Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

The rate of reaction depends upon concentration of only $$A.$$

$${t_{\frac{1}{2}}} \propto \frac{1}{{{{\left( p \right)}^{n - 1}}}}$$    where $$n$$ is the order of reaction
$$\eqalign{ & \frac{2}{4} = {\left( {\frac{{50}}{{100}}} \right)^{n - 1}}\,\,{\text{or}}\,\,\frac{1}{2} = {\left( {\frac{1}{2}} \right)^{n - 1}} \cr & n - 1 = 1;n = 2 \cr} $$

Given, order of reaction = 1
Rate of reaction at $$10\,s = 0.04\,mol\,{L^{ - 1}}{s^{ - 1}}$$
Rate of reaction at $$20\,s = 0.03\,mol\,{L^{ - 1}}{s^{ - 1}}$$
∴ Half-life period $$\left( {{t_{\frac{1}{2}}}} \right) = ?$$
We have the equation for rate-constant $$'k'$$  in first order reaction.
$$\eqalign{ & k = \frac{{2.303}}{t}{\text{log}}\frac{{{A_t}}}{{{A_0}}} = \frac{{2.303}}{{105}}{\text{log}}\frac{{0.04}}{{0.03}} \cr & \,\,\,\,\, = \frac{{2.303}}{{105}} \times 0.124 \cr & k = 0.028\,{s^{ - 1}} \cr & {\text{We know that,}} \cr & {t_{\frac{1}{2}}} = \frac{{0.693}}{k} = \frac{{0.693}}{{0.028773391\,{s^{ - 1}}}} \cr & \,\,\,\,\,\,\, = 24.14\,s \approx 24.1\,s \cr} $$

In a reaction $$A \to B,$$   concentration of reactant decreases as concentration of product increases during the course of a reaction.

$$\eqalign{ & \frac{{{k_2}}}{{{k_1}}} \cr & = \frac{{A{e^{ - \frac{{{E_{{a_2}}}}}{{RT}}}}}}{{A{e^{ - \frac{{{E_{{a_1}}}}}{{RT}}}}}} \cr & = {e^{\frac{{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)}}{{RT}}}} \cr & 2.303\,\log \frac{{{k_2}}}{{{k_1}}} \cr & = \frac{{{E_{{a_1}}} - {E_a}}}{{RT}} \cr & = \frac{{\left( {83.314 - 75} \right) \times {{10}^3}}}{{8.314 \times 500}} = 2 \cr & \log \,{k_2} = \frac{2}{{2.303}} = 0.868 \cr & {\text{Taking Antilog}} \cr & {k_2} = 7.38 \cr} $$

$$30\% \,\,{\text{decomposition means}}$$     $$X = 30\% \,\,{\text{of}}\,\,{R_0}$$
\[\begin{align} & \text{or}\,\,R={{R}_{0}}-0.3{{R}_{0}}=0.7{{R}_{0}} \\ & \text{For first order,} \\ & k=\frac{2.303}{t}\log \frac{\left[ {{R}_{0}} \right]}{\left[ R \right]} \\ & \,\,\,=\frac{2.303}{40}\log \frac{10}{7}{{\min }^{-1}} \\ & \,\,\,=8.918\times {{10}^{-3}}\,\min {{.}^{-1}} \\ & {{t}_{\frac{1}{2}}}=\frac{0.693}{k} \\ & \,\,\,\,\,\,=\frac{0.693}{8.918\times {{10}^{-3}}\min {{.}^{-1}}} \\ & \,\,\,\,\,\,=77.7\,\min . \\ \end{align}\]

$$\eqalign{ & {\text{For first order reaction,}} \cr & A \to B \cr & {\text{rate}} = k \times \left[ A \right] \cr & {\text{Rate}} = 2.0 \times {10^{ - 5}}mol\,{L^{ - 1}}{s^{ - 1}} \cr & \left[ A \right] = 0.01\,M \cr & {\text{So,}}\,\,{\text{2}}{\text{.0}} \times {\text{1}}{{\text{0}}^{ - 5}} = k \times 0.01 \cr & k = \frac{{2.0 \times {{10}^{ - 5}}}}{{0.01}}\,{s^{ - 1}} \cr & = 2.0 \times {10^{ - 3}}{s^{ - 1}} \cr & {\text{For first order reaction,}} \cr & {t_{\frac{1}{2}}} = \frac{{0.693}}{k} = \frac{{0.693}}{{2.0 \times {{10}^{ - 3}}}} \cr & = 346.5 \approx 347\,s \cr} $$

$$\eqalign{ & {t_{\frac{1}{2}}} = \frac{{0.693}}{k} = 2100\,s = 35\,\min \cr & {t_{75\% }} = 2{t_{\frac{1}{2}}} = 2 \times 35 = 70\,\min \cr} $$

NOTE: The rate of photochemical process varies with the intensity of absorption.
Since greater the intensity of absorbed light more photons will fall at a point, and further each photon causes one molecule to undergo reaction.

$${\text{Rate}} = k{\left[ A \right]^2}{\left[ B \right]^3} = a$$
When volume is reduced to one half then cone. of reactants will be doubled.
$$\eqalign{ & {\text{Rate}} = k{\left[ {2A} \right]^2}{\left[ {2B} \right]^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 32\,k{\left[ A \right]^2}{\left[ B \right]^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 32a \cr} $$