NCERT Solutions for Class 12 Chemistry Chapter 4 - Chemical Kinetics

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

$$\textbf{(i) 3NO(g)}\xrightarrow{}\textbf{N}_2\textbf{O(g)}+\textbf{NO}_2\textbf{(g)}\\\textbf{Rate = k[NO]}^{2}\\\textbf{(ii) H}_2\textbf{O}_2(aq)+3\textbf{I}^{\normalsize-}(aq)+2\textbf{H}^{\normalsize+}\xrightarrow{}2\textbf{H}_2\textbf{O(l)}+\textbf{I}_{3}^{\normalsize-}\\\textbf{Rate = k[H}_2\textbf{O}_2][I^{\normalsizeโ€“}]\\\textbf{(iii) CH}_3\textbf{CHO(g)}\xrightarrow{}\textbf{CH}_4\textbf{(g)}+\textbf{CO(g)}\\\textbf{Rate = k}[\textbf{CH}_3\textbf{CHO}]^{3/2}\\\textbf{(iv) C}_2\textbf{H}_5\textbf{Cl(g)}\xrightarrow{}\textbf{C}_2\textbf{H}_4\textbf{(g)}+\textbf{HCl(g)}$$

Ans. (i) Order of reaction = 2

Dimension of rate constant

$$k=\frac{\text{Rate}}{[\text{NO}]^{2}}=\frac{\text{mol L}^{\normalsize-1}s^{\normalsize-1}}{(\text{mol L}^{\normalsize-1})^{2}}=\text{mol L}^{\normalsize-1}s^{\normalsize-1}$$

(ii) Order or reaction = 2

Dimension of rate constant

$$k=\frac{\text{Rate}}{[\text{H}_2\text{O}_2][\text{I}^{\normalsize-}]}=\frac{\text{mol L}^{\normalsize-1}s^{\normalsize-1}}{(\text{mol}\space L^{\normalsize-1})(\text{mol L}^{\normalsize-1})}=\text{mol L}^{\normalsize-1}s^{\normalsize-1}$$

(iii) Order of reaction = 3/2

$$k=\frac{\text{Rate}}{[\text{CH}_3\text{CHO}]^{3/2}}=\frac{\text{mol}^{\normalsize-1}s^{\normalsize-1}}{(\text{mol L}^{\normalsize-1})(\text{mol L}^{\normalsize-1})}=\text{mol}^{\normalsize-1/2 }\text{L}^{1/2}s^{\normalsize-1}\\\text{(iv) Order of reaction = 1}\\\text{Dimension of rate constant}\\k=\frac{\text{Rate}}{\text{C}_2\text{H}_5 \text{Cl}}=\frac{\text{mol L}^{\normalsize-1}s^{\normalsize-1}}{\text{mol L}^{\normalsize-1}}=s^{\normalsize-1}$$

2. For the reaction; $$2\textbf{A + B}\xrightarrow{}\textbf{A}_2\textbf{B},$$the rate = k [A][B]2 with k = 2ยท0 ร— 10โ€“6 molโ€“2 L2 sโ€“1. Calculate the initial rate of the reaction when [A] = 0ยท1 mol Lโ€“1; [B] = 0ยท2 mol Lโ€“1. Calculate the rate of reaction after [A] is reduced to 0ยท06 mol Lโ€“1.

Ans. Given : k = 2 ร— 10โ€“6 molโ€“2 L2 sโ€“1

To find : Rate when [A] = 0.1 mol Lโ€“1, [B] = 0.2 mol Lโ€“1

Rate when [A] = 0.1 mol Lโ€“1, [B] = 0.06 mol Lโ€“1

Formula : Rate = k[A][B]2

First case :

Rate = k[A][B]2

= (2.0 ร— 10โ€“6 molโ€“2 L2 sโ€“1) ร— (0.1 mol Lโ€“1) ร— (0.2 mol Lโ€“1)2

= 8 ร— 10โ€“9 mol Lโ€“1 sโ€“1 = 8 ร— 10โ€“9 Msโ€“1.

Second case :

The concentration of A after taking part in the reaction = 0.06 mol Lโ€“1.

Amount of A reacted = (0.1 โ€“ 0.06) = 0.04 mol Lโ€“1

$$\text{Amount of B reacted =}\frac{1}{2}ร—0.04\space\text{mol L}^{\normalsize-1}=0.02\space\text{mol L}^{\normalsize-1}.$$

The concentration of B after taking part in the reaction = (0.2 โ€“ 0.02) = 0.18 mol Lโ€“1

Rate = k[A][B]2

= (2.0 ร— 10โ€“6 molโ€“2 L2 sโ€“1) ร— (0.06 mol Lโ€“1) ร— (0.18 mol Lโ€“1)2.

= 3.89 ร— 10โ€“9 mol Lโ€“1 sโ€“1 = 3.90 ร— 10โ€“9 Msโ€“1.

Hence, the rate of reaction is 3.90 ร— 10โ€“9 Msโ€“1.

