Question

Decomposition of NH$_3$ on the surface of platinum is a zero–order reaction. What will be the rate of formation of N$_2$ and H$_2$? \; (Given: $k=2.5\times10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}}$)

Hint:

For any reaction \(aA\to\) products, the common rate is \(-\frac{1}{a}\frac{d[A]}{dt}\).
If \(-\frac{d[A]}{dt}=k\) (zero order), multiply the common rate by each stoichiometric coefficient to get individual product formation rates.

Answer 1

Overall reaction on Pt: \(\;2\,\mathrm{NH_3}\;\rightarrow\;\mathrm{N_2}+3\,\mathrm{H_2}\;\).
For a reaction \(2\,\mathrm{NH_3}\to \mathrm{N_2}+3\,\mathrm{H_2}\): \(-\frac{1}{2}\frac{d[\mathrm{NH_3}]}{dt}=\frac{d[\mathrm{N_2}]}{dt}=\frac{1}{3}\frac{d[\mathrm{H_2}]}{dt}=r\).
Zero order in NH$_3$ on Pt means the rate of disappearance of NH$_3$ is \( -\frac{d[\mathrm{NH_3}]}{dt}=k\).
Hence \(r=\dfrac{k}{2}\).
\(\Rightarrow\) Rate of formation of nitrogen: \(\displaystyle \frac{d[\mathrm{N_2}]}{dt}=r=\frac{k}{2}=\frac{2.5\times10^{-4}}{2}=1.25\times10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}}\).
\(\Rightarrow\) Rate of formation of hydrogen: \(\displaystyle \frac{d[\mathrm{H_2}]}{dt}=3r=\frac{3k}{2}=3.75\times10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}}\).
\[ \boxed{\frac{d[\mathrm{N_2}]}{dt}=1.25\times10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}}, \frac{d[\mathrm{H_2}]}{dt}=3.75\times10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}}} \]