A and B react as per the reaction :

$\mathrm{A}\left(g\right)+2\mathrm{B}\left(g\right)⟶\mathrm{C}\left(g\right)$

From the data given below identify the Rate L aw.

[A] (M)	[B] (M)	$\text{ ROR }M{\sec }^{-1}$
0.3	0.4	$2\times {10}^{-3}$
0.6	0.8	$8\times {10}^{-3}$
0.6	0.4	$4\times {10}^{-8}$

 

Rate $R=k\left[A{]}^m\right[B{]}^n$

From Exp. (i) $R_1=2\times {10}^{-3}=k\left[0.3{]}^m\right[0.4{]}^n$

From Exp. (ii) $R_2=8\times {10}^{-3}=k\left[0.6{]}^m\right[0.8{]}^n$

From Exp. (iii) $R_3=4\times {10}^{-3}=k\left[0.6{]}^m\right[0.4{]}^n$

$\begin{array}{l}\frac{R_3}{R_1}\Rightarrow \frac{4\times {10}^{-3}}{2\times {10}^{-3}}=\frac{k[0.6{]}^m{0.4}^n}{k{\left[{0.3}^{m}0.\right]}^n}\\ 2^{\mathrm{\prime }}=2^m\therefore m=1\\ \frac{R_2}{R_3}\Rightarrow \frac{8\times {10}^{-3}}{4\times {10}^{-3}}=\frac{{\left.k[0.6{]}^{m}0.8\right]}^n}{k[0.6{]}^m{0.4}^n}\\ 2=2^n\therefore n=1\end{array}$

$\therefore$Rate law $\left(\mathrm{R}\right)=\mathrm{k}\left[\mathrm{A}\right]\left[\mathrm{B}\right]$