Question

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?

Answer 1

The decomposition of NH₃ is

2NH₃ → N₂ + 3H₂

Rate of reaction,

\(\frac{dx}{dt}=\frac{1}{2}\,\frac{d[NH_3]}{dt}\) \(=\frac{d[N_2]}{dt}=\frac{1}{3}\,\frac{d[H_2]}{dt}\) = k

Where k is the rate constant. Since, reaction is of zero order,

Rate of reaction = \(\frac{dx}{dt}=\) \(\frac{d[N_2]}{dt}\) = k = 2.5 x 10^-4 M s^-1

^But, \(\frac{d[N_2]}{dt}\) = \(\frac{1}{3}\,\frac{d[H_2]}{dt}\)

∴ \(\frac{d[H_2]}{dt}\) = \(3\frac{d[N_2]}{dt}\) = 3 x 2.5 x 10^-4 mol^-1 L s^-1

= 7.5 x 10^-4 M s^-1