CBSE Class 12 Chemistry Chemical Kinetics NEET JEE Questions Set 02

Single Correct Answer

Question 1. t_1/4 can be taken as the time taken for the concentration of a reactant to drop to \( \frac{3}{4} \) of its initial value. If the rate constant for a first order reaction is K, then t_1/4 can be written as
(a) 0.10/K
(b) 0.29/K
(c) 0.69/K
(d) 0.75/K
Answer:
Ans. (b)
\( t_{1/4} = \frac{2.303}{K}\log\frac{a}{(3/4)a} = \frac{2.303}{K}\log\frac{4}{3} = \frac{2.303\times0.125}{K} = \) \( \frac{0.29}{K} \)

Teacher's Note:
1. t_1/4 means one-fourth has reacted, so three-fourths remain; do not confuse it with one-fourth remaining.

Question 2. A plot of log t_1/2 versus log C₀ is given in the adjoining fig. The conclusion that can be drawn from this graph is
(a) Order = 1, \( t_{1/2} = \frac{1}{ka} \)
(b) Order = 1, \( t_{1/2} = \frac{2.303}{k}\log 2 \)
(c) Order = 0, \( t_{1/2} = \frac{1}{2ka} \)
(d) Order = 2, \( t_{1/2} = \frac{1}{a} \)

Answer:
Ans. (b)
t_1/2 ∝ (C₀)¹⁻ⁿ or t_1/2 = K C₀¹⁻ⁿ (K = constant)
∴ log t_1/2 = log K + (1 − n) log C₀ (y = mx + c)
Thus, plot of log t_1/2 vs log C₀ will be linear with slope = 1 − n. But as graph is parallel to log C₀ axis, slope = 0, i.e., 1 − n = 0 or n = 1. Hence, it is a reaction of 1st order and for 1st order reactions,
\( t_{1/2} = \frac{2.303}{k}\log 2 \)

Teacher's Note:
1. A horizontal log t_1/2 vs log C₀ line means t_1/2 does not depend on C₀, which is true only for first order.

Question 3. For a reaction A + B → Products, rate law is \( -\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]_0 \). The concentration of A left after time \( t = \frac{1}{k} \) is
(a) \( \frac{[\mathrm{A}]_0}{e} \)
(b) [A]₀ × e
(c) \( \frac{[\mathrm{A}]_0}{e^2} \)
(d) \( \frac{1}{[\mathrm{A}]_0} \)
Answer:
Ans. (a)
The given rate law shows that it is a reaction of 1st order. For 1st order reactions, [A] = [A]₀ e^−kt
When \( t = \frac{1}{k} \), \( [\mathrm{A}] = [\mathrm{A}]_0\,e^{-k\cdot\frac{1}{k}} = [\mathrm{A}]_0\,e^{-1} = \) \( \frac{[\mathrm{A}]_0}{e} \)

Teacher's Note:
1. t = 1/k is the average life of a first order reaction; at this time about 37% (1/e) of the reactant is left.

Question 4. Decomposition of H₂O₂ follows a first order reaction. In fifty minutes, the concentration of H₂O₂ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H₂O₂ reaches 0.05 M, the rate of formation of O₂ will be (JEE Main 2016)
(a) 6.93 × 10⁻² mol min⁻¹
(b) 6.93 × 10⁻⁴ mol L⁻¹ min⁻¹
(c) 2.66 L min⁻¹ at STP
(d) 1.34 × 10⁻² mol min⁻¹
Answer:
Ans. (b)
Decomposition of H₂O₂ takes place as: \( \mathrm{H_2O_2 \longrightarrow H_2O + \frac{1}{2}O_2} \)
Decrease of concentration of H₂O₂ from 0.5 to 0.125 in 50 minutes means two half lives (0.5 → 0.25 → 0.125), i.e., 2 × t_1/2 = 50 min or t_1/2 = 25 min
For a 1st order reaction, \( k = \frac{0.693}{t_{1/2}} = \frac{0.693}{25}\ \text{min}^{-1} \)
Rate of reaction = Rate of decomposition of H₂O₂, i.e., \( -\frac{d[\mathrm{H_2O_2}]}{dt} = k[\mathrm{H_2O_2}] = \frac{0.693}{25}\times0.05 \) = 1.386 × 10⁻³ mol L⁻¹ min⁻¹
Rate of formation of O₂ \( = \frac{d[\mathrm{O_2}]}{dt} = -\frac{1}{2}\frac{d[\mathrm{H_2O_2}]}{dt} = \frac{1}{2}\times1.386\times10^{-3} \) = 6.93 × 10⁻⁴ mol L⁻¹ min⁻¹

Teacher's Note:
1. Spot the number of half-lives first (0.5 → 0.125 is two), then remember that O₂ forms at half the rate at which H₂O₂ decomposes.

