Free Energy and Phase equilibria Thermodynamic integration 7.1 - PowerPoint PPT Presentation

Free Energy and Phase equilibria Thermodynamic integration 7.1 Chemical potentials 7.2 Overlapping distributions 7.2 Umbrella sampling 7.4 Application: Phase diagram of Carbon

Why free energies? • Reaction equilibrium constants A ↔ B K = [ B ] [ A ] = p B [ ] = exp − β ( G B − G A ) p A • Examples: – Chemical reactions, catalysis, etc.... – Protein folding, binding: free energy gives binding constants • Phase diagrams – Prediction of thermodynamic stability of phases – Coexistence lines – Critical points – Triple points – First order/second order phase transitions

Phase diagrams Along the liquid gas coexistence line increasing the pressure and Critical point: no difference between temperature at constant volume the liquid and vapor liquid density becomes lower and the Triple point: liquid, vapor and solid in vapor density higher. equilibrium. Carbon Phase Diagram How do we compute these lines?

Phase equilibrium Criteria for equilibrium (for single component) T I = T II P I = P II µ I = µ II Chemical potential # & # & µ = ∂ F = ∂ G = G m % ( % ( $ ∂ N ' $ ∂ N ' V , T P , T If µ I > µ II : transport of particles from phase I to phase II. Stable phase: Lowest chemical potential (for single phase: lowest Gibbs free energy)

Relation thermodynamic potentials Helmholtz free energy: F = U - TS Gibbs free energy: G = F + PV Suppose we have F(n,V,T) Then we can find G from F from: # & G = F − ∂ F $ ' P = − ∂ F V % ( & ) $ ∂ V ' ∂ V % ( n , T n , T All thermodynamic quantities can be derived from F and its derivatives

Phase equilibria from F(V,T) Common tangent construction liquid Equal tangents F $ ' P = − ∂ F & ) % ∂ V ( n , T gas Connecting line: equal ! V

Common tangent construction Helmholtz Free Energy Perspective F ∂ & # G F V = − $ ! liquid V F ∂ % " n , T gas V

Common tangent construction Gibbs Free Energy Perspective $ ' G = F − V ∂ F & ) G ! ∂ V % ( n , T Both liquid and vapor G equal and minimal liquid gas V ! Only equilibrium when P,T is on coexistence line.

We need F or µ • So equilibrium from F(V) alone or from P and µ # & ∂ F V ∫ ∫ F ( V ) = F ( V 0 ) + dV = F ( V 0 ) − P dV % ( ∂ V $ ' V 0 N,T P ( # ρ ) ρ ∫ F ( ρ ) = F ( ρ 0 ) + N d # ρ ρ 2 ρ 0 • So in fact for only 1 point of the equation of state the F is needed • For liquid e.o.s even from ideal gas β P ( $ ρ ) - $ ρ β F ( ρ )/ N = β F id ( ρ )/ N + ρ ∫ d $ ρ ρ 2 0

Equation of state ( ) P = P ρ , T # & ∂ F = − P % ( ∂ V $ ' N , T P ( # ρ ) ρ ∫ F ( ρ ) = F ( ρ 0 ) + N d # ρ ρ 2 ρ 0 β P ( $ ρ ) - $ ρ β F ( ρ )/ N = β F id ( ρ )/ N + ρ ∫ d $ ρ ρ 2 0

Free Energies and Phase Equilibria General Strategies • Determine free energy of both phases relative to a reference state Free energy difference calculation General applicable: Gas, Liquid, Solid, Inhomogeneous systems, … • Determine free energy difference between two phases Gibbs Ensemble (Lecture Thijs Vlugt) Specific applicable: Gas, Liquid

Statistical Thermodynamics Probability to find a particular configuration 1 1 δ r' N − r N [ ] ∝ exp − β U r N [ ] ( ) = ( ) exp − β U r' N ( ) ( ) P r N ∫ dr' N Λ 3 N N ! Q NVT Partition function 1 [ ] ( ) ∫ dr N exp − β U r N Q NVT = Λ 3 N N ! Free energy ( ) F ln Q β = − NVT Ensemble average 1 1 [ ] ( ) exp − β U r N ( ) ∫ dr N A r N A NVT = Λ 3 N N ! Q NVT

