The rate of the reaction is proportional to the
first power of the concentration of the reactant R. For example,
R → P
Again, C is the constant of integration and its value can be determined
When t = 0, A = [A]0, where [A]0 is the initial concentration of the reactant.
Therefore,above equation can be written as
ln [A]0 = –k × 0 + C
ln [A]0 = C
putting the value of C in equation (1), we get
This reaction follows second order kinetics.
So that, the rate equation for this reaction will
Rate, R = k[X]2 .............(1)
Let initial concentration is x mol L−1,
Plug the value in equation (1)
Rate, R1 = k .(a)2
Given that concentration is increasing by 3 times so new concentration will 3a mol L−1
Plug the value in equation (1) we get
Rate, R2 = k (3a)2
We have already get that R1 = ka2 plus this value we get
R2 = 9 R1
So that, the rate of formation will increase by 9 times.
Rate = k[A]2
If concentration of X is increased to three times,
Rate = k[3A]2
or Rate = 9 k A2
Thus, rate will increase 9 times.