JEE Chemistry Chemical Kinetics MCQs Set 25

Answer: (c) 0.72 M

Explanation:
1. The units of k (mol L\(^{-1}\) s\(^{-1}\)) show that the reaction is of zero order.
2. For zero order, [B] formed \( = kt = 0.6 \times 10^{-3} \times 1200 = 0.72 \) M.

Teacher's Note:
1. Identify the order from the units of k first. Zero-order products form linearly with time.

Answer: (b) k' = 2k, k'' = k/2

Explanation:
1. From the stoichiometry, \( \frac{d[NO_2]}{dt} = 2 \times \left(-\frac{d[N_2O_5]}{dt}\right) \), so k' = 2k.
2. \( \frac{d[O_2]}{dt} = \frac{1}{2}\left(-\frac{d[N_2O_5]}{dt}\right) \), so k'' = k/2.

Teacher's Note:
1. Rates of formation and disappearance are related through the stoichiometric coefficients.

Answer: (a) 0

Explanation:
1. Rate \( = k[A]^n \). Rate = k only if \( [A]^n = 1 \) for all concentrations, which means n = 0.
2. The rate is independent of concentration.

Teacher's Note:
1. For zero order, rate = k and the units of k are mol L\(^{-1}\) time\(^{-1}\).

Answer: (b) A plot of ln [A] against time is a straight line with a negative slope.

Explanation:
1. For a first-order reaction \( \ln[A] = \ln[A]_0 - kt \), so ln [A] against t is a straight line with slope \(-k\).
2. The half-life \( 0.693/k \) is independent of concentration, k has units s\(^{-1}\), and [A] falls exponentially.

Teacher's Note:
1. First order: constant half-life, log-linear decay, k in time\(^{-1}\).

Answer: (d) \( \frac{dc}{dt} = k[A] \)

Explanation:
1. Experiments 1 and 2: doubling [B] leaves the rate unchanged, so the order in B is 0.
2. Experiments 1 and 3: doubling [A] doubles the rate, so the order in A is 1.
3. Rate \( = k[A] \).

Teacher's Note:
1. Compare two experiments in which only one concentration changes.

Answer: (c) 0.25 h

Explanation:
1. For zero order \( t_{1/2} = \frac{[A]_0}{2k} \), so \( k = \frac{2.0}{2 \times 1} = 1.0 \) mol L\(^{-1}\) h\(^{-1}\).
2. Time \( = \frac{0.50 - 0.25}{1.0} = 0.25 \) h.

Teacher's Note:
1. In zero-order kinetics the concentration falls by the same amount in equal times, so the time depends on the change in concentration.

Answer: (b) the energy below which colliding molecules will not react

Explanation:
1. E is the activation energy, the minimum (threshold) energy that colliding molecules must have to react.
2. Molecules with less energy do not react. The fraction of molecules with enough energy is \( e^{-E/RT} \), not E.

Teacher's Note:
1. The factor \( e^{-E_a/RT} \) is the fraction of molecules with energy equal to or greater than \( E_a \).

Answer: (c) remains unchanged

Explanation:
1. A catalyst changes only the pathway (lowering the activation energy).
2. The energies of reactants and products are the same, so \( \Delta H \) does not change.

Teacher's Note:
1. A catalyst does not change \( \Delta H \), \( \Delta G \) or the equilibrium constant.

Answer: (b) determining the rate constants at two temperatures

Explanation:
1. From \( \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \), \( E_a \) can be found if k is known at two temperatures.
2. A single temperature is insufficient.

Teacher's Note:
1. Two values of k at two temperatures give \( E_a \), assuming that A is constant.

Answer: (d) increases exponentially with decreasing activation energy and increasing temperature

Explanation:
1. k increases when the exponent \( \frac{E_a}{RT} \) becomes smaller.
2. That happens when \( E_a \) decreases or when T increases, so k increases exponentially in these cases.

Teacher's Note:
1. Lower barrier and higher temperature both speed up a reaction.

Answer: (b) It alters \( \Delta G \) of the reaction.