3. 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 ร— 10โ€“4 molโ€“1 Lsโ€“1? $$\textbf{Ans.}\space\text{NH}_3\xrightarrow{}\frac{1}{2}\text{N}_2+\frac{3}{2}\text{H}_2\\\therefore\text{Rate}=-\frac{d[\text{NH}_3]}{dt}=\frac{2d[\text{N}_2]}{dt}=\frac{2d[\text{H}_2]}{3dt}\\= k = 2.5 ร— 10^{\normalsizeโ€“4} \text{mol L}^{\normalsizeโ€“1}\text{s}^{\normalsizeโ€“1}.\\\therefore\text{Rate of production of N}_2=\frac{d[\text{N}_2]}{dt}=\frac{k}{2}=\frac{2.5ร—10^{\normalsize-4}\text{mol L}^{\normalsize-1}s^{\normalsize-1}}{2}\\= 1.25 ร— 10^{\normalsizeโ€“4} \text{mol L}^{\normalsizeโ€“1} s^{\normalsizeโ€“1}\\\text{Rate of production of H}_2=\frac{d[\text{H}_2]}{dt}=\frac{3k}{2}=\frac{3ร—2.5ร—10^{\normalsize-4}\text{mol L}^{\normalsize-1}s^{\normalsize-1}}{2}\\= 3.75 ร— 10^{\normalsizeโ€“4} \text{mol L}^{\normalsizeโ€“1} s^{\normalsizeโ€“1}.$$ Hence, the rates of production of N2 and H2 is 1.25 ร— 10โ€“4 mol Lโ€“1 sโ€“1 and 3.75 ร— 10โ€“4 mol Lโ€“1 sโ€“1 respectively.

4. The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction, rate is given by Rate = k [CH3OCH3]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 (PCH3OCH3)3/2

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

Ans. As the concentration in the rate law equation is given in terms of pressure,

โˆด Unit of rate = bar minโ€“1

$$\therefore\space\text{Unit of k =}\frac{\text{Rate}}{[\text{P}_{\text{CH}_3\text{OCH}_3}]^{3/2}}=\frac{\text{bar min}^{\normalsize-1}}{\text{ bar}^{3/2}}\\\text{= bar}^{-1/2}\text{min}^{\normalsize-1}.$$

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

Ans. The factors that affects the rate of chemical reaction are as follows :

(i) Nature of reactants: Nature or reactivity of reactants affects the rate of chemical reaction.

E.g. Al is more reactive than Zn. Therefore, the rate of reaction of Al with HCl is higher than that of Zn.

(ii) Size of the particles of the reactants: Smaller the size of reactant particles, higher the rate of reaction.

E.g. When HCl is added to pieces of Sahabad tiles, the CO2 effervescence is formed slowly. However, when HCl is added to sahabad powder, the CO2 effervescence is formed at a faster rate.
(iii) Concentration of reactants: The rate of chemical reaction increases with increase in the concentration of reactants.

E.g. Dilute HCl reacts slowly with CaCO3 whereas concentrated HCl reacts rapidly with CaCO3.

(iv) Temperature of the reaction: Higher the temperature, faster the rate of chemical reaction.

E.g. The temperature in the refrigerator is low. So, the rate of decomposition of perishable foodstuffs is low and it remains fresh for a longer time.

(v) Catalyst: Presence of catalyst increase the rate of a chemical reaction.

E.g. The decomposition of H2O2 into H2O and O2 takes place slowly at room temperature. However, the same reaction occurs at a faster rate on adding MnO2 powder in it.

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 ?

Ans. Rate = k[A]2 = ka2

(i) If [A] = 2a, rate = k(2a)2 = 4 ka2 = 4ka2 = 4 times.

Rate of reaction becomes 4 times.

$$\text{(ii) If [A] = a/2, rate = k}\bigg(\frac{a}{2}\bigg)=\frac{1}{4}\text{ka}^{2}=\frac{1}{4}\text{th.}$$

Rate of reaction reduced to one fourth.

7. What is the effect of temperature on the rate constant of reaction? How can this effect of on rate constant be represented quantitatively?

Ans. The rate constant (k) of a reaction depends on temperature and it increases with rise in temperature and becomes nearly double with about every 10ยฐ rise in temperature. The effect of temperature on the rate constant is expressed quantitatively by Arrhenius equation.

k = Aeโˆ’Ea/Rt

$$\text{log}\frac{k_2}{k_1}=\frac{\text{E}_a}{2.303 \text{R}}\bigg[\frac{1}{\text{T}_1}-\frac{1}{\text{T}_2}\bigg]=\frac{0}{2.303\space \text{R}}\bigg[\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\bigg]=0\\\frac{k_2}{k_1}=\text{Antilog 0 = 1 or}\space k_2=k_1=1.6ร—10^{6}s^{\normalsize-1}$$

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

t/s 0 30 60 90
[A]/mol Lโ€“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.

