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

AI Thread Summary
The discussion centers on calculating the average kinetic energy of ethane at 410K, focusing on the degrees of freedom needed for the calculation. The participant initially calculated the degrees of freedom as 18 using the formula 3n-6, but later questioned whether to include all degrees of freedom or just the rotational ones. They received guidance on converting energy units from joules to kilojoules per mole and were advised to double-check their arithmetic. The professor provided a formula for total internal energy, emphasizing the importance of correctly identifying degrees of freedom. The conversation concludes with the participant expressing gratitude for the clarification.
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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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