Question

Given that Delta. G for the reaction below is –957. 9 kJ, what is Delta. Gf of H2O? 4NH3(g) 5O2(g) Right arrow. 4NO(g) 6H2O(g) Delta. Gf,NH3 = -16. 66 kJ/mol Delta. Gf,NO = 86. 71 kJ/mol –228. 6 kJ/mol –206. 4 kJ/mol 46. 7 kJ/mol 90. 7 kJ/mol.

Answer 1

ΔG for the formation of H₂O is -228.6 kJ.

How we calculate Gibb's free energy of the reaction?

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:

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