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

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(B) The rate of reaction is given by the expression: $Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$ Given

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$1.25 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$

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(B) The rate of reaction is given by the expression: $Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$ Given that the rate of formation of $NH_3$ is $\frac{d[NH_3]}{dt} = 2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$. Equating the rate of disappearance of $N_2$ to the rate of formation of $NH_3$: $-\frac{d[N_2]}{dt} = \frac{1}{2} 	imes \frac{d[NH_3]}{dt}$ $-\frac{d[N_2]}{dt} = \frac{1}{2} 	imes (2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1})$ $-\frac{d[N_2]}{dt} = 1.25 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$

For the reaction $N_2 + 3H_2 \rightarrow 2NH_3$ in the Haber process,the rate of formation of $NH_3$ is $2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$. What is the rate of disappearance of $N_2$?

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