Chemistry - Chemical Kinetics Question with Solution | TestHub

Let the initial rate be $R$ and order with respect to $A$ be $x$ and $B$ be $y$. Thus, rate lawcan be written as, rate, $R=[A{]}^x[B{]}^y$...(i) After doubling the concentration of $A$, rate becomes $4R$ $$4R=[2A{]}^x[B{]}^y...(ii)$$ After doubling the concentration of $B$, rate remains $R$, $R=[A{]}^x[2B\}$...(iii) From Eq. (i) and (ii), we get $$\frac{R}{4R}={\left(\frac{1}{2}\right)}^x\Rightarrow {\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^x$$ So, $x=2$ From Eq. (i) and (iii), we get $$\frac{R}{R}={\left[\frac{1}{2}\right]}^y\Rightarrow {\left(\frac{1}{1}\right)}^0={\left(\frac{1}{2}\right)}^y$$ So, $Y=0$ Hence, the rate lawis, rate $\mathrm{R}=[A{]}^2[B{]}^0$ This clearly shows that the order of this reaction is 2 and for second order reaction units of rate constant are ${\mathrm{mol}}^{-1}{\mathrm{Ls}}^{-1}$.