The decomposition of NH_3 on the platinum sur | vedclass

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Question

(B) The balanced chemical equation for the decomposition of ammonia is: $2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}$For a zero-order reaction,the rat

Vedclass · Accepted Answer

$7.5 	imes 10^{-4}$

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(B) The balanced chemical equation for the decomposition of ammonia 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 $K$.Rate $= K = 2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$From the stoichiometry of the reaction,the rate of reaction is related to the rate of production of $H_2$ as: $	ext{Rate} = \frac{1}{3} \frac{d[H_2]}{dt}$Therefore,$\frac{d[H_2]}{dt} = 3 	imes 	ext{Rate} = 3 	imes (2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}) = 7.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$

Time $(s)$	Concentration of the reactant $(mol \ L^{-1})$
$0$	$0.1$
$15$	$0.05$
$20$	$0.1 - x$

The decomposition of $NH_3$ on the platinum surface is a zero-order reaction. If $K = 2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$,what will be the rate of production of $H_2$ in $mol \ L^{-1} \ s^{-1}$ unit?

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