Q1 
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3 NO(g) → N_{2}O(g) Rate = k[NO]^{2}
(ii) H_{2}O_{2} (aq) + 3 I^{  } (aq) + 2 H^{+}→ 2 H_{2}O (l) + I_{3}^{} Rate = k[H_{2}O_{2}][I^{  }]
(iii) CH_{3}CHO(g) → CH_{4}(g) + CO(g) Rate = k [CH_{3}CHO]^{3/2}
(iv) C_{2}H_{5}Cl(g) → C_{2}H_{4}(g) + HCl(g) Rate = k [C_{2}H_{5}Cl] 
Ans: 
(i) Given rate = k[NO]^{2}
Therefore, order of the reaction = 2
Dimension of k = Rate / [NO]^{2}
= mol L^{1} s^{1} / (mol L^{1})^{2}
= mol L^{1} s^{1} / mol^{2} L^{2}
= L mol^{1} s^{1}
(ii) Given rate = k[H_{2}O_{2}][I^{  }]
Therefore, order of the reaction = 2
Dimension of k = Rate / [H_{2}O_{2}][I^{  }]
= mol L^{1} s^{1 }/ (mol L^{1}) (mol L^{1})
= L mol^{1} s^{1}
(iii) Given rate =k [CH_{3}CHO]^{3/2}
Therefore, order of reaction = 3/2
Dimension of k = Rate / [CH_{3}CHO]^{3/2}
= mol L^{1} s^{1 }/ (mol L^{1})^{3/2}
= mol L^{1} s^{1 }/ mol^{3/2} L^{3/2}
= L^{½} mol^{½ }s^{1}
(iv) Given rate = k [C_{2}H_{5}Cl]
Therefore,order of the reaction = 1
Dimension of k = Rate / [C_{2}H_{5}Cl]
= mol L^{1} s^{1 }/ mol L^{1}
= s^{1} 

Q2 
For the reaction:
2A + B → A_{2}B
the rate = k[A][B]^{2}with k= 2.0 x 10^{6}mol^{2}L^{2}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 reduced to 0.06 mol L^{1}. 
Ans: 
The initial rate of the reactionis
Rate = k [A][B]^{2}
= (2.0 × 10^{  6}mol^{  2}L^{2}s^{  1}) (0.1 mol L^{  1}) (0.2 mol L^{  1})^{2}
= 8.0 × 10^{  9}mol^{  2}L^{2}s^{  1}
When [A] is reduced from 0.1 mol L^{  1}to 0.06 mol^{  1}, the concentration of A reacted = (0.1  0.06) mol L^{  1} = 0.04 mol L^{  1}
Therefore, concentration of B reacted= 1/2 x 0.04 mol L^{1} = 0.02 mol L^{  1}
Then, concentration of B available, [B] = (0.2  0.02) mol L^{  1}
= 0.18 mol L^{  1}
After [A] is reduced to 0.06 mol L^{  1}, the rate of the reaction is given by,
Rate = k [A][B]^{2}
= (2.0 × 10^{  6}mol^{  2}L^{2}s^{  1}) (0.06 mol L^{  1}) (0.18 mol L^{  1})^{2}
= 3.89 mol L^{  1}s^{  1} 

Q3 
The decomposition of NH_{3}on platinum surface is zero order reaction. What are the rates of production of N_{2}and H_{2}if k = 2.5 x 10^{4}mol^{1}L s^{1}? 
Ans: 
The decomposition of NH_{3}on platinum surface is represented by the following equation.
= 7.5 × 10^{  4}mol L^{  1}s^{  1} 

Q4 
The decomposition of dimethyl ether leads to the formation of CH_{4}, H_{2}and CO and the reaction rate is given by
Rate = k [CH_{3}OCH_{3}]^{3/2}
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 = k (P_{CH3OCH3})^{3/2}
If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants? 
Ans: 
If pressure is measured in bar and time in minutes, then
Unit of rate = bar min^{  1}
Rate = k (P_{CH3OCH3})^{3/2}
⇒ k = Rate / (P_{CH3OCH3})^{3/2}
Therefore, unit of rate constants (k) = bar min^{1 }/ bar^{3/2}
= bar^{½} min^{  1}


