Question

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.

Answer 1

∆G =  ∆H - T∆S

NH3: ∆G = -46.11x10^3 - (298)(192.45) = -103460.1 J
O2: ∆G = 0 - (298)(205) = -61090 J
NH4NO3: ∆G = -365.56x10^3 - (298)(151.08) = -410581.84 J
H2O: ∆G = -285.830x10^3 - (298)69.91) = -306663.18 J

∆Grex = ∆Gproducts - ∆Greactants
∆Grex = (-410581.84 +  -306663.18) - (-103460.1/2 + -61090/2)
∆Grex =-634969.97 J/mol = -634.97 kJ/mol