Compounds $$'A'$$ and $$'B'$$ react according to the following chemical equation.
$${A_{\left( g \right)}} + 2{B_{\left( g \right)}} \to 2{C_{\left( g \right)}}$$
Concentration of either $$'A'$$ or $$'B'$$ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.

Experiment	Initial concentration of $$\left[ A \right]/mol\,{L^{ - 1}}$$	Initial concentration of $$\left[ B \right]/mol\,{L^{ - 1}}$$	Initial rate of formation of $$\left[ C \right]/mol\,{L^{ - 1}}\,{s^{ - 1}}$$
1.	0.30	0.30	0.10
2.	0.30	0.60	0.40
3.	0.60	0.30	0.20

Solution :
Let order with respect to $$A$$  and $$B$$  are $$x$$  and $$y$$  respectively.
$$\eqalign{ & \therefore \,{\text{Rate}} = k{\left( A \right)^x}{\left( B \right)^y} \cr & \,\,\,\,\,0.1 = k{\left( {0.3} \right)^x}{\left( {0.3} \right)^y}...\left( {\text{i}} \right) \cr & \,\,\,\,\,0.4 = k{\left( {0.3} \right)^x}{\left( {0.6} \right)^y}...\left( {{\text{ii}}} \right) \cr & \,\,\,\,\,0.2 = k{\left( {0.6} \right)^x}{\left( {0.3} \right)^y}...\left( {{\text{iii}}} \right) \cr & {\text{Dividing}}\left( {{\text{ii}}} \right){\text{by}}\left( {\text{i}} \right)\,\frac{{0.4}}{{0.1}} = \frac{{{{\left( {0.6} \right)}^y}}}{{{{\left( {0.3} \right)}^y}}} \cr & \therefore \,y = 2 \cr & {\text{Dividing}}\left( {{\text{iii}}} \right){\text{by}}\left( {\text{i}} \right)\,\frac{{0.2}}{{0.1}} = \frac{{{{\left( {0.6} \right)}^x}}}{{{{\left( {0.3} \right)}^x}}} \cr & \therefore x = 1 \cr & {\text{Rate law will be}}\,{\text{:}}\,{\text{Rate}} = k{\left[ A \right]^1}{\left[ B \right]^2} \cr} $$