Q20 / 150 · MHT CET 2025 (26 Apr Shift 1) · Chemistry
Rate of the reaction $\mathrm{A}+\mathrm{B} \rightarrow$ product is $3.6 \times 10^{-2} \mathrm{~mol~} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ and rate law is $\mathrm{r}=\mathrm{k}[\mathrm{A}][\mathrm{B}]^2$. What is rate constant of the reaction if $[\mathrm{A}]=0 \cdot 2 \mathrm{M}$ and $[\mathrm{B}]=0 \cdot 1 \mathrm{M}$ ?
- A$10 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}$
- Correct answerB$18 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}$
- C$24 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}$
- D$4.8 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}$
Solution
The rate law is given by $r = k[A][B]^2$.
Solving for the rate constant yields $k = \frac{r}{[A][B]^2}$.
Substituting the values $r = 3.6 \times 10^{-2} \> \text{mol} \> \text{dm}^{-3} \> \text{s}^{-1}$, $[A] = 0.2 \> \text{mol} \> \text{dm}^{-3}$, and $[B] = 0.1 \> \text{mol} \> \text{dm}^{-3}$ gives
$k = \frac{3.6 \times 10^{-2}}{0.2 \times (0.1)^2} = \frac{3.6 \times 10^{-2}}{0.2 \times 0.01} = \frac{3.6 \times 10^{-2}}{2 \times 10^{-3}} = 1.8 \times 10^{1} = 18 \> \text{mol}^{-2} \> \text{dm}^6 \> \text{s}^{-1}$.
This corresponds to option B.
Hence, the correct answer is option: 'B'.