Consider the following reaction: N_{2(g)} + 3 | vedclass

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(C) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is expressed as: Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3}

Vedclass · Accepted Answer

$0.06$

Vedclass · Answer

(C) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is expressed as: Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$. Given that $-\frac{d[N_2]}{dt} = 0.02 \ mol \ L^{-1} \ s^{-1}$. Equating the terms for $N_2$ and $H_2$: $-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt}$. Therefore,$-\frac{d[H_2]}{dt} = 3 	imes (-\frac{d[N_2]}{dt})$. $-\frac{d[H_2]}{dt} = 3 	imes 0.02 \ mol \ L^{-1} \ s^{-1} = 0.06 \ mol \ L^{-1} \ s^{-1}$.

$1$. Burning candle	$i$. Line spectrum
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Consider the following reaction: $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$. The rate of this reaction in terms of $N_2$ at $T \ K$ is $-\frac{d[N_2]}{dt} = 0.02 \ mol \ L^{-1} \ s^{-1}$. What is the value of $-\frac{d[H_2]}{dt}$ (in units of $mol \ L^{-1} \ s^{-1}$) at the same temperature?

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