Question

The reaction of butadiene gas (C4H6) with itself produces C8H12 gas as follows: The reaction is second order with a rate constant equal to 5.76 × 10−2 L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration in molarity remaining after 10.0 min? Report your answer to 3 decimal places.

Answer 1

`C_(C_4H_6)=0.179M`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`2C_4H_6\rightarrow C_8H_(12)`

And the rate law is:

`(dC_(C_4H_6))/(dt)=kC_(C_4H_6)^2`

Which integrated is:

`(1)/(C_(C_4H_6)) =(1)/(C_(C_4H_6)^0)+kt`

In such a way, the concentration after 10.0 min is:

`(1)/(C_(C_4H_6)) =(1)/(0.200M)}+5.76x10^(-2)(L)/(mol*min)*10.0min\\ \\(1)/(C_(C_4H_6))=5.58(L)/(mol) \\\\C_(C_4H_6)=(1)/(5.58(L)/(mol) ) \\\\C_(C_4H_6)=0.179M`

Regards.