Q44 / 150 · MHT CET 2023 (09 May Shift 2) · Chemistry
Find the average rate of formation $\mathrm{O}_{2(\mathrm{~g})}$ in the following reaction.
$\begin{aligned}
& 2 \mathrm{NO}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \\
& {\left[-\frac{\Delta[\mathrm{NO}]}{\Delta \mathrm{t}}\right]=\mathrm{x} \mathrm{moldm}^{-3} \mathrm{~s}^{-1}}
\end{aligned}$
- Correct answerA$\frac{x}{2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
- B$\mathrm{x} \mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
- C$2 \times \mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
- D$4 \times \mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
Solution
Average rate of reaction
$=-\frac{1}{2} \frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{O}_2\right]}{\Delta \mathrm{t}}$
Rate of formation of $\mathrm{O}_2=\frac{\Delta\left[\mathrm{O}_2\right]}{\Delta \mathrm{t}}$
$\begin{aligned}
& =\frac{1}{2} \frac{-\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}} \\
& =\frac{1}{2} \times \mathrm{x}=\frac{\mathrm{x}}{2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}
\end{aligned}$
Hence, the correct answer is option: 'A'.