Calculating free energy change (Topscore test)

[soxinabox90]

Hello,

Question #50 on topscore test 1 is:

Given that the standard enthalpy of formation for NO(g) is 90.25 kJ/Mole, calculate the free energy change for the following reaction at 25C.

N2(g) + O2(g) <-> 2NO(g)

The correct answer is 180.5 - (298)(25)(1x10^-3)

Their solution simply says "Determine the entropy change for the reaction"

Can somebody please help explain how to find the answer for this problem. I understand it has to do with the Gibbs free energy equation G = H - TS, but I'm not sure exactly how each piece fits into the puzzle.

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[Lnguyen7]

ΔSrxn = Sum mole*S(products) - Sum mole*S(reactants)
ΔS = 2(210.6) - (191.5 + 205.0) = 24.7 J/(K*mol)

ΔHrxn = Sum mole*ΔHf(products) - Sum mole*ΔHf(reactants)
ΔH = 2(90.25) - (0 + 0) = 180.5 kJ/mol

ΔG = ΔH - TΔS = 180.5 kJ/mol - (25+273)K*24.7 J/(K.mol)*( 1 kJ/1000 J)

Kinda funny that they gave you ΔHf of NO and not a single S of NO/N2/O2?

You have to look up S of NO/N2/O2.