Question 5. If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by (NEET 2019)
(a) t = 2.303/k
(b) t = 0.693/k
(c) t = 6.909/k
(d) t = 4.606/k
Answer:
Ans. (d)
For 1st order reaction, \( t = \frac{2.303}{k}\log\frac{a}{a-x} = \frac{2.303}{k}\log\frac{100}{100-99} = \frac{2.303}{k}\log10^2 = \frac{2.303}{k}\times2\log10 = \) \( \frac{4.606}{k} \)

Teacher's Note:
1. Useful shortcuts for first order: 90% needs 2.303/k, 99% needs 4.606/k and 99.9% needs 6.909/k.

Question 6. The half-life of a reaction is halved as the initial concentration of the reactant is doubled. The order of reaction is
(a) 0.5
(b) 1
(c) 2
(d) 0
Answer:
Ans. (c)
\( t_{1/2} \propto \frac{1}{[\mathrm{A}]_0^{\,n-1}} \), \( \frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{[\mathrm{A}_0]_2^{\,n-1}}{[\mathrm{A}_0]_1^{\,n-1}} = \left\{\frac{[\mathrm{A}_0]_2}{[\mathrm{A}_0]_1}\right\}^{n-1} \)
\( \frac{t}{t/2} = \left(\frac{2a}{a}\right)^{n-1} \) or 2 = (2)ⁿ⁻¹ or n − 1 = 1 or n = 2.

Teacher's Note:
1. t_1/2 inversely proportional to [A]₀ is the signature of a second order reaction.

Question 7. A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as A → B (k₁) and A → C (k₂), with k₁ = 1.26 × 10⁻⁴ s⁻¹, k₂ = 3.8 × 10⁻⁵ s⁻¹. The percentage distribution of B and C are
(a) 75% B and 25% C
(b) 80% B and 20% C
(c) 60% B and 40% C
(d) 90% B and 10% C
(e) 76.83% B and 23.17% C
Answer:
Ans. (e)
% of B \( = \frac{k_1}{k_1+k_2}\times100 = \frac{1.26\times10^{-4}}{1.26\times10^{-4}+3.8\times10^{-5}}\times100 = \frac{1.26\times10^{-4}}{10^{-4}(1.26+0.38)}\times100 \) = 76.83%
% of C = 100 − 76.83 = 23.17%

Teacher's Note:
1. In parallel first order reactions the products form in the ratio of the rate constants, k₁ : k₂, at every instant.

Question 8. For a second order reaction, 2A → Products, a plot of log t_1/2 vs log a (where a is the initial concentration) will give an intercept equal to which of the following? (SCRA 2007)
(a) 1/k
(b) log (1/2k)
(c) log (1/k)
(d) log k
Answer:
Ans. (c)
For a second order reaction, 2A → Products, \( k = \frac{1}{t}\left[\frac{1}{[\mathrm{A}]}-\frac{1}{[\mathrm{A}]_0}\right] \)
When t = t_1/2, \( [\mathrm{A}] = \frac{[\mathrm{A}]_0}{2} \). Hence, \( t_{1/2} = \frac{1}{k}\left[\frac{2}{[\mathrm{A}]_0}-\frac{1}{[\mathrm{A}]_0}\right] = \frac{1}{k[\mathrm{A}]_0} \)
∴ log t_1/2 = − log k − log [A]₀ = − log k − log a
Plot of log t_1/2 vs log a will be linear (slope = −1) with intercept = − log k = log (1/k).

Teacher's Note:
1. Write the half-life equation in y = mx + c form; the intercept is the constant term, −log k, not log k.

