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 x 10^-4 mol^-1Ls^-1?

Answer 1

The decomposition of NH₃ on platinum surface is represented by the following equation.
\(2NH_3(g) → N_2(g) + 3H_2(G)\)
Therefore, 
\(Rate = -\frac 12 \frac {d[NH_3]}{dt} = \frac {d[N_2]}{dt} = \frac 13 \frac {d[H_2]}{dt}\)
However,it is given that the reaction is of zero order
Therefore
\(-\frac 12 \frac {d[NH_3]}{dt} = \frac {d[N_2]}{dt} = \frac 13 \frac {d[H_2]}{dt}\) = \(k = 2.5 \times 10^{-4} mol L^{-1} s^{-1}\)
Therefore, the rate of production of N₂ is
\(\frac { d[N_2]}{dt} = 2.5 \times 10^{-4} mol L^{-1} s^{-1}\)
And, the rate of production of H₂ is
\(\frac {d[H_2]}{dt} = 3\times 2.5\times 10^{-4} mol L^{-1} s^{-1}\)
= \(7.5\times 10^{-4} mol L^{-1} s^{-1}\)