In a catalytic conversion of N_2 to NH_3 by H | vedclass

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(A) The balanced chemical equation for Haber's process is: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.The rate of reaction is expressed as: $-\frac{d[N_2

Vedclass · Accepted Answer

$60 	imes 10^{-3}$

Vedclass · Answer

(A) The balanced chemical equation for Haber's process is: $N_2(g) + 3H_2(g) ightarrow 2NH_3(g)$.The rate of reaction is expressed as: $-\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$.Given that the rate of formation of ammonia is $\frac{d[NH_3]}{dt} = 40 	imes 10^{-3} \ mol \ L^{-1} s^{-1}$.To find the rate of disappearance of hydrogen $(-\frac{d[H_2]}{dt})$,we use the relation: $-\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$.$-\frac{d[H_2]}{dt} = \frac{3}{2} 	imes \frac{d[NH_3]}{dt} = \frac{3}{2} 	imes 40 	imes 10^{-3} \ mol \ L^{-1} s^{-1} = 60 	imes 10^{-3} \ mol \ L^{-1} s^{-1}$.

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