Question 9. Under the same reaction conditions, initial concentration of 1.386 mol dm⁻³ of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics respectively. Ratio (k₁/k₀) of the rate constants for first order (k₁) and zero order (k₀) of the reactions is (IIT 2008)
(a) 0.5 mol⁻¹ dm³
(b) 1.0 mol dm⁻³
(c) 1.5 mol dm⁻³
(d) 2.0 mol⁻¹ dm³
Answer:
Ans. (a)
For 1st order reaction, \( t_{1/2} = \frac{0.693}{k} \) or \( k_1 = \frac{0.693}{t_{1/2}} = \frac{0.693}{40\ \text{s}} \)
For zero order reaction, \( t_{1/2} = \frac{[\mathrm{A}]_0}{2k} \) or \( k_0 = \frac{[\mathrm{A}]_0}{2t_{1/2}} = \frac{1.386\ \text{mol dm}^{-3}}{2\times20\ \text{s}} \)
\( \therefore \frac{k_1}{k_0} = \frac{0.693}{40\ \text{s}}\times\frac{2\times20\ \text{s}}{1.386\ \text{mol dm}^{-3}} \) = 0.5 mol⁻¹ dm³

Teacher's Note:
1. Keep the units: k₁ is in s⁻¹ and k₀ in mol dm⁻³ s⁻¹, so their ratio has units mol⁻¹ dm³.

Question 10. In the catalysed decomposition of benzene diazonium chloride, C₆H₅N₂Cl → C₆H₅Cl + N₂ (Δ/Cu), half-life period is found to be independent of the initial concentration of the reactant. After 10 minutes, the volume of N₂ gas collected is 10 L and after the reaction is complete, it is 50 L. Hence, rate constant of the reaction (in min⁻¹) is
(a) \( \frac{2.303}{10}\log\frac{10}{50} \)
(b) \( \frac{2.303}{10}\log\frac{50}{50-10} \)
(c) \( \frac{2.303}{10}\log\frac{50}{10} \)
(d) \( \frac{2.303}{10}\log\frac{10}{50-10} \)
Answer:
Ans. (b)
As half-life period is independent of initial concentration, hence it is a reaction of 1st order. \( \therefore k = \frac{2.303}{t}\log\frac{a}{a-x} \)
But V_∞ ∝ a and V_t ∝ x. Hence, \( k = \frac{2.303}{10}\log\frac{50}{50-10} \)

Teacher's Note:
1. The final volume of N₂ measures the initial amount (a); the volume collected at time t measures the amount reacted (x).

Question 11. Kinetics of the reaction A(g) → 2B(g) + C(g) is followed by measuring the total pressure at different times. It is given that
Initial pressure of A = 0.5 atm.
Total pressure of A after 2 hours = 0.7 atm.
Rate constant of the reaction = 1 × 10⁻³ s⁻¹
What is the rate of reaction \( -\frac{d[\mathrm{A}]}{dt} \) when the total pressure is 0.7 atm? (IAS Prelim 2010)
(a) 2.0 × 10⁻⁴ atm s⁻¹
(b) 4.0 × 10⁻⁴ atm s⁻¹
(c) 5.0 × 10⁻⁴ atm s⁻¹
(d) 7.0 × 10⁻⁴ atm s⁻¹
Answer:
Ans. (b)
A(g) → 2B(g) + C(g). Initial: 0.5 atm. After 2 hrs: 0.5 − p, 2p, p
Total pressure after 2 hours = (0.5 − p) + 2p + p = 0.5 + 2p ∴ 0.5 + 2p = 0.7 or p = 0.1 atm
∴ Pressure of A after 2 hours = 0.5 − 0.1 = 0.4 atm
Rate constant in s⁻¹ shows that it is a reaction of 1st order.
∴ Rate after 2 hours = k [A] = k P = (10⁻³ s⁻¹)(0.4 atm) = 4.0 × 10⁻⁴ atm s⁻¹

Teacher's Note:
1. Use the partial pressure of A (0.4 atm), not the total pressure (0.7 atm), in the rate law.

Question 12. The rate of a first order reaction is 0.04 mol L⁻¹ s⁻¹ at 10 seconds and 0.03 mol L⁻¹ s⁻¹ at 20 seconds after initiation of the reaction. The half-life period of the reaction is (NEET Phase I 2016)
(a) 44.1 s
(b) 54.1 s
(c) 24.1 s
(d) 34.1 s
Answer:
Ans. (c)
For a first order reaction, A → Products, and for concentration of the reactant at two different times, \( k = \frac{2.303}{(t_2-t_1)}\log\frac{[\mathrm{A}]_1}{[\mathrm{A}]_2} \)
Also for first order reaction, rate ∝ [A]. Hence, \( k = \frac{2.303}{t_2-t_1}\log\frac{(\text{Rate})_1}{(\text{Rate})_2} = \frac{2.303}{(20-10)}\log\left(\frac{0.04}{0.03}\right) \) = 0.0287 s⁻¹
\( t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0287\ \text{s}^{-1}} \) = 24.14 s

Teacher's Note:
1. For first order, the ratio of rates at two times equals the ratio of concentrations, so rates can be put directly into the integrated equation.

