Chapter 3 Chemical Kinetics – NCERT Class 12 Chemistry

Question 3.1

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) $3NO(g) \to N_2O(g)$ Rate = $k[NO]^2$
(ii) $H_2O_2 + 3I^- + 2H^+ \to 2H_2O + I_3^-$ Rate = $k[H_2O_2][I^-]$
(iii) $CH_3CHO \to CH_4 + CO$ Rate = $k[CH_3CHO]^{3/2}$
(iv) $C_2H_5Cl \to C_2H_4 + HCl$ Rate = $k[C_2H_5Cl]$

Concept

Order ($n$) is the sum of powers of concentration terms in the rate law.
Units of $k = (\text{mol L}^{-1})^{1-n} \text{ s}^{-1}$.

Solutions

(i) Order = 2. Unit: $(\text{mol L}^{-1})^{-1} \text{s}^{-1} = \text{L mol}^{-1} \text{s}^{-1}$.
(ii) Order = 1 + 1 = 2. Unit: $\text{L mol}^{-1} \text{s}^{-1}$.
(iii) Order = 1.5 (3/2). Unit: $(\text{mol L}^{-1})^{-0.5} \text{s}^{-1} = \text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$.
(iv) Order = 1. Unit: $\text{s}^{-1}$.

Question 3.2

For the reaction: $2A + B \to A_2B$, rate = $k[A][B]^2$ with $k = 2.0 \times 10^{-6} \text{ mol}^{-2} \text{ L}^2 \text{ s}^{-1}$. Calculate the initial rate when $[A] = 0.1 \text{ M}, [B] = 0.2 \text{ M}$. Calculate rate after $[A]$ is reduced to $0.06 \text{ M}$.

1. Initial Rate

$$\text{Rate}_1 = k[A][B]^2 = 2.0 \times 10^{-6} \times (0.1) \times (0.2)^2$$ $$= 2.0 \times 10^{-6} \times 0.1 \times 0.04 = 8.0 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1}$$

2. Rate after reduction of A

Reacted A = $0.1 – 0.06 = 0.04 \text{ M}$.
From stoichiometry ($2A + B$), B reacted = $0.04 / 2 = 0.02 \text{ M}$.
Remaining $[B] = 0.2 – 0.02 = 0.18 \text{ M}$.

$$\text{Rate}_2 = 2.0 \times 10^{-6} \times (0.06) \times (0.18)^2$$ $$= 2.0 \times 10^{-6} \times 0.06 \times 0.0324 = 3.89 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1}$$

Initial Rate = $8.0 \times 10^{-9} \text{ M/s}$
Final Rate = $3.89 \times 10^{-9} \text{ M/s}$

Question 3.3

Decomposition of $NH_3$ on platinum surface is zero order. What are rates of production of $N_2$ and $H_2$ if $k = 2.5 \times 10^{-4} \text{ M s}^{-1}$?

Analysis

Reaction: $2NH_3 \to N_2 + 3H_2$.
For zero order, Rate = $k$.

$$\text{Rate} = -\frac{1}{2}\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3}\frac{d[H_2]}{dt} = k$$

Rate of production of $N_2 = k = 2.5 \times 10^{-4} \text{ M s}^{-1}$.

Rate of production of $H_2 = 3k = 3 \times 2.5 \times 10^{-4} = 7.5 \times 10^{-4} \text{ M s}^{-1}$.

Question 3.4

Decomposition of dimethyl ether: Rate = $k [P_{ether}]^{3/2}$. If pressure is in bar and time in minutes, what are units of rate and k?

Rate: Change in pressure per unit time = bar min⁻¹.

Rate Constant (k):
Rate = $k (P)^{3/2}$ $\implies$ bar min⁻¹ = $k (\text{bar})^{3/2}$.
$k = \text{bar} \text{ min}^{-1} \text{ bar}^{-1.5} = \textbf{bar}^{-1/2} \textbf{ min}^{-1}$.

Question 3.5

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

Concentration of reactants: Higher concentration usually increases rate.
Temperature: Rate generally increases with temperature.
Nature of reactants: Ionic reactions are faster than covalent ones.
Catalyst: Increases rate by lowering activation energy.
Surface Area: For solids, larger surface area increases rate.
Exposure to light: Affects photochemical reactions.

Question 3.6

A reaction is second order wrt a reactant. How is rate affected if concentration is (i) doubled, (ii) reduced to half?

