Question

The decomposition of NH₃ on platinum surface, 

\[\ce{2NH3_{(g)} ->[Pt] N2_{(g)} + 3H2_{(g)}}\]

is a zero order reaction with k = 2.5 × 10⁻⁴ Ms⁻⁴. What are the rates of production of N₂ and H₂?

Answer 1

The given reaction is \[\ce{2NH3_{(g)} ->[Pt] N2_{(g)} + 3H2_{(g)}}\]

It’s a zero-order reaction, so Rate = k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹

From stoichiometry \[\ce{2NH3 -> 1N2 + 3H2}\]

This means

Rate of formation of N₂ = \[\ce{\frac{1}{2} \times Rate of decomposition of NH3}\]

Rate of formation of H₂ = \[\ce{\frac{3}{2} \times Rate of decomposition of NH3}\]

Rate of production of N₂ = \[\ce{\frac{1}{2} \times 2.5 \times 10^{-4}}\]

= 1.25 × 10⁻⁴ mol L⁻¹ s⁻¹

Rate of production of H₂ = \[\ce{\frac{3}{2} \times 2.5 \times 10^{-4}}\]

= 3.75 × 10⁻⁴ mol L⁻¹ s⁻¹