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Question: Answered & Verified by Expert
Rate of reaction $\mathrm{x}+\mathrm{y} \rightarrow$ product is $5.4 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{sec}^{-1}$. When $[\mathrm{x}]=0.2 \mathrm{~mol} \mathrm{dm}^{-3}$ and $[\mathrm{y}]=0.1 \mathrm{~mol} \mathrm{dm}^{-3}$, calculate rate constant of reaction if it is first order in $\mathrm{X}$ and second order in $\mathrm{Y}$
ChemistryChemical KineticsMHT CETMHT CET 2022 (06 Aug Shift 1)
Options:
  • A $18 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{sec}^{-1}$
  • B $27 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{sec}^{-1}$
  • C $32 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{sec}^{-1}$
  • D $12 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{sec}^{-1}$
Solution:
1789 Upvotes Verified Answer
The correct answer is: $27 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{sec}^{-1}$
$\begin{aligned} & \mathrm{r}=-\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}=-\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}} \\ & \mathrm{r}=5.4 \times 10^{-2} \frac{\mathrm{mol}}{\mathrm{dm}^3-\mathrm{sec}} \\ & \mathrm{r}=\mathrm{k}[\mathrm{x}] \cdot[\mathrm{y}]^2 \\ & 5.4 \times 10^{-2}=\mathrm{K}(0.2)(0.1)^2 \\ & \mathrm{~K}=27 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{sec}^{-1}\end{aligned}$

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