$$\eqalign{ & {\text{By Arrhenius equation}} \cr & K = A{e^{ - \,\frac{{Ea}}{{RT}}}} \cr & {\text{where,}}\,{E_a} = {\text{energy of activation}} \cr & {\text{Applying log on both the side,}} \cr & {\text{ln}}\,k = {\text{ln}}A - \frac{{{E_a}}}{{RT}}\,\,\,...\left( {\text{i}} \right) \cr & {\text{or}}\,\,\,{\text{log}}\,k = - \frac{{{E_a}}}{{2.303\,RT}} + {\text{log}}\,A\,\,\,...\left( {{\text{ii}}} \right) \cr} $$
Compare the above equation $$w.r.t.$$  straight line equation of $$y = mx + c.$$   Thus, if a plote of $${\text{ln}}\,k\,\,{\text{vs}}\,\frac{1}{T}$$   is a straight line, the validity of the equation is confirmed.
Slope of the line $$ = - \frac{{{E_a}}}{R}$$
Thus, measuring the slope of the line, the value of $${{E_a}}$$ can be calculated.

$$A + 2B \to C$$
$$1\,mole$$  of $$A$$ reacts with $$2\,moles$$  of $$B$$  to give $$1\,mole$$  of $$C.$$
∴ $$5\,moles$$  of $$A$$ would react with $$10\,moles$$   of $$B$$  to give $$5\,moles$$  of $$C.$$
But, only $$8\,moles$$  of $$B$$  are available
∴ $$B$$  acts as a limiting reagent.
$$2\,moles$$  of $$B$$  gives $$1\,mole$$  of C
∴ $$8\,moles$$  of $$B$$  will give $$\frac{1}{2} \times 8 = 4\,moles$$    of $$C.$$

Arrhenius equation is given by
$$k = A{e^{ - \frac{{{E_a}}}{{RT}}}}$$
Taking log on both sides, we get
$$\log \,k = \log A - \frac{{{E_a}}}{{2.303RT}}$$
Arrhenius plot a graph between $$\log k$$  and $$\frac{1}{T}$$  whose slope is $$\frac{{ - {E_a}}}{{2.303R}}.$$

$$\eqalign{ & \log k = \log A - \frac{{{E_a}}}{{2.303RT}}\,\,...(1) \cr & {\text{Also}}\,{\text{given}}\,{\text{log}}k = 6.0 - \left( {2000} \right)\frac{1}{T}\,\,\,...(2) \cr & {\text{On comparing equations, (1) and (2)}} \cr & \log A = 6.0 \Rightarrow \,A = {10^6}{s^{ - 1}} \cr & and\,\frac{{{E_a}}}{{2.303\,R}} = 2000\,; \cr & \Rightarrow {E_a} = 2000 \times 2.303 \times 8.314 = 38.29\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {\text{From first order reaction,}} \cr & {\text{Rate constant,}} \cr & k = \frac{{2.303}}{t}{\log _{10}}\frac{a}{{\left( {a - x} \right)}} \cr & {k_1} = \frac{{2.303}}{{{t_1}}}{\text{log}}\frac{{{a_1}}}{{{a_1} - {x_1}}}\,\,\,...{\text{(i)}} \cr & {k_2} = \frac{{2.303}}{{{t_2}}}{\text{log}}\frac{{{a_2}}}{{{a_2} - {x_2}}}\,\,\,...{\text{(ii)}} \cr & {x_1} = \frac{{60}}{{100}}{a_1},{t_1} = 60 \cr & {x_2} = \frac{{50}}{{100}}{a_2},{t_2} = ? \cr & {\text{From Eqs}}{\text{. (i) and (ii)}} \cr} $$
$$\frac{{2.303}}{{{t_1}}}{\text{log}}\frac{{{a_1}}}{{{a_1} - {x_1}}}$$     $$ = \frac{{2.303}}{{{t_2}}}{\text{log}}\frac{{{a_2}}}{{{a_2} - {x_2}}}$$
$$\frac{{2.303}}{{60}}{\text{log}}\frac{{{a_1}}}{{\left( {{a_1} - \frac{{60}}{{100}}{a_1}} \right)}}$$      $$ = \frac{{2.303}}{{{t_2}}}{\text{log}}\frac{{{a_2}}}{{\left( {{a_2} - \frac{{50}}{{100}}{a_2}} \right)}}$$
$$\eqalign{ & \frac{{2.303}}{{60}}{\text{log}}\frac{{100{a_1}}}{{40{a_1}}} = \frac{{2.303}}{{{t_2}}}{\text{log}}\frac{{100{a_2}}}{{50{a_2}}} \cr & \frac{1}{{60}}{\text{log}}\frac{{100}}{{40}} = \frac{1}{{{t_2}}}{\text{log}}\frac{{100}}{{50}} \cr & {t_2} = \frac{{60\,{\text{log}}\frac{{100}}{{50}}}}{{{\text{log}}\frac{{100}}{{40}}}} = \frac{{60\left( {{\text{log}}\,10 - {\text{log}}\,5} \right)}}{{\left( {{\text{log}}\,10 - {\text{log}}\,4} \right)}} \cr & \,\,\,\,\,\, = \frac{{60\left( {1 - 0.69} \right)}}{{\left( {1 - 0.60} \right)}} = \frac{{60 \times 0.31}}{{0.40}} \cr & \,\,\,\,\,\, = 1.5 \times 31 = 46.5 \approx 45\min \cr} $$

