Question

The conversion of molecules A to B follow second order kinetics. If concentration of A is increased to three times, how will it affect the rate of formation of B?

Answer 1

For the reaction, A → B as it follows second order kinetics, therefore the rate law equation will be 

Rate = k[A]² = kx²

if [A] = x mol^-1  

if the concentration of A is increased three times, then 

[A] = 3x mol L^-1 

∴ Rate = k (3x)² = 9 kx²  

Thus, the rate of the reaction will become 9 times. 

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