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

The order of reaction is defined as the sum of the powers of concentrations In the rate law.

The rate of second order reaction can be expressed as     rate = k[A]²

The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k[X]^{2                                                    ________________________}   (1)

Let [X] = a mol L^-1 , then equation (1) can be written as:

Rate = k [a]²

= ka²

If the concentration of X is increased to three times, then [X] = 3a mol L^-1

Now, the rate equation will be:

Rate = k [3a]²

= 9(ka²)

Hence, the rate of formation will increase by 9 times.