During the kinetic study of the reaction, 2 A+B arrow C+D ,...

Let the order of reaction with respect to $\mathrm{A}$ is $x and with respect to$B$is$y$. Thus,$\text{ rate }=\mathrm{k}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$For the given cases, (I) rate$=\mathrm{k}(0.1)^{\mathrm{x}}(0.1)^{\mathrm{y}}=6.0 \times 10^{-3}$(II) rate$=\mathrm{k}(0.3)^{\mathrm{x}}(0.2)^{\mathrm{y}}=7.2 \times 10^{-2}$(III) rate =$\mathrm{k}(0.3)^{\mathrm{x}}(0.40)^{\mathrm{y}}$=2.88$\times$10^{-1} (IV) rate$=\mathrm{k}(0.4)^{\mathrm{x}}(0.1)^{\mathrm{y}}=2.40 \times 10^{-2}$On dividing Eq. (I) by (IV), we get$\left(\frac{0.1}{0.4}\right)^$x$\left(\frac{0.1}{0.1}\right)^$y=$\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}$or$\left(\frac{1}{4}\right)^$x=$\left(\frac{1}{4}\right)^$1 \therefore \quad$\mathrm{x}=$1 On dividing Eq. (II) by (III), we get$$\begin{aligned}& & \left(\frac{0.3}{0.3}\right)^x\left(\frac{0.2}{0.4}\right)^y & =\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}\\ \text{ or }& & \left(\frac{1}{2}\right)^y & =\frac{1}{4}\\ \text{ or }& & \left(\frac{1}{2}\right)^y & =\left(\frac{1}{2}\right)^2 \\ \therefore & & y & =2 \end{aligned}$$\therefore$ Thus, rate law is, \text{ rate }=\mathrm{k}[\mathrm{A}]^1[\mathrm{~B}]^2 \text{\text{ or } }=\mathrm{k}[\mathrm{A}][\mathrm{B}]^2