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?

\(4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}\)
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, \(\frac{\mathrm{d}\left[\mathrm{H}_{2} \mathrm{O}\right]}{\mathrm{dt}}=\frac{6}{4} \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}\)
\(
=\frac{6}{4} \times 3.6 \times 10^{-3}=5.4 \times 10^{-3} \mathrm{M} / \mathrm{s}
\)