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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?
Solution
The decomposition of NH3 on platinum surface is represented by the following equation.
\[\ce{2NH3_{(g)}->[Pt]N2_{(g)} +3H2_{(g)}}\]
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 × 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 xx 2.5 xx 10^(-4) "mol L"^(-1) "s"^(-1)`
= 7.5 × 10−4 mol L−1 s−1
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