NCERT Solutions for Exam Preparation Class 12 Chemistry Chapter 4 Chemical Kinetics are provided here with simple stepbystep explanations. These solutions for Chemical Kinetics are extremely popular among Exam Preparation Class 12 students for Chemistry Chemical Kinetics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Exam Preparation Class 12 Chemistry Chapter 4 are provided here for you for free. You will also love the adfree experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class Exam Preparation Class 12 Chemistry are prepared by experts and are 100% accurate.
Page No 98:
Question 4.1:
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 of reaction
= 6.67 × 10^{−6} M s^{−1}
Page No 98:
Question 4.2:
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:
Average rate
= 0.005 mol L^{−1 }min^{−1}
= 5 × 10^{−3} M min^{−1}
Page No 103:
Question 4.3:
For a reaction, A + B → Product; the rate law is given by,. What is the order of the reaction?
Answer:
The order of the reaction
= 2.5
Page No 103:
Question 4.4:
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Answer:
The reaction X → Y follows second order kinetics.
Therefore, the rate equation for this reaction will be:
Rate = k[X]^{2} (1)
Let [X] = a mol L^{−1}, then equation (1) can be written as:
Rate_{1} = k .(a)^{2}
= ka^{2}
If the concentration of X is increased to three times, then [X] = 3a mol L^{−1}
Now, the rate equation will be:
Rate = k (3a)^{2}
= 9(ka^{2})
Hence, the rate of formation will increase by 9 times.
Page No 111:
Question 4.5:
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:
From the question, we can write down the following information:
Initial amount = 5 g
Final concentration = 3 g
Rate constant = 1.15 10^{−3} s^{−1}
We know that for a 1^{st} order reaction,
= 444.38 s
= 444 s (approx)
Page No 111:
Question 4.6:
Time required to decompose SO_{2}Cl_{2} to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Answer:
We know that for a 1^{st} order reaction,
It is given that t_{1/2} = 60 min
Page No 116:
Question 4.7:
What will be the effect of temperature on rate constant?
Answer:
The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,
Where,
A is the Arrhenius factor or the frequency factor
T is the temperature
R is the gas constant
E_{a} is the activation energy
Page No 116:
Question 4.8:
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_{a}.
Answer:
It is given that T_{1} = 298 K
∴T_{2} = (298 + 10) K
= 308 K
We also know that the rate of the reaction doubles when temperature is increased by 10°.
Therefore, let us take the value of k_{1} = k and that of k_{2} = 2k
Also, R = 8.314 J K^{−1} mol^{−1}
Now, substituting these values in the equation:
We get:
= 52897.78 J mol^{−1}
= 52.9 kJ mol^{−1}
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
Page No 116:
Question 4.9:
The activation energy for the reaction
2HI_{(}_{g}_{)} → H_{2} + I_{2}_{(}_{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:
In the given case:
E_{a} = 209.5 kJ mol^{−1} = 209500 J mol^{−1}
T = 581 K
R = 8.314 JK^{−1} mol^{−1}
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:
$x={e}^{{E}_{a}/\mathrm{R}T}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{In}x=\frac{{E}_{a}}{\mathrm{R}T}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}x=\frac{{E}_{a}}{2.303\mathrm{R}T}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}x=\frac{209500\mathrm{J}{\mathrm{mol}}^{1}}{2.303\times 8.314{\mathrm{JK}}^{1}{\mathrm{mol}}^{1}\times 581}=18.8323\phantom{\rule{0ex}{0ex}}Now,x=\mathrm{Antilog}\left(18.8323\right)\phantom{\rule{0ex}{0ex}}=1.471\times {10}^{19}$
Page No 117:
Question 4.1:
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) + 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]
Answer:
(i) Given rate = k [NO]^{2}
Therefore, order of the reaction = 2
Dimension of
(ii) Given rate = k [H_{2}O_{2}] [I^{−}]
Therefore, order of the reaction = 2
Dimension of
(iii) Given rate = k [CH_{3}CHO]^{3/2}
Therefore, order of reaction =
Dimension of
(iv) Given rate = k [C_{2}H_{5}Cl]
Therefore, order of the reaction = 1
Dimension of
Page No 117:
Question 4.2:
For the reaction:
2A + B → A_{2}B
the rate = k[A][B]^{2} with k = 2.0 × 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}.
Answer:
The initial rate of the reaction is
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 = 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}
Page No 117:
Question 4.3:
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 × 10^{−4} mol^{−1} L s^{−1}?
Answer:
The decomposition of NH_{3} on platinum surface is represented by the following equation.
Therefore,
However, it is given that the reaction is of zero order.
Therefore,
Therefore, the rate of production of N_{2} is
And, the rate of production of H_{2} is
= 7.5 × 10^{−4} mol L^{−1} s^{−1}
Page No 117:
Question 4.4:
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.,
If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?
Answer:
If pressure is measured in bar and time in minutes, then
Unit of rate = bar min^{−1}
Therefore, unit of rate constants
Page No 117:
Question 4.5:
Mention the factors that affect the rate of a chemical reaction.
Answer:
The factors that affect the rate of a reaction are as follows.
(i) Concentration of reactants (pressure in case of gases)
(ii) Temperature
(iii) Presence of a catalyst
Page No 118:
Question 4.6:
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:
Letthe 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
= 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. , then the rate of the reaction would be
Therefore, the rate of the reaction would be reduced to
Page No 118:
Question 4.7:
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Answer:
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,
where, k is 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
Page No 118:
Question 4.8:
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.
Answer:
(i) Average rate of reaction between the time interval, 30 to 60 seconds,
= 4.67 × 10^{−3} mol L^{−1} s^{−1}
(ii) For a pseudo first order reaction,
For t = 30 s,
= 1.911 × 10^{−2} s^{−1}
For t = 60 s,
= 1.957 × 10^{−2} s^{−1}
For t = 90 s,
= 2.075 × 10^{−2} s^{−1}
Then, average rate constant,
Page No 118:
Question 4.9:
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?
Answer:
(i) The differential rate equation will be
(ii) If the concentration of B is increased three times, then
Therefore, the rate of reaction will increase 9 times.
(iii) When the concentrations of both A and B are doubled,
Therefore, the rate of reaction will increase 8 times.
Page No 118:
Question 4.10:
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?
Answer:
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,
Dividing equation (i) by (ii), we obtain
Dividing equation (iii) by (ii), we obtain
= 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.
Page No 118:
Question 4.11:
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.
Answer:
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,
According to the question,
Dividing equation (iv) by (i), we obtain
Dividing equation (iii) by (ii), we obtain
Therefore, the rate law is
Rate = k [A] [B]^{2}
From experiment I, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
From experiment II, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
From experiment III, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
From experiment IV, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
Therefore, rate constant, k = 6.0 L^{2} mol^{−2} min^{−1}
Page No 118:
Question 4.12:
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}
Answer:
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 × 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 × 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} × 0.4 mol L^{−1}
= 0.08 mol L^{−1} min^{−1}
From experiment IV, we obtain
2.0 × 10^{−2} mol L^{−1} min^{−1} = 0.2 min^{−1} [A]
⇒ [A] = 0.1 mol L^{−1}
Page No 119:
Question 4.13:
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}
Answer:
(i) Half life,
= 3.47 $\times $10 ^{3} s (approximately)
(ii) Half life,
= 0.35 min (approximately)
(iii) Half life,
= 0.173 years (approximately)
Page No 119:
Question 4.14:
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.
Answer:
Here,
It is known that,
= 1845 years (approximately)
Hence, the age of the sample is 1845 years.
Page No 119:
Question 4.15:
The experimental data for decomposition of N_{2}O_{5}
in gas phase at 318K are given below:

t(s)
0
400
800
1200
1600
2000
2400
2800
3200
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).
Answer:
(ii) Time corresponding to the concentration, _{ }is the half life. From the graph, the half life is obtained as 1450 s.
(iii)

t(s)
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, v/s t, is a straight line. Therefore, the rate law of the reaction is
(v) From the plot, v/s t, we obtain
Again, slope of the line of the plot v/s t is given by
.
Therefore, we obtain,
(vi) Halflife is given by,
This value, 1438 s, is very close to the value that was obtained from the graph.
Page No 119:
Question 4.16:
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?
Answer:
It is known that,
Hence, the required time is 4.6 × 10^{−2} s.
Page No 119:
Question 4.17:
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.
Answer:
Here,
It is known that,
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.
Page No 119:
Question 4.18:
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 a first order reaction, the time required for 99% completion is
For a first order reaction, the time required for 90% completion is
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.
Page No 119:
Question 4.19:
A first order reaction takes 40 min for 30% decomposition. Calculate t_{1/2}.
Answer:
For a first order reaction,
Therefore, t_{1/2} of the decomposition reaction is
= 77.7 min (approximately)
Page No 119:
Question 4.20:
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.
Answer:
The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.
After time, t, total pressure,
= 2P_{0} − P_{t}
For a first order reaction,
When t = 360 s,
= 2.175 × 10^{−3} s^{−1}
When t = 720 s,
= 2.235 × 10^{−3} s^{−1}
Hence, the average value of rate constant is
= 2.21 × 10^{−3} s^{−1}
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
Page No 119:
Question 4.21:
The following data were obtained during the first order thermal decomposition of SO_{2}Cl_{2} at a constant volume.

