$$\eqalign{ & {\text{For a first order reaction,}}\,A \to {\text{Products}} \cr & r = k\left[ A \right]\,{\text{or}}\,k = \frac{1}{{\left[ A \right]}} \cr & \Rightarrow k = \frac{{1.5 \times {{10}^{ - 2}}}}{{0.5}} = 3 \times {10^{ - 2}} \cr & {\text{Further}},\,{t_{\frac{1}{2}}} = \frac{{0.693}}{k} = \frac{{0.693}}{{3 \times {{10}^{ - 2}}}} = 23.1 \cr} $$

According to Arrhenius equation,
$$\eqalign{ & k = A{e^{ - \,\frac{{Ea}}{{RT}}}} \cr & \therefore \,\,{\text{when}}\,{E_a} = 0,k = A \cr} $$
Also $${\text{ln}}\,k\,\,{\text{vs}}\,\,\frac{1}{T}$$    is a straight line with slope $$ = - \frac{{{E_a}}}{R}.$$
∴ Statements (ii) and (v) are correct.

The catalyst decreases the activation energy of the reaction and thus increases the rate of reaction.

$$k = \frac{{2.303}}{t}\log \frac{a}{{a - x}}$$
For $$\frac{7}{8}$$  of the reaction to complete, $$t = \frac{7}{{8t}}$$
$$\eqalign{ & a - x = a - \frac{{7a}}{8} = \frac{a}{8} \cr & \therefore {t_{\frac{7}{8}}} = \frac{{2.303}}{k}\log \frac{a}{{\frac{a}{8}}} = \frac{{2.303}}{k}\log 8 \cr} $$
For half of the reaction to complete, $$t = {t_{\frac{1}{2}}},$$
$$\eqalign{ & x = a - \frac{a}{2} = \frac{a}{2} \cr & \therefore {t_{\frac{1}{2}}} = \frac{{2.303}}{k}\log \frac{a}{{\frac{a}{2}}} = \frac{{2.303}}{k}\log 2 \cr & \therefore \frac{{{t_{\frac{7}{8}}}}}{{{t_{\frac{1}{2}}}}} = \frac{{\log 8}}{{\log 2}} = \frac{{\log {2^3}}}{{\log 2}} = \frac{{3\log 2}}{{\log 2}} = 3 \cr} $$

$$_{11}^{23}Na \to _{10}^{23}X + _1^0\beta $$

$$A \to B\,\,{\text{For a first order reaction}}$$
$${\text{Given}}\,\,a = 0.8\,mol,$$     $$\left( {a - x} \right) = 0.8 - 0.6 = 0.2$$
$$k = \frac{{2.303}}{1}{\text{log}}\frac{{0.8}}{{0.2}}\,{\text{or}}\,k = 2.303\,{\text{log}}\,4$$
$${\text{again}}\,a = 0.9,$$   $$a - x = 0.9 - 0.675 = 0.225$$
$$\eqalign{ & k = \frac{{2.303}}{t}{\text{log}}\frac{{0.9}}{{0.225}} \cr & 2.303\,{\text{log}}\,4 = \frac{{2.303}}{t}{\text{log}}4 \cr & {\text{Hence}}\,\,t = 1\,{\text{hour}} \cr} $$

$$\eqalign{ & k = A.{e^{ - \,\frac{{{E_a}}}{{\left( {RT} \right)}}}} \cr & \therefore \,\,{\text{Effective overall energy of activation}} \cr & {E_a} = {E_a}\left( 2 \right) - {E_a}\left( 3 \right) + \frac{1}{2}{E_a}\left( 1 \right) - \frac{1}{2}{E_a}\left( 5 \right) \cr & = 60 - 50 + \frac{1}{2} \times 40 - \frac{1}{2} \times 10 \cr & = 25\,kJ/mol \cr} $$

$$\eqalign{ & t = \frac{{2.303}}{k}{\text{log}}\frac{a}{{a - x}} \cr & {\text{or}}\,\,t = \frac{{2.303}}{k}{\text{log}}a - \frac{{2.303}}{k}{\text{log}}\left( {a - x} \right) \cr & {\text{log}}\left( {a - x} \right) = \frac{{ - k}}{{2.303}}t + {\text{log}}\,a \cr} $$

$${\text{Rate}} = k{\left[ X \right]^0}{\left[ Y \right]^0}\,\,{\text{or}}\,\,{\text{rate = }}k$$

rate of appearance of $$HI = \frac{1}{2}\frac{{d\left[ {HI} \right]}}{{dt}}$$
rate of disappearance of $${H_2} = \frac{{ - d\left[ {{H_2}} \right]}}{{dt}}$$
rate of disappearance of $${I_2} = \frac{{ - d\left[ {{I_2}} \right]}}{{dt}}$$
hence $$\frac{{ - d\left[ {{H_2}} \right]}}{{dt}} = - \frac{{d\left[ {{I_2}} \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ {HI} \right]}}{{dt}}$$
or $$ - \frac{{2d\left[ {{H_2}} \right]}}{{dt}} = - \frac{{2d\left[ {{I_2}} \right]}}{{dt}} = \frac{{d\left[ {HI} \right]}}{{dt}}$$