$$\textbf{Ans.}\space\text{(i) Average rate of reaction between the time interval 30 โ€“ 60 sec =}\space\frac{\text{C}_2-\text{C}_1}{t_2-t_1}\\=\frac{0.31-0.17}{60-30}=\frac{0.14}{30}=4.6ร—10^{\normalsize-3}\text{mol L}^{\normalsize-1}s^{\normalsize-1}\\\text{(ii) For a pseudo first order reaction, kโ€™ =}\frac{2.303}{t}\text{log}\frac{[\text{A}_0]}{\text{A}}\\\text{when initial concentration of A, [A}_0] = 0.55\text{M}\\\text{when t = 30 sec k}_1โ€ฒ =\frac{2.303}{30}\text{log}\frac{0.55}{0.31}=1.92ร—10^{\normalsize-2}s^{\normalsize-1}\\\text{when t = 60 sec k}_2=\frac{2.303}{60}\text{log}\frac{0.55}{0.17}=1.92ร—10^{\normalsize-2}s^{\normalsize-1}\\\therefore\text{Average rate constant,}\space k'=\frac{k'_1+k'_2+k'_3}{3}=\frac{(1.91+1.96+2.07)ร—10^{\normalsize-2}}{3}$$

= 1.98 ร— 10โ€“2 sโ€“1.

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 is doubled?

$$\textbf{Ans.}(i)\space\text{Rate of reaction =}\frac{-d[\text{A}]}{dt}=\frac{1}{2}=\frac{-d[B]}{dt}$$

= k[A][B]2

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

When [B] becomes three times, rate

= k[A][3B]2 = 9k[A][B]2

= 9 times

Hence, when concentration of B is tripled, rate of reaction increased by 9 times.

(iii) When both [A] and [B] are doubled, rate = k[2A][2B]2 = 8[A][B]2 = 8 times.

Hence, rate of reaction increases by 8 times.

10. 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 Lโ€“1 0.20 0.20 0.40
B/mol Lโ€“1 0.30 0.10 0.05
r0/mol Lโ€“1 sโ€“1 5.07ร— 10โ€“5 5.07 ร— 10โ€“5 1.43 ร— 10โ€“4

What is the order of the reaction with respect to A and B?

Ans. Rate = [A]ฮฑ[B]ฮฒ

r1 = 5.07 ร— 10โ€“5 = (0.20)ฮฑ(0.30)ฮฒ ...(i)

r2 = 5.07 ร— 10โ€“5 = (0.20)ฮฑ(0.10)ฮฒ ...(ii)

r3 = 1.43 ร— 10โ€“4 = (0.40)ฮฑ(0.05)ฮฒ ...(iii)

$$\frac{r_1}{r_2}=\frac{5.07ร—10^{\normalsize-5}}{5.07ร—10^{\normalsize-5}}=\frac{(0.2)^{\alpha}}{(0.2)^{\alpha}}\frac{(0.3)^{\beta}}{(0.10)^{\beta}} =1=(3)^{\beta}\\\beta=0\\\frac{r_3}{r_2}=\frac{1.43ร—10^{\normalsize-4}}{5.06ร—10^{\normalsize-5}}=\frac{(0.4)^{\alpha}}{(0.20)^{\alpha}}\frac{(0.05)^{\beta}}{(0.10)^{\beta}}\\=2.826=2^\alpha\bigg(\frac{1}{2}\bigg)^\beta$$

2.826 = 2ฮฑ (as ฮฒ = 0)

Taking log on both the sides, we get

log 2.826 = ฮฑ log 2

โ‡’ 0.4511 = ฮฑ ร— 0.3010


โˆด Hence, Order with respect to A = 1.5 and order with respect to B = 0

11. The following results have been obtained during the kinetic studies of the reaction.

$$2\text{A + B}\xrightarrow{}\text{C + D}$$

Experiment [A] mol L-1 [B] mol L-1 Initial rate of formation of D/mol Lโ€“1 minโ€“1
I 0.1 0.1 6.0 ร— 10โ€“3
II 0.3 0.2 7.2 ร— 10โ€“2
III 0.3 0.4 2.88 ร— 10โ€“1
IV 0.4 0.1 2.40 ร— 10โ€“2

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

Ans. Rate = k[A]ฮฑ[B]ฮฒ

r1 = k[0.1]ฮฑ[0.1]ฮฒ = 6.0 ร— 10โ€“3

r2 = k[0.3]ฮฑ[0.2]ฮฒ = 7.2 ร— 10โ€“2

r3 = k[0.3]ฮฑ[0.4]ฮฒ = 2.88 ร— 10โ€“1

r4 = k[0.4]ฮฑ[0.1]ฮฒ = 2.40 ร— 10โ€“2


โˆด ย ฮฒย = 2

โˆด Rate law expression, is Rate = k[A][B]2

โˆด Order with respect to A = 1 and

Order with respect to B = 2

Overall order of the reaction = 1 + 2 = 3

Calculation of rate constant k
Substituting values of experiment I, we get

6.0 ร— 10โ€“3 = k(0.1)(0.1)2

= k ร— 1 ร— 10โ€“3

โˆด k = 6 molโ€“2 L2 minโ€“1

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 Lโ€“1 minโ€“1
I 0.1 0.1 2.0ร—10-2
II โ€“ 0.2 4.0 ร— 10โ€“2
III 0.4 0.4 โ€“
IV โ€“ 0.2 2.0 ร— 10โ€“2

Ans. Rate law expression :

Rate = k[A]1[B]0 = k[A]

