The decomposition of NH_{3} on platinum surface is zero order reaction.

What are the rates of production of N_{2} and H_{2}. if k = 2.5 × 10^{-4} mol L^{-1} sec^{-1}?

2NH_{3}(g) → N_{2}(g) + 3H_{2}(g)

Rate of zero order reaction is equal to rate constant. i.e. Rate = 2.5 × 10^{-4}mol L^{-1}sec^{-1}.

According to rate law,

2.5 × 10^{-4}mol L^{-1}sec^{-1} =

i.e. the rate of production of N_{2} is 2.5 × 10^{-4}mol L^{-1} sec^{-1}.

According to rate law,

i.e. rate of formation of H_{2} is 3 times rate of reaction = 3 × 2.5 × 10^{-4}mol L^{-1}sec^{-1}

= 7.5 × 10^{-4}mol L^{-1}sec^{-1}

__Rate of formation of N _{2} and H_{2} is 2.5 × 10^{-4} mol L^{-1}sec^{-1} and 7.5 × 10^{-4} mol L^{-1}sec^{-1} respectively.__

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