Chemical kinetics practice · Medium

Option A ($- \frac{{dc}}{{dt}} = - \frac{1}{V}\frac{{dn}}{{dt}} = - \frac{1}{{RT}}\frac{{dP}}{{dt}}$) is correct because it correctly relates the rate of change of concentration, moles, and pressure for a gaseous reactant.

Option B ($1000\;K$) is correct because it is the temperature at which the rate constants k1 and k2 are equal, found by equating the given expressions and solving for T.

Option C is correct because Statement I is true, as the reciprocal of the time for 66% conversion in a first-order reaction is related to the rate constant, and Statement II is false, since the rate constant for a first-order reaction is independent of the initial concentration of reactants.

Option B is correct because the rate law is determined by the slow step in the reaction mechanism.

Option D (10) is correct because the ratio of times for 99.9% completion to half completion in a first-order reaction is approximately 10.

Option B (0.72 M) is correct because, for a zero-order reaction, the concentration of product B is directly proportional to the rate constant and time.

Option B is correct because both the Assertion and Reason are correct statements, but the Reason does not explain the Assertion.

Option C ($0.11\,{\text{mol}}\,{{\text{L}}^{ - 1}}{{\text{S}}^{ - 1}}$) is correct because it is the average rate of the reaction calculated using the change in concentration over the change in time.

Rate of the reaction involving the thermal decomposition of acetaldehyde = $k{\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}} \right]^{3/2}}$ is correct because the thermal decomposition of acetaldehyde follows a 3/2 order kinetics.

Option A (1) is correct because the half-life is constant, indicating a first-order reaction.

Option C (Both A and R are correct and R is correct explanation of A) is correct because for a free radical combination, the activation energy ($E_a$) is approximately zero, which makes the rate constant (K) equal to the pre-exponential factor (A).

The correct answer is B (2828 second) because the reaction is first order, and 75% completion takes twice the time for 50% completion.

The correct answer is 0.32 M because after 4 half-lives, the initial concentration is reduced to 0.02 M, so the initial concentration is 0.02 * 2⁴.

Option A (1) is correct because the rate of formation of NO2 is twice the rate of consumption of N2O4, and the calculation yields approximately 1 mol lit⁻¹ min⁻¹.

Option D (All of these) is correct because the relative rates depend on the concentration of the reactant [A]. If [A] > 1, then r3 > r2 > r1; if [A] < 1, then r1 > r2 > r3; and if [A] = 1, then r1 = r2 = r3.

Option D (48 {~min}) is correct because for a first-order reaction, the time required for a certain percentage completion is proportional to the logarithm of the remaining reactant.

Option D (-20) is correct because the enthalpy change is the difference between the activation energies of the forward and reverse reactions, and the catalyst lowers both by the same amount.

Option B (4.92 K) is correct because it accurately calculates the temperature increase required for reaction B based on the given activation energies and the Arrhenius equation.

Option A (A is correct but R is wrong) is incorrect. Both A and R are incorrect. Catalytic reactions do not always show zero order for the catalyst, and the rate of reaction *does* depend on the concentration of the catalyst.

Option B ($4.6 \times 10^{-2} \mathrm{~s}$) is correct because it's calculated using the first-order kinetics formula with the given rate constant and the fraction of reactant remaining.

Option C is correct because the half-life is inversely proportional to the initial concentration, indicating a second-order reaction.

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 $t_{75\%} = 2 \times t_{50\%}$.

Option D (low $T$ and high $E_{\text {a }}$) is correct because the rate constant $k$ decreases exponentially with increasing activation energy ($E_a$) and decreasing temperature ($T$) according to the Arrhenius equation.

Option D ($0.1\;M\;\mathrm{N}{\mathrm{a}_3}\mathrm{P}{\mathrm{O}_4}$) is correct because it has the highest van't Hoff factor (i=4), leading to the greatest lowering of vapor pressure.

The correct answer is B ($3{K_1}{\left[ A \right]^2}{\left[ B \right]^3} - 3{k_2}{\left[ C \right]^2}$) because it correctly represents the rate of disappearance of B in terms of the forward and reverse reaction rates, considering the stoichiometric coefficient of B.

Option B (1/3) is correct because the rate is proportional to the concentration raised to the power of the order of the reaction, and solving for the order gives 1/3.

Option C (12) is correct because the activation energy is calculated using the Arrhenius equation.

Option A (${\text{mol}}{{\text{e}}^{ - 2/3}}{\text{li}}{{\text{t}}^{2/3}}{\text{tim}}{{\text{e}}^{ - 1}}$) is correct because it is derived from the rate law expression and the overall order of the reaction.

Option B is correct because the Arrhenius equation shows that a larger activation energy results in a greater change in the rate constant for a given temperature change.

Option C (One) is correct because when X is in large excess, its concentration remains nearly constant, making the reaction's rate dependent only on the concentration of Y, thus first order.