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## The free energy surface

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**The free energy surface**dG1 = V1dp1 – S1dT1 dG2 = V2dp2 – S2dT2 CHEM 471: Physical Chemistry**Chapter 4: the exact Clapeyron equation**ΔfG = G1 – G2 (blue surface) dΔfG = ΔfVdp – ΔfSdT At a reversible phase transition, the free energy change is zero: ΔfG = 0 (Grey surface) dΔfG = 0 = ΔfVdp – ΔfSdT ΔfVdp = ΔfSdT = ΔfHdT/T This is the exact Clapeyron equation CHEM 471: Physical Chemistry**The Clausius-Clapeyron equation**Assumption 3: ΔvH is constant: We can do either a definite or an indefinite integral: first, definite Now, indefinite: At some standard pressure p° (usually 1 bar), ln p° = 0. We call the boiling point at this standard pressure Tband the molar enthalpy of vaporization ΔvH° CHEM 471: Physical Chemistry**Chapter 4: phase transitions**There are 2 degrees of freedom in a closed system, and one equation of state CHEM 471: Physical Chemistry At a phase boundary, there is one degree of freedom, and a phase-change equation At the triple point, there are no degrees of freedom**Inexact Clapeyron**We make three assumptions for liquid-gas phase transitions: (1) The volume change of vaporization is just the volume of the vapor At 1 bar and 298 K, Vm, water = Mwater/ρwater = (18.017 g/mol)/(0.9984 g/mL) = 18.045 mL/mol Vm, water vap. = RT/p = 24.79 L/mol >> Vm, water (2) The vapor is an ideal gas (3) The enthalpy of vaporization is temperature independent This is the weakest of the three, but we can remove it later Let’s take the exact Clapeyron equation, and work on the volume CHEM 471: Physical Chemistry**The Clausius-Clapeyron equation**Assumption 3: ΔvH is constant: We can do either a definite or an indefinite integral: first, definite Now, indefinite: At some standard pressure p° (usually 1 bar), ln p° = 0. We call the boiling point at this standard pressure Tband the molar enthalpy of vaporization ΔvH° CHEM 471: Physical Chemistry**Clausius-Clapeyron equation:example**Original Improved CHEM 471: Physical Chemistry Improved version**Clausius-Clapeyron equation:example**Fit to Clausius-Clapeyron equation gives ΔvH° = 41.300 kJ/mol ΔvS° = 109.0 J mol–1 K–1 CHEM 471: Physical Chemistry