Answer:
Rate law: [tex]k[C_4H_6]^2[/tex]
Integrated Rate Law: [tex]\frac{1}{[C_4H_6]}=\frac{1}{[C_4H_6]_0}+kt[/tex]
[tex]k = 1.4 \times 10^{-2}[/tex]
Explanation:
We can see that the graph of time is linear compared to [tex]\frac{1}{[C_4H_6]}[/tex] and the reaction is second order hence we get the rate law from [tex]k[A]^n[/tex].
The integrated rate law for second order is [tex]\frac{1}{[A]}=\frac{1}{[A]_0} +kt[/tex] where A is [tex]C_4H_6[/tex].
The slope of the graph [tex]\frac{1}{[C_4H_6]}[/tex] w.r.t time is equal to k. The slope of the graph from the table is 0.014 which is equal to k.