Question 13. At 518°C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 torr, was 1.00 torr s⁻¹ when 5% had reacted and 0.5 torr s⁻¹ when 33% had reacted. The order of the reaction is (JEE Main 2018)
(a) 2
(b) 3
(c) 1
(d) 0
Answer:
Ans. (a)
Suppose order of reaction = x. Then
1st case: Rate = k [CH₃CHO]^x, 1 = k (363 × 0.95)^x (∵ 95% CH₃CHO is present) or 1 = k (344.85)^x ...(i)
2nd case: 0.5 = k (363 × 0.67)^x (∵ 67% CH₃CHO is present) or 0.5 = k (243.21)^x ...(ii)
Dividing eqn. (i) by eqn. (ii), \( \frac{1}{0.5} = \left(\frac{344.85}{243.21}\right)^x \) or 2 = (1.414)^x or x = 2

Teacher's Note:
1. Use the amount still left (95% and 67%), not the amount reacted, in the rate law.

Question 14. Decomposition of X exhibits a rate constant of 0.05 μg/year. How many years are required for the decomposition of 5 μg of X into 2.5 μg? (JEE Main 2019)
(a) 50
(b) 25
(c) 20
(d) 40
Answer:
Ans. (a)
The units of the rate constant show that it is a reaction of zero order. Half-life for a reaction of zero order is given by \( t_{1/2} = \frac{[\mathrm{A}]_0}{2k} = \frac{5\ \mu\text{g}}{2\times0.05\ \mu\text{g/year}} \) = 50 years

Teacher's Note:
1. Always read the order from the units of k: amount per time means zero order.

Question 15. For a first order reaction, the time required for the completion of 90% reaction is 'x' times the half-life of the reaction. The value of 'x' is (Given ln 10 = 2.303 and log 2 = 0.3010) (JEE Main 2022)
(a) 1.12
(b) 2.43
(c) 3.32
(d) 33.31
Answer:
Ans. (c)
\( t_{1/2} = \frac{0.693}{k} \) or \( k = \frac{0.693}{t_{1/2}} \)
\( t_{90\%} = \frac{2.303}{(0.693/t_{1/2})}\log\frac{100}{100-90} = \frac{2.303\,t_{1/2}}{0.693} \) = 3.32 t_1/2

Teacher's Note:
1. t_90%/t_1/2 = log 10/log 2 = 1/0.301 ≈ 3.32 for every first order reaction.

Question 16. At 300°C, the half-life for the decomposition of AB₂ is 200 s and is independent of the initial concentration of AB₂. The time required for 80% of the AB₂ to decompose is (Given: log 2 = 0.30, log 3 = 0.48) (JEE Main 2022)
(a) 200 s
(b) 323 s
(c) 467 s
(d) 532 s
Answer:
Ans. (c)
As t_1/2 is independent of initial concentration, it is a reaction of 1st order. Hence, \( k = \frac{0.693}{t_{1/2}} = \frac{0.693}{200}\ \text{s}^{-1} \)
\( t_{80\%} = \frac{2.303}{0.693/200}\log\frac{100}{100-80} = \frac{2.303\times200}{0.693}\log 5 \)
With the given data, 0.693 = 2.303 log 2, so \( t_{80\%} = 200\times\frac{\log 5}{\log 2} = 200\times\frac{0.70}{0.30} \) = 467 s

Teacher's Note:
1. Writing t = t_1/2 × log(a/(a − x))/log 2 avoids decimals: here log 5 = 1 − log 2 = 0.70.

