For the reaction,NO_{2(g)} + CO_{(g)} rightar | vedclass

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(A) The given chemical reaction is: $NO_{2(g)} + CO_{(g)} \rightarrow NO_{(g)} + CO_{2(g)}$.The rate of reaction can be expressed as: $Rate = -\frac{d

Vedclass · Accepted Answer

$Y$

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(A) The given chemical reaction is: $NO_{2(g)} + CO_{(g)} \rightarrow NO_{(g)} + CO_{2(g)}$.The rate of reaction can be expressed as: $Rate = -\frac{d[NO_2]}{dt} = -\frac{d[CO]}{dt} = \frac{d[NO]}{dt} = \frac{d[CO_2]}{dt}$.Given that the rate of formation of $NO_{(g)}$ is $\frac{d[NO]}{dt} = Y \ mol \ dm^{-3} \ s^{-1}$.Since the stoichiometric coefficients of $NO$ and $CO$ are both $1$,the rate of disappearance of $CO_{(g)}$ is equal to the rate of formation of $NO_{(g)}$.Therefore,$-\frac{d[CO]}{dt} = Y \ mol \ dm^{-3} \ s^{-1}$.

For the reaction,$NO_{2(g)} + CO_{(g)} \rightarrow NO_{(g)} + CO_{2(g)}$,the rate of formation of $NO_{(g)}$ is $Y$ (in $mol \ dm^{-3} \ s^{-1}$). Find the rate of disappearance of $CO_{(g)}$ (in $mol \ dm^{-3} \ s^{-1}$)?

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