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

Use the formula for ΔHrxn= (sum product) - (sum reactants)
and formula for ΔSrxn= (sum product) - (sum reactants)
and ΔG= ΔH-TΔS

there are different ways to solve this question, but I think the simplest is to find the ΔG of each one of the substance and solve for ΔG as (sum product) - (sum reactants). also because the answers are in Kj, you need to convert the ΔS to Kj by simply dividing the value by 1000. 

ΔG= ΔH-TΔS

ΔGNH4NO3= (-365.56) - (298 X 151.08x10-3)= -410.58
ΔG NH3= (-46.11) - (298 X 192.45x10-3)= -103.46
ΔG H2O= (-285.830) - (298 X 69.91x10-3)= -306.66
ΔG O2= (0) - (298 X 205x10-3)= -61.09

ΔGrxn= product- reactants

ΔGrxn= [∆Gf(NH4NO3) + ∆Gf(H2O)] - [2 ∆Gf(NH3) + 2 ∆Gf(O2)] 
ΔGrxn= [ -410.58 + -306.66] - [ (2 x -103.46) + (2 x -61.09)]= -388.14 kj

the answers is closest to E