For the decomposition reaction N_2O_{4(g)} ri | vedclass

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(B) The rate of reaction is given by: $Rate = -\frac{d[N_2O_4]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$ Given,change in pressure of $N_2O_4$ is $\Delta

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$1.2 	imes 10^{-2} \ atm \ min^{-1}$

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(B) The rate of reaction is given by: $Rate = -\frac{d[N_2O_4]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$ Given,change in pressure of $N_2O_4$ is $\Delta P = 0.28 - 0.46 = -0.18 \ atm$ over $\Delta t = 30 \ min$. The rate of disappearance of $N_2O_4$ is $-\frac{\Delta P}{\Delta t} = -\frac{-0.18}{30} = 0.006 \ atm \ min^{-1}$. Equating the rates: $\frac{1}{2} \frac{d[NO_2]}{dt} = 0.006 \ atm \ min^{-1}$. Therefore,the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = 2 	imes 0.006 = 0.012 \ atm \ min^{-1} = 1.2 	imes 10^{-2} \ atm \ min^{-1}$.

$t \ (s)$	$0$	$30$	$60$	$90$
$[\text{Ester}] \ (mol \ L^{-1})$	$0.55$	$0.31$	$0.17$	$0.085$

For the decomposition reaction $N_2O_{4(g)} \rightarrow 2NO_{2(g)}$,the initial pressure of $N_2O_4$ decreases from $0.46 \ atm$ to $0.28 \ atm$ in $30 \ minutes$. What is the rate of formation of $NO_2$?

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