The decomposition of NH₃ on the platinum surface is zero order reaction. If k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹ the rate of production of H₂ is

7.5 × 10⁻⁴ mol L⁻¹ s⁻¹

Given rate constant = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹

The decomposition of NH₃ on the platinum surface takes place as

\(2NH_3 ————→ N_2 + 3H_2\)

So, Rate of reaction = \(−\frac{1}{2}\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3}\frac{d[H_2]}{dt}\)

The rate of production of \(H_2 = \frac{d[H_2]}{dt}\)

                                           \(= 3 × 2.5 × 10^{− 4} \text{ mol L}^{−1}\text{ s}^{−1}\)

                                           \( = 7.5 × 10^{− 4} \text{ mol L}^{−1}\text{ s}^{−1}\)