Question

The decomposition of $N{H}_3$ on platinum surface is zero order reaction. What are the rates of production of N_2 and H_2 if $k=2.5\times {10}^{-4}mo{l}^{-1}L{s}^{-1}$ ?

Answer 1

$2N{H}_3\to N_2+3H_2$

By dividing the equation by 2

$N{H}_3\to \frac{1}{2}{N}_2+\frac{3}{2}{H}_2$

Rate $=-\frac{d\left[N{H}_3\right]}{dt}=+\frac{2d\left[N_2\right]}{dt}=\frac{2}{3}\frac{d\left[H_2\right]}{dt}$

For zero order reaction, rate = k

So, $\frac{-d\left[N{H}_3\right]}{dt}=\frac{2d\left[N_2\right]}{dt}=\frac{2}{3}\frac{d\left[H_2\right]}{dt}$

$=2.5\times {10}^{-}4mol{L}^{-1}{s}^{-1}$

$\therefore$ Rate of production of $N_2=\frac{d\left[N_2\right]}{dt}$

$=\frac{(2.5\times {10}^{-4}mol{L}^{-1}{s}^{-1})}{2}$

$=1.25\times {10}^{-4}mol{L}^{-1}{s}^{-1}$

$\therefore$ Rate of production of

$H_2=\frac{d\left[H_2\right]}{dt}=\frac{3}{2}\times (2.5\times {10}^{-4}mol{L}^{-1}{s}^{-1})$

$=3.75\times {10}^{-4}mol{L}^{-1}{s}^{-1}$