–1

× (0.18 mol L–1)2

= 3.88 × 10–9 mol L–1 s–1

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#30 SUB 2M 🖼 6 ✎ Edit ⚠ Log Error

Question

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

The decomposition reaction of ammonia is

2NH3(g) → N2(g) + 3H2(g),

This is zero order reaction,

Rate = =

= = k[NH3]0

Rate of appearance of N2 =

= k[NH3]0 = k

= 2.5 × 10–4 mol L–1 s–1

Rate of appearance of H2 =

= k[NH3]0 = k

∴ = 3k = 3 × 2.5 × 10– 4 mol L–1 s–1

= 7.5 × 10–4 mol L–1 s–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

The decomposition reaction of ammonia is

2NH3(g) → N2(g) + 3H2(g),

This is zero order reaction,

Rate = =

= = k[NH3]0

Rate of appearance of N2 =

= k[NH3]0 = k

= 2.5 × 10–4 mol L–1 s–1

Rate of appearance of H2 =

= k[NH3]0 = k

∴ = 3k = 3 × 2.5 × 10– 4 mol L–1 s–1

= 7.5 × 10–4 mol L–1 s–1

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#31 SUB 0M 🖼 5 ✎ Edit ⚠ Log Error

Question

The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k[CH3OCH3]

The 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 =

If the pressure is measured in bar and time in minutes, then what are the units of rate and rate costants ?

Answer

The pressure is measured in bar and time in minutes,

So, the unit of rate = Bar min–1

Unit of K =

=

= min–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k[CH3OCH3]

The 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 =

If the pressure is measured in bar and time in minutes, then what are the units of rate and rate costants ?

Answer

The pressure is measured in bar and time in minutes,

So, the unit of rate = Bar min–1

Unit of K =

=

= min–1

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#32 SUB 0M ✎ Edit ⚠ Log Error

Question

Mention the factors that affect the rate of a chemical reaction.

Answer

Factors affecting the rate of a chemical reaction:

(i) Nature 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.

(ii) Concentration of reactant: More is the concentration of reactant, more will be the rate of reaction.

(iii) Pressure: On increasing pressure, the collision between reactant molecule increases, which results in to increase in the rate of reaction.

(iv) Temperature: In most of the cases, on increasing temperature the rate of reaction and rate constant both increase.

For endothermic reaction the rate of reaction increases on increasing temperature, whereas for exothermic reaction the rate of reaction decreases on increasing temperature.

(v) Catalyst: 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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Mention the factors that affect the rate of a chemical reaction.

Answer

Factors affecting the rate of a chemical reaction:

(i) Nature 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.

(ii) Concentration of reactant: More is the concentration of reactant, more will be the rate of reaction.

(iii) Pressure: On increasing pressure, the collision between reactant molecule increases, which results in to increase in the rate of reaction.

(iv) Temperature: In most of the cases, on increasing temperature the rate of reaction and rate constant both increase.

For endothermic reaction the rate of reaction increases on increasing temperature, whereas for exothermic reaction the rate of reaction decreases on increasing temperature.

(v) Catalyst: The energy of activation decreases on using catalyst. Thus the value of energy barrier decreases for reaction, hence the rate of reaction increases.

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#33 SUB 2M 🖼 4 ✎ Edit ⚠ Log Error

Question

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

Reaction → Products

Rate r1 = k[A]2

(i) When concentration of reactant is doubled;

Rate r2 = k[2A]2 = 4k[A]2 = 4r1

∴ When concentration of the reactant is doubled, the rate of reaction will become 4 times of the initial rate.

(ii) When concentration of reactant is reduced to half;

Rate r2 = = k[A]2 =

∴ when concentration of the reactant is reduced by half, the rate of reaction will be reduce to of the initial rate.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Reaction → Products

Rate r1 = k[A]2

(i) When concentration of reactant is doubled;

Rate r2 = k[2A]2 = 4k[A]2 = 4r1

∴ When concentration of the reactant is doubled, the rate of reaction will become 4 times of the initial rate.

(ii) When concentration of reactant is reduced to half;

Rate r2 = = k[A]2 =

∴ when concentration of the reactant is reduced by half, the rate of reaction will be reduce to of the initial rate.

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#34 SUB 2M 🖼 2 ✎ Edit ⚠ Log Error

Question

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

Most of the chemical reactions are accelerated by increase in temperature.

Increasing the temperature will result in an exponential increase in the rate constant.

It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.

Rate = (2)

According to Swedish chemist Arrhenius, the quantitative effect of temperature on the rate constant can be express by following equation.

k = A

Where, A = Arrhenius factor or frequency factor.

R = gas constant

Ea = activation energy in terms of
joules/mole (J mol–1).

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Most of the chemical reactions are accelerated by increase in temperature.

Increasing the temperature will result in an exponential increase in the rate constant.

It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.