For 1st experiment R1 = 2.0 ร— 10โ€“2 mol Lโ€“1 minโ€“1

= k[0.1] mol Lโ€“1

k = 0.2 minโ€“1

For IInd experiment R2 = 4.0 ร— 10โ€“2 mol Lโ€“1 minโ€“1

= k[0.2 minโ€“1] [A]

[A] = 0.2 mol Lโ€“1

For IIInd experiment R3 = Rate = k[A]

= (0.2 minโ€“1) (0.4 mol Lโ€“1)

= 0.08 mol Lโ€“1 minโ€“1

For IVth experiment R4 = 2.0 ร— 10โ€“2 mol Lโ€“1 minโ€“1

= k[A] = 0.2 minโ€“1[A]

โˆด [A] = 0.1 mol Lโ€“1

[A] = 0.2 mol Lโ€“1

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

(i) 200 โ€“1 (ii) 2 minโ€“1 (iii) 4 yearsโ€“1

Ans. Given : k1 = 200 sโ€“1, k2 = 2 minโ€“1, k3 = 4 yearsโ€“1

To find t1/2 in all three cases

$$\text{Formula : Half life for a first order reaction is, t}_{1/2} =\frac{0.693}{k}\\\text{Calculation :}\\\text{(i)}\space\text{t}_{1/2}=\frac{0.693}{200 s ^{\normalsize-1}}=0.346ร—10^{\normalsize-2}s=3.46ร—10^{\normalsize-3}s\\\text{(ii)}\space\text{t}_{1/2}=\frac{0.693}{2\space\text{min}^{\normalsize-1}}=0.346\space\text{min}=3.46ร—10^{\normalsize-1}\space\text{min.}\\\text{(iii)}\space t_{1/2}=\frac{0.693}{4 \text{Year}^{\normalsize-1}}=0.173\space\text{year}=1.73ร—10^{\normalsize-1}\text{year}$$

14. The half-life for radioactive decay of HC 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.

Ans. Radioactive decay follows first order kinetics.

Given: [A0] = 100, [A] = 80, t1/2 = 5730 year

To Find: Age of sample?

$$\text{Formula}:\space k=\frac{0.639}{t_{1/2}}\\t=\frac{2.303}{k}\text{log}\frac{[\text{A}_0]}{[\text{A}]}\\\text{Calculation: Radioactive decay follows first order kinetics. Therefore,}\\\text{Decay constant (k)}\space t=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730}\text{Year}^{\normalsize-1}\\\text{t}=\frac{2.303}{k}\text{log}\frac{[\text{A}_0]}{[\text{A}]}\\=\frac{2.303}{(0.693/5730\space \text{years}^{\normalsize-1})}\text{log}\frac{100}{8}\\=\frac{2.303ร—5730}{0.693}ร—0.0969=1845\space\text{years}$$

Hence, the age of wood will be 1845 years.

15. The experimental data for decomposition of N2O5:


in gas phase at 318 K are given below :

t/s 0 400 800 1200 1600 2000 2400 2800 3200
102 ร— [N2O5]/mol Lโ€“1 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35

(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).

Ans. (i) Plot of [N2O5] Vs time

Chemical Kinetics_ans15

(ii) Initial concentration of N2O5 = 1.63 ร— 102 M

Time corresponding to this concentration

$$\text{N}_2\text{O}_5=\frac{1.63ร—10^{2}}{2}=81.5\space\text{molL}^{\normalsize-1}\text{is half life.}$$

From the plot, half-life corresponding to this concentration is, t2 = 1450 S.

(iii) Plot of log [N2O5] Vs time.

t(s) 0 400 800 1200 1600 2000 2400 2800 3200
102 ร— [N2O5]/mol Lโ€“1 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35
log[N2O5] โ€“1.79 โ€“1.87 โ€“1.94 โ€“2.03 โ€“2.11 โ€“2.19 โ€“2.28 โ€“2.37 โ€“2.46

(iv) As the plot obtained for log [N2O5] Vs time is a straight line, hence it is a first order reaction. Therefore, rate law for the reaction is:

โˆด Rate = k[N2O5]

$$\text{(v)}\space\text{Slope of the line =}\frac{k}{2.303}\\\text{Slope}=\frac{-2.46-(-1.79)}{3200-0}=-\frac{0.67}{3200}\\k=\frac{0.67ร—2.303}{3200}=4.82ร—10^{\normalsize-4}\text{mol L}^{\normalsize-1}\text{s}^{\normalsize-1}\\\text{(vi)}\space\text{Half life period}\space t_{1/2}=\frac{0.693}{k}=\frac{0.693}{4.82ร—10^{\normalsize-4}}=1438\space\text{sec.}$$

Half life period (t1/2) is calculated from the formula and slope are approximately the same.

16. The rate constant for a first order reaction is 60 sโ€“1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

$$\textbf{Ans.}\space \text{Given: k = 60 s}^{\normalsize-1},\text{R}=\frac{\text{R}_0}{16}\\\text{To find: time t}\\\text{For 1}^{\text{st}} \text{order reaction}\\\text{Formula:}\space t=\frac{2.303}{k}\text{log}\frac{\text{R}_0}{\text{R}}\\\text{Calculation:}\space t=\frac{2.303}{60}\text{log}\frac{\text{R}_0}{\text{R}_{0/16}}=\frac{2.303}{60}\text{log}\space16=4.62ร—10^{\normalsize-2}s$$

17. During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1 ฮผg 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?

