The decomposition of $NH_{3}$ on a platinum surface is a zero-order reaction. What are the rates of production of $N_{2}$ and $H_{2}$ if $k = 2.5 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$?

The decomposition of $NH_{3}$ on a platinum surface is represented by the following equation:
$2NH_{3(g)} \xrightarrow{Pt} N_{2(g)} + 3H_{2(g)}$
The rate of reaction is given by:
$Rate = -\frac{1}{2} \frac{d[NH_{3}]}{dt} = \frac{d[N_{2}]}{dt} = \frac{1}{3} \frac{d[H_{2}]}{dt} = k$
Given that the reaction is zero order,the rate of reaction is equal to the rate constant $k = 2.5 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$.
Therefore,the rate of production of $N_{2}$ is:
$\frac{d[N_{2}]}{dt} = k = 2.5 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$
And the rate of production of $H_{2}$ is:
$\frac{d[H_{2}]}{dt} = 3 \times k = 3 \times 2.5 \times 10^{-4} \, mol \, L^{-1} \, s^{-1} = 7.5 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$