Chemistry - Chemical Kinetics Question with Solution | TestHub

Let the order of reaction with respect to $A$ and $B$ is $x$ and $y$ respectively. So, the rate law can be given as $R=k[A{]}^x[B{]}^y$ When the concentration of only $B$ is doubled, the rate is doubled, so $R_1=k[A{]}^x[2B{]}^y=2R$ If concentrations of both the reactants $A$ and $B$ are doubled, the rate increases by a factor of 8 , so $\begin{array}{rl}{R}^{\prime \prime }=k[2A{]}^x[2B{]}^y& =8R\\ \Rightarrow k{2}^x{2}^y[A{]}^x[B{]}^y& =8R\end{array}$ From Eq. (i) and (ii), we get $\begin{array}{rlrl}& \Rightarrow & \frac{2R}{R}& =\frac{[A{]}^x[2B{]}^y}{[A{]}^x[B{]}^y}\\ & \therefore & =2^y& \\ & y& =1& \end{array}$ From Eq. (i) and (iv), we get $\Rightarrow \frac{8R}{R}=\frac{2^x{2}^y[A{]}^x[B{]}^y}{[A{]}^x[B{]}^y}$ or $8=2^x{2}^y$ Substitution of the value of $y$ gives, $\begin{array}{rl}& 8=2^x{2}^1\\ & 4=2^x\end{array}$ $\begin{array}{rlrl}& & (2{)}^2& =(2{)}^x\\ \therefore & & x& =2\end{array}$ Substitution of the value of $x$ and $y$ in Eq. (i) gives, $R=k[A{]}^2[B]$