$$\eqalign{ & {\text{For a zero order reaction}}{\text{.}} \cr & {\text{rate}} = k{\left[ A \right]^ \circ }\,i.e.,\,{\text{rate}} = k \cr & {\text{hence unit of}}\,k = M.{\sec ^{ - 1}} \cr & {\text{For a first order reaction}}{\text{.}} \cr & {\text{rate}} = k\left[ A \right] \cr & k = M.{\sec ^{ - 1}}/M = {\sec ^{ - 1}} \cr} $$

\[\begin{align} & \text{For the first order reaction,} \\ & \text{Rate}\,\left( \frac{dx}{dt} \right)=k\left[ A \right] \\ & \left[ A \right]=\text{concentration of reactant} \\ & k=\text{rate constant} \\ & \text{Given that,} \\ & \frac{dx}{dt}=1.5\times {{10}^{-2}}mol\,{{L}^{-1}}{{\min }^{-1}} \\ & k=?\,\text{and}\,\left[ A \right]=0.5\,M \\ & \therefore \,\,1.5\times {{10}^{-2}}=k\times 0.5 \\ & \therefore \,k=\frac{1.5\times {{10}^{-2}}}{0.5} \\ & =3\times {{10}^{-2}}{{\min }^{-1}} \\ & \text{For first order reaction,} \\ & \text{half-life period,} \\ & {{t}_{\frac{1}{2}}}=\frac{0.693}{k}=\frac{0.693}{3\times {{10}^{-2}}} \\ & \,\,\,\,\,\,=23.1\,\min \\ \end{align}\]

$$\eqalign{ & {\text{For the reaction,}}\,2N{O_2} \to 2NO + {O_2} \cr & - \frac{1}{2}\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ {NO} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}} \cr & - \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = 6 \times {10^{ - 12}}\,mol\,{L^{ - 1}}\,{s^{ - 1}} \cr & \frac{{d\left[ {{O_2}} \right]}}{{dt}} = 3 \times {10^{ - 12}}\,mol\,{L^{ - 1}}\,{s^{ - 1}} \cr} $$

Since for every $${10^ \circ }C$$  rise in temperature rate doubles for $${50^ \circ }C$$  rise in temp increase in reaction rate $$ = {2^5} = 32\,{\text{times}}$$

Rate of forward reaction decreases and rate of backward reaction increases with passage of time. At equilibrium both the rates become equal.

Average rate during the time interval 30-60 $$s.$$
$$\eqalign{ & {\text{Rate}} = - \frac{{{C_2} - {C_1}}}{{{t_2} - {t_1}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = - \frac{{\left( {0.17 - 0.31} \right)}}{{60 - 30}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.14}}{{30}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 4.67 \times {10^{ - 3}}\,mol\,{L^{ - 1}}\,{s^{ - 1}} \cr} $$

$$\eqalign{ & {H_2}{O_2}\left( {aq} \right) \to {H_2}O\left( {aq} \right) + \frac{1}{2}{O_2}\left( g \right) \cr & {\text{For a first order reaction}} \cr & k = \frac{{2.303}}{t}\log \frac{a}{{\left( {a - x} \right)}} \cr} $$
$${\text{Given}}\,a = 0.5,\left( {a - x} \right) = 0.125,$$       $$t = 50\,\min $$
\[\begin{align} & \therefore \,\,k=\frac{2.303}{50}\log \frac{0.5}{0.125} \\ & =2.78\times {{10}^{-2}}{{\min }^{-1}} \\ & r=k\left[ {{H}_{2}}{{O}_{2}} \right]=2.78\times {{10}^{-2}}\times 0.05 \\ & =1.386\times {{10}^{-3}}\text{mol}\,\text{mi}{{\text{n}}^{-1}} \\ & \text{Now} \\ & -\frac{d\left[ {{H}_{2}}{{O}_{2}} \right]}{dt}=\frac{d\left[ {{H}_{2}}O \right]}{dt}=\frac{2d\left[ {{O}_{2}} \right]}{dt} \\ & \therefore \,\,\frac{2d\left[ {{O}_{2}} \right]}{dt}=-\frac{d\left[ {{H}_{2}}{{O}_{2}} \right]}{dt} \\ & \therefore \,\frac{d\left[ {{O}_{2}} \right]}{dt}=-\frac{1}{2}\times \frac{d\left[ {{H}_{2}}{{O}_{2}} \right]}{dt} \\ & =\frac{1.386\times {{10}^{-3}}}{2} \\ & =6.93\times {{10}^{-4}}\text{mol}\,\text{mi}{{\text{n}}^{-1}} \\ \end{align}\]

For the reaction,
$$2A + B \to 3C + D$$
The reaction rate is written as follows :
The reaction rate with respect to $$A = - \frac{1}{2}\frac{{d\left[ A \right]}}{{dt}}$$
The reaction rate with respect to $$B = - \frac{{d\left[ B \right]}}{{dt}}$$
The reaction rate with respect to $$C = + \frac{1}{3}\frac{{d\left[ C \right]}}{{dt}}$$
The reaction rate with respect to $$D = \frac{{d\left[ D \right]}}{{dt}}$$
Hence, the answer (A) is not correct expression to represent the rate of the reaction.

From slow reaction
$${\text{Rate}} = k\left[ A \right]\left[ B \right]$$

Arrhenius equation, $$k = A{e^{ - \frac{{{E_a}}}{{RT}}}}$$
Given equation is $$k = \left( {4.5 \times {{10}^{11}}\,{s^{ - 1}}} \right){e^{ - \frac{{28000\,K}}{T}}}$$
Comparing both the equations, we get $$ - \frac{{{E_a}}}{{RT}} = - \frac{{28000\,K}}{T}$$
$$\eqalign{ & {E_a} = 28000\,K \times R \cr & \,\,\,\,\,\,\,\, = 28000\,K \times 8.314\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr & \,\,\,\,\,\,\,\, = 232.79\,kJ\,mo{l^{ - 1}} \cr} $$