How will the rate of reaction
`2SO_(2)(g) + O_(2)(g) rarr 2SO_(3)(g)` change if the volume of the reaction vessel is halved?

Options

A. (a) It will br `1//16th` of its initial value.
B. (b) it will be `1//4th` of its initial value.
C. It will be `8` times of its initial value.
D. It will be `4` times of its initial value.

Answer

C

Explanation

`V = (1)/(2)`, then concentration `= 2` times,
Assuming the rate law as:
`r_(1) = k[SO_(2)]^(2)[O_(2)]`
`r_(2) = k[2SO_(2)]^(2)[2O_(2)]`
`r_(2) = 8r_(1)`