Ans. Given : t1/2 = 28.1 years

[R0] = 1 mg

To find: (i) Amount left after 10 years

Amount left after 60 years

$$\text{Formula}:\space\lambda=\frac{0.693}{t_{1/2}}\\\lambda=\frac{2.303}{t}\text{log}\frac{[\text{R}_0]}{[\text{R}]}\\\text{Calculation: As radioactive disintegration follows first order kinetics. Hence}\\\text{Decay constant of}\space^{90}\text{Sr},(\lambda)=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.1}$$

= 2.466 ร— 10โ€“2 yrโ€“1

(i) To calculate the amount left after 10 year

$$\text{Using formula,}\space\lambda=\frac{2.303}{t}\text{log}\frac{[\text{R}_0]}{[\text{R}]}$$

$$\text{Or}\qquad 2.466ร—10^{\normalsize-2}=\frac{2.303}{10}\text{log}\frac{1}{[\text{R}]}\\\frac{2.466ร—10^{\normalsize-2}ร—10}{2.303}=-\text{log[R]}$$

Or, log [R] = โ€“ 0.1071

Or, [R] = Antilog (โ€“ 0.1071) = 0.7814 ยตg

Amount left after 10 years = 0.7814 ยตg

(ii) To calculate the amount left after 60 years,

$$\text{Or},\space 2.466ร—10^{\normalsize-2}=\frac{2.303}{60}\text{log}\frac{1}{[\text{R}]}\\\text{Or,}\qquad\frac{2.466ร—10^{\normalsize-2}ร—60}{2.303}=-\text{log}[\text{R}]$$

Or, log [R] = โ€“ 0.6425

Or, [R] = Antilog (โ€“0.6425) = 0.2278 ยตg

Amount left after 60 years = 0.2278 ยตg

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

$$\textbf{Ans.}\space\text{For a first order reaction;}\space t=\frac{2.303}{k}\text{log}\frac{[\text{R}_0]}{[\text{R}]}\\\textbf{I}^{\textbf{st}} \textbf{case:}\space \text{R}_0 = 100\%, \text{R} = 100 โ€“ 99 = 1\%\\t_{99\%}=\frac{2.303}{k}\text{log}\frac{100}{1}=\frac{2.303}{k}\text{log}10^{2}\\=\frac{2.303ร—2}{k}=\frac{4.606}{k}\\\textbf{II}^{\textbf{nd}}\textbf{case:}\space \text{R}_0 = 100\%; \text{R} = 100 โ€“ 90 = 10\%\\t_{90\%}=\frac{2.303}{k}\text{log}\frac{100}{10}=\frac{2.303}{k}\text{log \space 10}=\frac{2.303}{k}\\\text{Dividing eq. (ii) by eqn. (i),}\\\frac{t_{(99\%)}}{t_{(90\%)}}=\frac{4.606}{k}ร—\frac{k}{2.303}=2$$

It means that time required for 99% completion of reaction is twice the time required to complete 90% of the reaction.

19. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.

Ans. Given: 30% decomposition means that x = 30% of [R0] or, [R] = [R0] โ€“3[R0] = 0.7[R0]

To Find: t1/2

$$\text{Formula: k}=\frac{2.303}{t}\text{log}\frac{[\text{R}_0]}{[\text{R}]}\\\text{Calculation: For reaction of 1}^{st} \text{order,}\\k=\frac{2.303}{t}\text{log}\frac{[\text{R}_0]}{[\text{R}]}=\frac{2.303}{40}\text{log}\frac{[\text{R}_0]}{0.70[\text{R}_0]}=\frac{2.303}{40}\text{log}\frac{10}{7}\text{min}^{\normalsize-1}\\=\frac{2.303}{40}ร—0.1549\space\text{min}^{\normalsize-1}=8.918ร—10^{\normalsize-3}\text{min}^{\normalsize-1}\\\text{For a 1}^\text{st} \text{order reaction half life is,}\\t_{1/2}=\frac{0.693}{k}=\frac{0.693}{8.918ร—10^{\normalsize-3}\text{min}^{\normalsize-1}}=77.7\space\text{min}$$

Hence, half-life is 77.7 min.

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.

$$\textbf{Ans.}\space\underset{\text{Azoisopropane}}{(\text{CH}_3)_2\text{CHN = NCH(CH}_3)_2(\text{g})}\xrightarrow{}\underset{\text{Hexane}}{\text{N}_2(g)+\text{C}_6\text{H}_{14}\text{(g)}}$$