Explanation:
1. A catalyst does not change \( \Delta G \) of a reaction because the free energies of reactants and products are fixed.
2. Statements A, C and D are correct.

Teacher's Note:
1. A catalyst changes the rate (kinetics), not the thermodynamics.

Answer: (b) A and D

Explanation:
1. For an elementary reaction the rate law follows the stoichiometry, so order = molecularity.
2. Molecularity is the number of molecules colliding in the step, so it is a positive whole number, never zero.

Teacher's Note:
1. Order can be zero or fractional; molecularity cannot.

Answer: (c) A and C

Explanation:
1. For a zero-order reaction, rate \( = k[NH_3]^0 = k \), so A is correct.
2. The rate does not depend on the concentration of ammonia (the catalyst surface is saturated), so B is false.
3. The rate stays constant until the reactant is used up, so C is correct.

Teacher's Note:
1. Zero order at high pressure arises because the catalyst surface is fully covered with reactant.

Answer: (b) A and D

Explanation:
1. The area under the curve represents the total number (or fraction, equal to 1) of molecules, which stays constant.
2. With higher temperature the curve flattens (broadens) and its peak moves to the right.

Teacher's Note:
1. Total molecules are fixed, so the area is constant at every temperature.

Answer: (d) B and D

Explanation:
1. A catalyst gives a path of lower activation energy, so A and C are correct.
2. It does not raise the activation energy and does not change the enthalpy change of a reaction, so B and D are incorrect.

Teacher's Note:
1. A catalyst changes the path and the barrier, but not the energies of the reactants and products.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. A catalyst provides an alternative path with a lower barrier.
2. The energies of the reactants and products are unaffected, so \( \Delta H \) is unchanged.
3. The Reason therefore explains the Assertion.

Teacher's Note:
1. Enthalpy is a state function, so it depends only on the initial and final states.

Answer: (c) 8

Explanation:
1. New rate \( = k(4[A])^{1/2}(2[B])^2 = 4^{1/2} \times 2^2 \times \) old rate.
2. Factor \( = 2 \times 4 = 8 \).

Teacher's Note:
1. Raise each concentration factor to its own order and multiply.

Answer: (b) 4

Explanation:
1. Rate \( = k[A]^2 \).
2. If [A] is doubled the rate becomes \( k(2[A])^2 = 4k[A]^2 \), i.e. 4 times.

Teacher's Note:
1. For order n, doubling the concentration multiplies the rate by \( 2^n \).

Answer: (b) 33 min

Explanation:
1. A unimolecular elementary reaction is first order.
2. \( k = \frac{0.693}{10} = 0.0693 \) min\(^{-1}\).
3. \( t = \frac{2.303}{k}\log\frac{100}{10} = \frac{2.303}{0.0693} = 33.2 \) min.

Teacher's Note:
1. Time to reach 10% remaining is 3.32 half-lives for a first-order reaction.

Answer: (a) 0

Explanation:
1. The half-life is proportional to the initial concentration (pressure): \( t_{1/2} \propto [A]_0^{1-n} \).
2. Halving the pressure halves the half-life, so \( 1 - n = 1 \) and n = 0.

Teacher's Note:
1. Zero order: half-life proportional to \( [A]_0 \). First order: independent. Second order: inversely proportional.

Answer: (c) 58 kcal

Explanation:
1. \( E_{a(forward)} - E_{a(reverse)} = \Delta H \).
2. \( 20 - E_{a(reverse)} = -38 \), so \( E_{a(reverse)} = 58 \) kcal.

Teacher's Note:
1. For an exothermic reaction the reverse reaction has the higher activation energy.

Answer: (b) 30

Explanation:
1. In 120 min X goes through 6 half-lives, so its activity becomes \( \frac{4a}{64} = \frac{a}{16} \) when Y started at a.
2. For the ratio to be 1 : 1, Y's activity must also fall to \( \frac{a}{16} \), which is 4 half-lives.
3. \( t_{1/2}(Y) = \frac{120}{4} = 30 \) min.

Teacher's Note:
1. Count half-lives for each nuclide separately and then equate the final activities.