Chapter 5: Gibbs Free Energy

Chapter 5-Gibbs Free Energy

Problems: 5.1,3,5,11; concept:5.9,19,21,22,23

1.G-energy to drive chemical reactions

-ΔpG- tabulated (in table)

-ΔrxnG- usually calculated

eq.1: ΔrG- ∑v*Δ_fG_products –∑v*Δ_fG_reactants

-if 0, reaction is at equilibrium

-if negative, reactions proceeds towards right (spontaneous)

-if positive, reaction will not proceed (non-spontaneous)

2.Phase Transitions

-Phase 1 --> Phase 2:

eq.2:ΔrxnG= n* ΔfG_{phase 2} – n* ΔfG_phase1 (ΔG is molar)

-if ΔGf phase 2 > phase 1, ΔGrxn is positive (non-spontaneous)

-if ΔGf phase 1 > phase 2, ΔGrxn is negative (spontaneous)

-if ΔGf phase 1 = phase 2, ΔGrxn is 0 (equilbrium)

-stable phase- phase with lowest ΔGf

3.Molar ΔGf vs Pressure- ΔGf,m or ΔGm = Vm*ΔP

-ΔG is directly proportional to changes in pressure

-larger molar volume- ΔG changes more as pressure changes for substances with larger molar volumes

eq.3:Vm=ΔG/ΔP

-solids,liquids: final ΔGm= ΔGm,i + Vm(Pf-Pi), where pressure is in bars

-gases: final ΔGm*(Pf)= ΔGm*(Pi)+RTln(Pf/Pi)

-reactions:  ΔGrxn= ∑V*ΔGm,final (products) – ∑V*ΔGm,final (reactants)

4.Molar ΔGf vs Temperature

-for small changes in T, entropy values are nearly constant

eq.4:ΔGm=-Sm*ΔT

5.Phase boundaries- on phase diagrams, show the pressure and temperature combinations at which 2+ phases are stable

-liquid-vapor boundary-liquid is in contact with and in equilbrium with a gas of that composition; pressure of the vapor is its vapor pressure, which substantially increases with temperature

-solid-vapor boundary-solid in contact with and in equilbrium with vapor; sublimination vapor pressure of the solid can be determined in the same way as the liquid vapor pressure

-slope- slope of boundary determined by thermodynamic properties

 6. Clapeyron equation- finds slope of boundary (ΔP/ΔT), ONLY fo small changes in P and T

-ΔtrsH- change of heat during transition between phases (where temperature and pressure stay constant)

eq.5: ΔP/ΔT=ΔtrsH/T*ΔtrsV

7.Clausius-Clapeyron equation: Δ(lnP)= ΔvapH/RT² *ΔT

-liquid-gas boundaries

eq.6:lnP’=lnP+ΔvapH/R*(1/T-1/T’), where P’ is final P, and T’ is final T

-Table 5.1- log(P in kPa)=A-(B/T), where A and B are constants (see derivation on p.113)

8.Critical point-terminal point on liquid-gas boundary curve

-highest pressure at which liquid can be condensed

-fluid- above critical point, state of matter is called “fluid”, no boundary

-observed between states of matter as seen between liquid and gas

9.Normal boiling point- boiling temperature at 1 atm

-Standard boiling point- boiling temperature at 1 bar

-Normal boiling point- melting temperature at 1 atm

-Standard melting point- melting temperature at 1 bar

10. Phase Rule- for a system at equilbrium, F=C-P+2

-F=degrees of freedom, C=number of component, P=number of phases, 2=T+P

-components- minimum number of species necessary to define all the phases present in a system

-degrees of freedom- number of intensive variables (P,T,mol fraction, etc.) that can be changed without disturbing equilbrium

-triple point- F=0 (C=1, P=3)

-phase boundary- F=1 (C=1, P=2)