The decomposition of NH_3 on Pt surface is a | vedclass

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(C) The decomposition reaction of $NH_3$ is: $2NH_3 \rightarrow N_2 + 3H_2$. Since it is a zero-order reaction,the rate of reaction is equal to the ra

Vedclass · Accepted Answer

$N_2 = 2 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$; $H_2 = 6 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$

Vedclass · Answer

(C) The decomposition reaction of $NH_3$ is: $2NH_3 ightarrow N_2 + 3H_2$. Since it is a zero-order reaction,the rate of reaction is equal to the rate constant $k = 2 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$. The rate of reaction is given by: $	ext{Rate} = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt} = k$. Therefore,the rate of appearance of $N_2$ is $\frac{d[N_2]}{dt} = k = 2 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$. The rate of appearance of $H_2$ is $\frac{d[H_2]}{dt} = 3k = 3 	imes (2 	imes 10^{-4}) = 6 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

The decomposition of $NH_3$ on $Pt$ surface is a zero order reaction. If the value of rate constant is $2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$,the rates of appearance of $N_2$ and $H_2$ are respectively:

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