Q5 
Mention the factors that affect the rate of a chemical reaction. 
Ans: 
The factors that affect the rate of a reaction areas follows.
(i) Concentration of reactants (pressure in case of gases)
(ii) Temperature
(iii) Presence of a catalyst 

Q6 
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? 
Ans: 
Let the concentration of the reactant be [A] = a
Rate of reaction, R = k [A]^{2} = ka^{2}
(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R = k(2a)^{2}
= 4ka^{2}
= 4 R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e. [A] = 1/2 a,then the rate of the reaction would be
R = k(1/2a)^{2}
= 1/4 Ka^{2}
= 1/4 R
Therefore, the rate of the reaction would be reduced to ¼^{th} 

Q7 
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively? 
Ans: 
The rate constant is nearly doubled with a rise in temperature by 10° for a chemical reaction.
The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,
K = Ae^{ Ea / RT}
where, kis the rate constant,
A is the Arrhenius factor or the frequency factor,
R is the gas constant,
T is the temperature, and
E_{a} is the energy of activation for the reaction 

Q8 
In a pseudo first order hydrolysis of ester in water, the following results were obtained:
t/s 
0 
30 
60 
90 
[Ester]mol L^{  1}

0.55 
0.31 
0.17 
0.085 
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester. 
Ans: 
(i) Average rate of reaction between the time interval, 30 to 60 seconds, = d[ester] / dt
= (0.310.17) / (6030)
= 0.14 / 30
= 4.67 × 10  3mol L^{  1}s ^{ 1}
(ii) For a pseudo first order reaction,
k = 2.303/ t log [R]_{º} / [R]
For t= 30 s, k_{1} = 2.303/ 30 log 0.55 / 0.31
= 1.911 × 10 ^{ 2}s^{  1}
For t= 60 s, k_{2} = 2.303/ 60 log 0.55 / 0.17
= 1.957 × 10^{  2}s ^{ 1}
For t= 90 s, k_{3} = 2.303/ 90 log 0.55 / 0.085
= 2.075 × 10 ^{ 2}s ^{ 1 }
Then, average rate constant, k = k_{1} + k_{2}+ k_{3 }/ 3
= 1.911 × 10 ^{ 2 }+ 1.957 × 10^{  2} + 2.075 × 10 ^{ 2} / 3
= 1.98 x 10^{2 }s ^{ 1} 

Q9 
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled? 
Ans: 


Q10 
In a reaction between A and B, the initial rate of reaction (r_{0}) 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}

What is the order of the reaction with respect to A and B? 
Ans: 
Let the order of the reaction with respect to A be xand with respect to B be y.
Therefore,
= 1.496
= 1.5 (approximately)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero. 

Q11 
The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D
Experiment 
A/ mol L^{  1}

B/ mol L^{  1}

Initial rate of formation of D/mol L^{  1} min^{  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}

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


Q12 
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
Experiment 
A/ mol L^{  1}

B/ mol L^{  1}

Initial rate/mol L^{  1} min^{  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}


Ans: 
The given reaction is of the first order with respect to A and of zero order with respect to B.
Therefore, the rate of the reaction is given by,
Rate = k [A]^{1}[B]^{0}
⇒ Rate = k [A]
From experiment I, we obtain
2.0 x 10^{2}mol L^{1}min^{1}= k (0.1 mol L^{1})
⇒ k= 0.2 min^{1}
From experiment II, we obtain
4.0 x 10^{2}mol L^{1}min^{1}= 0.2 min^{1}[A]
⇒ [A] = 0.2 mol L^{1}
From experiment III, we obtain
Rate = 0.2 min^{1} x 0.4 mol L^{1}
= 0.08 mol L^{1}min^{1}
From experiment IV, we obtain
2.0 x 10^{2}mol L^{1}min^{1}= 0.2 min^{1}[A]
⇒ [A] = 0.1 mol L^{1} 

Q13 
Calculate the halflife of a first order reaction from their rate constants given below:
(i) 200 s^{1}
(ii) 2 min^{1}
(iii) 4 years^{1} 
Ans: 
(i) Half life, t_{½} = 0.693 / k
= 0.693 / 200 s^{1}
= 3.47 ×10 ^{3} s (approximately)
(ii) Half life, t_{½} = 0.693 / k
= 0.693 / 2 min^{1}
= 0.35 min (approximately)
(iii) Half life,t_{½} = 0.693 / k
= 0.693 / 4 years^{1}
= 0.173 years (approximately) 

Q14 
The halflife for radioactive decay of ^{14}C is 5730 years. An archaeological artifact containing wood had only 80% of the ^{14}C found in a living tree. Estimate the age of the sample. 
Ans: 
Here,
K = 0.693 / t_{½}
= 0.693 / 5730 years^{1}
It is known that,
= 1845 years (approximately)
Hence, the age of the sample is 1845 years. 