Rate = (2)

According to Swedish chemist Arrhenius, the quantitative effect of temperature on the rate constant can be express by following equation.

k = A

Where, A = Arrhenius factor or frequency factor.

R = gas constant

Ea = activation energy in terms of
joules/mole (J mol–1).

✓ Apply Cancel

#35 SUB 2M 🖼 4 ▦ 1 ✎ Edit ⚠ Log Error

Question

In a pseudo first order reaction in water, the following results were obtained:

calculate the average rate of reaction between the time interval 30 to 60 second.

Time/s	0	30	60	90
[A]/mol L–1	0.55	0.31	0.17	0.085

Answer

Hydrolysis of ester is pseudo first order reaction.

RCOOR' + H2O RCOOH + R'OH

Ester Water Acid Alcohol

Average rate of reaction rav =

rav = –

[R]2 = concentration at 60 s = 0.17 M

[R]1 = concentration at 30 s = 0.31 M

rav =

rav = = 4.67 × 10–3 mol L–1 s–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

In a pseudo first order reaction in water, the following results were obtained:

calculate the average rate of reaction between the time interval 30 to 60 second.

Answer

Hydrolysis of ester is pseudo first order reaction.

RCOOR' + H2O RCOOH + R'OH

Ester Water Acid Alcohol

Average rate of reaction rav =

rav = –

[R]2 = concentration at 60 s = 0.17 M

[R]1 = concentration at 30 s = 0.31 M

rav =

rav = = 4.67 × 10–3 mol L–1 s–1

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#36 SUB 2M ✎ Edit ⚠ Log Error

Question

A reaction is first order in A and second
order in B.

(i) Write the differential rate equation. [June 2025]

(ii) How is the rate affected on increasing the concentration of B three times ?

(iii) How is the rate affected when the concentration of both A and B are doubled ?

[Topic 3.2] [June 2025] [2 Marks]

Answer

The given reaction is first order in A and second order in B.

(i) Rate = k[A][B]2

(ii) Initial rate r1 = k[A][B]2

When concentration of B is increased to three times,

Rate r2 = k[A][3B]2 = 9k [A][B]2 = 9r1

∴ Rate becomes nine times.

(iii) Initial rate r1 = k[A][B]2

When concentration of both A and B is increased to two times,

Rate r2 = k[2A][2B]2 = 8k [A][B]2 = 8r1

∴ Rate becomes eight times.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

A reaction is first order in A and second
order in B.

(i) Write the differential rate equation. [June 2025]

(ii) How is the rate affected on increasing the concentration of B three times ?

(iii) How is the rate affected when the concentration of both A and B are doubled ?

[Topic 3.2] [June 2025] [2 Marks]

Answer

The given reaction is first order in A and second order in B.

(i) Rate = k[A][B]2

(ii) Initial rate r1 = k[A][B]2

When concentration of B is increased to three times,

Rate r2 = k[A][3B]2 = 9k [A][B]2 = 9r1

∴ Rate becomes nine times.

(iii) Initial rate r1 = k[A][B]2

When concentration of both A and B is increased to two times,

Rate r2 = k[2A][2B]2 = 8k [A][B]2 = 8r1

∴ Rate becomes eight times.

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#37 SUB 4M 🖼 3 ▦ 1 ✎ Edit ⚠ Log Error

Question

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:

A/mol L–1	0.20	0.20	0.40
B/mol L–1	0.30	0.10	0.05
r0/mol L–1 s–1	5.07 × 10–5	5.07 × 10–5	1.43 × 10–4

Answer

Suppose,

Order of reaction with respect to A = x

Order of reaction with respect to B = y

Then,

Rate = k[A]x[B]y

∴ r1 = k[0.20]x[0.30]y = 5.07 ×10–5 ... ... (1)

r2= k[0.20]x[0.10]y = 5.07 ×10–5 ... ... (2)

r3 = k[0.40]x [0.05]y = 1.43 ×10– 4 ... ... (3)

Dividing (1) and (2) we get,

∴ = [3]y = 1

∴ y = 0, because [3]0 = 1 = [3]y

Dividing (3) and (2) we get,

= 2.82

∴ = 2.82

∴ [2]x = 2.82

Taking log of both the side we get,

log[2]x = log2.82

∴ x log2 = log2.82

x × 0.3010 = 0.4503

∴ x = = 1.5

∴ order of reaction with respect to A = 1.5

order of reaction with respect to B = 0

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Suppose,

Order of reaction with respect to A = x

Order of reaction with respect to B = y

Then,

Rate = k[A]x[B]y

∴ r1 = k[0.20]x[0.30]y = 5.07 ×10–5 ... ... (1)

r2= k[0.20]x[0.10]y = 5.07 ×10–5 ... ... (2)

r3 = k[0.40]x [0.05]y = 1.43 ×10– 4 ... ... (3)

Dividing (1) and (2) we get,

∴ = [3]y = 1

∴ y = 0, because [3]0 = 1 = [3]y

Dividing (3) and (2) we get,

= 2.82

∴ = 2.82

∴ [2]x = 2.82

Taking log of both the side we get,

log[2]x = log2.82

∴ x log2 = log2.82

x × 0.3010 = 0.4503

∴ x = = 1.5

∴ order of reaction with respect to A = 1.5

order of reaction with respect to B = 0

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#38 SUB 4M 🖼 2 ▦ 1 ✎ Edit ⚠ Log Error

Question

The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D

Determine the rate law and the rate constant for the reaction.

