The rate of a gaseous reaction is given by the expression `k[A]^(2)[B]^(3)`. The volume of the reaction vessel is reduced to one half of the initial volume. What will be the reaction rate as compared to the original rate a?

Options

A. (a) `1//24`
B. (b) `1//32`
C. ( c) `32`
D. (d) `24`

Answer

C

Explanation

(c ) Rate `= k[A]^(2)[B]^(3)`
When the volume is halved, the concentration will become double.
`:.` Rate `= k[2A]^(2)[2B]^(3)`
`= 32k[A]^(2)[B]^(3)`
`= 32 xx` Original rate form Eq. (i).