Question 17. The given graph is a representation of kinetics of a reaction. The y and x axes for zero and first order reactions, respectively are (NEET 2022)
(a) zero order (y = concentration and x = time), first order (y = t_1/2 and x = concentration)
(b) zero order (y = concentration and x = time), first order (y = rate constant and x = concentration)
(c) zero order (y = rate and x = concentration), first order (y = t_1/2 and x = concentration)
(d) zero order (y = rate and x = concentration), first order (y = rate and x = t_1/2)

Answer:
Ans. (c)
For zero order reaction, rate = k [A]⁰ or rate = k (constant). Hence, plot of rate (y-axis) vs concentration (x-axis) will give a horizontal line.
For 1st order reaction, \( t_{1/2} = \frac{0.693}{k} \), i.e., t_1/2 of a 1st order reaction is independent of concentration. Hence, plot of t_1/2 (y-axis) vs concentration (x-axis) will be a horizontal line.

Teacher's Note:
1. A horizontal line means y does not depend on x: the rate of a zero order reaction and the t_1/2 of a first order reaction are both independent of concentration.

Question 18. For a first order reaction, t_av (average life time), t_50% and t_75% are in the order:
(a) t₅₀ < t_av < t₇₅
(b) t₅₀ < t₇₅ < t_av
(c) t_av < t₅₀ < t₇₅
(d) t_av = t₅₀ < t₇₅
Answer:
Ans. (a)
\( t_{av} = \frac{1}{k} = \frac{1}{0.693/t_{1/2}} = \frac{t_{1/2}}{0.693} \) = 1.44 t_50%
Also, we know that t_75% = 2 × t_50%. Hence, t_50% < t_av < t_75%.

Teacher's Note:
1. Express all three in terms of t_1/2: 1, 1.44 and 2 times respectively.

Question 19. In the study of inversion of sucrose in presence of acid, if r₀, r_t and r_∞ represent the polarimetric readings at times 0, t and ∞ respectively, then at the 50% inversion, which of the following relationship will hold good?
(a) r_t = r₀ + r_∞
(b) \( r_t = \frac{1}{2}(r_0 + r_\infty) \)
(c) r_t = r₀ − r_∞
(d) \( r_t = \frac{1}{2}(r_0 - r_\infty) \)
Answer:
Ans. (b)
Rotation in time t = (r₀ − r_t) ∝ x; Rotation in time ∞ = (r₀ − r_∞) ∝ a
50% inversion means x = a/2
i.e., \( r_0 - r_t = \frac{r_0 - r_\infty}{2} \) or 2r₀ − 2r_t = r₀ − r_∞ or 2r_t = r₀ + r_∞ or \( r_t = \frac{1}{2}(r_0 + r_\infty) \)

Teacher's Note:
1. Change in rotation is proportional to the amount of sucrose inverted; at half inversion the reading is midway between the initial and final readings.

Question 20. Number of natural life times (t_av) required for a first order reaction to complete 99.9% is
(a) 6.93
(b) 2.31
(c) 9.2
(d) infinite
Answer:
Ans. (a)
t_99.9 = 10 × t_1/2 \( = 10\times\frac{0.693}{k} \) (∵ t_1/2 = 0.693/k) = 10 × 0.693 × t_av (∵ t_av = 1/k)
= 6.93 × t_av, i.e., 6.93 natural life times.

Teacher's Note:
1. 99.9% completion takes about 10 half-lives (2¹⁰ ≈ 1000), and t_av = 1/k.

Question 21. Half-life period of a first order reaction is 100 min. After 144.3 min, concentration of reactant is reduced to ... of the original concentration
(a) 40%
(b) 30%
(c) 1/e
(d) 1/e²
Answer:
Ans. (c)
\( t_{av} = \frac{1}{k} = \frac{t_{1/2}}{0.693} \) = 1.443 t_1/2 = 1.443 × 100 min = 144.3 min
Average life is the time in which the concentration is reduced to 1/e of the original concentration.

Teacher's Note:
1. At t = 1/k, ln([A]₀/[A]) = 1, so [A] = [A]₀/e (about 37%).

Question 22. The rate constant of a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy of the reaction (R = 8.314 J K⁻¹ mol⁻¹) (AIIMS 2014)
(a) 4.17 kJ mol⁻¹
(b) 41.7 kJ mol⁻¹
(c) 417.0 kJ mol⁻¹
(d) 417.0 kJ mol⁻¹
Answer:
Ans. (b)
According to Arrhenius equation, \( \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1T_2}\right) \)
\( \log\frac{6}{1} = \frac{E_a}{2.303\times8.314}\left(\frac{400-350}{350\times400}\right) \) or \( 0.7782 = \frac{E_a}{2.303\times8.314}\times\frac{50}{350\times400} \)
or \( E_a = \frac{0.7782\times2.303\times8.314\times350\times400}{50} \) = 41720 J = 41.72 kJ

Teacher's Note:
1. Keep T in kelvin and R in J K⁻¹ mol⁻¹; the answer comes in J mol⁻¹, so divide by 1000 for kJ.