Experiment	[A]	[B]	Initial rate of formation of D/mol L–1 min–1
	mol L–1	mol L–1
I	0.1	0.1	6.0 × 10–3
II	0.3	0.2	7.2 × 10–2
III	0.3	0.4	2.88 × 10–1
IV	0.4	0.1	2.40 × 10–2

Answer

Suppose,

Order of reaction with respect to A = x

Order of reaction with respect to B = y

∴ Rate = k[A]x[B]y

∴ r1 = k[0.1]x[0.1]y = 6.0 × 10–3 ... ... (1)

r2 = k[0.3]x[0.2]y = 7.2 × 10–2 ... ... (2)

r3 = k[0.3]x[0.4]y = 2.88 × 10–1... ... (3)

r4 = k[0.4]x[0.1]y = 2.40 × 10–2... ... (4)

Dividing (4) and (1) we get,

∴ = [4]x = [4]1

Order of reaction with respect to A = 1

Dividing (3) and (2) we get,

∴ = [2]y = 4 = [2]2

∴ y = 2

∴ order of reaction with respect to A = 1

order of reaction with respect to B = 2

overall order of reaction = 3

∴ rate equation,

Rate = k[A]1[B]2

Calculation of rate constant k

From equation (1)

r1 = k[0.1]1[0.1]2 = 6.0 × 10–3

k =

=

= 6.0 mol–2 L–2 min–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D

Determine the rate law and the rate constant for the reaction.

Answer

Suppose,

Order of reaction with respect to A = x

Order of reaction with respect to B = y

∴ Rate = k[A]x[B]y

∴ r1 = k[0.1]x[0.1]y = 6.0 × 10–3 ... ... (1)

r2 = k[0.3]x[0.2]y = 7.2 × 10–2 ... ... (2)

r3 = k[0.3]x[0.4]y = 2.88 × 10–1... ... (3)

r4 = k[0.4]x[0.1]y = 2.40 × 10–2... ... (4)

Dividing (4) and (1) we get,

∴ = [4]x = [4]1

Order of reaction with respect to A = 1

Dividing (3) and (2) we get,

∴ = [2]y = 4 = [2]2

∴ y = 2

∴ order of reaction with respect to A = 1

order of reaction with respect to B = 2

overall order of reaction = 3

∴ rate equation,

Rate = k[A]1[B]2

Calculation of rate constant k

From equation (1)

r1 = k[0.1]1[0.1]2 = 6.0 × 10–3

k =

=

= 6.0 mol–2 L–2 min–1

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#39 SUB 3M 🖼 3 ▦ 1 ✎ Edit ⚠ Log Error

Question

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:

Experiment	[A]	[B]	Initial rate/ mol L–1 min–1
	mol L–1	mol L–1
I	0.1	0.1	2.0 × 10–2
II	–	0.2	4.0 × 10–2
III	0.4	0.4	–
IV	–	0.2	2.0 × 10–2

Answer

Reaction A → B

Rate = k[A]1[B]0

Tate constant in Experiment-I:

Rate = k[A]

∴ k =

∴ k = 2.0 × 10–1 = 0.2 min–1

[A] in Experiment-II:

Rate = k[A]

∴ [A] =

∴ [A] = 0.2 mol L–1

Initial rate in Experiment-III:

Rate = k[A]

∴ rate = 0.2 × 0.4 = 0.08 mol L–1 min–1

[A] in Experiment-IV:

Rate = k[A]

∴ [A] =

∴ [A] = 0.1 mol L–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Reaction A → B

Rate = k[A]1[B]0

Tate constant in Experiment-I:

Rate = k[A]

∴ k =

∴ k = 2.0 × 10–1 = 0.2 min–1

[A] in Experiment-II:

Rate = k[A]

∴ [A] =

∴ [A] = 0.2 mol L–1

Initial rate in Experiment-III:

Rate = k[A]

∴ rate = 0.2 × 0.4 = 0.08 mol L–1 min–1

[A] in Experiment-IV:

Rate = k[A]

∴ [A] =

∴ [A] = 0.1 mol L–1

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#40 SUB 1M 🖼 8 ✎ Edit ⚠ Log Error

Question

Calculate the half-life of a first order reaction from their rate constants given below:

(1) 200 s–1 (ii) 2 min–1 (iii) 4 years–1

Answer

For first order reaction =

(i) = 3.465 × 10–3 s

(ii) = 0.3465 min

(iii) = 0.17325 year

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Calculate the half-life of a first order reaction from their rate constants given below:

(1) 200 s–1 (ii) 2 min–1 (iii) 4 years–1

Answer

For first order reaction =

(i) = 3.465 × 10–3 s

(ii) = 0.3465 min

(iii) = 0.17325 year

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#41 SUB 4M 🖼 9 ✎ Edit ⚠ Log Error

Question

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

Half-life of 14C = 5730 years.

