The enthalpy of formation of liquid water at 25°C is -286 kJ. Given C P = 75.4 J/K-mol for H 2 O(l) and 33.4 J/K-mol for H 2 O(g) and that the molar enthalpy of vaporization of liquid water at 125°C i

Question

The enthalpy of formation of liquid water at 25°C is -286 kJ. Given C P = 75.4 J/K-mol for H 2 O(l) and 33.4 J/K-mol for H 2 O(g) and that the molar enthalpy of vaporization of liquid water at 125°C is 40.8 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

TestHub · Accepted Answer

The enthalpy of vaporization at 298 K, Δ H_vap(298), can be calculated using the enthalpy of vaporization at 398 K, Δ H_vap(398), and the heat capacities of the gaseous and liquid phases. The formula is:

Δ H_vap(298) = Δ H_vap(398) + [C_P(g) - C_P(l)](298 - 398)

Substituting the given values:

Δ H_vap(298) = 40.8 + (33.4 - 75.4) × 10^⁻³ × (-100)

Δ H_vap(298) = 40.8 + 4.2 = 45.0 kJ/mol

Next, we calculate the standard enthalpy of formation of gaseous water, Δ_f H(H₂O,g).

Δ_f H(H₂O,g) = -286 + 45 = -241 kJ/mol

Therefore, the dissociation of H_₂O(g) requires +241 kJ/mol.

Answer: 0241.

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