Let the order of reaction with respect to $A$ be $x$ and with respect to B be $y$. $\therefore \text{ Rate of reaction } = k[A]^{x}[B]^{y}$ According to details given , $4.2 \times 10^{-2} = k[0.2]^{x}[0.3]^{y}$ .....(1) $6.0 \times 10^{-3} = k[0.1]^{x}[0.1]^{y}$ .......(2) $1.68 \times 10^{-1} = k[0.4]^{x}[0.3]^{y}$ .......(3) $2.40 \times 10^{-2} = k[0.1]^{x}[0.4]^{y}$ ........(4) Dividing equation (4) by (2), we get $\frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = \frac{k[0.1]^{x}[0.4]^{y}}{k[0.1]^{x}[0.1]^{y}}$ $4 = \frac{[0.4]^{y}}{[0.1]^{y}}$ $(4)^1 = (4)^{y}$ $y = 1$ Dividing equation (1) by (3), we get $\frac{4.2 \times 10^{-2}}{1.68 \times 10^{-1}} = \frac{k[0.2]^{x}[0.3]^{y}}{k[0.4]^{x}[0.3]^{y}}$ $0.25 = \frac{[0.1]^{x}}{[0.2]^{x}}$ $(0.25) = (0.5)^{x}$ $(0.5)^2 = (0.5)^{x} \quad x = 2$ (a) So the rate of reaction with respect to $A$ is 2 and with respect to $B$ is 1 . (b) Rate law $= k[A]^2[B]$ Overall order of reaction is 3 . (c) Rate constant, $K = \frac{\text{Rate}}{[A]^2[B]}$ $= \frac{6.0 \times 10^{-3}}{(0.1)^2(0.1)}$ $= 6.0 \text{ mol}^{-2} \text{ L}^2 \text{ min}^{-1}$