All radioactive decay follows first order kinetics.

k =

k = year–1

For first order reaction.

t = log

t = log

t = log1.25

t = × 0.0969

t = 1845.2 years.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Half-life of 14C = 5730 years.

All radioactive decay follows first order kinetics.

k =

k = year–1

For first order reaction.

t = log

t = log

t = log1.25

t = × 0.0969

t = 1845.2 years.

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#42 SUB 4M 🖼 13 ▦ 1 ✎ Edit ⚠ Log Error

Question

The experimental data for decomposition of N2O5; [2N2O5 → 4NO2 + O2]

in gas phase at 318K are given below:

(i) Plot [N2O5] against t.

(ii) Find the half-life period for the reaction.

(iii) Draw a graph between log [N2O5] and t

(iv) What is the rate law?

(v) Calculate the rate constant.

(vi) Calculate the half-life period from k and compare it with (ii).

Time/s	0	400	800	1200	1600	2000	2400	2800	3200
102× [N2O5] mol L–1	1.63	1.36	1.14	0.93	0.78	0.64	0.53	0.43	0.35

Answer

(i)

(ii) Initial concentration of N2O5

= 1.63 × 10–2 M

Half of this concentration = 0.815 × 10–2 M

From graph, time corresponding to this concentration is 1400 s.

Hence, = 1400 s.

(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

(iv) As plot of log [N2O5] → t is straight line. Hence, it is first order reaction.

∴ Rate = k [N2O5]

(v) From plot of log [N2O5] → t

Slope =

∴ k = – slope × 2.303

= × 2.303

= × 2.303

= 4.82 × 10–4 s–1

OR

From formula of integrated rate law for first order reaction,

k = log

∴ k = log

∴ k = log4.6571

∴ k = × 0.6681

∴ k = 4.81 × 10– 4 s–1

(vi) = =

= 0.1437 × 104 s = 1437 s

Hence, half-life from both (ii) and (vi) are nearly same.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

The experimental data for decomposition of N2O5; [2N2O5 → 4NO2 + O2]

in gas phase at 318K are given below:

(i) Plot [N2O5] against t.

(ii) Find the half-life period for the reaction.

(iii) Draw a graph between log [N2O5] and t

(iv) What is the rate law?

(v) Calculate the rate constant.

(vi) Calculate the half-life period from k and compare it with (ii).

Answer

(i)

(ii) Initial concentration of N2O5

= 1.63 × 10–2 M

Half of this concentration = 0.815 × 10–2 M

From graph, time corresponding to this concentration is 1400 s.

Hence, = 1400 s.

(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

(iv) As plot of log [N2O5] → t is straight line. Hence, it is first order reaction.

∴ Rate = k [N2O5]

(v) From plot of log [N2O5] → t

Slope =

∴ k = – slope × 2.303

= × 2.303

= × 2.303

= 4.82 × 10–4 s–1

OR

From formula of integrated rate law for first order reaction,

k = log

∴ k = log

∴ k = log4.6571

∴ k = × 0.6681

∴ k = 4.81 × 10– 4 s–1

(vi) = =

= 0.1437 × 104 s = 1437 s

Hence, half-life from both (ii) and (vi) are nearly same.

✓ Apply Cancel

#43 SUB 3M 🖼 7 ✎ Edit ⚠ Log Error

Question

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

Rate constant k = 60 s–1

If, initial concentration = [R]0 M

Then, concentration at time t = M

For first order reaction,

t = log

= log

= log16

= × 1.2044

= 4.62 × 10-2 Sec.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Rate constant k = 60 s–1

If, initial concentration = [R]0 M

Then, concentration at time t = M

For first order reaction,

t = log

= log

= log16

= × 1.2044

= 4.62 × 10-2 Sec.

✓ Apply Cancel

#44 SUB 3M 🖼 21 ✎ Edit ⚠ Log Error

Question

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

Nuclear explosion is first order reaction.

= 28.1 years and [R]0 = 1 µg.

For first order reaction,

k = = = 2.466 × 10–2 year–1

Calculation for amount of 90Sr remain after 10 years:

t = log

∴ log =

∴ log =

∴ log = 0.1070

∴ = anti log0.1070 = 1.2794

∴ [R]t = = 0.7816 µg 90Sr is left after 10 years.

