JEE Chemistry Chemical Kinetics MCQs Set 05

Answer: (b) t₁/₂ does not depend upon concentration of reactant

Explanation:
1. r = k means the reaction is zero order.
2. For zero order t₁/₂ = a/(2k), which does depend on the initial concentration.
3. So statement (b) is incorrect.

Teacher's Note:
1. Zero order: rate constant, half-life proportional to a.
2. First order is the only order with a concentration-independent half-life.

Answer: (c) 2

Explanation:
1. Doubling the pressure halves t₁/₂, so t₁/₂ ∝ 1/p.
2. That corresponds to second order.

Teacher's Note:
1. Check each doubling: 3.64 → 1.82 → 0.91.
2. t₁/₂ ∝ p^{1−n} gives n = 2.

Answer: (a) 3

Explanation:
1. Unit s⁻¹ means first order, so rate = kx[A].
2. Tripling [A] triples the rate.

Teacher's Note:
1. Identify the order from the unit of k.
2. First-order rate is proportional to concentration.

Answer: (b) A second-order reaction with molecularity two

Explanation:
1. Rate = k[ester][OH⁻], so it is second order.
2. An ester molecule and a hydroxide ion collide: molecularity two.

Teacher's Note:
1. Here the nucleophile is not in excess in water as a solvent, so no pseudo-order appears.
2. Order and molecularity agree for this elementary-type step.

Answer: (d) \( \frac{2.303}{k}\log\frac{4}{3} \)

Explanation:
1. After one quarter of the reaction, x = a/4, so a − x = 3a/4.
2. t = (2.303/k) log[a/(a − x)] = (2.303/k) log(4/3).

Teacher's Note:
1. Be careful: the fraction decomposed is ¼, the fraction left is ¾.
2. The ratio inside the log is always initial/left.

Answer: (a) 100 minutes

Explanation:
1. First case: 75% decomposed in 100 min. Second case: 150/200 = 75% as well.
2. The fraction decomposed is the same, so the time is the same, 100 minutes.

Teacher's Note:
1. For first order only the fraction matters.
2. Do not compare masses directly.

Answer: (b) Zero order

Explanation:
1. Rate = k[A]⁰ = k.
2. So both have units mol L⁻¹ s⁻¹.

Teacher's Note:
1. Check using units: rate = k × (conc)ⁿ.
2. Only n = 0 gives identical units.

Answer: (d) 0.034 min⁻¹

Explanation:
1. k = 0.693/t₁/₂ = 0.693/20 = 0.0347 min⁻¹ ≈ 0.034 min⁻¹.

Teacher's Note:
1. The first-order constant has unit of reciprocal time.
2. Initial concentration is not needed.

Answer: (d) Rate of reaction is directly proportional to collision frequency

Explanation:
1. Only effective collisions (enough energy and proper orientation) give products.
2. The rate is proportional to the collision frequency multiplied by the fraction of effective collisions.

Teacher's Note:
1. Not all collisions are effective.
2. Rate = Z × e^{−Ea/RT} × P.

Answer: (c) The minimum extra amount of energy needed to overcome the potential barrier of the reaction

Explanation:
1. Activation energy is the extra energy reactants must gain over their average energy to reach the activated complex.
2. It is the barrier height, not an energy released.

Teacher's Note:
1. Ea = threshold energy − average energy of reactants.
2. A lower barrier means a faster reaction.

Answer: (b) E₁ < E₂

Explanation:
1. k = A e^{−Ea/RT}; a larger k at the same T (with similar A) means a smaller Ea.
2. So E₁ < E₂.

Teacher's Note:
1. Lower activation energy gives a larger rate constant.
2. This assumes similar frequency factors.

Answer: (c) The activation energy of 2NO + O₂ → 2NO₂ being less than that of 2CO + O₂ → 2CO₂

Explanation:
1. Speed of a reaction at a given temperature is controlled by the activation energy.
2. A lower Ea for the NO reaction makes it fast.

Teacher's Note:
1. Rates differ because of different energy barriers.
2. Size and toxicity are irrelevant to kinetics.