Q15 
The experimental data for decomposition of N_{2}O_{5}
[2N_{2}O_{5} → 4NO_{2} + O_{2}]
in gas phase at 318K are given below:
t/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 
(i) Plot [N_{2}O_{5}] against t.
(ii) Find the halflife period for the reaction.
(iii) Draw a graph between log[N_{2}O_{5}] and t.
(iv) What is the rate law ?
(v) Calculate the rate constant.
(vi) Calculate the halflife period from k and compare it with (ii). 
Ans: 
(i)
(ii) Time corresponding to the concentration, 1630x10^{2} / 2 mol L^{1} = 81.5 mol L^{1} is the half life. From the graph, the half life is obtained as 1450 s.
(iii)
t/s 
10^{2} × [N_{2}O_{5}] 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) The given reaction is of the first order as the plot, log[N_{2}O_{5}] v/s t, is a straight line. Therefore, the rate law of the reaction is
Rate = k [N_{2}O_{5}]
(v) From the plot, log[N_{2}O_{5}]
v/s t, we obtain
Slope = 2.46 (1.79) / 32000
= 0.67 / 3200
Again, slope of the line of the plot log[N_{2}O_{5}] v/s t is given by
 k / 2.303
.Therefore, we obtain,
 k / 2.303 =  0.67 / 3200
⇒ k = 4.82 x 10^{4} s^{1}
(vi) half life is given by,
t_{½} = 0.693 / k
= 0.639 / 4.82x10^{4 }s
=1.438 x 10^{3 }s
Or we can say
1438 S
Which is very near to what we obtain from graph.


Q16 
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 1/16^{th} value? 
Ans: 


Q17 
During nuclear explosion, one of the products is ^{90}Sr with halflife of 28.1 years. If 1μg of ^{90}Sr 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. 
Ans: 
Therefore, 0.7814 μg of ^{90}Sr will remain after 10 years.
Again,
Therefore, 0.2278 μg of ^{90}Sr will remain after 60 Years 

Q18 
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. 
Ans: 
For a first order reaction, the time required for 99% completionis
t_{1 }= 2.303/k Log 100/10099
= 2.303/k Log 100
= 2x 2.303/k
For a first order reaction, the time required for 90% completion is
t_{2 }= 2.303/k Log 100/10090
= 2.303/k Log 10
= 2.303/k
Therefore, t_{1 }= 2t_{2}
Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction. 

Q19 
A first order reaction takes 40 min for 30% decomposition. Calculate t_{1/2}. 
Ans: 
For a first order reaction,
t_{ }= 2.303/k Log [R] _{º }/ [R]
k = 2.303/40min Log 100_{ }/ 10030
= 2.303/40min Log 10_{ }/ 7
= 8.918 x 10^{3} min^{1}
Therefore, t_{1/2} of the decomposition reaction is
t_{1/2 }= 0.693/k
= 0.693 / 8.918 x 10^{3} min
= 77.7 min (approximately) 

Q20 
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
t (sec) 
P(mm of Hg) 
0 
35.0 
360 
54.0 
720 
63.0 
Calculate the rate constant 
Ans: 
The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.
After time, t, total pressure, P_{t} = (P_{º}  p) + p + p
⇒ P_{t} = (P_{º} + p)
⇒ p = P_{t }  P_{º}
therefore, P_{º}  p = P_{º } P_{t }  P_{º}
= 2 P_{º } P_{t}
For a first order reaction,
k = 2.303/t Log P_{º}_{ }/ P_{º } p
= 2.303/t Log P_{º}_{ }/ 2 P_{º } P_{t}
When t = 360 s, k = 2.303 / 360s log 35.0 / 2x35.0  54.0
= 2.175 × 10 ^{ 3} s ^{ 1}
When t = 720 s, k = 2.303 / 720s log 35.0 / 2x35.0  63.0
= 2.235 × 10 ^{ 3} s ^{ 1}
Hence, the average value of rate constant is
k = (2.175 × 10 ^{ 3 }+ 2.235 × 10 ^{ 3} ) / 2 s ^{ 1}
= 2.21 × 10 ^{ 3 }s^{  1} 

