Given that g for the reaction below is –957.9 kj, what is gf of h2o? 4nh3(g) + 5o2(g) =4no(g) + 6h2o(g)

gf,

nh3 = -16.66 kj/mol

gf, no = 86.71 kj/mol –228.6 kj/mol –206.4 kj/mol 46.7 kj/mol 90.7 kj/mol

- 228.6 kJ/mol.

Explanation:

∵ ΔGrxn = ∑ΔGf, products - ∑ΔGf, reactants

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O,  

∴ ΔGrxn = [6(ΔGf, H₂O) + 4(ΔGf, NO)] - [4(ΔGf, NH₃) + 5(ΔGf, O₂)]

ΔGrxn = – 957.9 kJ, ΔGf, H₂O = ??? KJ/mol, ΔGf, NO = 86.71 kJ/mol, ΔGf, NH₃ = - 16.66 kJ/mol, ΔGf, O₂ = 0.0 kJ/mol.

∴ – 957.9 kJ = [6(ΔGf, H₂O) + 4(86.71 kJ/mol)] - [4(- 16.66 kJ/mol) + 5(0.0 kJ/mol)].

– 957.9 kJ = 6(ΔGf, H₂O) + 346.8 kJ/mol + 66.64 kJ/mol

– 957.9 kJ = 6(ΔGf, H₂O) + 413.8 kJ/mol

6(ΔGf, H₂O) = – 957.9 kJ - 413.8 kJ/mol = - 1372 kJ/mol.

∴ ΔGf, H₂O = (- 1372 kJ/mol)/6 = - 228.6 kJ/mol.

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.