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.

Therefore

`Rate = -1/2 (d[NH_3])/dt = (d[N_2])/dt = 1/3(d[H_2])/dt`

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

Therefore,

`-1/2 (d[NH_3])/dt = (d[N_2])/dt = 1/3 (d[H_2])/dt = k` 

= 2.5 x 10^-4 mol L^-1 s^-1

Therefore, the rate of production of N₂ is

`(d[N_2])/dt = 2.5xx10^(-4) mol L^(-1) s^(-1)`

 And, the rate of production of H₂ is

`(d[H_2])/dt = 3 x 2.5 xx 10^(-4) mol L^(-1) s^(-1)`

= 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