The rate constant for the reaction,2N_2O_5 to | vedclass

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(D) The stoichiometry of the reaction is $2N_2O_5 	o 4NO_2 + O_2$.From the stoichiometry,$1 \ mol$ of $O_2$ is formed from $2 \ mol$ of $N_2O_5$.Ther

Vedclass · Accepted Answer

$9.6 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$

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(D) The stoichiometry of the reaction is $2N_2O_5 	o 4NO_2 + O_2$.From the stoichiometry,$1 \ mol$ of $O_2$ is formed from $2 \ mol$ of $N_2O_5$.Therefore,the concentration of $N_2O_5$ reacted when $[O_2] = 0.1 \ mol \ L^{-1}$ is $2 	imes 0.1 = 0.2 \ mol \ L^{-1}$.The concentration of $N_2O_5$ remaining is $[N_2O_5] = 1.0 - 0.2 = 0.8 \ mol \ L^{-1}$.The rate of reaction is given by $Rate = k[N_2O_5] = 3.0 	imes 10^{-4} \ s^{-1} 	imes 0.8 \ mol \ L^{-1} = 2.4 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.From the rate expression,$Rate = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.Thus,the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = 4 	imes Rate = 4 	imes 2.4 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1} = 9.6 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

Observation	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	Rate $(mol \ L^{-1} \ sec^{-1})$
$1$	$0.1$	$0.1$	$2 \times 10^{-3}$
$2$	$0.4$	$0.1$	$3.2 \times 10^{-3}$
$3$	$0.1$	$0.2$	$8 \times 10^{-3}$

$[A_0]$	$0.1 \ M$	$0.025 \ M$
$t_{1/2} \ (s)$	$100 \ s$	$50 \ s$

The rate constant for the reaction,$2N_2O_5 \to 4NO_2 + O_2$ is $3.0 \times 10^{-4} \ s^{-1}$. If the reaction starts with $1.0 \ mol \ L^{-1}$ of $N_2O_5$,calculate the rate of formation of $NO_2$ at the moment when the concentration of $O_2$ is $0.1 \ mol \ L^{-1}$.

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