For the reaction $4 \mathrm{NH}_{3}+50_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$ if the…

For the reaction $4 \mathrm{NH}_{3}+50_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$ if the rate of disappearance of $\mathrm{NH}_{3}$ is $3.6 \times 10^{-3} \mathrm{M} / \mathrm{s}$. What is the rate of formation of water?
  1. $4.0 \times 10^{4} \mathrm{M} / \mathrm{S}$
  2. $3.6 \times 10^{-3} \mathrm{M} / \mathrm{S}$
  3. $6.0 \times 10^{-4} \mathrm{M} / \mathrm{S}$
  4. $5.4 \times 10^{-3} \mathrm{M} / \mathrm{S}$

Solution

$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$ Rate of reaction $=-\frac{1}{4} \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=-\frac{1}{5} \frac{\mathrm{d}\left[\mathrm{O}_{2}\right]}{\mathrm{dt}}=\frac{1 \mathrm{~d}[\mathrm{NO}]}{4}=\frac{1}{6} \frac{\mathrm{d}\left[\mathrm{H}_{2} \mathrm{O}\right]}{\mathrm{dt}}$ $\therefore$ Rate of formation of water, $\frac{\mathrm{d}\left[\mathrm{H}_{2} \mathrm{O}\right]}{\mathrm{dt}}=\frac{6}{4} \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}$ $=\frac{6}{4} \times 3.6 \times 10^{-3}=5.4 \times 10^{-3} \mathrm{M} / \mathrm{s}$

Asked in: MHT CET 2020 (14 Oct Shift 1)

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