For the reaction,N_2 + 3H_2 rightarrow 2NH_3, | vedclass

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(D) For the reaction,$N_2 + 3H_2 \rightarrow 2NH_3,$ the rate of reaction is given by:Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \f

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$3 	imes 10^{-4}$

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(D) For the reaction,$N_2 + 3H_2 ightarrow 2NH_3,$ the rate of reaction is given by:Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$Given that $\frac{d[NH_3]}{dt} = 2 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}.$Equating the terms for $H_2$ and $NH_3$:$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$$-\frac{d[H_2]}{dt} = \frac{3}{2} 	imes \frac{d[NH_3]}{dt}$$-\frac{d[H_2]}{dt} = \frac{3}{2} 	imes (2 	imes 10^{-4}) = 3 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}.$

For the reaction,$N_2 + 3H_2 \rightarrow 2NH_3,$ if $\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1},$ the value of $-\frac{d[H_2]}{dt}$ would be .................. $mol \ L^{-1} \ s^{-1}$.

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