Question

Ammonia and oxygen react at high temperature as:
$4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$
In an experiment , the rate of formation of $NO$ is $3.6\times {10}^{-3}M{s}^{-1}$. Calculate (a) the rate of disappearance of ammonia and (b) the rate of formation of water.

Answer 1

In this reaction, $-\frac{1}{4}\frac{dN{H}_3}{dt}=+\frac{1}{4}\frac{dNO}{dt}=+\frac{1}{6}\frac{d{H}_{2}O}{dt}$

(i)$\frac{dN{H}_3}{dt}=\frac{dNO}{dt}=3.6\times {10}^{-3}M{s}^{-1}$ i.e. rate of disappearance of $N{H}_3=3.6\times {10}^{-3}M{s}^{-1}$

(ii) $\frac{d{H}_{2}O}{dt}=\frac{3}{2}\frac{dNO}{dt}=\frac{3}{2}\times 3.6\times {10}^{-3}=5.4\times {10}^{-3}M{s}^{-1}$

i.e. rate of formation of $H_{2}O=5.4\times {10}^{-3}M{s}^{-1}$