Question

The rate for the decomposition of $N{H}_3$ on platinum surface is zero order. What are the rate of production of $N_2$ and $H_2$ if $K=2.5\times {10}^{-4}mollitr{e}^{-1}{s}^{-1}$.

Answer 1

$2N{H}_3\to N_2+3H_2$
Rate of reaction $=-\frac{1}{2}\frac{d}{dt}\left[N{H}_3\right]=\frac{d}{dt}\left[N_2\right]=\frac{1}{3}\frac{d}{dt}\left[H_2\right]=k$

For a zero order reaction,
Rate $=k=2.5\times {10}^{-4}M/s$

Rate of production of $N_2=\frac{d}{dt}\left[N_2\right]=2.5\times {10}^{-4}M/s$
Rate of production of $H_2=\frac{d}{dt}\left[H_2\right]=3\times 2.5\times {10}^{-4}M/s=7.5\times {10}^{-4}M/s$