Question

For a reaction $\mathrm{A}+\mathrm{B}\to \mathrm{C}$, the rate law is written as $\mathrm{r}=\mathrm{k}\left[\mathrm{A}{]}^2\right[\mathrm{B}]$. Doubling the concentrations of both of ' A ' and 'B' increases the rate of reaction by

• A. 2 times
• B. 4 times
• C. 8 times
• D. 16 times

Answer 1

The correct option is C

8 times

The explanation for the correct option C):

1. Given reaction is $\mathrm{A}+\mathrm{B}\to \mathrm{C}$, the rate law is written as $\mathrm{r}=\mathrm{k}\left[\mathrm{A}{]}^2\right[\mathrm{B}]$
2. Because the reaction is one of $2^{\text{nd}}$order with relation to ${\mathrm{A}}_{\mathrm{t}}$ and $1^{\text{st}}$ order in relation to B.
3. The rate is proportional to the reactant concentration.
4. On doubling the concentration of both A and B, the new rate is expressed as-

$$\begin{gathered} {\mathrm{r}}_1={\left[2\mathrm{A}\right]}^2\times \left[2\mathrm{B}\right]=4{\left[\mathrm{A}\right]}^2\times 2\left[\mathrm{B}\right] \\ =8{\left[\mathrm{A}\right]}^2\left[\mathrm{B}\right] \\ \therefore {\mathrm{r}}_1=8\mathrm{r} \end{gathered}$$

As a result, doubling the concentrations of A and B increases the rate by 8 times.

Conclusion: The correct option is (C) .