Chem Ch 23/#2 Today’s To Do List

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Presentation transcript:

Chem 300 - Ch 23/#2 Today’s To Do List Gibbs & Phase Stability Chemical Potential Phase Equilibrium Clapeyron Equation Clausius- Clapeyron Equation

Another Look at G(P) dGT = +VmdP Vm(g) >> Vm(l)  Vm(s)

Gm(P): (a) most substances (Vms < Vml), (b) H2O (Vms > Vml)

How does G(P) look at different Temperatures?

Gm(P) vs P (a) T<Ttr (b) T=Ttr (c) T<Tcr (d) T>Tcr

Phase Equilibrium & Chem Potential dG = 0 for 2 phases in equilib. For 2 phases 1 & 2 in general: dG = dG1 + dG2 If dn mols are moved [G = f(n1, n2)]: dG = (G1/  n1)P,T dn1 + ( G2/  n2)P,T dn2 But dn2 = - dn1 dG = [( G1/  n1)P,T - ( G2/  n2)P,T ]dn1

Continued Chemical Potential (m): m1 = ( G1/  n1)P,T dG = (m1 - m2)dn1 (const T, P) For 2 phases to be in equilibrium, the m’s must be equal.

The Clapeyron Equation For a single substance: m = ( G/  n)P,T = G/n = Gm For 2 phases in equilibrium (s, l, g): ma = mb  Ga = Gb dGa = dGb -Sa dt + Va dP = -Sb dt + Vb dP dP/dT = (Sb – Sa)/(Vb –Va) dP/dT = DtrS/DtrV  DtrS = DtrH/Ttr dP/dT = DtrH/ TtrDtrV Clapeyron Equation

High-Pressure Phase Diagram of H2O (Again)

Clapeyron & Clausius-Clapeyron eqs. dP/dT = DtrH/ TtrDtrV Clapeyron Eq. Any 2 phases V/L or V/S phases: DtrV = Vg – Vl or s  Vg (Vg >> Vl or s ) Vapor  Ideal Gas  Vg = RT/P (1/P)dP/dT = DvapH/RT2 d lnP/dT = DvapH/RT2 Clausius- Clapeyron eq.

C-C Eq. Integrated d lnP = (DvapH/RT2) dT Assume: Integrate: DvapH is constant Integrate: ln (P2/P1) = (DvapH/R)(1/T2 – 1/T1) In general: ln P = (DvapH/R)(1/T) + const

LnP vs 1/T for benzene

L/V line in H2O

Next Time Start Chapter 24 Partial Molar Quantities Gibbs-Duhem Equation Raoult’s Law & The Ideal Solution