Chemical kinetics practice · Mixed difficulty

Option B (0.042 M sec⁻¹) is incorrect. Option D (0.125 M sec⁻¹) is correct because the rate of disappearance of Br⁻ is calculated based on the stoichiometry of the reaction and the rate of appearance of Br₂.

Option B (1) is correct because molecularity is the number of reactant molecules involved in an elementary step, which is one in this case.

Option D (Rate is independent of the surface coverage) is correct because a first-order reaction's rate depends only on the concentration of the reactant, not the surface coverage in this specific scenario.

Option D (2) is correct because the rate law, derived from the rate-determining step and equilibrium, shows a second-order dependence on $\mathrm{NO}$.

Option D (Unimolecular reaction) is correct because a reaction involving two different reactants must have at least two molecules participating in the rate-determining step, making it impossible for it to be unimolecular.

Option B ($\ln \;A$) is correct because in the Arrhenius equation, the intercept of the Arrhenius plot corresponds to the natural logarithm of the pre-exponential factor A.

Option A ($36.6\;\mathrm{kJ}/\mathrm{mol}$) is correct because it is calculated using the Arrhenius equation relating rate constant and temperature.

Option A ($\frac{{200}}{{3.506}}\mathrm{K}$) is incorrect. Option B ($\frac{{300}}{{2.303}}\mathrm{K}$) is correct because equating the rate constants and solving for T using the Arrhenius equation leads to this result.

Option C (If Assertion is True but the Reason is False) is correct because while increasing temperature generally increases the rate of reaction, it doesn't always shift the equilibrium in the forward direction.

Option B is correct because the Assertion is wrong (activation energy does not *always* increase potential energy) but the Reason is correct (definition of activation energy).

Option B (105) is correct because the activation energy for the backward reaction is the sum of the activation energy for the forward reaction and the enthalpy change.

Option A ($4k_3$) is correct because the rate constants are related by the stoichiometry of the reaction.

Option A ($2{K_1} = {K_2} = 4{K_3}$) is correct because it correctly relates the rate constants based on the stoichiometry of the reaction.

Option B (0.5112) is correct because it is calculated using the integrated rate laws for parallel first-order reactions, considering the ratio of rate constants and the reaction time.

Option A (Rate is proportional to the surface coverage) is correct because for a first-order reaction on a surface, the rate is directly proportional to the surface coverage of the reactant.

Option D ((A)Only) is correct because only mechanism A is consistent with the given rate equation, rate = k[Cl2][H2S], as the slow step in mechanism A directly reflects this rate law.

The correct answer is B because the rate-determining step is the second step, where NOBr2 reacts with NO, consistent with the observed rate law.

Option D (decrease the time required for the reaction) is correct because catalysts lower the activation energy, speeding up the reaction.

Option D (1.0 M) is correct because it is calculated using the integrated rate law for a zero-order reaction.

Option D is correct. Statement II is correct because a catalyst always decreases the activation energy but does not alter Gibb's energy, while statement I is incorrect because a catalyst can only increase the rate of reaction, not decrease it.

Option C (${t_{1/2}} = \frac{{2.303}}{k}\log \frac{4}{3}$) is correct because it represents the time taken for one-fourth of the reaction to complete, derived from the first-order kinetics equation.

Option B (46.06 min) is correct because it is the time required for 99% completion of a first-order reaction, calculated using the integrated rate law and the given half-life.

Option B (1) is correct because when reactant A is in large excess, its concentration remains nearly constant, making the reaction's rate dependent only on the concentration of reactant B, thus making it a first-order reaction.

Option C ($100 \mathrm{~M}^{-2} \mathrm{~min}^{-1}$) is correct because it accurately calculates the true rate constant K based on the given rate law, pH, and half-life.

Option A (310 min) is correct because it is the calculated time for the radioactive substance to decay from $5 \times 10^{20}$ to $5 \times 10^{17}$ atoms.

Option D ($\mathrm{C}{\mathrm{H}_3}\mathrm{COOC}{\mathrm{H}_3} + \mathrm{NaOH} \to \mathrm{C}{\mathrm{H}_3}\mathrm{COONa} + {\mathrm{H}_2}\mathrm{O}$) is correct because the saponification of an ester by NaOH is a second-order reaction.

Option D (2) is correct because the ratio of the time for 99% completion to the time for 90% completion in a first-order reaction is approximately 2.

Option B (32 times) is correct because the rate doubles for every 10°C increase, so for a 50°C increase, the rate increases by 2^(50/10) = 2⁵ = 32 times.

Option A (1) is correct because the reaction being 50% complete in 2 hours and 75% complete in 4 hours indicates a first-order reaction, where $t_{75\%} = 2 \times t_{50\%}$.

The correct answer is B ($r=k^{\prime}\left[{O}_3\right]^2\left[{O}_2\right]^{-1}$) because it correctly represents the rate law derived from the given mechanism, considering the fast equilibrium step and the slow rate-determining step.