The following results were obtained during kinetic studies of the reaction ;

$$2A + B\,\,\,\,{\text{Products}}$$
Experiment	$$\left[ A \right]\left( {{\text{in}}\,mol\,{L^{ - 1}}} \right)$$	$$\left[ B \right]\left( {{\text{in}}\,mol\,{L^{ - 1}}} \right)$$	Initial Rate of reaction $$\left( {{\text{in}}\,mol\,{L^{ - 1}}\,{{\min }^{ - 1}}} \right)$$
I	0.10	0.20	$$6.93 \times {10^{ - 3}}$$
II	0.10	0.25	$$6.93 \times {10^{ - 3}}$$
III	0.20	0.30	$$1.386 \times {10^{ - 2}}$$

The time (in minutes) required to consume half of $$A$$ is :

Solution :
From experiment I and II, it is observed that order of reaction $$w.r.t.$$  $$(c)$$ is zero.
From experiment II and III, $$\alpha $$ can be calculated as :
$$\eqalign{ & \frac{{1.386 \times {{10}^{ - 2}}}}{{6.93 \times {{10}^{ - 3}}}} = {\left( {\frac{{0.2}}{{0.1}}} \right)^\alpha }\,\,\therefore \,\,\alpha = 1 \cr & {\text{Now,}}\,\,{\text{Rate}} = \,K{\left[ A \right]^1} \cr & {\text{or}},\,\,6.93 \times {10^{ - 3}} = K\left( {0.1} \right) \cr & K = 6.93 \times {10^{ - 2}} \cr & {\text{For the reaction,}}\,{\text{2A + B}} \to {\text{Products}} \cr & {\text{2}}Kt = \ln \frac{{{{\left[ A \right]}_0}}}{{\left[ A \right]}} \cr & \therefore \,\,{t_{\frac{1}{2}}} = \frac{{0.693}}{{2K}} = \frac{{0.693}}{{0.693 \times {{10}^{ - 2}} \times 2}} \cr & {t_{\frac{1}{2}}} = 5 \cr} $$