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Chemistry Question on Chemical Kinetics

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

The decomposition of NH3 on platinum surface is represented by the following equation.
2NH3(g)→N2(g)+3H2(G)2NH_3(g) → N_2(g) + 3H_2(G)
Therefore,
Rate=−12d[NH3]dt=d[N2]dt=13d[H2]dtRate = -\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
−12d[NH3]dt=d[N2]dt=13d[H2]dt-\frac 12 \frac {d[NH_3]}{dt} = \frac {d[N_2]}{dt} = \frac 13 \frac {d[H_2]}{dt} = k=2.5×10−4molL−1s−1k = 2.5 \times 10^{-4} mol L^{-1} s^{-1}
Therefore, the rate of production of N2 is
d[N2]dt=2.5×10−4molL−1s−1\frac { d[N_2]}{dt} = 2.5 \times 10^{-4} mol L^{-1} s^{-1}
And, the rate of production of H2 is
d[H2]dt=3×2.5×10−4molL−1s−1\frac {d[H_2]}{dt} = 3\times 2.5\times 10^{-4} mol L^{-1} s^{-1}
= 7.5×10−4molL−1s−17.5\times 10^{-4} mol L^{-1} s^{-1}