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^{−1} L s^{−1}?

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#### Solution

The decomposition of NH_{3} 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 N_{2} is

`(d[N_2])/dt = 2.5xx10^(-4) mol L^(-1) s^(-1)`

And, the rate of production of H_{2} 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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