Question 23. A reactant (A) forms two products: A → B (k₁), Activation Energy E_a1; A → C (k₂), Activation Energy E_a2. If E_a2 = 2E_a1, then k₁ and k₂ are related as
(a) k₁ = 2k₂ e^E_a2/RT
(b) k₂ = k₁ e^E_a1/RT
(c) k₂ = k₁ e^E_a2/RT
(d) k₁ = A k₂ e^E_a1/RT
Answer:
Ans. (d)
k₁ = A₁ e^−E_a1/RT, k₂ = A₂ e^−E_a2/RT
\( \therefore \frac{k_1}{k_2} = \frac{A_1}{A_2}e^{(E_{a2}-E_{a1})/RT} = \frac{A_1}{A_2}e^{(2E_{a1}-E_{a1})/RT} \) (∵ E_a2 = 2E_a1)
or k₁ = k₂ A e^E_a1/RT (A = A₁/A₂)

Teacher's Note:
1. The path with the lower activation energy (A → B) is faster, so k₁ > k₂ when the A factors are similar.

Question 24. In the presence of a catalyst, the activation energy of a reaction is lowered by 2 kcal at 27°C. The rate of reaction will increase by
(a) 2 times
(b) 14 times
(c) 28 times
(d) 20 times
Answer:
Ans. (c)
k₁ = A e^−E_a/RT, k₂ = A e^{−(E_a − 2)/RT} = A e^−E_a/RT e^2/RT
\( \therefore \frac{k_2}{k_1} = e^{2/RT} \) or \( \ln\frac{k_2}{k_1} = \frac{2}{RT} \)
Putting R = 2 × 10⁻³ kcal K⁻¹ mol⁻¹ and T = 300 K, we get \( \log\frac{k_2}{k_1} = \frac{2}{2.303\times2\times10^{-3}\times300} \) = 1.4473
\( \therefore \frac{k_2}{k_1} = 28 \) or k₂ = 28 × k₁

Teacher's Note:
1. Use R in the same energy unit as E_a (here kcal: R ≈ 2 × 10⁻³ kcal K⁻¹ mol⁻¹).

One or More Correct Answers

Question 25. For the first order reaction 2N₂O₅(g) → 4NO₂(g) + O₂(g) (IIT 2011)
(a) the concentration of the reactant decreases exponentially with time
(b) the half-life of the reaction decreases with increasing temperature
(c) the half-life of the reaction depends on the initial concentration of the reactant
(d) the reaction proceeds to 99.6% completion in eight half-life duration
Answer:
Ans. (a, b, d)
[A] = [A]₀ e^−kt. Hence, (a) is correct.
\( t_{1/2} = \frac{0.693}{k} \). As k increases with increasing temperature, therefore, t_1/2 decreases. Hence, (b) is correct.
(c) is incorrect because t_1/2 of a first order reaction is independent of initial concentration.
Amount left after 8 half-lives \( = \frac{[\mathrm{A}]_0}{2^8} = \frac{1}{256} \) of [A]₀ = 0.004 of [A]₀ ∴ Amount reacted = 0.996 of [A]₀, i.e., 99.6% of [A]₀. Hence, (d) is correct.

Teacher's Note:
1. After n half-lives the fraction left is (1/2)ⁿ; 8 half-lives leave 1/256.

Question 26. The increase in rate constant of a chemical reaction with increasing temperature is (are) due to the fact(s) that (West Bengal JEE 2015)
(a) the number of collisions among the reactant molecules increases with increasing temperature
(b) the activation energy of the reaction decreases with increasing temperature
(c) the concentration of the reactant molecules increases with increasing temperature
(d) the number of reactant molecules acquiring the activation energy increases with increasing temperature
Answer:
Ans. (a, d)
Both (a) and (d) are the correct reasons. E_a does not change with temperature, and concentration does not increase on heating.

Teacher's Note:
1. The main reason is (d): the fraction of molecules with energy ≥ E_a rises sharply with temperature.