Calculation for amount of 90Sr remain after
60 years:

t = log

∴ log =

∴ log =

∴ log = 0.6425

∴ = antilog 0.6425 = 4.39

∴ [R]t = = 0.228 µg 90Sr is left after 60 years.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Nuclear explosion is first order reaction.

= 28.1 years and [R]0 = 1 µg.

For first order reaction,

k = = = 2.466 × 10–2 year–1

Calculation for amount of 90Sr remain after 10 years:

t = log

∴ log =

∴ log =

∴ log = 0.1070

∴ = anti log0.1070 = 1.2794

∴ [R]t = = 0.7816 µg 90Sr is left after 10 years.

Calculation for amount of 90Sr remain after
60 years:

t = log

∴ log =

∴ log =

∴ log = 0.6425

∴ = antilog 0.6425 = 4.39

∴ [R]t = = 0.228 µg 90Sr is left after 60 years.

✓ Apply Cancel

#45 SUB 3M 🖼 13 ✎ Edit ⚠ Log Error

Question

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

For first order reaction,

t99% = log ... ... (1)

t90% = log ... ... (2)

Taking ratio of equation (1) and (2)

=

∴ =

∴ =

∴ =

∴ = 2

∴ t99% = 2 × t90%

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

For first order reaction,

t99% = log ... ... (1)

t90% = log ... ... (2)

Taking ratio of equation (1) and (2)

=

∴ =

∴ =

∴ =

∴ = 2

∴ t99% = 2 × t90%

✓ Apply Cancel

#46 SUB 2M 🖼 14 ✎ Edit ⚠ Log Error

Question

A first order reaction takes 40 min for 30% decomposition. Calculate the half-life period for the reaction .

Answer

uppose initial concentration [R]0 = 100

∴ Concentration at time t [R]t = 70, (30% decomposition).

t = 40 min.

Comparing k = log and

k = we get,

= log

∴ = log = × 0.1549

∴ = 0.0089185

∴ = = 77.7 min

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

A first order reaction takes 40 min for 30% decomposition. Calculate the half-life period for the reaction .

Answer

uppose initial concentration [R]0 = 100

∴ Concentration at time t [R]t = 70, (30% decomposition).

t = 40 min.

Comparing k = log and

k = we get,

= log

∴ = log = × 0.1549

∴ = 0.0089185

∴ = = 77.7 min

✓ Apply Cancel

#47 SUB 4M 🖼 12 ▦ 1 ✎ Edit ⚠ Log Error

Question

For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.

Calculate the rate constant.

Time (sec)	0	360	720
P(mm Hg)	35.0	54.0	63.0

Answer

Decomposition reaction of azoisopropane,

(CH3)2CHN = NCH(CH3)2 → C6H14(g) + N2(g)

Pressure at time t = 0	35 mm Hg	0	0
Change in pressure	– x	+x	+x
Pressure after t = 360 s	35 – x	+x	+x

∴ Total pressure after 360 s = 35 – x + x + x = 54 mm Hg

∴ Pressure after t = 360 s; 35 – 19 mm Hg = 16 mm Hg

Rate constant at 360 s:

Initial pressure [R]0 = pi = 35 mm Hg

Pressure at time t [R]t = pt = 16 mm Hg

For first order reaction,

k = log

= log

= log2.1875

= × 0.3399

∴ k = 2.1747 × 10–3 sec–1

k at t = 720 s:

Total pressure at t = 720 s,

Total pressure at equilibrium = 35 – x + x + x

= 63 mm Hg

Initial pressure at time t = 0 s, [R]0 = pi = 35 mm Hg

Pressure at time t = 720 s, [R]t = pt = 35 – 28

= 7 mm Hg

For first order reaction,

k = log

= log

= log5

= × 0.6990

∴ k = 2.21 × 10–3 sec–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.

Calculate the rate constant.

Answer

Decomposition reaction of azoisopropane,

(CH3)2CHN = NCH(CH3)2 → C6H14(g) + N2(g)

Pressure at time t = 0	35 mm Hg	0	0
Change in pressure	– x	+x	+x
Pressure after t = 360 s	35 – x	+x	+x

∴ Total pressure after 360 s = 35 – x + x + x = 54 mm Hg

∴ Pressure after t = 360 s; 35 – 19 mm Hg = 16 mm Hg

Rate constant at 360 s:

Initial pressure [R]0 = pi = 35 mm Hg

Pressure at time t [R]t = pt = 16 mm Hg

For first order reaction,

k = log

= log

= log2.1875

= × 0.3399

∴ k = 2.1747 × 10–3 sec–1

k at t = 720 s:

Total pressure at t = 720 s,

Total pressure at equilibrium = 35 – x + x + x

= 63 mm Hg

Initial pressure at time t = 0 s, [R]0 = pi = 35 mm Hg

Pressure at time t = 720 s, [R]t = pt = 35 – 28

= 7 mm Hg

For first order reaction,

k = log

= log

= log5

= × 0.6990

∴ k = 2.21 × 10–3 sec–1

✓ Apply Cancel

#48 SUB 4M 🖼 5 ▦ 1 ✎ Edit ⚠ Log Error

Question

The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

SO2Cl2(g) → SO2(g) + Cl2(g)

Calculate the rate of reaction when total pressure is 0.65 atm.

