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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.
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 x 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 x 2.5 xx 10^(-4) mol L^(-1) s^(-1)`
= 7.5 × 10−4 mol L−1 s−1
Concept: Integrated Rate Equations - Zero Order Reactions
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