For the reaction,2N_2O_{5(g)} rightarrow 4NO_ | vedclass

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(A) The stoichiometric equation is $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$.According to the rate law,the rate of reaction is given by: $-\fra

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$0.22$

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(A) The stoichiometric equation is $2N_2O_{5(g)} ightarrow 4NO_{2(g)} + O_{2(g)}$.According to the rate law,the rate of reaction is given by: $-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{d[O_2]}{dt}$.Therefore,the rate of disappearance of $N_2O_5$ is $-\frac{d[N_2O_5]}{dt} = 2 	imes \frac{d[O_2]}{dt}$.Given the rate of formation of $O_2$ is $0.032 \ g \ h^{-1}$,we convert this to molar rate: $	ext{Rate in mol } h^{-1} = \frac{0.032 \ g \ h^{-1}}{32 \ g \ mol^{-1}} = 0.001 \ mol \ h^{-1}$.The rate of disappearance of $N_2O_5$ in $mol \ h^{-1}$ is $2 	imes 0.001 = 0.002 \ mol \ h^{-1}$.Converting this back to mass rate: $0.002 \ mol \ h^{-1} 	imes 108 \ g \ mol^{-1} = 0.216 \ g \ h^{-1}$,which is approximately $0.22 \ g \ h^{-1}$.

For the reaction,$2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$,the rate of formation of $O_2$ is $0.032 \ g \ h^{-1}$. What is the rate of conversion of $N_2O_5$ in $g \ h^{-1}$?

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