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For the reaction system $2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g)$ volume is suddenly reduced to half its volume by increasing the pressure on it. If the reaction is first order with respect to $O_{2}$ and second order with respect to NO, the rate of reaction will

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increase to eight times of its initial value

increase to four times of its initial value

diminish to one-fourth of its initial value

diminish to one-eighth of its initial value

answer is D.

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Detailed Solution

$r_{1} = k [O_{2}] [NO]; r_{2} = k (2 [O_{2}]) (2 [NO])^{2} = 8 k [O_{2}] [NO]^{2}$ . Hence, $\frac{r_{2}}{r_{1}} = 8$ .

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