Experiment	Time/s–1	Total pressure/atm
1	0	0.5
2	100	0.6

Answer

Decomposition reaction of SO2Cl2,

SO2Cl2(g) → SO2(g) + Cl2(g)

Pressure at time t = 0 s pi atm 0 atm 0 atm

Pressure at time t pi – x atm x atm x atm

pi is initial pressure at time t = 0 s

∴ Total pressure at time t,

pt = pi – x + x + x = pi + x

∴ x = pt – pi

∴ p(SO2Cl2 ) = pi – x = pi – (pt – pi) = 2pi – pt atm.

For experiment - 1

At time t = 0 s, Initial pressure pi = 0.5atm

At time t = 100 s, p(SO2Cl2 ) = 2pi – pt

= 2 × 0.5 – 0.6

= 0.4 atm

For first order reaction,

∴ k = log

= log

= × 0.0969

∴ k = 2.2318 × 10–3 sec–1

For experiment-2

At time t = 0 s, Initial pressure pi = 0.5 atm

pSO2Cl2 = 2pi – pt = 2 × 0.5 – 0.65 = 0.35 atm

∴ rate = k ∙ pSO2Cl2

= 2.2318 × 0.35 × 10–3 sec–1

= 0.78113 × 10–3

= 7.8113 × 10–4 atm sec–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

SO2Cl2(g) → SO2(g) + Cl2(g)

Calculate the rate of reaction when total pressure is 0.65 atm.

Answer

Decomposition reaction of SO2Cl2,

SO2Cl2(g) → SO2(g) + Cl2(g)

Pressure at time t = 0 s pi atm 0 atm 0 atm

Pressure at time t pi – x atm x atm x atm

pi is initial pressure at time t = 0 s

∴ Total pressure at time t,

pt = pi – x + x + x = pi + x

∴ x = pt – pi

∴ p(SO2Cl2 ) = pi – x = pi – (pt – pi) = 2pi – pt atm.

For experiment - 1

At time t = 0 s, Initial pressure pi = 0.5atm

At time t = 100 s, p(SO2Cl2 ) = 2pi – pt

= 2 × 0.5 – 0.6

= 0.4 atm

For first order reaction,

∴ k = log

= log

= × 0.0969

∴ k = 2.2318 × 10–3 sec–1

For experiment-2

At time t = 0 s, Initial pressure pi = 0.5 atm

pSO2Cl2 = 2pi – pt = 2 × 0.5 – 0.65 = 0.35 atm

∴ rate = k ∙ pSO2Cl2

= 2.2318 × 0.35 × 10–3 sec–1

= 0.78113 × 10–3

= 7.8113 × 10–4 atm sec–1

✓ Apply Cancel

#49 SUB 4M 🖼 11 ▦ 2 ✎ Edit ⚠ Log Error

Question

The rate constant for the decomposition of N2O5 at various temperatures is given below:

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.

T/°C	0	20	40	60	80
105 × k/s–1	0.0787	1.70	25.7	178	2140

T/°C	T/K	k/s–1	lnk	1/T
0	273	0.0787 × 105 = 78700	11.27	3.663 × 10–3
20	293	1.70 × 105 = 170000	12.04	3.412 × 10–3
40	313	25.7 × 105 = 2570000	14.76	3.195 × 10–3
60	333	178 × 105 = 17800000	16.69	3.003 × 10–3
80	353	2140 × 105 = 214000000	19.18	2.832 × 10–3

Answer

Ea from the graph:

Slope =

∴ Ea = – slope × R

∴ Ea = × 8.314

∴ Ea = 94523.47 J mol-1 = 94.52347 kJ mol–1

Arrhenius constant from graph:

Intercept of graph shows value of ln A.

Intercept = 21.5

∴ ln A = 21.5

∴ 2.303 log A = 21.5

∴ log A = = 9.3356

∴ A = antilog 9.3356

∴ A = 2.1657 × 109

The rate constant at 30°C and 50°C temperature:

T = 30°C = 303K

∴ = = 3.3 × 10–3K–1

From graph, at = 3.3 × 10–3K–1, ln k = 13.8

∴ 2.303 log k = 13.8

∴ log k = = 6

∴ k = antilog 6

∴ k = 1 × 106 s–1

T = 50°C = 323K

∴ = = 3.09 × 10–3K–1

From graph, at = 3.09 × 10–3K–1, ln k = 16

∴ 2.303 log k = 16

∴ log k = = 6.9474

∴ k = antilog 6.9474

∴ k = 8.86 × 106 s-1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

The rate constant for the decomposition of N2O5 at various temperatures is given below:

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

Ea from the graph:

Slope =

∴ Ea = – slope × R

∴ Ea = × 8.314

∴ Ea = 94523.47 J mol-1 = 94.52347 kJ mol–1

Arrhenius constant from graph:

Intercept of graph shows value of ln A.

