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

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(D) The rate of reaction is given by: $Rate = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt}$.First,calculate the rate of disappea

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$4 	imes 10^{-3}$

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(D) The rate of reaction is given by: $Rate = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt}$.First,calculate the rate of disappearance of $N_2O_5$: $-\frac{d[N_2O_5]}{dt} = -\frac{1.4 - 2.0}{300} = \frac{0.6}{300} = 2 	imes 10^{-3} \ mol \ L^{-1} \ min^{-1}$.Since the rate of production of $NO_2$ is twice the rate of disappearance of $N_2O_5$ (from the stoichiometry $2:4$ or $1:2$): $\frac{d[NO_2]}{dt} = 2 	imes (-\frac{d[N_2O_5]}{dt}) = 2 	imes 2 	imes 10^{-3} = 4 	imes 10^{-3} \ mol \ L^{-1} \ min^{-1}$.

For the reaction $2 N_2O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$,the initial concentration of $N_2O_5$ is $2.0 \ mol \ L^{-1}$ and after $300 \ min$,it is reduced to $1.4 \ mol \ L^{-1}$. The rate of production of $NO_2$ (in $mol \ L^{-1} \ min^{-1}$) is

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