For a reaction between gaseous compounds, `2A + B rarrC + D` , the reaction rate law is rate k[A][B]. If the volume of the container is made `1//4^(th)` of the initial, then what will be the rate of reaction as compared to the initial rate?

Options

A. `(1)/(16)`
B. `(1)/(8)`
C. `16` times
D. 8 times

Answer

C

Explanation

Concentration will increase by 4 times of each reactant
`therefore ` Rate will become ` (4)^(2)` I .e 16 times