Intercept = 21.5

∴ ln A = 21.5

∴ 2.303 log A = 21.5

∴ log A = = 9.3356

∴ A = antilog 9.3356

∴ A = 2.1657 × 109

The rate constant at 30°C and 50°C temperature:

T = 30°C = 303K

∴ = = 3.3 × 10–3K–1

From graph, at = 3.3 × 10–3K–1, ln k = 13.8

∴ 2.303 log k = 13.8

∴ log k = = 6

∴ k = antilog 6

∴ k = 1 × 106 s–1

T = 50°C = 323K

∴ = = 3.09 × 10–3K–1

From graph, at = 3.09 × 10–3K–1, ln k = 16

∴ 2.303 log k = 16

∴ log k = = 6.9474

∴ k = antilog 6.9474

∴ k = 8.86 × 106 s-1

✓ Apply Cancel

#50 SUB 3M 🖼 3 ✎ Edit ⚠ Log Error

Question

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

Arrhenius equation,

logk = logA –

∴ logA = logk +

Here, T = 546K, k = 2.418× 10–5 s–1

Ea = 179.9 kJ/mol = 179900 J/mol

A = ?

∴ logA = log2.418 × 10–5 +

∴ logA = log2.418 + log 10–5 + 17.2082

∴ logA = 0.3834 – 5.0 + 17.2082

∴ logA = 12.5917

∴ A = antilog12.5917 = 3.9057 × 1012 sec–1.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Arrhenius equation,

logk = logA –

∴ logA = logk +

Here, T = 546K, k = 2.418× 10–5 s–1

Ea = 179.9 kJ/mol = 179900 J/mol

A = ?

∴ logA = log2.418 × 10–5 +

∴ logA = log2.418 + log 10–5 + 17.2082

∴ logA = 0.3834 – 5.0 + 17.2082

∴ logA = 12.5917

∴ A = antilog12.5917 = 3.9057 × 1012 sec–1.

✓ Apply Cancel

#51 SUB 3M 🖼 10 ✎ Edit ⚠ Log Error

Question

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

k = 2.0 × 10–2 sec–1

[R]0 = 1 mol L–1, t = 100 sec

[R]t = ?

k = log

∴ log =

∴ log =

∴ log = 0.8684

∴ = anti log0.8684 = 7.3863

∴ [R]t = = = 0.1354 mol L–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

k = 2.0 × 10–2 sec–1

[R]0 = 1 mol L–1, t = 100 sec

[R]t = ?

k = log

∴ log =

∴ log =

∴ log = 0.8684

∴ = anti log0.8684 = 7.3863

∴ [R]t = = = 0.1354 mol L–1

✓ Apply Cancel

#52 SUB 3M 🖼 13 ✎ Edit ⚠ Log Error

Question

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

For first order reaction,

k = = = 0.231 hour–1

Fraction of sample of sucrose remains after 8 hours:

Taking initial concentration [R]0 = 1 M,

Concentration after 8 hours [R]t = ?

k = log

∴ log =

∴ log =

∴ log = 0.8024

∴ = antilog 0.8024 = 6.3445

∴ [R]t = = = 0.1576 M

∴ The fraction of sample of sucrose that remains after 8 hours is 0.1576 M.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

For first order reaction,

k = = = 0.231 hour–1

Fraction of sample of sucrose remains after 8 hours:

Taking initial concentration [R]0 = 1 M,

Concentration after 8 hours [R]t = ?

k = log

∴ log =

∴ log =

∴ log = 0.8024

∴ = antilog 0.8024 = 6.3445

∴ [R]t = = = 0.1576 M

∴ The fraction of sample of sucrose that remains after 8 hours is 0.1576 M.

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#53 SUB 3M 🖼 4 ✎ Edit ⚠ Log Error

Question

The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s–1) . Calculate
activation energy Ea.

Answer

Arrhenius equation,

k = A ... ... (I)

k = 4.5 × 1011 s–1 ... ... (II)

Taking ratio of Eq. (I) and Eq. (II) we get,

Ea = 28000K × R

Ea = 28000K × 8.314 J K–1 mol–1

Ea = 232792 J mol-1 = 232.792 kJ mol–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s–1) . Calculate
activation energy Ea.

Answer

Arrhenius equation,

k = A ... ... (I)

k = 4.5 × 1011 s–1 ... ... (II)

Taking ratio of Eq. (I) and Eq. (II) we get,

Ea = 28000K × R

Ea = 28000K × 8.314 J K–1 mol–1

Ea = 232792 J mol-1 = 232.792 kJ mol–1

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#54 SUB 4M 🖼 5 ✎ Edit ⚠ Log Error

Question

The rate constant for the first order decomposition of H2O2 is given by the following equation:

log k = 14.34 – 1.25 × 104 K/T

Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes ?

