Question

In the first step of the Ostwald process for synthesis of nitric acid, ammonia is converted into nitric oxide by high-temperature reaction.
$4N{H}_3\left(g\right)+5O_2\left(g\right)⟶4NO\left(g\right)+6H_{2}O\left(g\right)$
(a) How is the rate of disappearance of $O_2$ related to the rate of disappearance of ${NH}_3$?
(b) How are the rates of appearance of $NO$ and $H_{2}O$ related to the rate of disappearance of ${NH}_3$?

Answer 1

a.) $\frac{-1}{4}\frac{d\left[N{H}_3\right]}{dt}=\frac{-1}{5}\frac{d\left[O_2\right]}{dt}=\frac{+1}{4}\frac{d\left[NO\right]}{dt}+\frac{1}{6}\frac{d\left[H_{2}O\right]}{dt}$

$\frac{-1}{5}\frac{d\left[O_2\right]}{dt}=\frac{-1}{4}\frac{d\left[N{H}_3\right]}{dt}$

$\frac{-d\left[O_2\right]}{dt}=\frac{-5}{4}\frac{d\left[N{H}_3\right]}{dt}$

b. $\frac{+1}{4}\frac{d\left[NO\right]}{dt}=\frac{-1}{4}\frac{d\left[N{H}_3\right]}{dt}$

$\frac{d\left[NO\right]}{dt}=\frac{-4}{4}\frac{d\left[N{H}_3\right]}{dt}=\frac{-d\left[N{H}_3\right]}{dt}$

$\frac{d\left[NO\right]}{dt}=\frac{-d\left[N{H}_3\right]}{dt}$

$\frac{1}{6}\frac{d\left[H_{2}O\right]}{dt}=\frac{-1}{4}\frac{d\left[N{H}_3\right]}{dt}\Rightarrow \frac{d\left[H_{2}O\right]}{dt}=\frac{6}{4}\frac{d\left[N{H}_3\right]}{dt}$

$\frac{d\left[H_{2}O\right]}{dt}=\frac{-3}{2}\frac{d\left[N{H}_3\right]}{dt}$