In a chemical reaction $$A$$ is converted into $$B.$$ The rates of reaction, starting with initial concentrations of $$A$$ as $$2 \times {10^{ - 3}}M$$   and $$1 \times {10^{ - 3}}M,$$   are equal to $$2.40 \times {10^{ - 4}}M{s^{ - 1}}$$    and $$0.60 \times {10^{ - 4}}M{s^{ - 1}}$$    respectively. The order of reaction with respect to reactant $$A$$ will be

Solution :
$$\eqalign{ & A \to B \cr & {\text{Initial concentration }}\,\,\,\,\,{\text{Rate of reaction}} \cr & {\text{2}} \times {10^{ - 3}}M\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2.40 \times {10^{ - 4}}M{s^{ - 1}} \cr & 1 \times {10^{ - 3}}M\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.60 \times {10^{ - 4}}M{s^{ - 1}} \cr & {\text{rate of reaction}} \cr & r = k{\left[ A \right]^x} \cr & {\text{where }}x = {\text{order of reaction}} \cr & {\text{hence}} \cr & 2.40 \times {10^{ - 4}} = k{\left[ {2 \times {{10}^{ - 3}}} \right]^x}\,\,\,......\left( {\text{i}} \right) \cr & 0.60 \times {10^{ - 4}} = k{\left[ {1 \times {{10}^{ - 3}}} \right]^x}\,\,\,.......\left( {{\text{ii}}} \right) \cr & {\text{On dividing }}eqn.{\text{(i) from }}eqn.{\text{ (ii) we get}} \cr & 4 = {\left( 2 \right)^x} \cr & \therefore \,\,x = 2 \cr & {\text{i}}{\text{.e}}{\text{. order of reaction = 2}} \cr} $$