The convertion of molecules X to Y follows second order kinetics. If concentration of X is increased by three times, how will it affect the rate of formation of Y.

Explanation

For the reaction `x rarr y`
Reaction rate `(r )=K[x]^(2)" "-(i)`
If the concentration of x is increased three times, then
Reaction rate `(r')=k[3x]^(2)=kxx[9x]^(2)" "-(ii)`
Dividing Eq. (ii) by Eq. (i)
`(r')/(r )=(kxx[9x]^(2))/(kxx[x]^(2))=9`
It means that time rate of formation of y will increase by nine times.