For the reaction 2 NO_{(g)} + O_{2(g)} righta | vedclass

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(A) The rate expression for the reaction $2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$ is given by: $-\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[O_2]}{

Vedclass · Accepted Answer

$0.026$

Vedclass · Answer

(A) The rate expression for the reaction $2 NO_{(g)} + O_{2(g)} ightarrow 2 NO_{2(g)}$ is given by: $-\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$ Given that the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$. Equating the rate of disappearance of $O_2$ to the rate of formation of $NO_2$: $-\frac{d[O_2]}{dt} = \frac{1}{2} 	imes \frac{d[NO_2]}{dt}$ $-\frac{d[O_2]}{dt} = \frac{1}{2} 	imes 0.052 \ mol \ dm^{-3} \ s^{-1} = 0.026 \ mol \ dm^{-3} \ s^{-1}$.

For the reaction $2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$,the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$. Calculate the rate of disappearance of $O_2$,i.e.,$-\frac{d[O_2]}{dt}$.

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