The experimental rate law for a reaction
`2A + 2B rarr` Product, is
`V prop C_(A) C_(B)^(1//2)`. If the concentration of both A and B are doubled the rate of reaction increases by a factor of

Options

A. `sqrt(2)`
B. 2
C. `2.sqrt(2)`
D. 4

Answer

C

Explanation

In first case ,
`V prop C_(A) C_(B)^(1//2)`
or `V =kC_(A)C_(B)^(1//2)`
In second case,
`V'=k2 C_(A)(2 C_(B))^(1//2)`
Dividing
`(V')/(V)=(2.C_(A)(2C_(B))^(1//2))/(C_(A)(C_(B))^(1//2))=2sqrt(2)`
`V'=2sqrt(2)V`
The rate of reaction increases by a factor of `2 sqrt(2)`.