Initial pressure P0 0 0
Pressure P0 โ€“ P p p

After time t

Total pressure after time t(Pt) = (P0โ€“p) + p + p = P0 + p or p = Pt โ€“ P0

[R]0 โˆ P0 and [R] โˆ P0 โ€“ p

On substituting the value of p,

[R] โˆ P0 โ€“ (Pt โ€“ P0), i.e. [R] โˆ 2P0 โ€“ Pt

As decomposition of azoisopropane is a first order reaction

$$\therefore\space k=\frac{2.303}{t}\text{log}\frac{[\text{R}_0]}{[\text{R}]}=\frac{2.303}{t}\text{log}\frac{\text{P}_0}{2\text{P}_0-\text{P}_t}\\\text{When t = 360 sec,}\\\text{Rate constant k}=\frac{2.303}{360}\text{log}\frac{35.0}{2ร—35.0-54.0}=\frac{2.303}{360}\text{log}\frac{35.0}{16}=2.175ร—10^{\normalsize-3}s^{\normalsize-1}\\\text{When t = 720 sec,}\\\text{Rate constant k =}\frac{2.303}{720}\text{log}\frac{35.0}{2ร—35.0-63}=\frac{2.303}{720}\text{log}5=2.235ร—10^{\normalsize-3}s^{\normalsize-1}\\\therefore\space\text{Average value of k =}\frac{2.175ร—2.235}{2}ร—10^{\normalsize-3}s^{\normalsize-1}=2.20ร—10^{\normalsize-3}s^{\normalsize-1}$$

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


Experiment Time/sโ€“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.

$$\textbf{Ans.} \space\text{SO}_2\text{Cl}_2\text{(g)}\xrightarrow{}\text{SO}_{2(g)}+\text{Cl}_{2(g)}$$

Let initial pressure P0 0 0
Pressure at time t P0 โ€“ p p p

Let initial pressure P0 โˆ R0

Pressure at time t, Pt = P0 โ€“ p + p + p = P0 + p

โˆด Pressure of reactant at time Pt = P0 โ€“p

= 2P0 โ€“ Pt โˆ R

$$\text{Using formula:}\qquad k=\frac{2.303}{t}\text{log}\frac{\text{P}_0}{2\text{P}_0-\text{P}_t}\\\text{When t = 100 s,}\space k=\frac{2.303}{100}\text{log}\frac{0.5}{2ร—0.5-0.6}=\frac{2.303}{100}\text{log}(1.25)\\=\frac{2.303}{100}(0.0969)=2.2316ร—10^{\normalsize-3}s^{\normalsize-1}$$

When Pt = 0.65 atm,

โˆด Pressure of SO2Cl2 at time t (PSO2Cl2),

R = 2P0 โ€“ pt = 2 ร— 0.50 โ€“ 0.65 atm = 0.35 atm

Rate at that time = k ร— PSO2Cl2

= (2.2316 ร— 10โ€“3) ร— (0.35)

= 7.8 ร— 10โ€“4 atm sโ€“1

Hence, the rate of the reaction when total pressure is 0.65 atm is 7.8 ร— 10โ€“4 atm sโ€“1.

22. The rate constant for the decomposition of N2O5 at various temperatures is given below :
T/ยฐC 105 ร— k/sโ€“1
0 0.0787
20 1.70
40 25.7
60 178
80 2140

Draw a graph between ln k and 1T and calculate the value of A and Ea. Predict the rate constant at 30ยฐC and 50ยฐC.

Ans. For the given data, we obtain,

T/ยฐC 0 20 40 60 80
T/(K) 273 293 313 333 353
$$\frac{1}{\text{T}}/\text{K}^{\normalsize-1}$$ 3.66 ร— 10โ€“3 3.41 ร— 10โ€“3 3.19 ร— 10โ€“3 3.0 ร— 10โ€“3 2.83 ร— 10โ€“3
105ร— k/sโ€“1 0.0787 1.70 25.7 178 2140
ln k โ€“7.147 โ€“ 4.075 โ€“ 1.359 โ€“ 0.577 3.063
Chemical Kinetics_ans22

$$\text{Slope of the line,}\\\frac{y_2-y_1}{x_2-x_1}=-12.301\space\text{K}\\\text{According to Arrhenius equation,}\\\text{Slope}=\frac{\text{E}_a}{\text{R}}$$

โ‡’ Ea = โ€“Slope ร— R

= โ€“(โ€“12.301 K) ร— (8.314 J Kโ€“1 molโ€“1)

= 102.27 kJ molโ€“1


$$\text{In k = In A =}\frac{\text{E}_a}{\text{RT}}\\\text{In A = In k =}\frac{\text{E}_a}{\text{RT}}\\\text{When}\space\text{ T = 273 K.}\\\text{In k = โ€“7.147}\\\text{Then, In}\space \text{A}=-7.147+\frac{102.27ร—10^{3}}{8.314ร—273}=37.911\\\text{Therefore, A = 2.91 ร— 10}^{6}\\\text{When T = 30 + 273 K = 303 K,}\\\frac{1}{\text{T}}=0.0033\space\text{K}=3.3ร—10^{\normalsize-3}\text{K}\\\text{Then,}\space\text{at}\space \frac{1}{\text{T}}=3.1ร—10^{\normalsize-3}\text{K}$$

In k = โ€“0.5

Therefore,ย  k = 0.607 sโ€“1

23. The rate constant for the decomposition of a hydrocarbon is 2ยท418 ร— 10โ€“5 sโ€“1 at 546 K. If the energy of activation is 179ยท9 kJ molโ€“1, what will be the value of pre-exponential factor?

