For the reaction N_{2}O_{5(g)} rightarrow 2NO | vedclass

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(B) The rate of reaction is given by the expression: $-\frac{d[N_{2}O_{5}]}{dt} = \frac{1}{2} \frac{d[NO_{2}]}{dt} = 2 \frac{d[O_{2}]}{dt}$.Given,$-\f

Vedclass · Accepted Answer

$1.25 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$ and $3.125 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$

Vedclass · Answer

(B) The rate of reaction is given by the expression: $-\frac{d[N_{2}O_{5}]}{dt} = \frac{1}{2} \frac{d[NO_{2}]}{dt} = 2 \frac{d[O_{2}]}{dt}$.Given,$-\frac{d[N_{2}O_{5}]}{dt} = 6.25 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$.For $NO_{2}$ formation: $\frac{d[NO_{2}]}{dt} = 2 	imes (-\frac{d[N_{2}O_{5}]}{dt}) = 2 	imes 6.25 	imes 10^{-3} = 1.25 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$.For $O_{2}$ formation: $\frac{d[O_{2}]}{dt} = \frac{1}{2} 	imes (-\frac{d[N_{2}O_{5}]}{dt}) = \frac{6.25 	imes 10^{-3}}{2} = 3.125 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

$t / s$	$0$	$30$	$60$	$90$
$[A] / mol \ L^{-1}$	$0.55$	$0.31$	$0.17$	$0.085$

For the reaction $N_{2}O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2} O_{2(g)}$,the rate of disappearance of $N_{2}O_{5}$ is given as $6.25 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$. The rate of formation of $NO_{2}$ and $O_{2}$ is given respectively as:

Similar Questions

In a pseudo first order reaction in water,the following results were obtained:

$t / s$ $0$ $30$ $60$ $90$

$[A] / mol \ L^{-1}$ $0.55$ $0.31$ $0.17$ $0.085$

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