$$\eqalign{ & k = A{e^{ - \,\frac{{{E_a}}}{{RT}}}} \cr & \ln k = \ln A - \frac{{{E_a}}}{{RT}} \cr & {\text{For}}\,\ln \,k\,Vs\,\frac{1}{T} \cr & \ln \,A = {\text{intercept}} \cr & - \,\frac{{{E_a}}}{R} = {\text{slope}} = - 2 \times {10^4}K \cr & \therefore \,\,{E_a} = 8.3 \times 2 \times {10^4}\,J\,mo{l^{ - 1}} \cr & = 16.6 \times {10^4}\,J\,mo{l^{ - 1}}\,{\text{or}}\,\,166\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {\text{For the reaction,}} \cr & {N_2} + 3{H_2} \to 2N{H_3} \cr} $$
$${\text{Rate}} = - \frac{{d\left[ {{N_2}} \right]}}{{dt}} = - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$       $$ = + \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$
$$\eqalign{ & {\text{or}}\,\, - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}} = + \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}} \cr & - \frac{{d\left[ {{H_2}} \right]}}{{dt}} = \frac{3}{2} \times 2 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$

$$\eqalign{ & {\text{rate of appearance}}\,{\text{of}}\,HI = \frac{1}{2}\frac{{d\left[ {HI} \right]}}{{dt}} \cr & {\text{rate of formation of}}\,{H_2} = \frac{{ - d\left[ {{H_2}} \right]}}{{dt}} \cr & {\text{rate of formation of}}\,{{\text{I}}_2} = \frac{{ - d\left[ {{I_2}} \right]}}{{dt}} \cr & {\text{hence}}\,\frac{{ - d\left[ {{H_2}} \right]}}{{dt}} = \frac{{d\left[ {{I_2}} \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ {HI} \right]}}{{dt}} \cr & or - \frac{{2d\left[ {{H_2}} \right]}}{{dt}} = - \frac{{2d\left[ {{I_2}} \right]}}{{dt}} = \frac{{d\left[ {HI} \right]}}{{dt}} \cr} $$

$$\eqalign{ & C{H_3}CHO \to C{H_4} + CO \cr & {\text{Generally}}\,\,r \propto \left( {a - x} \right)m \cr & {r_1} = 1\,torr\,{s^{ - 1}},\,{\text{when}}\,{\text{5}}\% \,{\text{reacted}} \cr & {r_2} = 0.5\,torr\,{s^{ - 1}},\,{\text{when}}\,33\% \,{\text{reacted}} \cr & \left( {a - {x_1}} \right) = 0.95\left( {{\text{unreacted}}} \right) \cr & \left( {a - {x_2}} \right) = 0.67\left( {{\text{unreacted}}} \right) \cr & \frac{{{r_1}}}{{{r_2}}} = {\left[ {\frac{{\left( {a - {x_1}} \right)}}{{\left( {a - {x_2}} \right)}}} \right]^m};\frac{1}{{0.5}} = {\left( {\frac{{0.95}}{{0.67}}} \right)^m} \cr & 2 = {\left( {1.41} \right)^m} \Rightarrow 2 = {\left( {\sqrt 2 } \right)^m} \cr & \Rightarrow m = 2 \cr} $$

According to Arrhenius equation,
$$k = A.{e^{ - \,\frac{{{E_a}}}{{RT}}}}$$
where $$k = $$  rate constant
or $$\ell n\,k = \ell n\,A - \frac{{{E_a}}}{{RT}}$$
$$A =$$  frequency factor
$${E_a} = $$  Energy of activation.

\[\begin{align} & \text{Average rate} \\ & =-\frac{1}{2}\frac{\Delta \left[ HI \right]}{\Delta t} \\ & =-\frac{1}{2}\times \frac{0.4-0.5}{10} \\ & =\frac{1}{2}\times \frac{0.1}{10} \\ & =5\times {{10}^{-3}}M\,{{\min }^{-1}} \\ \end{align}\]

NOTE : Order is the sum of the power of the concentrations terms in rate law expression.
Hence the order of reaction is = 1 + 2 = 3

$$\eqalign{ & A + B \to {\text{Product}} \cr & {\text{Rate,}}\,\,r \propto {\left[ A \right]^x}{\left[ B \right]^y}\,\,\,...{\text{(i)}} \cr} $$
The rate decreases by a factor 4 if the concentration of reactant $$B$$ is doubled
$$\eqalign{ & \frac{r}{4} \propto {\left[ A \right]^x}{\left[ {2B} \right]^y}\,\,\,...{\text{(ii)}} \cr & {\text{From Eqs}}{\text{. (i) and (ii}}) \cr & 4 = {\left( {\frac{1}{2}} \right)^y} \cr & y = - 2 \cr} $$
Hence, order of reaction with respect to $$B$$ is $$-2.$$

$$\eqalign{ & {t_{\frac{1}{2}}} = \frac{{0.693}}{k} \cr & \,\,\,\,\,\,\, = \frac{{0.693}}{{200}} \cr & \,\,\,\,\,\,\, = 3.46 \times {10^{ - 3}}\,s \cr} $$

$$\eqalign{ & {\text{Rate of reaction}} \cr & = \frac{{{\text{Rate of disappearance / appearance}}}}{{{\text{Stoichiometric coefficient}}}} \cr & = - \frac{{d\left[ A \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ B \right]}}{{dt}} \cr & = 2.6 \times {10^{ - 2}} \cr & = - \frac{1}{2}\frac{{d\left[ B \right]}}{{dt}} \cr & 5.2 \times {10^{ - 2}} = - \frac{{d\left[ B \right]}}{{dt}} \cr} $$