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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
Options:
$$
4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \text {. }
$$
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
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Verified Answer
The correct answer is:
$3.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
$$
4 \mathrm{NH}_{3(\mathrm{~g})}+5O_{2(\mathrm{~g})} \longrightarrow 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})
$$
$\frac{\text { Rate of disappearance of } \mathrm{NH}_3}{4}=\frac{\text { Rate of formation of NO }}{4}$
Rate of disappearance of $\mathrm{NH}_3=3.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~S}^{-1}$
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 $\mathrm{NH}_3=3.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~S}^{-1}$
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