Material Science. Equilibrium

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1 Lecture 4: Phase diagrams in Material Science. Equilibrium Lecture plan: phase diagrams in material science: microstructures t in isomorphous binary systems microstructures in eutectic alloys liquid crystals equilibrium equilibrium and Gibbs free energy description of equilibrium response of equilibrium to conditions (P, T, ph) problems

2 Phase diagrams and Microstructure

3 Binary phase diagrams Phase diagram with total t solubility in both liquid id and solid state: isomorphous system T( C) L (liquid) homogeneous liquid solution of Cu and Ni. liquidus L + solidus (FCC solid l ti ) Cu-Ni phase diagram 2 phases: L (liquid) id) (FCC solid solution) 3 phase fields: L L + solution) homogeneous solid solution of Cu and Ni. wt% Ni

4 C =35 wt% Ni Cu-Ni phase diagram at T A : Only liquid, composition of liquid is given by the overall composition Information we can extract from the diagram: the phases present; composition of the phases percentage of fraction of the phases T( C) TA (C =35 wt% Ni) 13 L (liquid) at T D : Only liquid, composition of liquid is given by the overall composition (C =35 wt% Ni) at T B : Both L and are present Composition at T B : Liquid phase (L) of 32% Ni Solid phase () of 43% Ni Weight ratio: B TB R S 12 TD WL S R (43 35) = ; W = = = 73% W R S + R (43 32) L + A tie line D CL C o liquidus L + solidus (solid) 43 C wt% Ni

5 Development of microstructure in a Cu-Ni alloy Equilibrium case (very slow cooling)

6 Development of microstructure in a Cu-Ni alloy Non-Equilibrium case (real) Fast cooling: Cored structure Slow cooling: Equilibrium structure First to solidfy: 46wt%Ni Last to solidfy: < 35wt%Ni Uniform C: 35wt%Ni How we can prevent coring and get equilibrium structure?

7 Binary Eutectic Systems: Sn-Pb Sn-Pb system: limited solubility in solid state 3 single phase regions (L, a, b); T E=183 C, no liquid below T E. Eutectic composition 61.9% At the eutectic temperature: LC ( ) ( C ) + β ( C ) E E βe T( C) L R L (liquid) 183 C L+β β β S Pb For a 4wt%Sn-6wt%Pb alloy at 15C, find... --the compositions of the phases: Ca = 11wt%Sn Cb = 99wt%Sn Co Sn Co, wt% Sn W = 59 = 67wt % 88 W β = = 33 wt %

8 Microstructures in binary systems C o < 2wt%Sn Result: --polycrystal of grains. T( C) 4 L: Cowt%Sn L 3 L 2 TE : Cowt%Sn L + (Pb-Sn System) 1 + β Co 2 (room T solubility limit) Co, wt% Sn

9 Microstructures in binary systems 4 T( C) L: Cowt%Sn 2wt%Sn < C o < 18.3wt%Sn Result: -- polycrystal with fine β crystals. 3 2 TE 1 L L + + β L : Cowt%Sn β Co (sol. limit at Troom) 18.3 (sol. limit at TE) 3 Co, wt% Sn

10 Microstructures in binary systems Eutectic composition

3 hypoeutectic: Co=5wt%Sn 6 eutectic 61.

11 Microstructures in binary systems: eutectic and around T( C) 3 L 2 TE 1 L + + β L + β β (Pb-Sn System) Co Co hypoeutectic hypereutectic hypoeutectic: Co=5wt%Sn 6 eutectic Co, o wt% Sn hypereutectic: (illustration only) 175μm eutectic: Co=61.9wt%Sn 16μm β β β β β β eutectic micro-constituent From: W.D. Callister, Materials Science and Engineering: An Introduction, 6e.

IRON-CARBON (Fe-C) PHASE DIAGRAM 2 important points -Eutectic (A): L γ + Fe 3 C -Eutectoid (B): γ +Fe 3 C T( C) 16 δ 14 12 1 8 6 γ γ+l (austenite) +γ R B γ γ γ γ 1148 C R L 727 C = Teutectoid A

12 IRON-CARBON (Fe-C) PHASE DIAGRAM 2 important points -Eutectic (A): L γ + Fe 3 C -Eutectoid (B): γ +Fe 3 C T( C) 16 δ γ γ+l (austenite) +γ R B γ γ γ γ 1148 C R L 727 C = Teutectoid A γ+fe3c S +Fe3C L+Fe3C S Fe3C (ce ementite e) 12μm Result: Pearlite = alternating ti layers of and Fe3C phases (Fe) (Adapted from Fig. 9.24, Callister 6e. (Fig from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) Co, wt% C Fe3C (cementite-hard) (ferrite-soft) toid Ceutec Adapted from Fig. 9.21,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 199.) 21

HYPOEUTECTOID STEEL γ γ γ γ 16 δ 14 12 T( C) γ γ+l (austenite) L 1148 C L+Fe3C Fe e3c (cem mentite) (Fe-C System) γ γ 1 Adapted from Figs. γ γ γ+fe3c 9.

