Part-9

The Gibbs Free Energy and equilibria

• “Gibbs Free Energy” is energy that is still useful. • A chemical reaction will occur if the Gibbs would decrease.

G = H - TS

Gibbs free energy is a measure of chemical energyGibbs free energy is a measure of chemical energyAll chemical systems tend naturally toward states of minimum

Gibbs free energyG = Gibbs Free EnergyH = Enthalpy (heat content)T = Temperature in KelvinsS = Entropy (can think of as

randomness)

Gibbs free energy also known as the free enthalpy

Is a thermodynamic potential that measures the maximum or reversible work that may be performed by a system at a constant temperature and pressure (Isothermal, Isobaric)

Spontaneity and Gibbs Free Energy

• Gibbs Free energy is a measure of the spontaneity of a process

• ΔG is the free energy change for a reaction under standard state

conditions

• At constant temperature and pressure: ΔG = ΔH – TΔS

– an increase in ΔS leads to a decrease in ΔG

– – if if ΔΔG < 0, the forward reaction is spontaneousG < 0, the forward reaction is spontaneous

– – if if ΔΔG > 0, the forward reaction is nonspontaneousG > 0, the forward reaction is nonspontaneous

– – if if ΔΔG = 0, the process is in equilibriumG = 0, the process is in equilibrium

• The Gibbs Free Energy is generally agreed to be the “weapon of choice” for describing (a) chemical reactions and (b) equilibria between phases. It is defined as:

• G = H – TS = U + PV – TS (1) Where H = Enthalpy• U = Total internal energy• T = [Absolute] Temperature• S = Entropy

• Obviously dG = dU + PdV +VdP – TdS – SdT

The Gibbs Free Energy and equilibria

• Remember that thermodynamic variables come in pairs One is “intrinsic” (does not depend on system size) The other is “extrinsic” (depends on system size)• Examples: P and V, T and S…

• Also G and n, the number of moles of stuff in the system.• Hence G is the appropriate variable when material is moving between

phases

Note:

From the First Law of Thermodynamics

• dU = TdS – PdV since dS = dQ/T and the mechanical work done on a system

when it expands is –PdV.• Substituting into • dG = dU + PdV +VdP – TdS – SdT

• Leaves: dG = -SdT + VdP

Clapeyron’s Equation

Closed System• Closed system contains pure substance

– vapor – condensed phase

• Phases co-exist in equilibrium.

Write the Free Energy Equation twice

• Once for each phase

• dGc = -ScdT + VcdP c refers to the condensed phase

• dGv = -SvdT + VvdP v refers to the vapor phase

Definition of chemical equilibrium between two phases

• Free energy is the same in both phases Gc = Gv

• Changes in free energy when some independent variable is changed must be the same if they are to remain in equilibrium

dGc = dGv

-ScdT + VcdP = -SvdT + VvdP (Sv - Sc )dT = (Vv- Vc)dP • (Sv - Sc ) is the entropy change that takes place when material moves from the condensed phase to the vapor•ΔS = ΔQ/T where ΔQ is the amount of heat required per mole of material moved between the phases•ΔQ is just the heat of vaporization!

• dP/dT = (Sv – Sc)/(Vv – Vc) = ΔHv/(TΔV)

This is the Clapeyron equation

• It relates the change in pressure of a vapor to the temperature in a closed, mono-component system to the heat of vaporization, system temperature and molar volume change of the material on vaporization.

dP SordT V

From the Clapeyron’s Equation we can calculate phase diagrams.

H=U+PV=Q

Creating of an Ideal Gas• For lack of a better model, we treat most vapors as ideal gases, whose

molar volume is given by:• V/n = RT/P• Alternatively, equation of state is needed• Molar volume of gas is typically factor of 500 larger than condensed phase• Hence Vc is negligible in comparison

Substituting and IntegratingdP = (ΔHv/Vv)dT/T = (PΔHv/RT)dT/T

dP/P = ΔHv/R)dT/T2

ln(P(T)/ P0) = -(ΔHv/R)(1/T – 1/T0)

P(T) = P0 exp(-ΔHv/R(1/T – 1/T0))

Integrating

• The vapor pressure in equilibrium with a condensed phase increases exponentially (sort of: exp(-1/T) isn’t exactly an exponential!) with temperature from zero up to the critical temperature.

• Deviations from linearity on the log-log plot– Temperature dependence of the heat of vaporization – exp (-1/T) isn’t really linear in the exponent.

Heat of Vaporization from CRC Data Log10p(Torr) = -0.2185*A/T + B

Vapor Pressure of Water

Temperature (C)-20 0 20 40 60 80 100 120

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"Normal boiling point"

1. Determine the vapor pressure at 77 K fora. Waterb. Carbon monoxide

2. What is the boiling point of water in a vacuum system at 10-6 Torr?

HW

3. In the chemical equation G = H - TS, the term G stands forA) entropyB) the reactantsC) enthalpyD) free energyE) the products