Q21 
The following data were obtained during the first order thermal decomposition of SO_{2}Cl_{2}at a constant volume.
SO_{2}Cl_{2(}g) → SO_{2(}g) + Cl_{2}(g)
Experiment 
Time/s^{  1}

Total pressure/atm 
1 
0 
0.5 
2 
100 
0.6 
Calculate the rate of the reaction when total pressure is 0.65 atm. 
Ans: 
The thermal decomposition of SO_{2}Cl_{2}at a constant volume is represented by the following equation.
After time, t, total pressure,P_{t} = (P_{º}  p) + p + p
⇒ P_{t} = (P_{º} + p)
⇒ p = P_{t }  P_{º}
therefore, P_{º}  p = P_{º } P_{t }  P_{º}
= 2 P_{º } P_{t}
For a first order reaction,
k = 2.303/t Log P_{º}_{ }/ P_{º } p
= 2.303/t Log P_{º}_{ }/ 2 P_{º } P_{t}
When t= 100 s,
k = 2.303 / 100s log 0.5 / 2x0.5  0.6
= 2.231 × 10^{  3}s^{  1 }
When Pt= 0.65 atm,
P_{0}+ p= 0.65
⇒ p= 0.65  P_{0}
= 0.65  0.5
= 0.15 atm
Therefore, when the total pressure is 0.65 atm, pressure of SOCl_{2} is
p_{SOCL2} = P_{0}  p
= 0.5  0.15
= 0.35 atm
Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,
Rate = k(p_{SOCL2})
= (2.23 × 10 ^{ 3}s ^{ 1}) (0.35 atm)
= 7.8 × 10 ^{ 4}atm s^{  1} 

Q22 
The rate constant for the decomposition of N_{2}O_{5} at various temperatures is given below:
T/°C

0 
20 
40 
60 
80 
10^{5} X K /S^{1}

0.0787 
1.70 
25.7 
178 
2140 
Draw a graph between ln k and 1/T and calculate the values of A and E_{a}.
Predict the rate constant at 30 º and 50 ºC.

Ans: 
From the given data, we obtain
T/°C 
0 
20 
40 
60 
80 
T/K

273 
293 
313 
333 
353 
1/T / k^{1} 
3.66×10^{  3}

3.41×10^{  3}

3.19×10^{  3}

3.0×10^{  3}

2.83 ×10^{  3}

10^{5} X K /S^{1} 
0.0787 
1.70 
25.7 
178 
2140 
In k 
7.147 
4.075 
1.359 
0.577 
3.063 
Slope of the line,
In k=  2.8
Therefore, k = 6.08x10^{2}s^{1}
Again when T = 50 + 273K = 323K,
1/T = 3.1 x 10^{3} K
In k =  0.5
Therefore, k = 0.607 s^{1}


Q23 
The rate constant for the decomposition of hydrocarbons is 2.418 x 10^{5} s^{1} at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of preexponential factor. 
Ans: 
k= 2.418 × 10^{5} s^{1}
T= 546 K
E_{a}= 179.9 kJ mol^{  1} = 179.9 × 10^{3}J mol^{  1}
According to the Arrhenius equation,
= (0.3835  5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 10^{12} s^{  1}(approximately) 

Q24 
Consider a certain reaction A → Products with k = 2.0 x 10^{2} s^{1}. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^{1}. 
Ans: 
k= 2.0 × 110^{2} s^{1}
T= 100 s
[A]_{o}= 1.0 moL ^{ 1}
Since the unit of kis s^{  1}, the given reaction is a first order reaction.
Therefore, k = 2.303/t Log [A]_{º }/ [A]
⇒2.0 × 110^{2} s^{1 }= 2.303/100s Log 1.0_{ }/ [A]
⇒2.0 × 110^{2} s^{1 }= 2.303/100s (  Log [A] )
⇒  Log [A] =  (2.0 x 10^{2} x 100) / 2.303
⇒ [A] = antilog [ (2.0 x 10^{2} x 100) / 2.303]
= 0.135 mol L^{  1 }(approximately)
Hence, the remaining concentration of A is 0.135 mol L ^{ 1}. 

