For the reaction,N_{2(g)} + 3 H_{2(g)} longri | vedclass

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Question

(D) The balanced chemical equation is: $N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$ The rate of reaction is given by: $	ext{Rate} = -\frac{d[N

Vedclass · Accepted Answer

$0.044 \ mol \ dm^{-3} \ s^{-1}$

Vedclass · Answer

(D) The balanced chemical equation is: $N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$ The rate of reaction is given by: $	ext{Rate} = -\frac{d[N_{2}]}{dt} = -\frac{1}{3} \frac{d[H_{2}]}{dt} = \frac{1}{2} \frac{d[NH_{3}]}{dt}$ Given that the rate of formation of $NH_{3}$ is $\frac{d[NH_{3}]}{dt} = 0.088 \ mol \ dm^{-3} \ s^{-1}$. Substituting this into the rate expression: $-\frac{d[N_{2}]}{dt} = \frac{1}{2} 	imes \frac{d[NH_{3}]}{dt}$ $-\frac{d[N_{2}]}{dt} = \frac{1}{2} 	imes 0.088 \ mol \ dm^{-3} \ s^{-1} = 0.044 \ mol \ dm^{-3} \ s^{-1}$ Therefore,the rate of consumption of $N_{2}$ is $0.044 \ mol \ dm^{-3} \ s^{-1}$.

$t / s$	$0$	$30$	$60$	$90$
$[A] / mol \ L^{-1}$	$0.55$	$0.31$	$0.17$	$0.085$

For the reaction,$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$,$NH_{3}$ is formed at a rate of $0.088 \ mol \ dm^{-3} \ s^{-1}$. Calculate the consumption rate of $N_{2(g)}$.

Similar Questions

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