Posted by **apoorva** on Monday, February 4, 2008 at 12:12am.

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.

- i forgot to write down another part of question -
**apoorva**, Monday, February 4, 2008 at 1:28am
The temperature of the reaction is 298 kelvin

- tried solving it, please check answer -
**apoorva**, Monday, February 4, 2008 at 2:12am
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

- chemistry, please help -
**drwls**, Monday, February 4, 2008 at 8:51am
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.

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