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 on platinum surface is represented by the following equation.

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

Therefore

Rate = \[\ce{-\frac{1}{2} \frac{d[NH3]}{dt} = \frac{d[N2]}{dt} = \frac{1}{3} \frac{d[H2]}{dt}}\]

However, it is given that the reaction is of zero order.

Therefore,

\[\ce{-\frac{1}{2} \frac{d[NH3]}{dt} = \frac{d[N2]}{dt} = \frac{1}{3} \frac{d[H2]}{dt}}\] = k

= 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹

Therefore, the rate of production of N₂ is

\[\ce{\frac{d[N2]}{dt}}\] = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹

 And, the rate of production of H₂ is

\[\ce{\frac{d[H2]}{dt}}\] = 3 × 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹

= 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