Question

For the reaction; $4N{H}_{3\left(g\right)}+5O_{2\left(g\right)}\to 4N{O}_{\left(g\right)}+6H_2{O}_{\left(g\right)}$, the rate of reaction in terms of disappearance of $N{H}_3$ is $-\frac{d\left[N{H}_3\right]}{dt}$, then write the rate expression in terms of concentration of $O_2,NO$ and $H_{2}O$.

Answer 1

The given reaction can be written as:

$N{H}_3+\frac{5}{4}{O}_2\to NO+\frac{6}{4}{H}_{2}O$

Then, rate of disappearance of

$N{H}_3=-\frac{d\left[N{H}_3\right]}{dt}$

Also, rate of reaction $=-\frac{d\left[N{H}_3\right]}{dt}$

$=-\frac{5}{4}\frac{d\left[O_2\right]}{dt}=\frac{d\left[NO\right]}{dt}=\frac{6}{4}\frac{d\left[H_{2}O\right]}{dt}$