The experimental data for the reaction $$2A + {B_2} \to 2\,AB$$    is

Exp.	$$\left[ A \right]$$	$$\left[ {{B_2}} \right]$$	Rate $$\left( {M\,{s^{ - 1}}} \right)$$
1.	0.50	0.50	$$1.6 \times {10^{ - 4}}$$
2.	0.50	1.00	$$3.2 \times {10^{ - 4}}$$
3.	1.00	1.00	$$3.2 \times {10^{ - 4}}$$

The rate equation for the above data is

Solution :
Consider the following rate law equation,
$$\eqalign{ & \frac{{dx}}{{dt}} = k{\left[ A \right]^m}{\left[ {{B_2}} \right]^n} \cr & 1.6 \times {10^{ - 4}} = k{\left[ {0.50} \right]^m}{\left[ {0.50} \right]^n}\,\,\,...{\text{(i)}} \cr & 3.2 \times {10^{ - 4}} = k{\left[ {0.50} \right]^m}{\left[ {1.0} \right]^n}\,\,\,...{\text{(ii)}} \cr & 3.2 \times {10^{ - 4}} = k{\left[ {1.00} \right]^m}{\left[ {1.0} \right]^n}\,\,\,...{\text{(iii)}} \cr & {\text{By dividing Eq}}{\text{. (iii) by (ii) we get,}} \cr & \frac{{3.2 \times {{10}^{ - 4}}}}{{3.2 \times {{10}^{ - 4}}}} = \frac{{k{{\left[ {1.00} \right]}^m}{{\left[ {1.0} \right]}^n}}}{{k{{\left[ {0.50} \right]}^m}{{\left[ {1.0} \right]}^n}}} \cr & 1 = {2^m}\,\,\,\,\,{\text{or}}\,\,\,\,{{\text{2}}^0} = {2^m} \cr & \therefore \,\,\,m = 0 \cr & {\text{By dividing Eq}}{\text{. (ii) by (i)}} \cr & \frac{{3.2 \times {{10}^{ - 4}}}}{{1.6 \times {{10}^{ - 4}}}} = \frac{{{{\left[ {0.50} \right]}^m}{{\left[ {1.0} \right]}^n}}}{{{{\left[ {0.50} \right]}^m}{{\left[ {0.50} \right]}^n}}} \cr & 2 = {2^n} \cr & {\text{or}}\,\,\,{{\text{2}}^1} = {2^n} \cr & \therefore \,\,\,n = 1 \cr & {\text{Hence rate,}} \cr & \left( {\frac{{dx}}{{dt}}} \right) = k{\left[ A \right]^0}{\left[ {{B_2}} \right]^1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = k\left[ {{B_2}} \right] \cr} $$