Rate $r = k[A]^2$.

(i) Doubled: $[A’] = 2[A]$. $r’ = k(2A)^2 = 4k[A]^2 = 4r$. (Increases 4 times).
(ii) Halved: $[A’] = [A]/2$. $r’ = k(A/2)^2 = \frac{1}{4}k[A]^2 = \frac{1}{4}r$. (Reduced to 1/4th).

Question 3.7

What is the effect of temperature on rate constant? How is it represented quantitatively?

The rate constant generally increases with an increase in temperature. For a chemical reaction, the rate constant nearly doubles for every $10^\circ$ rise in temperature.

Quantitative representation (Arrhenius Equation):

$$k = A e^{-E_a/RT}$$

Where $k$ = rate constant, $A$ = Arrhenius factor, $E_a$ = Activation energy, $R$ = Gas constant, $T$ = Temperature.

Question 3.8

In a pseudo first order reaction in water, calculate the average rate between 30 to 60 seconds. Data: t=30, [A]=0.31; t=60, [A]=0.17.

$$R_{avg} = -\frac{\Delta [A]}{\Delta t} = -\frac{0.17 – 0.31}{60 – 30}$$ $$= -\frac{-0.14}{30} = 4.67 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}$$

Average Rate = $4.67 \times 10^{-3} \text{ M s}^{-1}$

Question 3.9

Reaction is first order in A and second order in B. (i) Differential rate eq? (ii) Effect if B tripled? (iii) Effect if both doubled?

(i) Equation: $\text{Rate} = k[A][B]^2$.

(ii) B Tripled: $r’ = k[A](3B)^2 = 9k[A][B]^2 = 9r$. (Increases 9 times).

(iii) Both Doubled: $r” = k(2A)(2B)^2 = k(2A)(4B^2) = 8k[A][B]^2 = 8r$. (Increases 8 times).

Question 3.10

Determine orders wrt A and B from initial rate data.
Exp 1: [A]=0.2, [B]=0.3, r=5.07e-5
Exp 2: [A]=0.2, [B]=0.1, r=5.07e-5
Exp 3: [A]=0.4, [B]=0.05, r=1.43e-4

Analysis

Comparing Exp 1 & 2: [A] is constant. [B] changes from 0.3 to 0.1. Rate stays $5.07 \times 10^{-5}$. Since rate is independent of [B], order wrt B is 0.

Comparing Exp 1 & 3: Rate law $r = k[A]^x$.
$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} = \left(\frac{0.4}{0.2}\right)^x$.
$2.82 \approx 2^x \implies x \approx 1.5$.

Order wrt A = 1.5, Order wrt B = 0.

Question 3.11

Determine rate law and rate constant for $2A + B \to C + D$ from data table.

1. Determine Orders

Compare I and IV: [B] is constant (0.1). [A] changes $0.1 \to 0.4$ (4x). Rate changes $6.0 \times 10^{-3} \to 2.40 \times 10^{-2}$ (4x). Since rate $\propto [A]$, order wrt A = 1.

Compare II and III: [A] is constant (0.3). [B] changes $0.2 \to 0.4$ (2x). Rate changes $7.2 \times 10^{-2} \to 2.88 \times 10^{-1}$ (4x). Since rate $\propto [B]^2$, order wrt B = 2.

Rate Law: $\text{Rate} = k[A][B]^2$.

2. Calculate k

Using Exp I: $6.0 \times 10^{-3} = k(0.1)(0.1)^2 = k(0.001)$.

$$k = \frac{6.0 \times 10^{-3}}{10^{-3}} = 6.0 \text{ L}^2 \text{ mol}^{-2} \text{ min}^{-1}$$

Rate = $k[A][B]^2$, $k = 6.0$.

Question 3.12

Reaction is first order in A and zero order in B. Fill in the blanks.

Rate = $k[A]$. From Exp I: $2.0 \times 10^{-2} = k(0.1) \implies k = 0.2 \text{ min}^{-1}$.

Exp II: Rate = $4.0 \times 10^{-2}$. Rate = $0.2 [A] \implies [A] = 0.2 \text{ M}$.
Exp III: Rate = $0.2 \times 0.4 = 8.0 \times 10^{-2} \text{ M min}^{-1}$.
Exp IV: Rate = $2.0 \times 10^{-2}$. Same as Exp I, so $[A] = 0.1 \text{ M}$.

