\[\begin{align} & k=\frac{2.303}{t}\log \frac{a}{a-x}\,\,\text{(for first order)} \\ & k=\frac{2.303}{12}\log \frac{1}{\frac{1}{8}} \\ & \,\,\,=\left( \frac{2.303}{12}\log 8 \right){{\min }^{-1}} \\ \end{align}\]

$$ - \frac{1}{4}\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = + \frac{1}{6}\frac{{d\left[ {{H_2}O} \right]}}{{dt}}$$
$$\frac{{d\left[ {{H_2}O} \right]}}{{dt}} = \frac{6}{4} \times 3.6 \times {10^{ - 3}}$$      $$ = 5.4 \times {10^{ - 3}}\,mol\,{L^{ - 1}}\,{s^{ - 1}}$$

$$\eqalign{ & {\text{Let rate ofreaction}} = \frac{{d\left[ C \right]}}{t} = k{\left[ A \right]^x}{\left[ B \right]^y} \cr & {\text{Now from the given data}} \cr & 1.2 \times {10^{ - 3}} = k{\left[ {0.1} \right]^x}{\left[ {0.1} \right]^y}\,\,\,......\left( {\text{i}} \right) \cr & 1.2 \times {10^{ - 3}} = k{\left[ {0.1} \right]^x}{\left[ {0.2} \right]^y}\,\,\,......\left( {{\text{ii}}} \right) \cr & 2.4 \times {10^{ - 3}} = k{\left[ {0.2} \right]^x}{\left[ {0.1} \right]^y}\,\,\,......\left( {{\text{iii}}} \right) \cr & {\text{Dividing equation (i) by (ii)}} \cr & \Rightarrow \,\,\frac{{1.2 \times {{10}^{ - 3}}}}{{1.2 \times {{10}^{ - 3}}}} = \frac{{k{{\left[ {0.1} \right]}^x}{{\left[ {0.1} \right]}^y}}}{{k{{\left[ {0.1} \right]}^x}{{\left[ {0.2} \right]}^y}}} \cr & {\text{We find, }}y = 0 \cr & {\text{Now dividing equation (i) by (iii)}} \cr & \Rightarrow \,\,\frac{{1.2 \times {{10}^{ - 3}}}}{{2.4 \times {{10}^{ - 3}}}} = \frac{{k{{\left[ {0.1} \right]}^x}{{\left[ {0.1} \right]}^y}}}{{k{{\left[ {0.2} \right]}^x}{{\left[ {0.1} \right]}^y}}} \cr & {\text{We find,}}\,x = 1 \cr & {\text{Hence}}\,\,\frac{{d\left[ C \right]}}{{dt}} = k{\left[ A \right]^1}{\left[ B \right]^0} \cr} $$

$$a - x = \frac{{80}}{{100}} \times a = 0.8a,$$     $$t = 10\,\min .$$
$$k = \frac{{2.303}}{t}\log \frac{a}{{a - x}}$$     $$ = \frac{{2.303}}{{10}}\log \frac{a}{{0.8a}}$$
$$k = 0.2303\,\log \frac{{10}}{8}$$     $$ = 0.2303\left( {1 - 0.9030} \right)$$
$$\,\,\,\, = 0.2303 \times 0.0970$$     \[=0.0223\,\min {{.}^{-1}}\]

The graph show that reaction is exothermic.
$$\log k = \frac{{ - \Delta H}}{{RT}} + I$$
For exothermic reaction $$\Delta H < 0$$
$$\therefore \,\,\log \,k\,\,{\text{Vs}}\,\,\frac{1}{T}$$    would be negative straight line with positive slope.

$$\eqalign{ & {t_{\frac{1}{2}}} = \frac{{0.693}}{k} = \frac{{0.693}}{{0.0693}} = 10\,\min \cr & {\text{Reactant after }}10\min = 5\,mol \cr} $$
\[\text{Rate}\left( \frac{dx}{dt} \right)=k\left[ A \right]=0.0693\times 5\,mol\,{{\min }^{-1}}\]

Overall order = sum of orders w.r.t each reactant.
Let the order be $$x$$ and $$y$$ for $$G$$ and $$H$$ respectively

Exp. No.	$$\left[ G \right]mole\,\,{\text{litr}}{{\text{e}} ^{ - 1}}$$	$$\left[ H \right]mole\,\,{\text{litr}}{{\text{e}} ^{ - 1}}$$	$${\text{rate}}\left( {mole\,{\text{litr}}{{\text {e}}^ - }{\text{tim}}{{\text{e}}^{ - 1}}} \right)$$
$$1$$	$$a$$	$$b$$	$$r$$
$$2$$	$$2a$$	$$2b$$	$$8r$$
$$3$$	$$2a$$	$$b$$	$$2r$$

Applying $$r = k{\left[ G \right]^x}{\left[ H \right]^y}$$   we get, $$x=1,$$ $$y=2$$
$$\because $$ For (1) and (3), the rate is doubled when conc. of $$G$$ is doubled keeping that of $$H$$ constant i.e., rate $$ \propto $$ $$[G]$$
∴ $$x=1$$
From (2) and (3), $$y=2$$
∴ Overall order is $$3.$$

$$\eqalign{ & pH = 2;\,{r_1} = k \times {\left( {{{10}^{ - 2}}} \right)^n}\,\,\left\{ {Q\left[ {{H^ + }} \right] = {{10}^{ - pH}}} \right\} \cr & pH = 1;{r_2} = k \times {\left( {{{10}^{ - 1}}} \right)^n} \cr & {\text{Given}}\,\,{r_2} = 100\,{r_1} \cr & \Rightarrow \left( {\frac{{{{10}^{ - 1}}}}{{{{10}^{ - 2}}}}} \right) = 100 \cr & {10^n} = 100 \cr & \therefore \,\,n = 2 \cr} $$

Plot given is for zero order reaction.

$$\eqalign{ & {\text{Suppose activity of safe working}} = A \cr & {\text{Given}}\,{A_0} = 10A \cr & \lambda = \frac{{0.693}}{{{t_{\frac{1}{2}}}}} = \frac{{0.693}}{{30}} \cr & {t_{\frac{1}{2}}} = \frac{{2.303}}{\lambda }\log \frac{{{A_0}}}{A} = \frac{{2.303}}{{\frac{{0.693}}{{30}}}}\log \frac{{10A}}{A} \cr & = \frac{{2.303 \times 30}}{{0.693}} \times \log 10 = 100\,{\text{days}}{\text{.}} \cr} $$