The decomposition of NH_3 on platinum surface is a zero order reactio…

The correct answer is Option (3) → 2.5 × 10⁻⁴, 7.5 × 10⁻⁴

The balanced equation for the decomposition is

    \( \ce{2NH3(g)} \xrightarrow{\ce{Pt}} \ce{N2(g) + 3H2(g)} \)

For a zero order reaction, rate = k, so the rate of reaction is 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹. The rate of a reaction is the rate of change of the concentration of any species divided by its stoichiometric coefficient:

    \( \displaystyle \text{Rate} = -\frac{1}{2}\frac{d[\mathrm{NH_3}]}{dt} = \frac{d[\mathrm{N_2}]}{dt} = \frac{1}{3}\frac{d[\mathrm{H_2}]}{dt} \)

    Rate of production of \( \ce{N2} \) = 1 × 2.5 × 10⁻⁴ = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹

    Rate of production of \( \ce{H2} \) = 3 × 2.5 × 10⁻⁴ = 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹

Note 1: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

Note 2: The NCERT book prints the unit of k in this exercise as mol^–1 L s^–1. For a zero order reaction the unit of k is mol L^–1 s^–1, as given in the question.

NCERT (2026-27 edition):
Exercises – “3.3 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?”