Question 27. According to Arrhenius equation, (JEE Advanced 2016)
(a) a high activation energy usually implies a fast reaction
(b) rate constant increases with increase in temperature. This is due to greater number of collisions whose energy exceeds the activation energy
(c) higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant
(d) the pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy
Answer:
Ans. (b, c, d)
(a) is incorrect because high activation energy implies a slow reaction. (b), (c) and (d) are correct and follow from Arrhenius equation, viz., k = A e^−E_a/RT.

Teacher's Note:
1. A larger E_a gives a steeper ln k vs 1/T line, i.e., k is more sensitive to temperature.

Passage-Based Questions

Question 28. Comprehension 1: Arrhenius studied the effect of temperature on the rate of a reaction and postulated that rate constant varies with temperature exponentially as k = Ae^−E_a/RT. For most of the reactions, it was found that the temperature coefficient of the reaction lies between 2 to 3. The method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied by studying how much the rate of reaction changes in the presence of catalyst. In most of the cases, it is observed that catalyst lowers the activation energy barrier and increases the rate of reaction.
If the rate of reaction grows 15.6 times on increasing the temperature by 30 K, the temperature coefficient of the reaction will be nearly
(a) 2
(b) 2.5
(c) 3.0
(d) 3.5
Answer:
Ans. (b)
If n is the temperature coefficient, after 10 K rise, rate constant will increase n times, after next 10 K rise to n² times and after further 10 K rise to n³ times.
Hence, n³ = 15.6 or n = (15.6)^1/3 = 2.5 (2.5³ = 15.625).

Teacher's Note:
1. The effect of each 10 K rise multiplies, so a 30 K rise multiplies the rate by n³, not 3n.

Question 29. Comprehension 1: Arrhenius studied the effect of temperature on the rate of a reaction and postulated that rate constant varies with temperature exponentially as k = Ae^−E_a/RT. For most of the reactions, it was found that the temperature coefficient of the reaction lies between 2 to 3. The method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied by studying how much the rate of reaction changes in the presence of catalyst. In most of the cases, it is observed that catalyst lowers the activation energy barrier and increases the rate of reaction.
If x is the fraction of molecules having energy greater than E_a, it will be given by
(a) \( x = -\frac{E_a}{RT} \)
(b) \( \ln x = -\frac{E_a}{RT} \)
(c) x = e^E_a/RT
(d) Any of these.
Answer:
Ans. (b)
Fraction of molecules having energy greater than E_a at temperature T = e^−E_a/RT (Boltzmann factor), i.e., x = e^−E_a/RT. Hence, \( \ln x = -\frac{E_a}{RT} \).

Teacher's Note:
1. A fraction must lie between 0 and 1, which rules out e^+E_a/RT.

Matrix Match Questions

Teacher's Note:
1. Only a zero order reaction goes to 100% completion in a finite time; first order reactions approach completion only as t → ∞.

Teacher's Note:
1. Simplify the denominator in each limit first: X_s + [X] ≈ X_s when [X] is small and ≈ [X] when [X] is large.

Integer and Numerical Answer Questions

Teacher's Note:
1. A constant rate while the concentration keeps falling is the test for zero order.

Question 33. A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non-reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is .......... min (Round off to the nearest integer). (JEE Main 2021)
Answer:
Ans. 108
Suppose after time t min, concentration of A, [A]_t = 16 times the concentration of B, i.e., [A]_t = 16 [B]_t
Half lives of A in time t = t/54; Half lives of B in time t = t/18
\( \therefore [\mathrm{A}]_t = \frac{[\mathrm{A}]_0}{2^{t/54}} \) and \( [\mathrm{B}]_t = \frac{[\mathrm{B}]_0}{2^{t/18}} \)
As [A]_t = 16 [B]_t and [A]₀ = [B]₀ \( \therefore \frac{1}{2^{t/54}} = \frac{16}{2^{t/18}} \) or 2^−t/54 = 2^4−t/18
or \( -\frac{t}{54} = 4-\frac{t}{18} \) or −t = 216 − 3t or 2t = 216 or t = 108 min

Teacher's Note:
1. Count half-lives: in 108 min A goes through 2 and B through 6 half-lives, so [A]/[B] = 2⁶/2² = 16.

