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