Ans. Given: k = 2.418 ร— 10โ€“5 sโ€“1; Ea = 179900 molโ€“1; R = 8.314 JKโ€“1 molโ€“1; T = 546 K

To find: Pre-exponential Factor (A)

Formula:ย  k = AeEa/RT

$$\text{log k = log A}-\frac{\text{E}_a}{2.303\space\text{RT}}\\\text{log A}=\text{log k}+\frac{\text{E}_a}{2.303\text{RT}}\\=\text{log}(2.418ร—10^{5}s^{\normalsize-1})+\frac{(179900 \text{J mol}^{\normalsize-1})}{2.303ร—(8.314\space\text{J K}^{\normalsize-1} \text{mol}^{\normalsize-1})ร—546\text{K}}$$

= โ€“ 4.6184 + 17.21 = 12.5916

A = Antilog 12.5916 = 3.9 ร— 1012 sโ€“1.

Hence, the value of pre-exponential factor is 3.9 ร— 1012 sโ€“1.

24. Consider a certain reaction A โ†’ Products with k = 2.0 ร— 10โ€“2 sโ€“1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol Lโ€“1.

Ans. Given: k = 2.10โ€“2 sโ€“1

A0 = 1.0 molโ€“1

t = 100 sec

To find: A = ?

$$\text{Formula: k =}\frac{2.303}{t}\text{log}\frac{[\text{A}_0]}{[\text{A}]}\\\text{Calculation : The unit of k show that the reaction is of first order.}\\\text{Hence, k}=\frac{2.303}{t}\text{log}\frac{[\text{A}_0]}{[\text{A}]}\\\text{Or,}\qquad 2.0ร—10^{\normalsize-2}=\frac{2.303}{100}\text{log}\frac{1.0}{[A]}$$

Or,ย  ย log [A] = โ€“0.8684

โˆด [A] = Antilog (โ€“0.8684) = 0.1354 mol Lโ€“1

Hence, the concentration of A remaining after 100 s is 0.1354 mol Lโ€“1.

25. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Ans. Given: t1/2 = 3 hours, t = 8 hours

$$\text{To find:}\qquad\frac{A}{A}=?\\\text{Formula:}\qquad k=\frac{0.693}{t_{1/2}}\\k=\frac{2.303}{t}\text{log}\frac{\text{[A}_0]}{[\text{A}]}\\\text{Calculation : Sucrose decomposes according to first order rate law , hence}\\k=\frac{0.693}{t_{1/2}}=\frac{0.693}{3}=0.231\space \text{hr}^{\normalsize-1}\\\text{Hence, 0.231}=\frac{2.303}{8}\text{log}\frac{[\text{A}_0]}{\text{A}}\\\text{Or,}\qquad\text{log}\frac{[\text{A}_0]}{[\text{A}]}=0.8024\\\text{Or,}\qquad\frac{[\text{A}_0]}{\text{[A]}}=\text{Antilog (0.8024)}=6.345\\\text{Or,}\qquad\frac{[\text{A}]}{[\text{A}_0]}=\frac{1}{6.345}=0.158$$

Hence, the fraction of sample of sucrose remains after 8 hours is 0.158.

26. The decomposition of a hydrocarbon follows the equation

k = (4.5 ร— 1011 sโ€“1)eโ€“28000K/T. Calculate Ea.

Ans. Given: k0 (4.5 ร— 1011 sโ€“1)eโ€“28000K/T

To Find: Ea

Formula: k = A eโ€“Ea/RT

Given equation is k = (4.5 ร— 1011 sโ€“1)eโ€“28000K/T

Arrhenius equation, k = Aeโ€“Ea/RT

Comparing both the equations, we get

$$-\frac{\text{E}_a}{\text{RT}}=\frac{28000 \text{K}}{\text{T}}$$

Or,ย  ย  Ea = 28000 K ร— R = 28000 ร— 8.314

= 232.79 kJ molโ€“1

27. The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.34 โ€“ 1.25 ร— 104 K/T

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

Ans. Given: log k = 14.34 โ€“ 1.25 ร— 104 K/T

To find: Ea

Formula:ย  k = Aeโ€“Ea/RT

Calculation: According to Arrhenius equation, k = Aeโ€“Ea/RT

$$\text{Or,}\qquad\text{ln k = In A โ€“}\frac{\text{E}_a}{\text{RT}}\\\text{Or,}\qquad\text{log K}=\text{log A}-\frac{\text{E}_a}{2.303\text{RT}}\qquad...\text(i)\\\text{Given equation is}\\\text{log k = 14.34 โ€“ 1.25 ร— 10}^{4} \text{K/T}\space...\text{(ii)}\\\text{Comparing (i) with (ii),}\space\frac{\text{E}_a}{2.303\text{RT}}=\frac{1.25ร—10^{4}\text{K}}{\text{T}}$$

or , Ea = 2.303 R ร— 1.25 ร— 104 K

= 2.303 ร— (8.314) ร— 1.25 ร— 104

= 239.34 kJ molโ€“1

$$\text{When}\qquad\text{t}_{1/2}=256\space\text{min},k=\frac{0.693}{256ร—60}=4.51ร—10^{\normalsize-5}s^{\normalsize-1}\\\text{Substituting this value in the given equation,}\\\text{log (4.51 ร— 10}^{\normalsizeโ€“5}) = 14.34-\frac{1.25ร—10^{4}\text{K}}{\text{T}}\\\text{i.e}\qquad(-5+0.6542)=14.34-\frac{1.25ร—10^{4}}{\text{T}}\\\text{Or}\qquad\frac{1.25ร—10^{4}\text{K}}{\text{T}}=18.6858$$

Or , T = 669 K

28. The decomposition of A into product has value of k as 4.5 ร— 103 sโ€“1 at 10ยฐC and energy of activation 60 kJ molโ€“1. At what temperature would k be 1.5 ร— 104 sโ€“1 ?

