Q32 / 155 · WBJEE 2020 · Chemistry
For a reaction $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{P}$, when concentration of $\mathrm{B}$ alone is doubled, rate does not change and when concentrations of both $\mathrm{A}$ and $\mathrm{B}$ is doubled, rate increases by a factor of 4 . The unit of rate constant is,
- A$\mathrm{s}^{-1}$
- Correct answerB$\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$
- C$\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$
- D$\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}$
Solution
Hint :
Rate law: $\mathrm{r}=\mathrm{k}[\mathrm{A}]^{\alpha}[\mathrm{B}]^{\beta} \quad \mathrm{B}$ follows first order kinetics. $\therefore \beta=1$
$\frac{r_{2}}{r_{1}}=\frac{[2 A]^{\alpha}[2 B]^{p}}{[A]^{\alpha}[B]^{\beta}}$
or, $4=2^{a+\beta}$
or, $\alpha+\beta=2$
or, $\alpha=1$
$\therefore$ overall order of reaction $=2$
Hence unit of rate constant $=\mathrm{L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{S}^{-1}$
Hence, the correct answer is option: 'B'.