Answer:
Option (b) is correct
Explanation:
For a second order of reaction of type:
[tex]2A\rightarrow products[/tex]
Rate law is - [tex]rate = k[A]^{2}[/tex], where k is rate constant and [A] is concentration of A
Integrated rate law is- [tex]\frac{1}{[A]}=\frac{1}{[A]_{0}}+kt[/tex], where [tex][A]_{0}[/tex] is initial concentration of reactant A and [A] is concentration of A after "t" time (k and [A]_{0}[/tex] are constants)
The integrated rate law can be compared with y= mx+c equation for a straight line (x vs y) where [tex]\frac{1}{[A]}[/tex] is y and t is x.
As the rate of reaction depends only upon concentration of X therefore-
[tex]rate = k[X]^{2}[/tex]