Following data is for a reaction between reactants A and B : The...

Let the rate equation is Rate $=\mathrm{k}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$ Therefore, we can write $2 \times 10^{-3}=k[0.1] \times\left[0.1 \rrbracket^y\right.$...(i) $4 \times 10^{-3}=k[0.2]^x[0.1]^y$...(ii) $1.6 \times 10^{-2}=\mathrm{k}[0.2]^{\times}[0.2]^{\mathrm{y}}$...(iii) (ii) $\div$ (i); $\frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\frac{k[0.2]^x[0.1]^y}{k[0.1]^x[0.1]^y}$ $\Rightarrow \quad \frac{2}{1}=\frac{(0.2)^x}{(0.1)^x}=\left(\frac{2}{1}\right)^x$ $\therefore \quad x=1$ (ii) $\div$ (iii); $\frac{4 \times 10^{-3}}{1.6 \times 10^{-2}}=\frac{k[0.2]^{\mathrm{x}}[0.1]^y}{\mathrm{k}[0.2]^{\mathrm{x}}[0.2]^{\mathrm{y}}}$ $\Rightarrow \quad \frac{1}{4}=\frac{(0.1)^y}{(0.2)^y}=\left(\frac{1}{2}\right)^y$ $\therefore \quad y=2$ $\therefore \quad$ Rate $=k[A]^1[B]^2$ First order with respect to A while second order with respect to B .