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Calculate the boiling point of the water .
Solution: It shall apply: H2O(l) ⇄ H2O(g).
For this chemical equilibrium is: K = c(H2O(g))/c(H2O(l)) = 1
With dG = -R·T·ln(K) followed by K = 1: dG = 0
Furthermore: dG = dH-T·dS (=0)
thus T = dH/dS- T = dH/dS
ΔHoR = ΔHof(H2O, g) – ΔHof(H2O, l)
ΔHoR = -241.82 – (-285.83) = 44.01 kJ/mol
ΔSoR = So(H2O, g) – So(H2O, l)
ΔSoR = 188.85 – 69.91 = 118.92 J/(mol·K) = 0.11892 kJ(mol·K) - T = dH/dS = 44.01 / 0.11892 = 370 K or Approx. 97℃
The calculated (!) seating temperature of the water is therefore approximately 97℃ The deviation from the usual value of 100℃ is therefore due to the fact that dH or dS actually also (little) depend on the temperature.
- T = dH/dS
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