JEE Chemistry Chemical Kinetics MCQs Set 07

Answer: (a) \( k = \dfrac1t\ln\dfrac{r_\infty}{r_\infty-r_t} \)

Explanation:
1. Rotation is zero initially and \( r_\infty \) at the end; \( r_t \) is proportional to the amount reacted.
2. Initial amount \( \propto r_\infty \) and remaining amount \( \propto r_\infty - r_t \).
3. So \( k = \dfrac1t\ln\dfrac{r_\infty}{r_\infty-r_t} \).

Teacher's Note:
1. Express a as the full change in the property and a minus x as the remaining change.

Answer: (d) 30 mL

Explanation:
1. 200 mL at 2 atm is 400 mL at 1 atm.
2. Fraction decomposed at 15 min = 100/400 = 1/4, so 3/4 remains.
3. \( V_{KMnO_4} = 40\times\tfrac34 = 30 \) mL.

Teacher's Note:
1. Convert both volumes to the same pressure before taking the ratio.

Answer: (c) 30 hours

Explanation:
1. Overall \( k = 0.693/12 \).
2. \( k_B = 0.4k \) since C takes 60%.
3. \( t_{1/2(B)} = 0.693/(0.4k) = 12/0.4 = 30 \) h.

Teacher's Note:
1. Partial half life is the overall half life divided by the fraction of the rate going to that path.

Answer: (b) \( B \to C \)

Explanation:
1. The slowest step has the smallest rate constant.
2. \( 8\times10^{-6} \) s\(^{-1}\) is the smallest, so \( B \to C \) is rate determining.

Teacher's Note:
1. In a series of steps, the slowest step controls the overall rate.

Answer: (a) 28 times

Explanation:
1. \( \dfrac{k_c}{k} = e^{\Delta E_a/RT} = e^{8314/(8.314\times300)} = e^{3.33} = 28 \).

Teacher's Note:
1. Do not forget to convert kJ to J.

Answer: (d) s\(^{-1}\), M s\(^{-1}\)

Explanation:
1. Rate has units M s\(^{-1}\). First order: \( k = \text{rate}/[A] \) gives s\(^{-1}\).
2. Zero order: \( k = \text{rate} \) gives M s\(^{-1}\).

Teacher's Note:
1. The unit of k is (concentration)^(1-n) per second.

Answer: (c) \( 1.5\times10^{-11} \) L mol\(^{-1}\) s\(^{-1}\)

Explanation:
1. [A] \( = 10^{-6} \) M and [B] \( = 10 \) M.
2. At equilibrium \( k_f[A] = k_b[B]^2 \): \( k_b = \dfrac{1.5\times10^{-3}\times10^{-6}}{100} = 1.5\times10^{-11} \).

Teacher's Note:
1. Use concentrations (mol per litre), not moles.

Answer: (b) 600 s

Explanation:
1. A and B are consumed equally, so rate \( = k[A] \): effectively first order.
2. \( t = \dfrac{2.303}{k}\log\dfrac{1}{0.25} = \dfrac{1.386}{2.31\times10^{-3}} = 600 \) s.

Teacher's Note:
1. With equal concentrations the two half-orders combine into first order.

Answer: (a) 4.65

Explanation:
1. \( k_1 = \dfrac{\ln2}{t} \) and \( k_2 = \dfrac{\ln25}{t} \).
2. \( \dfrac{k_2}{k_1} = \dfrac{\ln25}{\ln2} = \dfrac{3.219}{0.693} = 4.65 \).

Teacher's Note:
1. 96% reacted means 4% remains, so the ratio is 25.

Answer: (d) 100 kJ mol\(^{-1}\)

Explanation:
1. Fraction \( = 3.8\times10^{-18} = e^{-E_a/RT} \).
2. \( E_a = RT\ln\dfrac{1}{3.8\times10^{-18}} = 8.314\times300\times40.1 \approx 1.0\times10^5 \) J.

Teacher's Note:
1. The percentage must be converted to a fraction by dividing by 100.

Answer: (c) many body collisions have a very low probability

Explanation:
1. The chance that more than three molecules collide simultaneously with the right orientation and energy is extremely small.
2. Hence most reactions proceed through a sequence of bimolecular or unimolecular steps.

Teacher's Note:
1. Molecularity is rarely more than 3.

Answer: (b) 75 min

Explanation:
1. \( k = \dfrac{2.303}{30}\log\dfrac{30-(-15)}{20-(-15)} = \dfrac{2.303}{30}\log\dfrac97 \).
2. \( \log\dfrac97 = 2(0.48)-0.84 = 0.12 \).
3. \( t_{1/2} = \dfrac{0.693\times30}{2.303\times0.12} = \dfrac{0.3\times30}{0.12} = 75 \) min.

Teacher's Note:
1. Use (r0 - r infinity) and (rt - r infinity) as proportional to a and a - x.

Answer: (a) 11

Explanation:
1. \( \ln\dfrac{k_{298}}{k_{273}} = \dfrac{65000}{8.314}\left(\dfrac1{273}-\dfrac1{298}\right) = 2.40 \).
2. Ratio \( = e^{2.40} \approx 11 \).

Teacher's Note:
1. A 25 degree rise increases the rate by roughly an order of magnitude for typical Ea.

Answer: (d) zero

Explanation:
1. Half life doubles when the initial concentration doubles, so half life is proportional to a.
2. This is the behaviour of a zero order reaction.

Teacher's Note:
1. Check how half life depends on the initial concentration to find the order.

Answer: (c) first order, \( k_1 \)

Explanation:
1. At low \( C_A \), \( k_2C_A \ll 1 \) so the denominator is nearly 1.
2. Rate \( \approx k_1C_A \): first order with constant \( k_1 \).

Teacher's Note:
1. At high concentration the same expression becomes zero order with constant k1/k2.