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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?
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Solution
The decomposition of NH3 on platinum surface is represented by the following equation.
\[\ce{2NH3_{(g)} ->[Pt] N2_{(g)} + 3H2_{(g)}}\]
Therefore
Rate = \[\ce{-\frac{1}{2} \frac{d[NH3]}{dt} = \frac{d[N2]}{dt} = \frac{1}{3} \frac{d[H2]}{dt}}\]
However, it is given that the reaction is of zero order.
Therefore,
\[\ce{-\frac{1}{2} \frac{d[NH3]}{dt} = \frac{d[N2]}{dt} = \frac{1}{3} \frac{d[H2]}{dt}}\] = k
= 2.5 × 10−4 mol L−1 s−1
Therefore, the rate of production of N2 is
\[\ce{\frac{d[N2]}{dt}}\] = 2.5 × 10−4 mol L−1 s−1
And, the rate of production of H2 is
\[\ce{\frac{d[H2]}{dt}}\] = 3 × 2.5 × 10−4 mol L−1 s−1
= 7.5 × 10−4 mol L−1 s−1
Notes
The answer in the textbook is incorrect.
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