Ensemble average versus free energy N ! ,r M { N ,r 2 N ,r 3 N ,r 4 N } r Generate configuration using MC: 1 dr N A r N M [ ] ( ) ( ) ∫ exp − β U r N A = 1 ( ) ∑ N A r i = A NVT ≈ dr N exp − β U r N M [ ] ( ) ∫ i = 1 N ! ,r M { N ,r 2 N ,r 3 N ,r 4 N } r Generate configuration using MD: 1 M T A = 1 ( ) ≈ 1 ∑ N ∫ ∫ A r i dtA ( t ) ≈ A M T NVT ergodicity i = 1 0 1 [ ] ( ) ∫ dr N exp −β U r N β F = − ln Q NVT = − ln Λ 3 N N ! F is difficult, because requires accounting of phase space volume

I - Thermodynamic integration • Known reference state λ =0 • Unknown target state λ =1 Reference System Coupling parameter Target System ( ) = 1 − λ ( ) U I + λ U II U λ 1 ∫ dr N [ ] ( ) = ( ) Q NVT λ exp − β U λ Λ 3 N N ! # & λ = 1 ( ) d λ ∂ F λ ∫ F ( λ = 1) − F ( λ = 0) = % ( ∂ λ $ ' λ = 0 N , V , T

Thermodynamic integration $ ' ( ) ∂ F λ = − 1 ( ) = − 1 1 ∂ Q ∂ ∂ λ ln Q & ) Q ∂ λ β β ∂ λ % ( N , T dr N ∂ U λ ∫ ( ) [ ] ( ) ∂ λ ( ) exp − β U λ = ∫ dr N [ ] ( ) exp − β U λ ( ) U ∂ λ = ∂ λ λ Free energy as ensemble average! ( ) ∫ λ ∂ U λ ( ) − F λ = 0 ( ) = F λ = 1 d ∂ λ λ

Example • In general ( ) = (1 − λ ) U I + λ U II U λ ( ) ∂ U λ = U II − U I λ ∂ λ λ • Specific example ( ) = U LJ + λ U dipole-dipole U λ ( ) = U LJ U 0 Lennard-Jones ( ) = U Stockm U 1 Stockmayer ( ) ∂ U λ = U dip − dip ∂ λ λ λ

Free energy of solid More difficult. What is reference? Not the ideal gas. Instead it is the Einstein crystal: harmonic oscillators around r 0 N ( ) = (1 − λ ) U r N ( ) + λ U r ( ) + λ U λ ; r N N ∑ i ) 2 α ( r i − r 0 i = 1 λ = 1 ( ) d λ ∂ U λ Note, here: _ ∫ F = F ein + λ = 1 Reference System ∂ λ λ = 0 Target System λ = 0 λ λ = 1 N _ ( ) + ( ) + U r N ∫ − U r N ∑ i ) 2 F = F ein + d λ α ( r i − r 0 λ = 0 i = 1 λ

Hard sphere freezing P Solid free energy from liquid free Einstein crystal energy from Ideal gas Equal µ/ µ/ P (and T) ρ

II - Thermodynamic perturbation Two systems: System 0: N, V, T, U 0 System 1: N, V, T, U 1 Q 0 = V N 1 = V N ∫ d s N ∫ d s N ( ) ( ) exp − β U 0 Q exp − β U 1 Λ 3 N N ! Λ 3 N N ! ( ) Δ β F = β F 1 − β F 0 = − ln Q 1 Q 0 d s N exp − β U 1 $ & ∫ % ' = − ln d s N ( ) exp − β U 0 ∫ $ & d s N ( ) $ & exp − β U 1 − U 0 ' exp − β U 0 ∫ % ' % = − ln d s N ( ) exp − β U 0 ∫ F ln exp U U ( ) Δ β = − − β − $ % & ' 1 0 0

Chemical potential Q NVT = V N [ ] ( ) ∫ ds N exp − β U s N ; L Λ 3 N N ! ( ) β F = − ln Q NVT $ ' V N ( ) [ ] ( ) ∫ ds N exp − β U s N ; L = − ln ) − ln & Λ 3 N N ! % ( % ( 1 ( ) [ ] ( ) ∫ ds N exp − β U s N ; L = − N ln * + N − ln β F = β F IG + β F ex } ' Λ 3 ρ & ) β µ = β µ IG + β µ ex IG ex F F & # & # ∂ β ∂ β $ ' µ ≡ ∂ F IG ex $ ! $ ! β µ ≡ β µ ≡ & ) $ ! $ ! N N ∂ ∂ % ∂ N ( % " % " V , T V , T V , T

