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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.

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Answer:

[tex]C_{C_4H_6}=0.179M[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2C_4H_6\rightarrow C_8H_{12}[/tex]

And the rate law is:

[tex]\frac{dC_{C_4H_6}}{dt}=kC_{C_4H_6}^2[/tex]

Which integrated is:

[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{C_{C_4H_6}^0}+kt[/tex]

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

[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{0.200M}}+5.76x10^{-2}\frac{L}{mol*min}*10.0min\\ \\\frac{1}{C_{C_4H_6}}=5.58\frac{L}{mol} \\\\C_{C_4H_6}=\frac{1}{5.58\frac{L}{mol} } \\\\C_{C_4H_6}=0.179M[/tex]

Regards.

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