Suppose order with respect to A is m and with respect to B is n.Then the rate law will be Rate $=k[A]^{m}[B]^{n}$ Substituting the value of experiments I to IV, we get Expt. I: Rate $=6.0 \times 10^{-3}=k(0.1)^{m}(0.1)^{n}\;\;\;\;\;\;\;$...(i) Expt. II: Rate $=7.2 \times 10^{-2}=k(0.3)^{m}(0.2)^{n}\;\;\;\;\;\;$...(ii) Expt. III: Rate $=2.88 \times 10^{-1}=k(0.3)^{m}(0.4)^{n}\;\;\;$...(iii) Expt. IV: Rate $=2.4 \times 10^{-2}=k(0.4)^{m}(0.1)^{n}\;\;\;\;$...(iv) Comparing equation (i) and equation (iv) $\therefore\; \frac{(\text{Rate})_{I}}{(\text{Rate})_{IV}} = \frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}$ $= \frac{k(0.1)^{m}(0.1)^{n}}{k(0.4)^{m}(0.1)^{n}}$ or, $\frac{1}{4} = \frac{(0.1)^{m}}{(0.4)^{m}} = \left(\frac{1}{4}\right)^{m} \;\;\;\therefore m=1$ Comparing equation(ii) and equation (iii) $\frac{(\text{Rate})_{II}}{(\text{Rate})_{III}} = \frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}$ $= \frac{k(0.3)^{m}(0.2)^{n}}{k(0.3)^{m}(0.4)^{n}}$ or, $\left(\frac{1}{2}\right)^{2} = \frac{(0.2)^{n}}{(0.4)^{n}} = \left(\frac{1}{2}\right)^{n} \;\;\;\therefore n=2$ Rate law expression is : Rate =$k[A][B]^{2}$. The rate constant can be calculated from the given data of any experimentusing expression: $\;\;\;\;\;k = \frac{\text{Rate}}{[A][B]^{2}}$ From Expt. I, $k = \frac{6.0 \times 10^{-3}}{0.1 \times (0.1)^{2}} = 6.0$ $\therefore\;$Rate constant $k = 6.0 \, \mathrm{mol}^{-2} \, \mathrm{L}^{2} \, \mathrm{min}^{-1}$ Unit of $k, k = \frac{\text{Rate}}{[A][B]^{2}}$ $= \frac{\mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{min}^{-1}}{(\mathrm{mol} \, \mathrm{L}^{-1})(\mathrm{mol} \, \mathrm{L}^{-1})^{2}}$ $= \mathrm{mol}^{-2} \, \mathrm{L}^{2} \, \mathrm{min}^{-1}$