Widom test particle insertion ( ) − β F N ) ( ) F β µ = β F N + 1) ∂ β & # β µ ≡ $ ! N N + 1 − N ∂ % " V , T ( ) = − ln Q N + 1 ( ) Q N $ ' V N + 1 & ) $ ' Λ 3 N + 3 N + 1 [ ] ( ) ∫ ds N + 1 exp − β U s N + 1 ; L ( ) ! & ) & ) = − ln − ln V N [ ] & ) ( ) & ) ∫ ds N exp − β U s N ; L % ( & ) Λ 3 N N ! % ( $ ' [ ] ( ) ∫ ds N + 1 exp − β U s N + 1 ; L $ ' V & ) = − ln ) − ln & Λ 3 N + 1 & [ ] ) ( ) ( ) ∫ ds N exp − β U s N ; L % ( % ( β µ = β µ IG + β µ ex % ( [ ] ( ) ∫ ds N + 1 exp − β U s N + 1 ; L β µ ex = − ln ' * ' [ ] * ( ) ∫ ds N exp − β U s N ; L & )

Widom test particle insertion % ( [ ] ( ) ∫ ds N + 1 exp − β U s N + 1 ; L β µ ex = − ln ' * [ ] ' * ( ) ∫ ds N exp − β U s N ; L & ) ) = Δ U + + U s N ; L ( ( ) U s N + 1 ; L & ) exp − β Δ U + + U s N ; L [ ] ( ) ( ) ∫ ds N ∫ ds N + 1 β µ ex = − ln ( + ( + [ ] ( ) ∫ ds N exp − β U s N ; L ' * & ) [ ] { [ ] } exp − β U s N ; L ( ) ∫ ∫ ds N exp − β Δ U + ds N + 1 ( + = − ln ds N exp − β U s N ; L [ ] ( + ( ) ∫ ' * ( ) Ghost particle! [ ] NVT ∫ exp − β Δ U + = − ln ds N + 1

Hard spheres β µ ex = − ln ( ) [ ] NVT ∫ exp − β Δ U + ds N + 1 % r ≤ σ ( ) = ∞ U r & 0 r > σ ' % ] = 0 if overlap [ exp − β Δ U + & 1 no overlap ' [ ] exp − β Δ U + probability to insert a test particle! But, … may fail at high density

III - Overlapping Distribution Method Two systems: System 0: N, V,T, U 0 System 1: N, V,T, U 1 V N 1 = V N d s N ∫ ∫ d s N ( ) Q 0 = exp − β U 0 ( ) Q exp − β U 1 Λ 3 N N ! Λ 3 N N ! # & d s N ∫ ( ) exp − β U 1 # & ( = − ln Q 1 % ( ( ) = − ln Δ β F = β F 1 − β F 0 = − ln Q 1 Q 0 % ( % ∫ d s N ( ) Q 0 exp − β U 0 $ ' $ ' = Δ U ( δ function) ∫ d s N ∫ d s N ( ) δ U 1 − U 0 − Δ U ( ) ( ) δ U 1 − U 0 − Δ U ( ) exp − β U 1 exp − β U 0 ( ) = ( ) = p 1 Δ U p 0 Δ U ∫ d s N ∫ d s N ( ) exp − β U 1 ( ) exp − β U 0 ∫ d s N [ ] exp − β U 0 ( ) [ ] δ U 1 − U 0 − Δ U ( ) exp − β U 1 − U 0 ( ) = p 1 Δ U ∫ d s N ( ) exp − β U 1 = 1 = Q 0 1 Q 0 ( ) = exp β Δ F d s N Q Q Q 0 ∫ [ ] δ U 1 − U 0 − Δ U ( ) exp − β U 0 = Q 0 1 1 Q ( ) exp − β Δ U 1 Q 1 Q 0 ) = Q 0 ( ( ) p 0 Δ U ( ) p 1 Δ U exp − β Δ U ( ) = β Δ F − Δ U ( ) + ln p 0 Δ U ( ) ln p 1 Δ U Q 1