Given below are two statements : Statement I: The rate law for the reaction \(A+B \rightarrow C\) is rate \((r)=k[A]^2[B]\). When the concentration of both A and B is doubled, the reaction rate is increased " \(x\) " times. Statement II :
The figure is showing "the variation in concentration against time plot" for a " \(y\) " order reaction. The Value of \(x+y\) is ______

Solution

$\mathrm{r}=\mathrm{K}[\mathrm{A}]^2|\mathrm{~B}|$ if conc. are doubled $\begin{aligned} & \mathrm{r}^{\prime}=\mathrm{K}[2 \mathrm{~A}]^2[2 \mathrm{~B}]^1 \\ & \mathrm{r}^{\prime}=8 \mathrm{r} \Rightarrow \mathrm{x}=8 \end{aligned}$
$\begin{aligned} & \Rightarrow \text { Zero order, } y=0 \\ & x+y=8\end{aligned}$

Hence, the correct answer is 8.