Ans. Given: k1 = 4.5 ร— 103 sโ€“1,

T1 = 10 + 273 K = 283 K; k2 = 1.5 ร— 104 sโ€“1

Ea = 60 kJ molโ€“1

To find : T2 = ?

Formula : k = Aeโ€“Ea/RT


Applying Arrhenius equation,

$$\text{log}\frac{k_2}{k_1}=\frac{\text{E}_a}{2.303\text{R}}\bigg(\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\bigg)\\\text{log}\frac{1.5ร—10^{4}}{4.5ร—10^{3}}=\frac{60000}{2.303ร—8.314}\bigg(\frac{\text{T}_2-283}{283\text{T}_2}\bigg)\\\text{Or,}\qquad\text{log}3.333=3133.63\bigg(\frac{\text{T}_2-283}{283\text{T}_2}\bigg)\\\text{Or,}\qquad\frac{0.5228}{3133.63}=\bigg(\frac{\text{T}_2-283}{283\text{T}_2}\bigg)\\\text{Or, 0.0472 T}_2 = \text{T}_2 โ€“ 283\\\text{Or}\qquad 0.9528 \text{T}_2 = 283\\\text{Or,}\qquad\text{T}_2=\frac{283}{0.9528}$$

= 297 โ€“ 273 = 24ยฐC

Hence, the temperature would be 24ยฐC.

29. The time required for 10% completion of the 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 ร— 1010sโ€“1, calculated at 318 K and Ea.
Ans. Given: T1 = 298 K T2 = 308 K

A = 4 ร— 1010 sโ€“1

To Find: Ea and k at 318 K

$$\text{Formula:}\qquad t=\frac{2.303}{k}\text{log}\frac{[\text{A}]_0}{\text{[A]}}\\k=\text{Ae}^{-\text{E}_a/\text{RT}}\\\text{(a) Calculation of activation energy (E}_a)\\\text{For I}^{\text{st}} \text{order reaction:}\space k=\frac{2.303}{t}\text{log}\frac{[\text{A}]_0}{[\text{A}]}\\\text{At 298 K;}\space k_1=\frac{2.303}{t}\text{log}\frac{100}{90}\qquad...\text{(i)}\\\text{At 308 K;}\space k_2=\frac{2.303}{t}\text{log}\frac{100}{75}\qquad...(\text{ii})\\\text{Dividing equation (ii) by (i)}\\\frac{k_2}{k_1}=\frac{\text{log}\frac{100}{75}}{\text{log}\frac{100}{90}}=\frac{0.1249}{0.0458}=2.73$$

According to Arrhenius theory;

$$\text{log}\frac{k_2}{k_1}=\frac{\text{E}_a}{2.303\text{R}}ร—\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\\\text{log}\space2.73=\frac{\text{E}_a}{2.303\text{R}}\bigg[\frac{308-298}{298ร—308}\bigg]\\\text{E}_a=\frac{0.4361ร—2.303ร—(8.314\text{J mol}^{\normalsize-1})ร—298ร—308}{10}\\\text{E}_a=76640\space\text{J mol}^{\normalsize-1}=76.640\space\text{kJ mol}^{\normalsize-1}\\\text{(b)Calculation of rate constant (k)}\\\text{According to Arrhenius equation}\\\text{logk}=\text{log A}\frac{\text{E}_a}{2.303\text{RT}}\\\text{log k}=\text{log(4ร—10}^{10})-\frac{76640\text{J mol}^{\normalsize-1}}{2.303ร—(8.314\text{J mol}^{\normalsize-1}\text{K}^{\normalsize-1})ร—(318\text{K})}$$

log k = 10.6021 โ€“ 12.5870 = โ€“1.9849

k = Antilog (โ€“1.9849)

Ea = 76.640 kJ molโ€“1

k = 1.035 ร— 10โ€“2 sโ€“1.

30. 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.

Ans. Given: r2 = 4r1, T1 = 293 K, T2 = 313 K,

To find: Ea

Formula: k = Aeโ€“Ea/RT

Using Arrhenius equation,

$$\text{log}\frac{k_2}{k_1}=\frac{\text{E}_a}{2.303\text{R}}\bigg(\frac{1}{\text{T}_1}-\frac{1}{\text{T}_2}\bigg)\\\text{log 4}=\frac{\text{E}_a}{2.303ร—8.314}\bigg(\frac{1}{293}-\frac{1}{313}\bigg)\\\text{Or,}\qquad\text{E}_a=\text{log 4ร—}\frac{2.303ร—8.314ร—293ร—313}{20}$$

Or , Ea = 52.864 kJ

Hence, activation energy is 52.864 kJ.