Q14 / 150 · MHT CET 2021 (24 Sep Shift 1) · Chemistry
Ammonia and oxygen react at high temperature as
$$
4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \text {. }
$$
If rate of formation of $\mathrm{NO}_{(\mathrm{g})}$ is $3.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ then rate of disappearance of ammonia is
- A$7.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
- B$1.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
- C$2.4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
- Correct answerD$3.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
Solution
$$
4 \mathrm{NH}_{3(\mathrm{~g})}+5O_{2(\mathrm{~g})} \longrightarrow 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})
$$
$\frac{\text { Rate of disappearance of } \mathrm{NH}_3}{4}=\frac{\text { Rate of formation of NO }}{4}$
Rate of disappearance of $\mathrm{NH}_3=3.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~S}^{-1}$
Hence, the correct answer is option: 'D'.