Question

The decomposition of $N{H}_3$ is studied on platinum surface. The rate of production of $N_2$ and $H_2$ are respectively $(k=2.5\times {10}^{-4}mol{L}^{-1}{s}^{-1})$:

• A. $5\times {10}^{-4}$ and $3.75\times {10}^{-4}$
• B. $1.67\times {10}^{-4}$ and $5\times {10}^{-4}$
• C. $1.25\times {10}^{-4}$ and $3.75\times {10}^{-4}$
• D. $3.75\times {10}^{-4}$ and $1.25\times {10}^{-4}$

Answer 1

The correct option is C $1.25\times {10}^{-4}$ and $3.75\times {10}^{-4}$

$N{H}_3\to \frac{1}{2}{N}_2+\frac{3}{2}{H}_2$
$\frac{-d\left[N{H}_3\right]}{dt}=k[N{H}_3{]}^0=2.5\times {10}^{-4}=2\frac{d\left[N_2\right]}{dt}=\frac{2}{3}\frac{d\left[H_2\right]}{dt}$.
$\frac{d\left[N_2\right]}{dt}=1.25\times {10}^{-4}$
$\frac{d\left[H_2\right]}{dt}=3.75\times {10}^{-4}$