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

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diminish to one-eighth of its initial value

diminish to one-fourth of its initial value

increase to eight times of its initial value

increase to four times of its initial value

answer is C.

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Detailed Solution

Rate law $= k [O_{2}] {[N O]}^{2}$

If volume is reduced to half. Concentrations will become double.

$∴$ Rate $= k [2 O_{2}] {[2 N O]}^{2} = 8 t i m e$

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