//M2//QN1//SUB//DL0//EQ

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]

//X

(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

//M3//QN2//SUB//DL0//EQ

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

//X

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 → A 2 B
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

//M2//QN3//SUB//DL0//EQ

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 ?

//X

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

//M0//QN4//SUB//DL0//EQ

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 ?

//X

The pressure is measured in bar and time in minutes,
So, the unit of rate = Bar min–1
Unit of K =
=
= min–1

//M0//QN5//SUB//DL0

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

//X

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.

//M2//QN6//SUB//DL0//EQ

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 ?

//X

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.

//M2//QN7//SUB//DL0//EQ

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?

//X

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).

//M2//QN8//SUB//DL0//EQ

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

//X

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

//M2//QN9//SUB//DL0

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]

//X

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.

//M4//QN10//SUB//DL0//EQ

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
r 0/mol L–1 s–1 | 5.07 × 10–5 | 5.07 × 10–5 | 1.43 × 10–4

//X

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

//M4//QN11//SUB//DL0//EQ

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

//X

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

//M3//QN12//SUB//DL0//EQ

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

//X

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

//M1//QN13//SUB//DL0//EQ

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

//X

For first order reaction =
(i) = 3.465 × 10–3 s
(ii) = 0.3465 min
(iii) = 0.17325 year

//M4//QN14//SUB//DL0//EQ

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.

//X

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.

//M4//QN15//SUB//DL0//EQ

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
10 2× [N 2 O 5] mol L–1 | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35

//X

(i)
(ii) Initial concentration of N 2 O 5
= 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 [N 2 O 5],

Time/(sec) | [N 2 O 5] × 10+2 mol L–1 | log[N 2 O 5]
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 [N 2 O 5] → t is straight line. Hence, it is first order reaction.
∴ Rate = k [N2O5]
(v) From plot of log [N 2 O 5] → 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.

//M3//QN16//SUB//DL0//EQ

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 ?

//X

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.

//M3//QN17//SUB//DL0//EQ

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.

//X

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.

//M3//QN18//SUB//DL0//EQ

For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

//X

For first order reaction,
t 99% = log ... ... (1)
t 90% = log ... ... (2)
Taking ratio of equation (1) and (2)
=
∴ =
∴ =
∴ =
∴ = 2
∴ t 99% = 2 × t 90%

//M2//QN19//SUB//DL0//EQ

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

//X

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

//M4//QN20//SUB//DL0//EQ

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

//X

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 = p i = 35 mm Hg
Pressure at time t [R]t = p t = 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 = p i = 35 mm Hg
Pressure at time t = 720 s, [R]t = p t = 35 – 28
= 7 mm Hg
For first order reaction,
k = log
= log
= log5
= × 0.6990
∴ k = 2.21 × 10–3 sec–1

//M4//QN21//SUB//DL0//EQ

The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO 2 Cl 2(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

//X

Decomposition reaction of SO2Cl2,
SO2Cl2(g) → SO2(g) + Cl2(g)
Pressure at time t = 0 s p i atm 0 atm 0 atm
Pressure at time t p i – x atm x atm x atm
p i is initial pressure at time t = 0 s
∴ Total pressure at time t,
p t = p i – x + x + x = p i + x
∴ x = p t – p i
∴ p(SO 2 Cl 2 ) = p i – x = p i – (p t – p i) = 2p i – p t atm.
For experiment - 1
At time t = 0 s, Initial pressure p i = 0.5atm
At time t = 100 s, p(SO 2 Cl 2 ) = 2p i – p t
= 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 p i = 0.5 atm
p SO 2 Cl 2 = 2p i – p t = 2 × 0.5 – 0.65 = 0.35 atm
∴ rate = k ∙ pSO 2 Cl 2
= 2.2318 × 0.35 × 10–3 sec–1
= 0.78113 × 10–3
= 7.8113 × 10–4 atm sec–1

//M4//QN22//SUB//DL0//EQ

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
10 5 × 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 × 10 5 = 78700 | 11.27 | 3.663 × 10–3
20 | 293 | 1.70 × 10 5 = 170000 | 12.04 | 3.412 × 10–3
40 | 313 | 25.7 × 10 5 = 2570000 | 14.76 | 3.195 × 10–3
60 | 333 | 178 × 10 5 = 17800000 | 16.69 | 3.003 × 10–3
80 | 353 | 2140 × 10 5 = 214000000 | 19.18 | 2.832 × 10–3

//X

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

//M3//QN23//SUB//DL0//EQ

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.

//X

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.

//M3//QN24//SUB//DL0//EQ

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.

//X

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

//M3//QN25//SUB//DL0//EQ

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?

//X

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.

//M3//QN26//SUB//DL0//EQ

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

activation energy Ea.

//X

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

//M4//QN27//SUB//DL0//EQ

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 ?

//X

E a 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

//M3//QN28//SUB//DL0//EQ

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 ?

//X

A → product
Here, T1 = 10°C = 283K
k 1 = 4.5 × 103 s–1
Ea = 60 kJ mol-1= 60000 J mol–1
k 2 = 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

//M4//QN29//SUB//DL0//EQ

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.

//X

(A) Calculation of E a:
(A) For first order reaction, k = log
At 298K temperature, k 1 = log
... ... (I)
At 308K temperature, k 2 = 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

//M3//QN30//SUB//DL0//EQ

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.

//X

At T1 = 293K, rate constant = k 1
At T2 = 313K, rate constant = k 2 = 4k 1
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