Apart from energy considerations, the colliding molecules should have proper orientation for effective collisions.

Unit of rate constant for the reaction is
$$\eqalign{ & k = \frac{{{\text{rate}}}}{{\left[ {{H_2}} \right]{{\left[ {NO} \right]}^2}}} \cr & \,\,\,\, = \frac{{mol\,{L^{ - 1}}\,{s^{ - 1}}}}{{\left( {mol\,{L^{ - 1}}} \right){{\left( {mol\,{L^{ - 1}}} \right)}^2}}} \cr & \,\,\,\, = mo{l^{ - 2}}\,{L^2}\,{s^{ - 1}} \cr} $$

$$\frac{{0.693}}{{{t_{\frac{1}{2}}}}} = \frac{{2.303}}{t}\log \frac{a}{{a - x}}$$
$$\frac{{0.693}}{{10}} = \frac{{2.303}}{{100}}\log \frac{{100}}{{100 - x}}$$     $$ \Rightarrow x = 99.9\% $$

Given : 75% reaction gets completed in $$32\,\min $$
$$\eqalign{ & {\text{Thus,}}\,k = \frac{{2.303}}{t}{\text{log}}\frac{a}{{\left( {a - x} \right)}} \cr & = \frac{{2.303}}{{32}}{\text{log}}\frac{{100}}{{\left( {100 - 75} \right)}} \cr & = \frac{{2.303}}{{32}}{\text{log}}4 \cr} $$
  \[=0.0433\,{{\min }^{-1}}\]
Now we can use this value of $$k$$  to get the value of time required for 50% completion of reaction
$$\eqalign{ & t = \frac{{2.303}}{k}{\text{log}}\frac{a}{{\left( {a - x} \right)}} = \frac{{2.303}}{{0.0433}}{\text{log}}\frac{{100}}{{50}} \cr & = \frac{{2.303}}{{0.0433}}{\text{log}}\,2 = 16\,\min \cr} $$

Since the reaction is 2nd order $$w.r.t$$  $$CO,$$ Thus, rate law is given as.
$$r = k{\left[ {CO} \right]^2}$$
Let initial concentration of $$CO$$  is a i.e. $$[CO]=a$$
$$\therefore \,\,{r_1} = k{\left( a \right)^2} = k{a^2}$$
when concentration becomes doubled, i.e.$$[CO] =2a$$
$$\therefore \,\,{r_2} = k{\left( {2a} \right)^2} = 4k{a^2}\,\,\,\therefore {r_2} = 4{r_1}$$
So, the rate of reaction becomes 4 times.