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

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(B) For the elementary reaction: $3 H_{2(g)} + N_{2(g)} \rightarrow 2 NH_{3(g)}$The rate of reaction is expressed as:Rate $= -\frac{1}{3} \frac{d[H_{2

Vedclass · Accepted Answer

$\frac{-2}{3} \frac{d[H_{2(g)}]}{dt} = \frac{d[NH_{3(g)}]}{dt}$

Vedclass · Answer

(B) For the elementary reaction: $3 H_{2(g)} + N_{2(g)} \rightarrow 2 NH_{3(g)}$The rate of reaction is expressed as:Rate $= -\frac{1}{3} \frac{d[H_{2(g)}]}{dt} = -\frac{d[N_{2(g)}]}{dt} = \frac{1}{2} \frac{d[NH_{3(g)}]}{dt}$To find the relation between $\frac{d[H_{2(g)}]}{dt}$ and $\frac{d[NH_{3(g)}]}{dt}$,we equate their respective parts:$-\frac{1}{3} \frac{d[H_{2(g)}]}{dt} = \frac{1}{2} \frac{d[NH_{3(g)}]}{dt}$Multiplying both sides by $2$:$-\frac{2}{3} \frac{d[H_{2(g)}]}{dt} = \frac{d[NH_{3(g)}]}{dt}$Thus,option $(B)$ is correct.

$t \ (s)$	$0$	$30$	$60$	$90$
$[\text{Ester}] \ (mol \ L^{-1})$	$0.55$	$0.31$	$0.17$	$0.085$

For the elementary reaction,$3 H_{2(g)} + N_{2(g)} \rightarrow 2 NH_{3(g)}$,identify the correct relation among the following relations:

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