For the reaction 2N_2O_{5(g)} to 4NO_{2(g)} + | vedclass

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(A) The rate of reaction is given by: Rate $= -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.The rate of deco

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$1:2$

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(A) The rate of reaction is given by: Rate $= -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.The rate of decomposition of $N_2O_5$ is defined as $-\frac{d[N_2O_5]}{dt}$.The rate of formation of $NO_2$ is defined as $\frac{d[NO_2]}{dt}$.From the rate expression: $-\frac{d[N_2O_5]}{dt} = \frac{2}{4} \frac{d[NO_2]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$.Therefore,the ratio of the rate of decomposition of $N_2O_5$ to the rate of formation of $NO_2$ is $\frac{-\frac{d[N_2O_5]}{dt}}{\frac{d[NO_2]}{dt}} = \frac{1}{2}$,which is $1:2$.

For the reaction $2N_2O_{5(g)} \to 4NO_{2(g)} + O_{2(g)}$,what is the ratio of the rate of decomposition of $N_2O_5$ to the rate of formation of $NO_2$?

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