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The free energy surfacePowerPoint Presentation

The free energy surface

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The free energy surface

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dG1 = V1dp1 – S1dT1

dG2 = V2dp2 – S2dT2

CHEM 471: Physical Chemistry

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

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

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

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

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

Original

Improved

CHEM 471: Physical Chemistry

Improved version

Fit to Clausius-Clapeyron equation gives ΔvH° = 41.300 kJ/mol

ΔvS° = 109.0 J mol–1 K–1

CHEM 471: Physical Chemistry