Q4 / 150 · MHT CET 2023 (13 May Shift 1) · Chemistry
Find the rate of formation of $\mathrm{NO}_{2(\mathrm{~g})}$ in the following reaction.
$$
\begin{aligned}
& 2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\
& {\left[\frac{-\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=0.02 \mathrm{~mol} \mathrm{dm}^{-3}\right]}
\end{aligned}
$$
- A$0.01 \mathrm{~mol} \mathrm{dm}^{-3}$
- B$0.02 \mathrm{~mol} \mathrm{dm}^{-3}$
- C$0.03 \mathrm{~mol} \mathrm{dm}^{-3}$
- Correct answerD$0.04 \mathrm{~mol} \mathrm{dm}^{-3}$
Solution
$\begin{aligned} \text { Rate of reaction } & =-\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \\ & =+\frac{1}{4} \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}\end{aligned}$
Rate of formation of $\begin{aligned} \mathrm{NO}_2 & =\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}} \\ & =-\frac{4}{2} \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \\ & =-2 \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \\ & =2 \times 0.02 \\ & =0.04 \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}$
Hence, the correct answer is option: 'D'.