Answer

Ea is the activation energy.

According to Arrhenius equation,

k = A

log k = logA – ... ... (I)

Now, log k = 14.34 – ... ... (II)

Comparing Eq. (I) and Eq. (II) we get,

Ea = 1.25 × 104K × 2.303 × R

Ea = 1.25 × 104K × 2.303 × 8.314 J K–1 mol–1

Ea = 23.93 × 104 J mol–1 = 239.3 kJ mol–1

For first order reaction

k =

k = = 4.51 × 10–5 s–1

For given equation,

log k = 14.34 –

log 4.51 × 10-5 = 14.34 –

– 4.35 = 14.34 –

= 14.34 + 4.35 = 18.69

T = = 669K

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

The rate constant for the first order decomposition of H2O2 is given by the following equation:

log k = 14.34 – 1.25 × 104 K/T

Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes ?

Answer

Ea is the activation energy.

According to Arrhenius equation,

k = A

log k = logA – ... ... (I)

Now, log k = 14.34 – ... ... (II)

Comparing Eq. (I) and Eq. (II) we get,

Ea = 1.25 × 104K × 2.303 × R

Ea = 1.25 × 104K × 2.303 × 8.314 J K–1 mol–1

Ea = 23.93 × 104 J mol–1 = 239.3 kJ mol–1

For first order reaction

k =

k = = 4.51 × 10–5 s–1

For given equation,

log k = 14.34 –

log 4.51 × 10-5 = 14.34 –

– 4.35 = 14.34 –

= 14.34 + 4.35 = 18.69

T = = 669K

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#55 SUB 3M 🖼 4 ✎ Edit ⚠ Log Error

Question

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

A → product

Here, T1 = 10°C = 283K

k1 = 4.5 × 103 s–1

Ea = 60 kJ mol-1= 60000 J mol–1

k2 = 1.5 × 104 s–1

T2 = ?

R = 8.314 J K–1 mol–1

log

log

0.5228 =

= 0.0001668

– 0.0001668

= 0.003534 – 0.0001668 = 0.003367

T2 = = 297.00 K

T2 = 297 – 273 = 24°C

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

A → product

Here, T1 = 10°C = 283K

k1 = 4.5 × 103 s–1

Ea = 60 kJ mol-1= 60000 J mol–1

k2 = 1.5 × 104 s–1

T2 = ?

R = 8.314 J K–1 mol–1

log

log

0.5228 =

= 0.0001668

– 0.0001668

= 0.003534 – 0.0001668 = 0.003367

T2 = = 297.00 K

T2 = 297 – 273 = 24°C

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#56 SUB 4M 🖼 12 ✎ Edit ⚠ Log Error

Question

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

(A) Calculation of Ea:

(A) For first order reaction, k = log

At 298K temperature, k1 = log
... ... (I)

At 308K temperature, k2 = log

... ... (II)

Dividing Eq. (II) by Eq.(I) we get,

= 2.73

According to Arrhenius equation,

log

log2.73 =

Ea =

Ea = 76640 J/mol = 76.640 kJ/mol

(B) According to Arrhenius equation,

log k = logA –

log k = log4 × 1010 –

log k = 10.6021 – 12.5870 = –1.9849 = .0151

k = antilog.0151

k = 1.035 × 10–2 s–1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

(A) Calculation of Ea:

(A) For first order reaction, k = log

At 298K temperature, k1 = log
... ... (I)

At 308K temperature, k2 = log

... ... (II)

Dividing Eq. (II) by Eq.(I) we get,

= 2.73

According to Arrhenius equation,

log

log2.73 =

Ea =

Ea = 76640 J/mol = 76.640 kJ/mol

(B) According to Arrhenius equation,

log k = logA –

log k = log4 × 1010 –

log k = 10.6021 – 12.5870 = –1.9849 = .0151

k = antilog.0151

k = 1.035 × 10–2 s–1

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#57 SUB 3M 🖼 1 ✎ Edit ⚠ Log Error

Question

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

At T1 = 293K, rate constant = k1

At T2 = 313K, rate constant = k2 = 4k1

R = 8.314 J K–1 mol–1

According to Arrhenius equation,

log

log

Ea =

Ea =

Ea = 52863.33 J/mol = 52.863 kJ/mol

Class 12 Chemistry (Part 1) 013

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

At T1 = 293K, rate constant = k1

At T2 = 313K, rate constant = k2 = 4k1

R = 8.314 J K–1 mol–1

According to Arrhenius equation,

log

log

Ea =

Ea =

Ea = 52863.33 J/mol = 52.863 kJ/mol

Class 12 Chemistry (Part 1) 013

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