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

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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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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.