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.
Answer:
The thermal decomposition of SO_{2}Cl_{2} at a constant volume is represented by the following equation.
After time, t, total pressure,
Therefore,
= 2 P_{0} − P_{t}
For a first order reaction,
When t = 100 s,
= 2.231 × 10^{−3} s^{−1}
When P_{t} = 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_{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()
= (2.23 × 10^{−3} s^{−1}) (0.35 atm)
= 7.8 × 10^{−4} atm s^{−1}
Page No 120:
Question 4.22:
The rate constant for the decomposition of N_{2}O_{5} at various temperatures is given below:

T/°C
0
20
40
60
80
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 30º and 50ºC.
Answer:
From the given data, we obtain

T/°C
0
20
40
60
80
T/K
273
293
313
333
353
3.66×10^{−3}
3.41×10^{−3}
3.19×10^{−3}
3.0×10^{−3}
2.83 ×10^{−3}
0.0787
1.70
25.7
178
2140
ln k
−7.147
− 4.075
−1.359
−0.577
3.063
Slope of the line,
According to Arrhenius equation,
Again,
When ,
Then,
Again, when ,
Then, at ,
Page No 120:
Question 4.23:
The rate constant for the decomposition of hydrocarbons is 2.418 × 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.
Answer:
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)
Page No 120:
Question 4.24:
Consider a certain reaction A → Products with k = 2.0 × 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}.
Answer:
k = 2.0 × 10^{−2} s^{−1}
T = 100 s
[A]_{o} = 1.0 moL^{−1}
Since the unit of k is s^{−1}, the given reaction is a first order reaction.
Therefore,
= 0.135 mol L^{−1} (approximately)
Hence, the remaining concentration of A is 0.135 mol L^{−1}.
Page No 120:
Question 4.25:
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?
Answer:
For a first order reaction,
It is given that, t_{1/2} = 3.00 hours
Therefore,
= 0.231 h^{−1}
Then, 0.231 h^{−1}
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.
Page No 120:
Question 4.26:
The decomposition of hydrocarbon follows the equation
k = (4.5 × 10^{11 }s^{−1}) e^{−28000 }^{K}^{/}^{T}
Calculate E_{a}.
Answer:
The given equation is
k = (4.5 × 10^{11 }s^{−1}) e^{−28000 }^{K}^{/}^{T} (i)
Arrhenius equation is given by,
(ii)
From equation (i) and (ii), we obtain
= 8.314 J K^{−1} mol^{−1} × 28000 K
= 232792 J mol^{−1}
= 232.792 kJ mol^{−1}
Page No 120:
Question 4.27:
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 × 10^{4 }K/T
Calculate E_{a} for this reaction and at what temperature will its halfperiod be 256 minutes?
Answer:
Arrhenius equation is given by,
The given equation is
From equation (i) and (ii), we obtain
= 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,
= 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)
Page No 120:
Question 4.28:
The decomposition of A into product has value of k as 4.5 × 10^{3} s^{−1} at 10°C and energy of activation 60 kJ mol^{−1}. At what temperature would k be 1.5 × 10^{4} s^{−1}?
Answer:
From Arrhenius equation, we obtain
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.
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
Page No 120:
Question 4.29:
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 × 10^{10 }s^{−1}. Calculate k at 318 K and E_{a}.
Answer:
For a first order reaction,
At 298 K,
At 308 K,
According to the question,
From Arrhenius equation, we obtain
To calculate k at 318 K,
It is given that,
Again, from Arrhenius equation, we obtain
Page No 120:
Question 4.30:
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.
Answer:
From Arrhenius equation, we obtain
Hence, the required energy of activation is 52.86 kJmol^{−1}.
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