Chemistry - Chemical Kinetics Question with Solution | TestHub

For the reaction, $\mathrm{A}+\mathrm{B}⟶$ Products On doubling the initial concentration of A only, the rate of reaction is also doubled, therefore $$\text{ Rate }\propto [\mathrm{A}{]}^1$$ Let initially rate law is $$\text{ Rate }=k[A][B{]}^y$$ If concentration of and both are doubled, the rate gets changed by a factor of 8 . $$\begin{array}{r}8\times \text{ Rate }=\mathrm{k}[2\mathrm{A}][2\mathrm{B}{]}^{\mathrm{y}}\dots (\text{ iii) }\\ \left[\because \text{ Rate }\propto [\mathrm{A}{]}^1\right]\end{array}$$ Dividing Eq. (iii) by Eq. (ii), $$\begin{array}{rl}& 8=2\times 2^y\\ & 4=2^y\\ & \therefore \mathrm{y}=2\\ & \end{array}$$ Hence, rate law is, rate $=\mathrm{k}[\mathrm{A}][\mathrm{B}{]}^2$