ΔG for the formation of H₂O is -228.6 kJ.
<h3>How we calculate
Gibb's free energy of the reaction?</h3>
Gibb's free energy of the reaction is calculated as:
ΔG = G for product - G for reactant
Given chemical reaction is:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
In the question, given that:
ΔG for the reaction = -957. 9 kJ
ΔGf of NH₃ = -16. 66 kJ/mol
ΔGf of NO = 86. 71 kJ/mol
Equation for ΔG will be written as:
ΔG = (ΔGf of NO + ΔGf of H₂O) - (ΔGf of NH₃+ ΔGf of O₂)
ΔGf of O₂ = 0
-957. 9 = (4×86. 71 + 6×ΔGf of H₂O) - (4×-16. 66 + 5×ΔGf of O₂)
-957. 9 = 346.84 + 6ΔGf of H₂O + 66.64
ΔGf of H₂O = (-957. 9 - 346.84 - 66.64) / 6
ΔGf of H₂O = -228.56 kJ ≅ -228.6 kJ
Hence, option (1) is correct i.e. -228.6 kJ is the ΔGf of H₂O.
To know more about Gibb's free energy, visit the below link:
brainly.com/question/14415025
