The decomposition of NH_3 on a platinum surfa | vedclass

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(C) The decomposition reaction is: $2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}$For a zero-order reaction,the rate of reaction is equal to the rate co

Vedclass · Accepted Answer

$2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$

Vedclass · Answer

(C) The decomposition reaction is: $2NH_{3(g)} ightarrow N_{2(g)} + 3H_{2(g)}$For a zero-order reaction,the rate of reaction is equal to the rate constant $k$.$	ext{Rate} = k = 2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.The rate of reaction is expressed as: $	ext{Rate} = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$.Therefore,the rate of production of $N_2$ is equal to the rate of the reaction.$\frac{d[N_2]}{dt} = 	ext{Rate} = 2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

The decomposition of $NH_3$ on a platinum surface is a zero-order reaction. What is the rate of production of $N_2$ if $k = 2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$?

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