Answer: (c) The total number of effective collisions

Explanation:
1. Only collisions with at least threshold energy and correct orientation lead to reaction.
2. Rate equals the number of effective collisions per unit volume per second.

Teacher's Note:
1. Distinguish total from effective collisions.
2. The rate is smaller than the collision frequency.

Answer: (b) An increase in the number of activated molecules

Explanation:
1. A small rise in temperature increases the collision frequency only slightly.
2. It greatly increases the fraction of molecules with energy ≥ threshold (e^{−Ea/RT}).

Teacher's Note:
1. The exponential term in the Arrhenius equation dominates.
2. Temperature does not change Ea.

Answer: (a) Of any reaction

Explanation:
1. Increasing temperature increases k = A e^{−Ea/RT} for any reaction with positive Ea.
2. This holds for both endothermic and exothermic reactions.

Teacher's Note:
1. Exothermic reactions also speed up on heating even though equilibrium shifts backwards.
2. Rate and equilibrium are different questions.

Answer: (c) \( \frac{1}{T} \)

Explanation:
1. log k = log A − Ea/(2.303RT).
2. So log k versus 1/T is linear with slope −Ea/(2.303R).

Teacher's Note:
1. The slope gives the activation energy.
2. The intercept gives log A.

Answer: (d) ln k versus 1/T

Explanation:
1. ln k = ln A − Ea/(RT).
2. The slope of ln k versus 1/T equals −Ea/R.

Teacher's Note:
1. Choose the plot that is linear in the Arrhenius form.
2. Ea = −slope × R.

Answer: (d) log A

Explanation:
1. As T → ∞, e^{−Ea/RT} → 1, so k → A.
2. Hence log k → log A.

Teacher's Note:
1. At infinite temperature every collision is effective.
2. k can never exceed A.

Answer: (b) 6 × 10¹⁴ s⁻¹

Explanation:
1. k = A e^{−Ea/RT}.
2. As T → ∞ the exponential factor tends to 1, so k → A = 6.0 × 10¹⁴ s⁻¹.

Teacher's Note:
1. The given k and Ea are distractors here.
2. The limiting value of k is the frequency factor.

Answer: (c) More than ΔH

Explanation:
1. For endothermic reactions, ΔH = Ea(f) − Ea(r) > 0.
2. Since Ea(r) cannot be negative, Ea(f) > ΔH.

Teacher's Note:
1. Products lie above reactants in energy, so the barrier is higher than the energy gap.
2. Compare with exothermic reactions where Ea may be small.

Answer: (d) k

Explanation:
1. The rate constant does not depend on the concentrations of reactants.
2. It is unchanged; only the rate changes.

Teacher's Note:
1. k depends only on temperature and catalyst.
2. Rate = k[A]ⁿ, so only the rate changes.

Answer: (c) Remain constant

Explanation:
1. For first order the unit of k is s⁻¹ and has no concentration term.
2. So its value is unchanged.

Teacher's Note:
1. Units of k for order n: (conc)^{1−n} time⁻¹.
2. Only n ≠ 1 gives concentration dependence in the unit.

Answer: (d) Ea = Ea′

Explanation:
1. Both equations describe the same process, only written with different coefficients.
2. The energy barrier is the same, so Ea = Ea′.

Teacher's Note:
1. Activation energy is per mole of reaction events (the energy barrier), not per balanced equation.
2. Rate constants are defined per the equation but Ea is not.

Answer: (d) Either greater or less than 50 kcal

Explanation:
1. Ea(reverse) = Ea(forward) − ΔH = 50 − ΔH.
2. ΔH is not given and can be positive or negative, so the reverse value may be larger or smaller.

Teacher's Note:
1. Never assume the sign of ΔH.
2. Exothermic: Ea(reverse) > 50; endothermic: < 50.

Answer: (c) 50 kJ

Explanation:
1. ΔE = Ea(f) − Ea(r), so −20 = 30 − Ea(r).
2. Ea(r) = 50 kJ.

Teacher's Note:
1. For exothermic reactions the reverse barrier is larger.
2. Activation energy is never negative.