Chemistry - Chemical Kinetics Question with Solution | TestHub

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