Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

The value of rate constant of a pseudo first order reaction depends not only on temperature but also on concentration of reactant present in excess.

$$\eqalign{ & {\text{Given}}\,\left[ A \right] = 0.01\,M \cr & {\text{Rate}} = 2.0 \times {10^{ - 5}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr & {\text{For a first order reaction}} \cr & {\text{Rate}} = k\left[ A \right] \cr & k = \frac{{2.0 \times {{10}^{ - 5}}}}{{\left[ {0.01} \right]}} = 2 \times {10^{ - 3}} \cr & {t_{\frac{1}{2}}} = \frac{{0.693}}{{2 \times {{10}^{ - 3}}}} = 347\,\sec \cr} $$

In equation $$k = A{e^{ - \,\frac{{{E_a}}}{{RT}}}};A = $$    Frequency factor
$$k =$$  velocity constant, $$R =$$  gas constant and
$${E_a} = $$  energy of activation

TIPS/Formulae :
The sum of mass number and atomic numbers of reactants = The sum of mass number and atomic no. of products in a nuclear reaction. The given nuclear fission reaction is
$$_{92}^{235}U + _0^1n \to _{54}^{139}Xe + _{38}^{94}Sr + 3_0^1n$$

In graph (i), In [Reactant vs time is linear with positive intercept and negative slope. Hence it is 1^st order In graph (ii), [Reactant] vs time is linear with positive intercept and negative slope. Hence, it is zero order.

The rate equation depends upon the rate determining step. The given rate equation is only consistent with the mechanism (i).

$$\left[ R \right] = \left[ {{R_0}} \right] - kt$$
For completion of reaction, $$\left[ R \right] = 0$$  or $$t = \frac{{\left[ {{R_0}} \right]}}{k}$$

$$\eqalign{ & {\text{Given, initial temperature,}} \cr & {T_1} = 20 + 273 \cr & \,\,\,\,\,\,\, = 293\,K \cr & {\text{Final temperature,}} \cr & {T_2} = 35 + 273 \cr & \,\,\,\,\,\,\, = 308\,K \cr & R = 8.314\,J\,mo{l^{ - 1}}{K^{ - 1}} \cr} $$
Since, rate becomes double on raising temperature,
$$\eqalign{ & \therefore \,\,{r_2} = 2{r_1}\,\,\,{\text{or}}\,\,\,\frac{{{r_2}}}{{{r_1}}} = 2 \cr & {\text{As rate constant,}}\,k \propto r \cr & \therefore \,\,\frac{{{k_2}}}{{{k_1}}} = 2 \cr & {\text{From Arrhenius equation, we know that}} \cr & {\text{log}}\frac{{{k_2}}}{{{k_1}}} = - \frac{{{E_a}}}{{2.303\,R}}\left[ {\frac{{{T_1} - {T_2}}}{{{T_1}{T_2}}}} \right] \cr & {\text{log}}2 = - \frac{{{E_a}}}{{2.303 \times 8.314}}\left[ {\frac{{293 - 308}}{{293 \times 308}}} \right] \cr & 0.3010 = - \frac{{{E_a}}}{{2.303 \times 8.314}}\left[ {\frac{{ - 15}}{{293 \times 308}}} \right] \cr & \therefore \,\,{E_a} = \frac{{0.3010 \times 2.303 \times 8.314 \times 293 \times 308}}{{15}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 34673.48\,J\,mo{l^{ - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 34.7\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {N_2}{O_5}\left( g \right) \to 2\,N{O_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \cr & - \frac{d}{{dt}}\left[ {{N_2}{O_5}} \right] = + \frac{1}{2}\frac{d}{{dt}}\left[ {N{O_2}} \right] = 2\frac{d}{{dt}}\left[ {{O_2}} \right] \cr & \frac{d}{{dt}}\left[ {N{O_2}} \right] = 1.25 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{s^{ - 1}}\,{\text{and}} \cr & \frac{d}{{dt}}\left[ {{O_2}} \right] = 3.125 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$

For a zero order reaction $${t_{\frac{1}{2}}}$$  is directly proportional to the initial concentration of the reactant $${\left[ R \right]_0}$$
$$\eqalign{ & {t_{\frac{1}{2}}} \propto {\left[ R \right]_0} \cr & {\text{For a first order reaction}} \cr & k = \frac{{2.303}}{t}{\text{log}}\frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}} \cr & {\text{at}}\,\,\,\,{t_{\frac{1}{2}}},\left[ R \right] = \frac{{{{\left[ R \right]}_0}}}{2} \cr & {\text{So, the above equation becomes}} \cr & K = \frac{{2.303}}{{{t_{\frac{1}{2}}}}}{\text{log}}\frac{{{{\left[ R \right]}_0}}}{{\frac{{{{\left[ R \right]}_0}}}{2}}} \cr & {t_{\frac{1}{2}}} = \frac{{2.303}}{K} = {\text{log}}2 = \frac{{2.303}}{K} \times .3010 \cr & {t_{\frac{1}{2}}} = \frac{{.693}}{K} \cr} $$
ie., half life period is independent of initial concentration of a reactant.