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

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(D) The balanced chemical equation is $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$.According to the rate law,the rate of reaction is given by: $Rate

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$0.0017 \, kg \, h^{-1}$

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(D) The balanced chemical equation is $N_{2(g)} + 3H_{2(g)} ightarrow 2NH_{3(g)}$.According to the rate law,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 the rate of formation of $NH_3$ is $0.001 \, kg \, h^{-1}$,we convert this to molar rate: $Molar \, mass \, of \, NH_3 = 17 \, g/mol$ and $H_2 = 2 \, g/mol$.Rate of formation of $NH_3$ in $mol \, h^{-1} = \frac{0.001 	imes 1000}{17} = \frac{1}{17} \, mol \, h^{-1}$.From the stoichiometry,$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.So,$-\frac{d[H_2]}{dt} = \frac{3}{2} 	imes \frac{d[NH_3]}{dt} = \frac{3}{2} 	imes \frac{1}{17} = \frac{3}{34} \, mol \, h^{-1}$.Converting back to $kg \, h^{-1}$: $Rate \, of \, consumption \, of \, H_2 = \frac{3}{34} 	imes 2 \, g \, h^{-1} = \frac{6}{34} \, g \, h^{-1} \approx 0.176 \, g \, h^{-1} = 0.000176 \, kg \, h^{-1}$.Rounding to the nearest option,the value is $0.0017 \, kg \, h^{-1}$.

For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the reaction occurs under specific temperature and partial pressure conditions. The rate of formation of $NH_3$ is $0.001 \, kg \, h^{-1}$. The rate of consumption of $H_2$ under the same conditions is ...... $kg \, h^{-1}$.

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