Only those collisions in which molecules collide with sufficient energy, called threshold energy and proper orientation are effective collisions. Rest of the molecules collide and bounce back.

$$\eqalign{ & {t_{90\% }} = \frac{{2.303}}{k}{\text{log}}\frac{{100}}{{100 - 90}}\,\,\,\left( {\text{I}} \right) \cr & {t_{50\% }} = \frac{{2.303}}{k}{\text{log}}\frac{{100}}{{100 - 50}}\,\,\left( {{\text{II}}} \right) \cr & {\text{Dividing}}\,\,\frac{{{t_{90\% }}}}{{{t_{50\% }}}} = \frac{{{\text{log}}\,10}}{{{\text{log}}2}} \cr & \therefore \,\,{t_{90\% }} = 3.3{t_{50\% }} \cr} $$

$$\eqalign{ & {\text{Using the relation}} \cr & \left[ A \right] = {\left[ A \right]_0}{\left( {\frac{1}{2}} \right)^n}\left[ {n = {\text{number of half - lives}}} \right] \cr & T = n \times {t_{\frac{1}{2}}} \cr & {\text{Here,}}\,n = \frac{{7.5}}{{2.5}} = 3 \cr & \therefore \,\,\left[ A \right] = 160 \times {\left( {\frac{1}{2}} \right)^3} \cr & = 160 \times \frac{1}{8} \cr & = 20\,g \cr} $$

$$\log \frac{{{k_2}}}{{{k_1}}} = \frac{{{E_a}}}{{2.303R}}\left[ {\frac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right]$$

Not only sufficient threshold energy of colliding atoms or molecules but also the proper orientation for the collision is required for the formation of products.

It is bimolecular first order reaction since $${\text{Rate}} \propto \left[ {{N_2}{O_5}} \right]$$

$$\eqalign{ & {\text{The rates of reactions for the reaction}} \cr & \frac{1}{2}A \to 2B \cr & {\text{can be written either as}} \cr & - 2\frac{d}{{dt}}\left[ A \right]\,\,\,{\text{with respect to }}'A' \cr & or\,\,\frac{1}{2}\frac{d}{{dt}}\left[ B \right]\,\,\,{\text{with respect to }}'B' \cr & {\text{From the above, we have}} \cr & - 2\frac{d}{{dt}}\left[ A \right] = \frac{1}{2}\frac{d}{{dt}}\left[ B \right] \cr & or\,\, - \frac{d}{{dt}}\left[ A \right] = \frac{1}{4}\frac{d}{{dt}}\left[ B \right] \cr & {\text{i}}{\text{.e}}{\text{., correct answer is (B)}} \cr} $$

$$_{12}M{g^{24}}{ \to _{11}}N{a^{23}}{ + _1}{H^1}$$

The values of rate constants $${k_0},{k_1}$$ for zero order and first order reaction, respectively, are given by the following equation :
$${k_0} = \frac{{{A_0}}}{{2 \times {t_{\frac{1}{2}}}}}$$     [ where $${{A_0} = }$$  initial concentration, and $${{t_{\frac{1}{2}}} = }$$  half - life period ]
and $${k_1} = \frac{{0.693}}{{{t_{\frac{1}{2}}}}}$$
substituting various given values, we get
$$\eqalign{ & {k_0} = \frac{{{\text{1}}{\text{.386}}\,{\text{mol litr}}{{\text{e}}^{ - 1}}}}{{2 \times 20\,\sec }}\,\,...{\text{(i)}} \cr & {\text{and}}\,{k_1} = \frac{{0.639}}{{40\,\sec }}\,\,...({\text{ii)}} \cr & {\text{Dividing (ii) by (i), we get}} \cr & \frac{{{k_1}}}{{{k_0}}} = \frac{{0.639}}{{40}} \times \frac{{2 \times 20}}{{1.386}}mo{l^{ - 1}}{\text{litre}} \cr & = \frac{{0.639}}{{1.386}}mo{l^{ - 1}}{\text{litre}} \cr & = 0.5\,mo{l^{ - 1}}{\text{litre}} \cr & = 0.5\,mo{l^{ - 1}}d{m^3}\,\,\,\,\,\,\,\,\,\left[ {1\,{\text{litre}} = 1d{m^3}} \right] \cr & {\text{Thus the correct answer is (A)}}{\text{.}} \cr} $$

Rate constant is independent of concentration.