Although a catalyst speeds up the reaction but it does not shift the position of equilibrium. This is due to the fact that the presence of catalyst reduces the height of barrier by providing an alternative path for the reaction and lowers the activation energy. However, the lowering in activation energy is to the same extent for the forward as well as the backward reaction.

$$\eqalign{ & {t_{\frac{1}{2}}} = \frac{{0.693}}{k} \cr & \,\,\,\,\,\,\, = \frac{{0.693}}{{2 \times {{10}^{ - 2}}}} \cr & \,\,\,\,\,\,\, = 34.65\,\min . \cr} $$

$$\eqalign{ & k = 1.2 \times {10^{14}}{e^{ - \frac{{25000}}{{RT}}}}{\sec ^{ - 1}}\,\,{\text{or}} \cr & {\text{log}}\,k = {\text{log}}\,1.2 \times {10^{14}} - \frac{{25000}}{R}.\frac{1}{T} \cr} $$

$$\eqalign{ & {\text{Equation of straight line}} \cr & {\text{slope}} = - \frac{{25000}}{R} \cr} $$

\[\begin{align} & \frac{E_{a}^{'}}{{{T}_{1}}}=\frac{E_{a}^{'}}{{{T}_{2}}}=\frac{10}{300}=\frac{20}{{{T}_{2}}} \\ & \therefore \,\,{{T}_{2}}=600\,K={{327}^{\circ }}C \\ \end{align}\]

$${t_{\frac{1}{4}}} = \frac{{2.303}}{K}\log \frac{1}{{\frac{3}{4}}} = \frac{{2.303}}{K}\log \frac{4}{3}$$
$$ = \frac{{2.303}}{K}\left( {{\text{log}}4 - \log 3} \right) = $$      $$\frac{{2.303}}{K}\left( {2\log 2 - \log 3} \right)$$
$$ = \frac{{2.303}}{K}\left( {2 \times 0.301 - 0.4771} \right) = \frac{{0.29}}{K}$$

\[\begin{align} & \text{Average rate} \\ & =-\frac{\Delta \left[ R \right]}{\Delta t} \\ & =\frac{\left( {{\left[ R \right]}_{2}}-{{\left[ R \right]}_{1}} \right)}{{{t}_{2}}-{{t}_{1}}} \\ & =-\frac{\left( 0.04-0.05 \right)}{30} \\ & =\frac{0.01}{30} \\ & =3.3\times {{10}^{-4}}M\,{{\min }^{-1}} \\ \end{align}\]

$$\eqalign{ & {\text{According to Arrhenius equation}} \cr & \log \frac{{{k_2}}}{{{k_1}}} = \frac{{{E_a}}}{{2.303\,R}}\left( {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right) \cr & \log \frac{{1.3 \times {{10}^{ - 3}}}}{{1.3 \times {{10}^{ - 4}}}} \cr & = \frac{{{E_a}}}{{2.303 \times 8.314}}\left[ {\frac{1}{{373}} - \frac{1}{{423}}} \right] \cr & 1 = \frac{{{E_a}}}{{2.303 \times 8.314}}\left[ {\frac{1}{{373}} - \frac{1}{{423}}} \right] \cr & {E_a} = 60\,kJ/mol \cr} $$

For a first order reaction $${t_{\frac{1}{2}}} = \frac{{0.693}}{K}$$   i.e. for a first order reaction $${t_{\frac{1}{2}}}$$ does not depend up on the concentration. From the given data, we can say that order of reaction with respect to $$B = 1$$  because change in concentration of $$B$$ does not change half life. Order of reaction with respect to $$A = 1$$  because rate of reaction doubles when concentration of B is doubled keeping concentration of A constant.
∴ Order of reaction $$= 1 + 1 = 2$$   and units of second order reaction are $$L\,mo{l^{ - 1}}\,{\sec ^{ - 1}}.$$

Radioactive decay follows first order kinetics.
$$\eqalign{ & {\text{Therefore,}} \cr & {\text{Decay constant}}\left( \lambda \right) = \frac{{0.693}}{{{t_{\frac{1}{2}}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.693}}{{5730}} \cr & {\text{Given,}}\left[ {{R_0}} \right] = 100 \cr & \therefore \,\,\left[ R \right] = 80 \cr & {\text{and}}\,\,t = \frac{{2.303}}{\lambda }\log \frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}} \cr & = \frac{{2.303}}{{\left( {\frac{{0.693}}{{5730}}} \right)}}\log \frac{{100}}{{80}} \cr & = \frac{{2.303 \times 5730}}{{0.693}} \times 0.0969 \cr & = 1845\,{\text{years}} \cr} $$

Diagram (C) represents exothermic reaction with
$$\eqalign{ & {E_a} = 200 - 150 \cr & \,\,\,\,\,\,\,\, = 50\,kJ\,mo{l^{ - 1}} \cr & {\text{and}}\,\,\Delta H = 50 - 150 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 100\,kJ\,mo{l^{ - 1}} \cr} $$