Question 3.13

Calculate half-life for first order reaction with k: (i) $200 \text{ s}^{-1}$, (ii) $2 \text{ min}^{-1}$, (iii) $4 \text{ yr}^{-1}$.

Formula: $t_{1/2} = \frac{0.693}{k}$.

(i) $0.693/200 = 3.465 \times 10^{-3} \text{ s}$.
(ii) $0.693/2 = 0.3465 \text{ min}$.
(iii) $0.693/4 = 0.173 \text{ years}$.

Question 3.14

Half-life of 14C is 5730 years. Sample has 80% of 14C found in living tree. Estimate age.

1. Decay Constant

$$k = \frac{0.693}{5730} \approx 1.21 \times 10^{-4} \text{ yr}^{-1}$$

2. Age Calculation

Formula: $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$.

Here $[R]_0 = 100, [R] = 80$.

$$t = \frac{2.303}{1.21 \times 10^{-4}} \log \frac{100}{80}$$ $$t = 19033 \times \log(1.25) = 19033 \times 0.0969 \approx 1845 \text{ years}$$

Age $\approx 1845$ years.

Question 3.15

Decomposition of $N_2O_5$. Analyze data (plot [N2O5] vs t, find rate law, half-life, rate constant).

vs t]

Analysis

(i) Plot [A] vs t: Curve shows exponential decay.
(ii) Half-life from graph: Initial = 1.63. Half = 0.815. Corresponds to $t \approx 1440 \text{ s}$.
(iii) Graph log[A] vs t: Yields a straight line, confirming First Order.
(iv) Rate Law: Rate = $k[N_2O_5]$.
(v) Rate Constant: Slope of log plot = $-k/2.303$. Slope $\approx -2.09 \times 10^{-4}$.
$k = 2.303 \times 2.09 \times 10^{-4} = 4.82 \times 10^{-4} \text{ s}^{-1}$.
(vi) Calc Half-life: $0.693 / 4.82 \times 10^{-4} \approx 1438 \text{ s}$. (Matches graph).

Question 3.16

Rate constant for first order is $60 \text{ s}^{-1}$. Time to reduce to 1/16th initial value?

Calculation

$[R] = [R]_0 / 16$.
Using $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$.

$$t = \frac{2.303}{60} \log(16) = \frac{2.303}{60} \times 1.204$$ $$t = 0.046 \text{ s} = 4.6 \times 10^{-2} \text{ s}$$

Time $\approx 4.6 \times 10^{-2} \text{ s}$

Question 3.17

Half-life of 90Sr is 28.1 years. If 1$\mu$g absorbed, amount remaining after 10 and 60 years?

1. Rate Constant

$k = 0.693 / 28.1 = 0.02466 \text{ yr}^{-1}$.

2. After 10 Years

$$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \implies 0.02466 = \frac{2.303}{10} \log \frac{1}{x}$$ $$\log(1/x) = 0.107 \implies 1/x = 1.279 \implies x = 0.78 \mu\text{g}$$

3. After 60 Years

$$\log(1/x) = \frac{0.02466 \times 60}{2.303} = 0.6425$$ $$1/x = 4.39 \implies x = 0.227 \mu\text{g}$$

After 10y: 0.78 $\mu$g. After 60y: 0.23 $\mu$g.

Question 3.18

Show that for a first order reaction, time for 99% completion is twice the time for 90% completion.

Proof

For 99% completion: $[R] = 0.01 [R]_0$.
$t_{99} = \frac{2.303}{k} \log \frac{100}{1} = \frac{2.303}{k} \times 2$.

For 90% completion: $[R] = 0.10 [R]_0$.
$t_{90} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{k} \times 1$.

Ratio: $t_{99} / t_{90} = 2 / 1 = 2$.

Hence, $t_{99\%} = 2 \times t_{90\%}$.

Question 3.19

First order reaction takes 40 min for 30% decomposition. Calculate $t_{1/2}$.

1. Rate Constant

$[R]_0 = 100, [R] = 70$. $t = 40 \text{ min}$.

$$k = \frac{2.303}{40} \log \frac{100}{70} = \frac{2.303}{40} \times 0.1549 = 8.918 \times 10^{-3} \text{ min}^{-1}$$

2. Half-life

$$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{8.918 \times 10^{-3}} \approx 77.7 \text{ min}$$

Half-life $\approx 77.7$ min.

