For the reaction {N_2}_{(g)} + 3{H_2}_{(g)} t | vedclass

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(B) For a general reaction $aA + bB 	o cC + dD$,the rate of reaction is given by: Rate $= -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} =

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$+\frac{d[N{H_3}]}{dt} = -\frac{2}{3}\frac{d[{H_2}]}{dt}$

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(B) For a general reaction $aA + bB 	o cC + dD$,the rate of reaction is given by: Rate $= -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$. For the given reaction ${N_2}_{(g)} + 3{H_2}_{(g)} 	o 2N{H_3}_{(g)}$,the rate expression is: Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[{H_2}]}{dt} = \frac{1}{2}\frac{d[N{H_3}]}{dt}$. Equating the terms for $N{H_3}$ and ${H_2}$: $\frac{1}{2}\frac{d[N{H_3}]}{dt} = -\frac{1}{3}\frac{d[{H_2}]}{dt}$. Multiplying both sides by $2$: $\frac{d[N{H_3}]}{dt} = -\frac{2}{3}\frac{d[{H_2}]}{dt}$.

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

For the reaction ${N_2}_{(g)} + 3{H_2}_{(g)} \to 2N{H_3}_{(g)}$,what is the relationship between $\frac{d[N{H_3}]}{dt}$ and $-\frac{d[{H_2}]}{dt}$?

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