What is the ∆g for the following reaction under standard conditions (t = 298 k) for the formation of nh4no3(s)?
2nh3(g) + 2o2(g) nh4no3(s) + h2o(l)
given:
nh4no3(s): ∆hf = -365.56 kj ∆sf = 151.08 j/k.
nh3(g): ∆hf = -46.11 kj ∆sf = 192.45 j/k.
h2o(l): ∆hf = -285.830 kj ∆sf = 69.91 j/k.
o2(g): ∆hf = 0.00 kj ∆sf = 205 j/k.
a. 186.6 kj
b. 6.9 kj
c. -10.4 kj
d. -126.3 kj
e. -382 kj

Under constant temperature and pressure the change in free energy (ΔG) equals the  difference between the change in enthalpy (ΔH) of the system and the product of the temperature, T, times the change in the entropy (ΔS).

This is: ΔG = ΔH - TΔS

The reaction is:

2NH3(g) + 2O2(g) -> NH4NO3(s) + H2O(l)

So,

ΔGf = ΔGf of the products - ΔGf of the reactants

ΔGf products = ΔHf products - TΔS products

ΔGf reactants = ΔHf reactants - TΔS reactants

T = 298 K

ΔHf products = ΔHf NH4NO3(s) + ΔHf H2O(l) = -365.56 kJ - 285.830 kJ =  -651.390 kJ

TΔSf products = TΔSf NH4NO3 + TΔSf H2O(l) = 298K (151.08 J/K + 69.91 J/k) =  65,855.02 J = 65.855 kJ

ΔGf products = -651.390kJ - 65.855kJ = -717.245 J

ΔHf reactants = 2ΔHf NH3(g) + 2ΔHf O2(g)  = 2*(-46.11kJ) + 2*(0.00kJ) = -92.22 kJ

TΔSf reactants = 298k*2*(ΔSf NH3) + 298k*2*(ΔSf O2) =298k*2(192.45 J/K + 205 J/K) = 236,880.2 J = 236.880 kJ

ΔG f reactants = -92.22 kJ - 236.880kJ = -329.100 kJ

ΔGf = ΔG f products - ΔG f reactants = -717.245kJ - (-329.100kJ) = -388.145 kJ

- 388.145 kJ