Ammonia and oxygen react at high temperature | vedclass

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Question

(A) For the reaction $4 NH_{3(g)} + 5 O_{2(g)} \rightarrow 4 NO_{(g)} + 6 H_2O_{(g)}$,the rate expression is given by: $-\frac{1}{4} \frac{d[NH_3]}{dt

Vedclass · Accepted Answer

$5.4 	imes 10^{-3} \ mol \ L^{-1} \ sec^{-1}$

Vedclass · Answer

(A) For the reaction $4 NH_{3(g)} + 5 O_{2(g)} ightarrow 4 NO_{(g)} + 6 H_2O_{(g)}$,the rate expression is given by: $-\frac{1}{4} \frac{d[NH_3]}{dt} = -\frac{1}{5} \frac{d[O_2]}{dt} = +\frac{1}{4} \frac{d[NO]}{dt} = +\frac{1}{6} \frac{d[H_2O]}{dt}$ Given that the rate of formation of $NO$ is $\frac{d[NO]}{dt} = 3.6 	imes 10^{-3} \ mol \ L^{-1} \ sec^{-1}$. Equating the terms for $NO$ and $H_2O$: $\frac{1}{4} \frac{d[NO]}{dt} = \frac{1}{6} \frac{d[H_2O]}{dt}$ $\frac{d[H_2O]}{dt} = \frac{6}{4} 	imes \frac{d[NO]}{dt} = 1.5 	imes 3.6 	imes 10^{-3} \ mol \ L^{-1} \ sec^{-1}$ $= 5.4 	imes 10^{-3} \ mol \ L^{-1} \ sec^{-1}$

Ammonia and oxygen react at high temperature as in the reaction,$4 NH_{3(g)} + 5 O_{2(g)} \rightarrow 4 NO_{(g)} + 6 H_2O_{(g)}$. If the rate of formation of $NO$ is $3.6 \times 10^{-3} \ mol \ L^{-1} \ sec^{-1}$,calculate the rate of formation of water.

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$N_2 + 3H_2 \to 2NH_3$. If the concentration of $NH_3$ changes from $0.01 \ M$ to $0.04 \ M$ in $20 \ seconds$,then the rate of reaction will be:

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