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

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(C) The rate of reaction $r$ is given by: $r = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.Given the rate of formation of $NH_3$

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$0.00017 \ kg \ L^{-1} \ h^{-1}$

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(C) The rate of reaction $r$ is given by: $r = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.Given the rate of formation of $NH_3$ is $0.001 \ kg \ L^{-1} \ h^{-1}$.To find the rate of consumption of $H_2$ in terms of mass,we use the stoichiometry: $\frac{d[H_2]}{dt} = \frac{3}{2} 	imes \frac{d[NH_3]}{dt}$.Since the rate is given in $kg \ L^{-1} \ h^{-1}$,we can directly use the stoichiometric coefficients: $	ext{Rate of consumption of } H_2 = \frac{3}{2} 	imes (	ext{Rate of formation of } NH_3) 	imes \frac{M_{H_2}}{M_{NH_3}}$.$M_{H_2} = 2 \ g/mol$ and $M_{NH_3} = 17 \ g/mol$.Rate of consumption of $H_2 = \frac{3}{2} 	imes 0.001 	imes \frac{2}{17} \approx 0.000176 \ kg \ L^{-1} \ h^{-1}$.Rounding to the nearest provided option,the answer is $0.00017 \ kg \ L^{-1} \ h^{-1}$.

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

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