Equilibrium and Stability Phase Diagrams of Binary Systems Second

Equilibrium and Stability Phase Diagrams of Binary Systems Second law for any change in chemical states says: (dS)total > 0 ↓ (dG)T,P < 0 This leads to fundamental requirement for equilibrium “The equilibrium state of a closed system is that state for which the total Gibbs energy is a minimum with respect to all possible changes at a given T and P”. Mathematically this is: (dG)T,P = 0 • this equation is necessary but not sufficient for chemical equilibrium 1 Criterion for Phase Separation This mixture with composition x1* will phase separate into two liquid phases, one with composition x1a and the other with x1b. The mathematical criterion for NO phase separation in a binary system at constant T and P is : d 2   ∆Gmix dx 2 1  RT   >0 const T,P Spinodal limit 2 Other criteria for phase stability GE d 2 ( RT > − 1 dx12 x1 x2 ) d ln γ 1 1 >− dx1 x1 ˆ df1 >0 dx1 dµ1 >0 dx1 dy1 >0 dx1 3 Regular Solution: Can It Predict Phase Separation If we have equations for free energy of a mixture we can calculate when d 2   ∆Gmix dx12  RT   >0 For 1 constant Margules model, phase separation can occur if the constant A > 2. Can you derive this? GE = Ax1 x2 RT ∆G = x1 ln x1 + x1 ln x2 + Ax1 x2 RT 4 Analysis of liquid/liquid mixtures Criterion for equilibrium, α and β are labels indicating a phase ˆ ˆ f1α = f1β ˆ f1 γ1 ≡ x1 f1 Substituting definition of activity coefficients α α x1 γ 1 f1 = x1β γ 1β f1 Rearranging gives α x1β γ1 = α β x1 γ1 5 LLE Calculation Consider a binary liquid system with two equilibrium phases. a) Write down all the variable required to describe the composition. b) Write down a family of equations to solve for variables assuming 1 constant Margules model. use x1α γ1α = x1β γ1β to describe equilibrium α β 6 The Simplest Case: A system that follows regular solution GE = Ax1 x 2 RT 2 ln γ 1 = Ax 2 = A(1 − x1 ) 2 α α x1 γ 1 = x1β γ 1β ln γ 2 = Ax12 = Ax12 α α β β x2 γ 2 = x2 γ 2 α β γ2 x2 ln β = ln α γ2 x2 γ 1α x1β ln β = ln α γ1 x1 x1β A((1 − x1 ) − (1 − x1 ) ) = ln α x1 α 2 β 2 1 − x1β A(( x1 ) − A( x1 ) ) = ln α 1 − x1 α 2 β 2 α In this special case, solubility curse is symmetrical about x 1 = 0.5, i.e., x1β = 1 − x1 A(1 − 2 x1 ) = ln 1 − x1 x1 A: x1α: 2.0, 2.03, 2.12, 2.31, 2.75 4.69, 6.92 0.5, 0.4, 0.3, 0.2, 0.1, 0.01, 0.001 7 Class Exercise Construct the phase diagram for a system that follows regular solution equation with A = -540/T+21.1-3lnT 8 Change of Gibbs Energy Recall : ∆G id = RT ∑ x i ln x i M i ∆G M = RT ∑ x i ln a i i = RT ∑ x i ln x i + RT ∑ x i ln γ i i i = ∆G id + G E M G E = RT ∑ x i ln γ i i 9 Phase Separation in Regular Solutions How big is alpha ? ∆H M = ∆H id + H xs \ M = Ωx A x B = αRTx A x B ∆G M = RT(x1 ln x1 + x1 ln x 2 ) + αRTx 1x 2 10 Where to separate ? Most comfortable for both A & B ∆G A (in