The conversion of molecules of X to Y follows second order kinetics. If concentration of X is increases to three how will it affect the rate of formation of Y?

Explanation

`(X) rarr (Y)`
Rate=`(k)[(X)]^(2)`
If concentration of `X` is increased to 3 xx `[X]=3[X]`
therefore new rate will be =`k[3 X]^(2)= k.9[X]^(2)=9.k[X]^(2)`…....(ii)
From (i) and (ii) new rate will be 9 times.