The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10−4 mol−1 L s−1?
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