Ammonia and oxygen react at high temperature | vedclass

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Question

(D) For the reaction: $4 NH_{3(g)} + 5 O_{2(g)} \longrightarrow 4 NO_{(g)} + 6 H_2 O_{(g)}$The rate of reaction is given by the expression:$Rate = -\f

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$3.6 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$

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(D) For the reaction: $4 NH_{3(g)} + 5 O_{2(g)} \longrightarrow 4 NO_{(g)} + 6 H_2 O_{(g)}$The rate of reaction is given by the expression:$Rate = -\frac{1}{4} \frac{d[NH_3]}{dt} = -\frac{1}{5} \frac{d[O_2]}{dt} = \frac{1}{4} \frac{d[NO]}{dt} = \frac{1}{6} \frac{d[H_2O]}{dt}$Given that the rate of formation of $NO$ is $\frac{d[NO]}{dt} = 3.6 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$.From the rate expression,we have:$-\frac{1}{4} \frac{d[NH_3]}{dt} = \frac{1}{4} \frac{d[NO]}{dt}$Therefore,the rate of disappearance of $NH_3$ is:$-\frac{d[NH_3]}{dt} = \frac{d[NO]}{dt} = 3.6 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

Ammonia and oxygen react at high temperature as $4 NH_{3(g)} + 5 O_{2(g)} \longrightarrow 4 NO_{(g)} + 6 H_2 O_{(g)}$. If the rate of formation of $NO_{(g)}$ is $3.6 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$,then the rate of disappearance of ammonia is:

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