Step-by-step explanation:
From the information of the graph plots, the reaction is a second order reaction.
a) What is the rate law for the reaction?
Rate = k [C4H6]²
The superscript 2 signifies it is a second order reaction.
b) How many half-lives will it take for the [C4H6] to decrease to 3.1% of its original concentration?
Using the formular below;
[A]t / [A]o = 0.50^n
[A]t = 100%
[A]o = 3.1%
100 / 3.1 = 0.50^n
32 = 0.50^n
2^5 = 2^-n
n = 5
It would take five half lives.