Question 3.20

Decomposition of azoisopropane to hexane and nitrogen. Calculate k from pressure data. P(0)=35, P(360)=54.

Calculation

$A \to B + C$. $P_0 = 35$.
At time t, $P_{total} = P_0 + p$. Reactant Pressure $P_A = P_0 – p = P_0 – (P_t – P_0) = 2P_0 – P_t$.

At $t=360$, $P_t = 54$.
$P_A = 2(35) – 54 = 70 – 54 = 16 \text{ mm}$.

$$k = \frac{2.303}{360} \log \frac{35}{16} = \frac{2.303}{360} \times 0.34 = 2.17 \times 10^{-3} \text{ s}^{-1}$$

k = $2.17 \times 10^{-3} \text{ s}^{-1}$

Question 3.21

Decomposition of $SO_2Cl_2$. Calculate rate when total pressure is 0.65 atm. Data: P(0)=0.5, P(100)=0.6.

1. Calculate k

$P_A = 2P_0 – P_t$. At t=100, $P_t = 0.6$.
$P_A = 2(0.5) – 0.6 = 0.4 \text{ atm}$.

$$k = \frac{2.303}{100} \log \frac{0.5}{0.4} = 2.23 \times 10^{-3} \text{ s}^{-1}$$

2. Rate at P_t = 0.65

$P_{SO2Cl2} = 2(0.5) – 0.65 = 0.35 \text{ atm}$.
Rate = $k \times P_{SO2Cl2} = 2.23 \times 10^{-3} \times 0.35 = 7.8 \times 10^{-4} \text{ atm s}^{-1}$.

Rate = $7.8 \times 10^{-4} \text{ atm s}^{-1}$

Question 3.22

Rate constants for decomposition of N2O5 at various temps. Draw ln k vs 1/T. Calc A and Ea.

[Image of Graph ln k vs 1/T]

Method

1. Convert T to Kelvin and k to ln k.
2. Plot ln k (y) vs 1/T (x).
3. Slope = $-E_a/R$. Intercept = ln A.
Using data points (e.g., T=0°C and T=20°C):
Slope $\approx -12300 \text{ K}$. $E_a = -\text{Slope} \times R = 12300 \times 8.314 \approx 102 \text{ kJ/mol}$.
Pre-exponential factor $A$ can be calculated from $k = A e^{-E_a/RT}$.

Question 3.23

k = $2.418 \times 10^{-5} \text{ s}^{-1}$ at 546 K. $E_a = 179.9 \text{ kJ/mol}$. Calculate A.

Calculation

$\log k = \log A – \frac{E_a}{2.303 RT}$.

$$\log(2.418 \times 10^{-5}) = \log A – \frac{179900}{2.303 \times 8.314 \times 546}$$ $$-4.6165 = \log A – 17.208$$ $$\log A = 12.59 \implies A = 3.9 \times 10^{12} \text{ s}^{-1}$$

A = $3.9 \times 10^{12} \text{ s}^{-1}$

Question 3.24

Consider reaction $A \to \text{Products}$ with $k = 2.0 \times 10^{-2} \text{ s}^{-1}$. Calc conc of A after 100 s if initial is 1.0 M.

Formula

$[A] = [A]_0 e^{-kt}$.

$$[A] = 1.0 \times e^{-(2.0 \times 10^{-2} \times 100)} = e^{-2}$$ $$[A] = 1 / 7.389 = 0.135 \text{ M}$$

Remaining Concentration = 0.135 M

Question 3.25

Sucrose decomposition (1st order), $t_{1/2} = 3.00 \text{ hours}$. What fraction remains after 8 hours?

Calculation

$k = 0.693 / 3 = 0.231 \text{ hr}^{-1}$.
Using $[R] = [R]_0 e^{-kt}$:

$$\frac{[R]}{[R]_0} = e^{-(0.231 \times 8)} = e^{-1.848}$$ $$\text{Fraction} = 0.158$$

Fraction remaining = 0.158

Question 3.26

Decomposition follows $k = (4.5 \times 10^{11} \text{s}^{-1}) e^{-28000 K/T}$. Calculate $E_a$.

