What Is the Average Kinetic Energy of Ethane at 410K?

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Homework Help Overview

The discussion revolves around calculating the average kinetic energy of ethane at a temperature of 410K, focusing on the degrees of freedom relevant to the problem.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of degrees of freedom for ethane and question the correct application of the relevant equations. There is a specific inquiry about whether to use all degrees of freedom or just the rotational ones.

Discussion Status

The discussion includes attempts to clarify the correct number of degrees of freedom to use in the kinetic energy equation. Some participants have provided feedback on arithmetic and conversion issues, while others seek confirmation on the approach to calculating degrees of freedom.

Contextual Notes

Participants reference an equation provided by the professor regarding total internal energy, which adds to the complexity of determining the appropriate degrees of freedom for ethane.

chemman218
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Homework Statement



Find the average kinetic energy in kJ/mol (including all degrees of freedom) of one mole of ethane at 410K.

Homework Equations



1/2Mu^s = kT/2
u= avg. velocity
E=n/2kT
N=degrees of freedom

The Attempt at a Solution



C2H6 degrees of freedom - 3n-6 = 18

(18/2)(1.380x10^-23 j/k)(410K)= 5.09e^20J

not sure how to get to KJ/mol from here.
 
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You should first check your arithmetics. Powers of 10, to be exact. Then comes the physics.
 
Okay, thanks for the note on converting j to kj. My main question is how do I know how many degrees of freedom to use in order to complete the equation.
My professor sent us an equation to use:

the total internal
energy is (3+J)/2RT where J is the number of rotational degrees of
freedom in the molecule.

I calculated the degrees of freedom to be 18 due to 3n-6. But that was not correct. Do I just use the rotational degrees of freedom or all degrees of freedom? Thanks.
 
Got it! Thanks for the help
 

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