The conversion of molecules `X` to `Y` follows second order kinetics. If the concenration of `X` is increased to three times, how will it affect the rate of formation of `Y` ?

Explanation

For the reaction, `XtoY`, as it follows second order kinetics, the rate law equation will be
Rate `=k[X]^(2)=ka^(2)" "("if "[X]=a" mol L"^(-1))`
If concentration of X is increased three times, now, `[X]=3" a mol L"^(-1)`
`:." ""Rate"=k(3" a")^(2)=9" ka"^(2)`
Thus, the rate of reaction will become 9 times. Hence, the rate of formation of Y will increase 9 times.