chemistry, please help
posted by apoorva on .
2NO2(g)> 2NO(g) + O2(g) , H=+114.2kJ
(Note: H, S, G all have a degree sign next to them)
NO: H(enthalpy)=90.3kJ/mol, S(entropy)=210.7J/mol*K, G(gibbs energy)=86.6
O2: H(enthalpy)=0, S(entropy)=?, G(gibbs energy)=0 kJ/mol
NO2: H(enthalpy)=33.2, S=239.9, G=51
For part a, I have to calcuate G for the reaction.
I used the Gibbs free standard reaction formula:where i have to find the sum of products minus the sum of reactants.
I got 71.2 kJ/mol. I know I got that right because my friend got the same answer too.
for part b, I have to calculate standard entropy, S for only O2.
There's a table in my book saying that entropy of O2 is 205 J/mol*K.
But I have to solve this using formulas and I was wondering if you can help on how to approach this problem.

The temperature of the reaction is 298 kelvin

This is what I did:
my teacher said that this is a two step calculation:
First step:equation of heat
q=mct
m=mass of O2=32 grams
T=298+273=573 Celsius
q=32 g of O2(4.184J/g C)(573 C)
q=76450J/mol
then I used the equation:
S(entropy)=q/T
S=76450J/mol / (298 K)
=257 J/mol*K
I looked at a table in my chemistry book, and it says that S of O2 is 205 J/mol*K.
I'm assuming that's the answer I should reach, but I didn't.
So please see what I did wrong.
Thank you 
298 K is not 573 C. It is 25 C
Entropy at 298 requires an integration of C dT/T from 0 K to 298, including solid and liquid phases. It cannot be calculated with a simple formula. The 205 J/mol K number for 298 K agrees with my JANNAF Tables.