Question

Ammonia and oxygen reacts at higher 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 concentration of NO increases by $1.08\ast {10}^{-2}mol{litre}^{-1}$ in 3 seconds.
Calculate,
(1) rate of reaction
(2) rate of disappearance of ammonia
(3) rate of formation of water

Answer 1

$4{NH}_3\left(g\right)+5O_2\left(g\right)⟶4NO\left(g\right)+6H_{2}O\left(g\right)$

rate $=\frac{1}{4}\frac{\left[△{NH}_3\right]}{△t}=\frac{1}{4}\frac{△\left[NO\right]}{△t}=\frac{1}{6}\frac{△\left[H_{2}O\right]}{△E}$

rate $=\frac{1}{4}\times \frac{1.08\times {10}^{-2}}{3}=0.09\times {10}^{-2}=9\times {10}^{-4}$

rate of dissappearance of ${NH}_3$ = rate of formation of $NO=\frac{1.08\times {10}^{-2}}{3}$

rate of formation of $H_{2}O\Rightarrow \frac{1}{6}\frac{\left[△H_{2}O\right]}{△t}=\frac{1}{4}\frac{△\left[NO\right]}{△t}$

$\frac{△\left[H_{2}O\right]}{△t}=\frac{6}{4}\frac{△\left[NO\right]}{△t}=\frac{6}{4}\times \frac{1.08\times {10}^{-2}}{3}=0.54\times {10}^{-2}$