For the reaction system: `2NO(g) + O_(2) (g) rarr 2NO_(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 be

Options

A. diminished to one-fourth of its initial value
B. diminished to one-eighth of its initial value
C. increased to eight times of its initial value
D. increased to four times of its initial value

Answer

C

Explanation

The rate law for the reaction is Rate `=k[O_(2)][NO]^(2)`
When volume is redued to half so the rate law becomes rate =`k[2O_(2)][2NO]^(2) rarr rate=8k [O_(2)[NO]^(2)`