Question 34. If the activation energy of a reaction is 80.9 kJ mol⁻¹, the fraction of molecules at 700 K having enough energy to react to form products is e^−x. The value of x is ...... (Round off to the nearest integer) [Use R = 8.31 J K⁻¹ mol⁻¹] (JEE Main 2021)
Answer:
Ans. 14
Fraction of molecules able to cross energy barrier = e^−E_a/RT = e^−x \( \therefore x = \frac{E_a}{RT} = \frac{80.9\times1000}{8.31\times700} \) = 13.91 = 14

Teacher's Note:
1. Convert E_a to J mol⁻¹ so that it matches R in J K⁻¹ mol⁻¹.

Question 35. Sucrose hydrolyses in acid solution into glucose and fructose following first order kinetics with half-life of 3.33 h at 25°C. After 9 h, the fraction of the sucrose remaining is f. The value of f is ......... × 10⁻² (Round off to the nearest integer) [Assume ln 10 = 2.303, ln 2 = 0.693] (JEE Main 2021)
Answer:
Ans. 15
No. of half-lives in 9 h \( = \frac{9}{3.33} = \frac{9}{10/3} \) = 2.7
\( C_t = \frac{C_0}{2^{2.7}} \) or \( \frac{C_t}{C_0} = 2^{-2.7} = x \) (say)
log x = −2.7 log 2 = −2.7 × 0.301 = −0.8127 = 1̅.1873
x = Antilog (1̅.1873) = 0.154 = 15.4 × 10⁻² ≈ 15 × 10⁻²

Teacher's Note:
1. Check with ln: ln f = −0.693 × 9/3.33 = −1.873, so f = e^−1.873 = 0.154.

Assertion–Reason Questions

Question 36. Statement-1: The rate constant of a pseudounimolecular reaction has the units of the rate constant of a second order reaction.
Statement-2: A pseudounimolecular reaction is a reaction of second order in which one of the reactant is present in large excess.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT the correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
Statement-2 is the correct explanation of Statement-1 (k′ = k [H₂O] so that k = k′/[H₂O] = time⁻¹/mol L⁻¹ = L mol⁻¹ time⁻¹).

Teacher's Note:
1. The observed (pseudo first order) constant k′ has units s⁻¹; the true constant k = k′/[excess reactant] has second order units.

Question 37. Assertion: In any reaction, the rate of disappearance of a reactant is same as the rate of reaction.
Reason: Both rate of reaction and rate of disappearance of a reactant represent decrease in the concentration of a reactant per unit time.
(a) If both assertion and reason are true and reason is the true explanation of the assertion.
(b) If both assertion and reason are true but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct A. Rate of reaction may or may not be equal to rate of disappearance of a reactant. It depends upon the stoichiometric coefficient of the reactant in the balanced chemical equation.
Correct R. Rate of disappearance represents decrease in concentration of a reactant per unit time but rate may or may not.

Teacher's Note:
1. The two are equal only when the reactant's coefficient is 1.

Question 38. Assertion: The molecularity of the reaction H₂ + Br₂ → 2HBr is 2.
Reason: The order of the reaction is 3/2. (AIIMS 2004)
(a) If both assertion and reason are true and reason is the true explanation of the assertion.
(b) If both assertion and reason are true but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. Molecularity is two because two molecules of the reactants are involved in the given elementary reaction.
The order (3/2) is found experimentally, so R is true but does not explain A.

Teacher's Note:
1. Molecularity is a whole number counted from the reacting species; order is experimental and may be fractional.

Question 39. Assertion: The order of a reaction can have fractional value.
Reason: The order of reaction cannot be written from the balanced equation of the reaction. (AIIMS 2008)
(a) If both assertion and reason are true and reason is the true explanation of the assertion.
(b) If both assertion and reason are true but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. The order of reaction depends upon the power to which the concentration is raised to express the rate of reaction as found experimentally.

Teacher's Note:
1. Both statements are true, but fractional orders arise because order is the experimental power in the rate law.

Question 40. Assertion: Half-life of a reaction is independent of the initial concentration of the reactant.
Reason: Half-life of a reaction is independent of the order of reaction.
(a) If both assertion and reason are true and reason is the true explanation of the assertion.
(b) If both assertion and reason are true but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct A. Half-life of a first order reaction only is independent of the initial concentration of the reactant.
Correct R. \( t_{1/2} \propto \frac{1}{[\mathrm{A}_0]^{n-1}} \), i.e., depends upon the order n of the reaction.

Teacher's Note:
1. Statements that hold only for first order reactions are false when written for 'a reaction' in general.