JEE Chemistry Chemical Kinetics MCQs Set 22

Answer: (a) 3

Explanation:
1. (A) true: k = Ae^−E_a/RT decreases as E_a increases. (B) true: reactions with larger E_a are more sensitive to temperature.
2. (C) true: d(ln k)/dT = E_a/RT² is larger at lower T.
3. (D) false: the slope is −E_a/R (negative). Hence 3 statements are correct.

Teacher's Note:
1. From the Arrhenius equation, ln k = ln A − E_a/RT.
2. The slope of ln k against 1/T is negative.

Answer: (b) 2

Explanation:
1. The half-life halves every time the pressure doubles, so t½ ∝ 1/p.
2. t½ ∝ [A]₀¹⁻ⁿ and 1 − n = −1, so n = 2.

Teacher's Note:
1. For order n, t½ ∝ [A]₀¹⁻ⁿ.
2. First-order half-life is independent of concentration; zero-order half-life is proportional to it.

Answer: (d) 60 min

Explanation:
1. 75 % completion means two half-lives: 100 → 50 → 25 % remaining.
2. t = 2 × 30 = 60 min.

Teacher's Note:
1. Each half-life reduces the remaining amount to half.
2. The first-order half-life is constant.

Answer: (c) 2

Explanation:
1. The unit of k for order n is (mol L⁻¹)¹⁻ⁿ s⁻¹.
2. The given unit L mol⁻¹ s⁻¹ corresponds to 1 − n = −1, so n = 2.

Teacher's Note:
1. Zero order: mol L⁻¹ s⁻¹; first order: s⁻¹; second order: L mol⁻¹ s⁻¹.
2. Read the order directly from the unit of k.

Answer: (b) 623 s

Explanation:
1. t = (2.303/k) log([A]₀/[A]) = (2.303/2.011 × 10⁻³) log(7/2).
2. log 3.5 = 0.845 − 0.301 = 0.544.
3. t = 1145 × 0.544 ≈ 623 s.

Teacher's Note:
1. Use the integrated first-order equation with the masses in place of concentrations.
2. Mass is proportional to concentration for the same volume.

Answer: (a) 1350 s

Explanation:
1. 60 % decomposed: [A]₀/[A] = 100/40 = 2.5, and log 2.5 = 1 − 2 × 0.3 = 0.4. So k = (2.3/540) × 0.4.
2. 90 % decomposed: [A]₀/[A] = 10, log 10 = 1, so t = 2.3/k.
3. t = 540/0.4 = 1350 s.

Teacher's Note:
1. Write t = (2.3/k) log([A]₀/[A]) for both cases and take the ratio.
2. t(90 %)/t(60 %) = 1/log 2.5.

Answer: (d) 2520 J mol⁻¹

Explanation:
1. \( \ln\dfrac{k_2}{k_1}=\dfrac{E_a}{R}\left(\dfrac1{T_1}-\dfrac1{T_2}\right) \) with k₂/k₁ = 5/3.
2. ln(5/3) = 2.3 × (0.70 − 0.48) = 0.506; 1/200 − 1/300 = 1/600.
3. E_a = 8.3 × 600 × 0.506 ≈ 2520 J mol⁻¹.

Teacher's Note:
1. Arrhenius two-point form: ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂).
2. Convert log to ln by multiplying by 2.3.

Answer: (c) 1 × 10¹³

Explanation:
1. Δn_g = 1 − 1.5 = −0.5, and K_p = K_c(RT)^Δn.
2. K_c = K_p(RT)^0.5 = 2 × 10¹² × (0.082 × 300)^0.5 = 2 × 10¹² × 4.96.
3. K_c ≈ 9.9 × 10¹² ≈ 1 × 10¹³.

Teacher's Note:
1. Because Δn is negative, K_c is larger than K_p.
2. T = 27 + 273 = 300 K.

Answer: (b) 0.173 min

Explanation:
1. t = (2.303/k) log 32 = (2.303/20) × 5 × 0.3010.
2. t = 0.1152 × 1.505 ≈ 0.173 min (17.3 × 10⁻² min).

Teacher's Note:
1. 1/32 = (1/2)⁵, so the reaction takes 5 half-lives.
2. t½ = 0.693/20 = 0.0347 min, and 5 × 0.0347 = 0.173 min.

Answer: (a) 15 min

Explanation:
1. Let [A]₀ = c. Then [A] = c(1/2)^t/15 and [B] = 4c(1/2)^t/5.
2. Setting [A] = [B]: 4 (1/2)^t/5 = (1/2)^t/15, i.e. 2 − t/5 = −t/15.
3. 2 = t/5 − t/15 = 2t/15, so t = 15 min.

Teacher's Note:
1. Write each concentration using its own half-life and equate.
2. Radioactive decay is first order.

Answer: (d) 75 min

Explanation:
1. For zero order, t½ = [A]₀/2k = 50 min, so [A]₀/k = 100 min.
2. Falling to one-fourth means 0.75[A]₀ consumed: t = 0.75[A]₀/k = 75 min.

Teacher's Note:
1. Zero-order half-life depends on the initial concentration; t = (amount consumed)/k.
2. Half the reactant is used in 50 min, the next quarter takes only 25 min.

Answer: (c) 1

Explanation:
1. (A) false: a first-order reaction never reaches 100 % completion in finite time. (B) false: t½ = 0.693/4.6 × 10⁻³ ≈ 151 s.
2. (C) false: t_{10 %} = 0.105/k and t_{90 %} = 2.303/k, so t_{90 %} is about 22 times t_{10 %}.
3. (D) true: [A] = [A]₀e^−kt, so the fraction reacted is 1 − e^−kt. (E) false: k has unit s⁻¹, rate has mol L⁻¹ s⁻¹. Only D is correct.

Teacher's Note:
1. For first order, t½ = 0.693/k does not depend on concentration.
2. Only zero-order reactions have rate and rate constant with the same unit.