The decomposition of

The decomposition of NH3 on platinum surface is represented by the following equation.

$2 \mathrm{NH}_{3(g)} \stackrel{\mathrm{P}_{1}}{\longrightarrow} \mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)}$

Therefore,

Rate $=-\frac{1}{2} \frac{d\left[\mathrm{NH}_{3}\right]}{d t}=\frac{d\left[\mathrm{~N}_{2}\right]}{d t}=\frac{1}{3} \frac{d\left[\mathrm{H}_{2}\right]}{d t}$

However, it is given that the reaction is of zero order.

Therefore,

$-\frac{1}{2} \frac{d\left[\mathrm{NH}_{3}\right]}{d t}=\frac{d\left[\mathrm{~N}_{2}\right]}{d t}=\frac{1}{3} \frac{d\left[\mathrm{H}_{2}\right]}{d t}=k$

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

Therefore, the rate of production of N2 is

$\frac{d\left[\mathrm{~N}_{2}\right]}{d t}=2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

And, the rate of production of H2 is

$\frac{d\left[\mathrm{H}_{2}\right]}{d t}=3 \times 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

= 7.5 × 10−4 mol L−1 s−1