In a reaction N_{2(g)} + 3H_{2(g)} longrighta | vedclass

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(C) The given reaction is: $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$The rate of appearance of $NH_3$ is given as $\frac{d[NH_3]}{dt} = 2.5 	i

Vedclass · Accepted Answer

$1.25 	imes 10^{-4}, 3.75 	imes 10^{-4}$

Vedclass · Answer

(C) The given reaction is: $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$The rate of appearance of $NH_3$ is given as $\frac{d[NH_3]}{dt} = 2.5 	imes 10^{-4} \ mol \ L^{-1} s^{-1}$.The rate of reaction $(ROR)$ is expressed as:$ROR = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$Calculating $ROR$:$ROR = \frac{1}{2} 	imes (2.5 	imes 10^{-4}) = 1.25 	imes 10^{-4} \ mol \ L^{-1} s^{-1}$.Calculating the rate of disappearance of $H_2$ $(-\frac{d[H_2]}{dt})$:$-\frac{d[H_2]}{dt} = \frac{3}{2} 	imes \frac{d[NH_3]}{dt} = \frac{3}{2} 	imes (2.5 	imes 10^{-4}) = 3.75 	imes 10^{-4} \ mol \ L^{-1} s^{-1}$.Thus,the rate of reaction is $1.25 	imes 10^{-4} \ mol \ L^{-1} s^{-1}$ and the rate of disappearance of $H_2$ is $3.75 	imes 10^{-4} \ mol \ L^{-1} s^{-1}$.Therefore,option $C$ is correct.

In a reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of appearance of $NH_3$ is $2.5 \times 10^{-4} \ mol \ L^{-1} s^{-1}$. The rate of reaction and rate of disappearance of $H_2$ will be (in $mol \ L^{-1} s^{-1}$):

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