solution m) = ∆G A (in solution q ) ∆G B (in solution m) = ∆G B (in solution q ) RT ln a A (in solution m) = RT ln a A (in solution q) RT ln a B(in solution m) = RT ln a B(in solution q) a A (in solution m) = a A (in solution q) a B(in solution m) = a B(in solution q) 11 Criteria for regular solutions ∆G M = RT (x A ln x A + x B ln x B ) + RTαx A x B  x  ∂∆G M = RT ln B + α(x A − x B ) ∂x B  xA   1  ∂ 2 ∆G M 1 = RT + − 2α  x  ∂x 2 xB B  A   1 ∂ 3 ∆G M 1   2 − 2 = RT 3  ∂x B  xA xB  Tcr = Ω RT 12 Effect of T on mixing Ω = 16,630 J ∆H M = ∆H id + H xs \ M = Ωx A x B = αRTx A x B ∆G M = RT (x1 ln x1 + x1 ln x 2 ) + αRTx 1x 2 13 Criteria for regular solutions ∆G B = RT ln a B ∆G B = ∆G M + x A ∂∆G M ∂x B ∂∆G B ∂ 2 ∆G M = xA ∂x B ∂x 2 B = RT ∂a B a B ∂x B ∂ 2 ∆G B ∂ 3∆G M ∂ 2 ∆G M = xA − ∂x 2 ∂x 3 ∂x 2 B B B RT ∂ 2a B RT  ∂a B   = − 2  a B ∂x 2 a B  ∂x B  B   2 14 Liquid and Solid Standard States Section 10.5 Mixing liquid with solid at T Tm ( A) < T < Tm ( B ) Go B ∆G o ( m ) = G o ( l ) − G o (s ) A A A ∆G o B( m ) l s =G o B( l) −G o B( s ) Go A s l Tm(A) T Tm(b) 15 Change of Gibbs Energies in Mixing Solid + Solid and Liquid + Liquid ∆G (M) = RT( x lA ln a lA + x lB ln a lB ) + x lB ∆G o ( B) l m ∆G M (s ) = RT( x ln a + x ln a ) − x ∆G s A s A s B s B s A o m(A) ∆G (M) = RT ( x lA ln x lA + x lB ln x lB ) + x lB ∆G o ( B) l m + RT( x lA ln γ lA + x lB ln γ lB ) 1444 24444 4 3 Ω l x lA x lB If regular ∆G (M) = RT ( x s ln x s + x s ln x s ) − x s ∆G o ( A ) s A A B B A m + RT ( x s ln γ s + x s ln γ s ) A B B 1444 24444 4A 3 Ωs x s x s A B 16 If regular Solid phase and Liquid phase ∆G : ∆G ∆G : ∆G M B M A M A (l) M B( l ) = ∆G = ∆G M A (s ) M B(s ) 17 Partial properties ∂∆G M ∆G = ∆G M + x B ∂x A M A ∂∆G M ∆G = ∆G + x A ∂x B M B M ∂ (n∆G M ) ∆G = ( ) T ,n j ∂n i M i M ∆G A ( l ) = RT ln a lA M ∆G A ( s ) = RT ln a s − ∆G o ( A ) A m ∆G M B( l) = RT ln a + ∆G l A o m ( B) M ∆G B(s ) = RT ln a s B 18 Relating activity to change of Gibbs energy in melting RT ln a lA = RT ln a s − ∆G o ( A ) A m RT ln a + ∆G l B o m ( B) = RT ln a s B ∆G ∆G o m(A) a = RT ln a s A l A ∆G a = exp( ) a RT s A l A o m(A) o m ( B) as = RT ln lB aB a ) = exp( a RT 19 s B l B ∆G o ( B) m Dependence of solubility on temperature ∆G o = ∆H o − T∆So m m m o o o ∆G m = ∆H m − T∆Sm = 0 at melting temperature: o ∆H m o if ∆Smo & ∆Hmo are independent of T ∆Sm = Tm ∆H o o o o Tm − T m ∆G m = ∆H m − T = ∆H m ( ) Tm Tm ∆G m ( i ) ∆H m ( i ) as T i = exp( ) = exp( (1 − )) l ai RT RT Tm ( i ) o o 20 If ideal solution  ∆G o ( A )  x m  = exp  RT  x   s A l A  ∆G o ( B)  xs m B  = exp l  RT  xB   question: x = s A xs A > or < x lA similarly xs B > or < x lB exp − ∆G o ( A ) RT − exp − ∆G o ( B) RT m m o m ( B) ( 1 − exp − ∆G o ( B) RT m ( ) ( ) ) ) x l A [1 − exp(− ∆G = exp(− ∆G o m(A) )] ( RT ) − exp(− ∆G RT exp − ∆G o ( A ) RT m o m ( B) RT ) 21 Example Germanium - Silicon ∆G o ,Si = 50,200 − 29.8T J m Tm,Si = 1685K o log PSi (atm) = −23,550 / T − 0.565 log T + 9.47 ∆G o ,Ge = 36,800 − 30.3T J m Tm,Si = 1213K o log PGe (atm) = −18,700 / T − 0.565 log T + 9.99 x l Si [1 − exp(− ∆G = exp(− ∆G o m ,Ge o m ,Si )] ( RT ) − exp(− ∆G RT exp − ∆G o ,Si RT m o m ,Ge RT ) ) x = s Si exp − ∆G o ,Si RT − exp − ∆G o ,Ge RT m m ( 1 − exp − ∆G o ,Ge RT m ( ) ( ) ) 22 Class Exercise A and B are mutually insoluble in the solid state and have the eutectic point of T = 650 K and xB = 0.20. The melting temperatures of A and B are, respectively, 1000 and 1600 K. The ∆H0m value of A is 30 kJ. A and B forms a regular solution. 1/ Estimate the ∆H0m value of B. 2/ Calculate the A liquidus composition at 800K. Neglect the differences between the liquid and solid heat capacities 23 Choose Standard Every road leads to Roman ∆G (M) = RT ( x lA ln a lA + x lB ln a lB ) + x lB ∆G o ( B) l m ∆G (M) = RT ( x s ln a s + x s ln a s ) − x s ∆G o ( A ) s A A B B A m Liquid A and Solid B Liquid A and Liquid B Solid A and Solid B 24 Phase Diagrams 25 Complete miscible in liquid Complete immiscible in solid G o (s ) = G A If ideal: A (l) = G o ( l ) + RT ln a lA A a lA = exp(− ∆G o ( A ) m RT )  ∆H o ( A )   ∆G o ( A )  T  m m l 1 −   = exp − x A = exp −  RT  RT  Tm ( A )        26 Solubility versus temperature Example: cadmium - bismuth ln x lCd = − 6400  T  1−   8.3145T  594K  ln x lBi = − 10900  T  1−   8.3145T  554K  27 If liquid phase is a regular solution ln γ = α x l A ( ) l 2 B = α 1− X ( l 2 A ) − ∆G o ( A ) = RT ln a lA = RT ln x lA + RTα(1 − x lA ) 2 m ∆G If o m(A) = ∆H o m(A) o m(A) (1 − T Tm ( A ) ) ∆H = 10 kJ, Tm ( A ) = 2000 K − 10000 + 5T = RT ln x lA + RTα(1 − x lA ) 2 28 − 10000 + 5T = RT ln x + RTα(1 − x ) l A l 2 A Ω = RTα kJ α cr = Ω cr 25,390 = =2 RTcr 8.3144 × 1413 29 What happens if Ω = 30 kJ? S-L-L phases 30 If both liquid and solid phases are regular solutions ∆G (M) = x lB ∆G o ( B) + RT ( x lA ln x lA + x lB ln x lB ) + Ω l x lA x lB l m ∆G (M) = − x s ∆G o ( A ) + RT ( x s ln x s + x s ln x s ) + Ω s x s x s s A m A A B B A B If ∆Gm ( A) = 8000 − 10T Ωs = 0 J, Ωl = -20000 J, ∆Gm ( B ) = 12000 − 10T Fig 10.20 Fig 10.21 Fig 10.22 31 Ωs = 10000 J, Ωl = -2000 J, Ωs = 30000 J, Ωl = 20000 J, Ωs = 0 J, Ωl = -20000 J, Fig 10.20 32 Ωs = 10000 J, Ωl = -2000 J, Fig 10.21 33 Ωs = 30000 J, Ωl = 20000 J, Fig 10.22 34 Various Ωl and Ωs 35 Class Exercise A binary system A-B forms ideal liquid solutions (Ωl = 0 J) Ω and regular solid solutions (Ωs = -15,000 J). The melting Ω temperatures of A and B are, respectively, 800 and 1200 K, and the molar Gibbs free energies of melting are ∆Gm(A) = 8,000 – 10 T (J) ∆Gm(B) = 12,000 – 10 T (J) 1/ What phase is the mixture of xB = 0.05 at T = 1100 K? 2/ What phase is the mixture of xB = 0.50 at T = 1300 K? 36