Comparison

Compare with $k = A e^{-E_a/RT}$.
$E_a/R = 28000 \text{ K}$.

$$E_a = 28000 \times R = 28000 \times 8.314 \text{ J/mol}$$ $$E_a = 232792 \text{ J/mol} = 232.79 \text{ kJ/mol}$$

Activation Energy = 232.79 kJ/mol

Question 3.27

$H_2O_2$ decomposition: $\log k = 14.34 – 1.25 \times 10^4 K/T$. Calculate $E_a$ and T for $t_{1/2} = 256 \text{ min}$.

1. Calculate Ea

Compare with $\log k = \log A – \frac{E_a}{2.303 RT}$.
$\frac{E_a}{2.303 R} = 1.25 \times 10^4$.
$E_a = 1.25 \times 10^4 \times 2.303 \times 8.314 = 239.34 \text{ kJ/mol}$.

2. Calculate T

$k = 0.693 / (256 \times 60) = 4.51 \times 10^{-5} \text{ s}^{-1}$.
$\log(4.51 \times 10^{-5}) = 14.34 – \frac{1.25 \times 10^4}{T}$.
$-4.35 = 14.34 – \frac{12500}{T} \implies \frac{12500}{T} = 18.69$.
$T = 12500 / 18.69 \approx 669 \text{ K}$.

$E_a = 239.34 \text{ kJ/mol}$, T = 669 K.

Question 3.28

$k_1 = 4.5 \times 10^3 \text{ s}^{-1}$ at $10^\circ \text{C}$. $E_a = 60 \text{ kJ/mol}$. At what T will $k_2 = 1.5 \times 10^4 \text{ s}^{-1}$?

Formula

$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 – T_1}{T_1 T_2} \right)$.
$T_1 = 283 \text{ K}$.

$$\log \frac{1.5 \times 10^4}{4.5 \times 10^3} = \log 3.33 = 0.522$$ $$0.522 = \frac{60000}{19.147} \left( \frac{T_2 – 283}{283 T_2} \right)$$ $$0.522 = 3133.6 \left( \frac{T_2 – 283}{283 T_2} \right)$$ $$0.000166 = \frac{T_2 – 283}{283 T_2} \implies 0.047 T_2 = T_2 – 283$$ $$0.953 T_2 = 283 \implies T_2 \approx 297 \text{ K} = 24^\circ \text{C}$$

Temperature = $24^\circ \text{C}$

Question 3.29

Time for 10% completion at 298K = Time for 25% completion at 308K. If A = $4 \times 10^{10} \text{ s}^{-1}$, calculate k at 318K and $E_a$.

1. Relation between k values

$t = \frac{2.303}{k} \log \frac{100}{100-x}$.
$t_1 (10\%) = \frac{2.303}{k_1} \log \frac{100}{90} = \frac{0.1054}{k_1}$.
$t_2 (25\%) = \frac{2.303}{k_2} \log \frac{100}{75} = \frac{0.2877}{k_2}$.
Since $t_1 = t_2 \implies k_2 / k_1 = 0.2877 / 0.1054 = 2.73$.

2. Calculate Ea

$$\log(2.73) = \frac{E_a}{2.303 R} \left( \frac{308 – 298}{308 \times 298} \right)$$ $$0.436 = \frac{E_a}{19.147} \left( \frac{10}{91784} \right)$$ $$E_a = \frac{0.436 \times 19.147 \times 91784}{10} \approx 76623 \text{ J/mol} = 76.6 \text{ kJ/mol}$$

3. Calculate k at 318K

$$\log k = \log(4 \times 10^{10}) – \frac{76623}{2.303 \times 8.314 \times 318}$$ $$\log k = 10.602 – 12.58 = -1.978$$ $$k = 10^{-1.978} \approx 1.05 \times 10^{-2} \text{ s}^{-1}$$

$E_a = 76.6 \text{ kJ/mol}$, $k_{318} \approx 1.05 \times 10^{-2} \text{ s}^{-1}$

Question 3.30

Rate quadruples when T changes from 293 K to 313 K. Calculate $E_a$.

Calculation

$k_2 / k_1 = 4$. $T_1 = 293, T_2 = 313$.

$$\log 4 = \frac{E_a}{2.303 \times 8.314} \left( \frac{313 – 293}{293 \times 313} \right)$$ $$0.602 = \frac{E_a}{19.147} \left( \frac{20}{91709} \right)$$ $$E_a = \frac{0.602 \times 19.147 \times 91709}{20} \approx 52863 \text{ J/mol}$$

Activation Energy $\approx 52.86 \text{ kJ/mol}$

NCERT Solutions