The enthalpy of formation of liquid water at 25°C is -286 kJ. Given C P = 75 J/K-mol for H 2 O(l) and 34 J/K-mol for H 2 O(g) and that the molar enthalpy of vaporization of liquid water at 100°C is 40

Question

The enthalpy of formation of liquid water at 25°C is -286 kJ. Given C P = 75 J/K-mol for H 2 O(l) and 34 J/K-mol for H 2 O(g) and that the molar enthalpy of vaporization of liquid water at 100°C is 40.6 kJ/mol. The enthalpy of dissociation (in kJ/mol) of H 2 O(g) into H 2 and O 2 gases at 25°C is

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ΔH vap (298) = ΔH vap (373) + [C P (g) - C P (l)](298-373) = 40.6 + (34-75)×10 -3 ×(-75) = 40.6 + 3.08 = 43.7 kJ/mol. Then Δ f H(H 2 O,g) = -286 + 43.7 = -242.3 kJ/mol, so dissociation of H 2 O(g) requires ≈ +242 kJ/mol. Answer: 0242.

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