What is the Reaction Order and Rate Constant for 2C4H6 --> C8H12 at 320°C?

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SUMMARY

The reaction 2C4H6 --> C8H12 at 320°C has been determined to be first-order, with the rate constant calculated based on pressure changes over time. The pressure data indicates a decrease from 632.0 torr to 474.6 torr as the reaction progresses, confirming that the consumption of C4H6 correlates with the formation of C8H12. The correct rate equation is rate = k[C4H6], not involving the product C8H12. This conclusion was reached by analyzing the pressure data and understanding the stoichiometry of the reaction.

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flybynight
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Homework Statement


I am considering the reaction 2C4H6 --> C8H12 at 320 degrees C. Both the product and the reactant are gases.
I have the data:
Time (minutes): 0.00 3.25 12.18 24.55 42.50 68.05
Total P (torr): 632.0 618.5 584.2 546.8 509.3 474.6

Find the reaction order and the rate constant.
Assume that only C4H6 is present at the start of the reaction.

Homework Equations


rate=k[C8H12]x
Perhaps PV=nRT?

The Attempt at a Solution


I tried to apply PV=nRT to the initial amount, hoping to find moles of reactant. However, I don't have a volume, so I don't know how to start. Any advice would be greatly appreciated.
 
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Any help?
 
Why does the pressure goes down? Can you use this change to calculate how much C4H6 reacted?

Your rate equation is wrong.
 
I got it. Every time the reactant is changed into product, the pressure is one half for that molecule. So I found the values of the reactants and the products, and found the order is 1.

Thanks for the help,
Peter
 

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