Q25 
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_{1/2} = 3.00 hours. What fraction of sample of sucrose remains after 8 hours? 
Ans: 
For a first order reaction,
k = 2.303/t Log [R]_{º }/ [R]
It is given that, t_{1/2 }= 3.00 hours
Therefore, k = 0.693 / t_{1/2 }
= 0.693 / 3 h^{1}
= 0.231 h^{  1}
Then, 0.231 h^{  1 }= 2.303 / 8h Log [R]_{º }/ [R]
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158. 

Q26 
The decomposition of hydrocarbon follows the equation
k = (4.5 x 10^{11} s^{1}) e^{28000 K/T}
Calculate E_{a}. 
Ans: 
The given equation is
k = (4.5 x 10^{11} s^{1}) e^{28000 K/T} (i)
Arrhenius equation is given by,
k= Ae^{ Ea/RT }(ii)
From equation (i) and (ii), we obtain
E_{a }/ RT = 28000K / T
⇒ E_{a }= R x 28000K
= 8.314 J K^{  1}mol^{  1}× 28000 K
= 232792 J mol ^{ 1 }
= 232.792 kJ mol^{  1} 

Q27 
The rate constant for the first order decomposition of H_{2}O_{2 }is given by the following equation:
log k = 14.34  1.25 x 10^{4 }K/T
Calculate E_{a }for this reaction and at what temperature will its halfperiod be 256 minutes? 
Ans: 
Arrhenius equation is given by,
k= Ae ^{Ea/RT }
⇒In k = In A  E_{a}/RT
⇒In k = Log A  E_{a}/RT
⇒ Log k = Log A  E_{a}/2.303RT (i)
The given equation is
Log k = 14.34  1.25 10^{4} K/T (ii)
From equation (i) and (ii), we obtain
E_{a}/2.303RT = 1.25 10^{4} K/T
⇒ E_{a }=1.25 × 10^{4}K × 2.303 × R
= 1.25 × 10^{4}K × 2.303 × 8.314 J K^{  1}mol^{  1}
= 239339.3 J mol^{  1} (approximately)
= 239.34 kJ mol  1
Also, when t_{1/2}= 256 minutes,
k = 0.693 / t_{1/2 }
= 0.693 / 256
= 2.707 × 10^{  3 }min ^{ 1}
= 4.51 × 10^{  5}s^{  1}
It is also given that, log k= 14.34  1.25 × 10^{4}K/T
= 668.95 K
= 669 K (approximately) 

Q28 
The decomposition of A into product has value of k as 4.5 x 10^{3} s^{1} at 10°C and energy of activation 60 kJ mol^{1}. At what temperature would k be 1.5 x 10^{4} s^{1}? 
Ans: 
From Arrhenius equation, we obtain
log k_{2}/k_{1 }= E_{a} / 2.303 R (T_{2}  T_{1}) / T_{1}T_{2}
Also, k_{1} = 4.5 × 10^{3} s^{  1}
T_{1} = 273 + 10 = 283 K
k_{2} = 1.5 × 10^{4} s^{  1}
E_{a} = 60 kJ mol ^{ 1} = 6.0 × 10^{4} J mol ^{ 1}
Then,
= 297 K = 24°C
Hence, k would be 1.5 × 10^{4} s^{  1} at 24°C. 

Q29 
The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 x 10^{10} s^{1}. Calculate k at 318 K and E_{a}. 
Ans: 
For a first order reaction,
t = 2.303 / k log a / a  x
At 298 K,
t = 2.303 / k log 100 / 90
= 0.1054 / k
At 308 K,
t' = 2.303 / k' log 100 / 75
= 2.2877 / k'
According to the question,
t = t'
⇒ 0.1054 / k = 2.2877 / k'
⇒ k' / k = 2.7296
From Arrhenius equation,we obtain
To calculate k at 318 K,
It is given that, A = 4 x 10^{10} s^{1}, T = 318K
Again, from Arrhenius equation, we obtain
Therefore, k = Antilog (1.9855)
= 1.034 x 10^{2} s ^{1}


Q30 
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. 
Ans: 
From Arrhenius equation, we obtain
Hence, the required energy of activation is 52.86 kJ mol^{  1}. 
