For the reaction system `2NO(g) + O_(2)(g) rarr 2NO(g)` volume is suddenly produced 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

Options

A. increase to four times of its initial value
B. increase to eight times of its initial value
C. diminish to one-eight of its initial value
D. diminish to one-fourth of its initial value

Answer

B

Explanation

We have
`Rate_(1)=k[NO]^(2)[2O_(2)]`
or, `(Rate_(2))/(Rate_(1))=(k[2NO]^(2)[2O_(2)])/(k[NO]^(2)[O_(2)])=8`
i.e., `Rate_(2)=8Rate_(1)`
The rate of reaction will increase to eight time of its initial rate if the volume of the reaction vessel is doubled