13 HYPOEUTECTOID STEEL γ γ γ γ 16 δ T( C) γ γ+l (austenite) L 1148 C L+Fe3C Fe e3c (cem mentite) (Fe-C System) γ γ 1 Adapted from Figs. γ γ γ+fe3c 9.21 and 9.26,Callister 6e. (Fig adapted γ γ 8 r s from Binary Alloy 727 C Phase Diagrams, 2nd γ γ R S ed., Vol. 1, T.B. w =s/(r+s) 6 Massalski (Ed.-in+Fe3C Chief), ASM wγ =(1-w) International, Materials 4 Park, OH, 199.) Co pearlite Co, wt% C w pearlite = wγ w =S/(R+S) w Fe3C =(1-w).77 1μm Hypoeutectoid steel Adapted from Fig. 9.27,Callister 6e. (Fig courtesy Republic Steel Corporation.) 22

14 Fe3C γ γ γ γ γ γ γ γ γ γ γ γ HYPEREUTECTOID STEEL T( C) 16 δ γ γ+l (austenite) R r s S L 1148 C γ+fe3c L+Fe3C3 w Fe3C =r/(r+s) 6 wγ =(1-w Fe3C ) +Fe3C pearlite w pearlite = wγ w =S/(R+S) w Fe3C =(1-w) Co Adapted from Fig. 9.3,Callister 6e. (Fig. 9.3 copyright 1971 by United States Steel Corporation.) Fe e3c (cem mentite e) Co, wt% C (Fe-C System) 6μm Hypereutectoid steel Adapted from Figs and 9.29,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 199.) 23

substance having a liquid-like imperfect

15 Liquid crystals Mesophase an intermedediate phase between solid and liquid. Example: liquid crystal Liquid crystal substance having a liquid-like imperfect order in at least one direction and longrange positional or orientational order in at least one another direction Nematic Smectic calamitic (rod-like) discotic Cholesteric

16 Nematic crystals in LCD

17 EQUILIBRIUM

18 Chemical Equilibrium A + B C+ D Chemical reaction tend to move towards a dynamic equilibrium in which both reactants and products are present but have no tendency to undergo net change The question: How to predict the composition of mixture at various condition

19 The Gibbs energy minimum Spontaneous change at const P and T happens towards lower values of the Gibbs energy Let s consider reaction dξ A B If some amount of A changed into B: dna dn B = dξ =+ dξξ extent of the reaction Reaction Gibbs energy (definition): G Δ rg = ξ P, T dg = μ dn + μ dn = μ dξ + μ dξ = ( μ μ ) dξ A A B B A B B A G Δ rg = = μb μa ξ PT, Difference between chemical potentials of the products and the reactants at the composition fo the reaction mixture At equilibrium Δ G = r

r Reverse reaction is spontaneous, reaction endergonic i.e. required work to go in forward reaction E i ti Exergonic reaction, e.

20 The Gibbs energy minimum Spontaneity reaction at const P, T Δ < Forward reaction is spontaneous, reaction exergonic (work-producing) r G Δ G = r Reaction at equilibrium Δ G > r Reverse reaction is spontaneous, reaction endergonic i.e. required work to go in forward reaction E i ti Exergonic reaction, e.g. glucose oxidation Endergonic reaction, e.g. protein synthesis

21 The description of equilibrium Perfect gas equilibrium Δ rg = μb μa = ( μb + RTln pb) ( μa + RTln pa) = pb = Δ r G + RT ln p At equilibrium: A r Q reaction quotient Δ G = =Δ G + RT p ln B r pa RT ln K = ΔG Why reaction doesn t go till the end: Δ G = nrt ( κ lnκ + κ ln κ ) mix A A B B r K- equilibrium constant

22 The description of equilibrium General case of a reaction 2A+ B 3C+ D Δ =Δ + rg rg RTln Q Δ = Δ Δ G ν G ν G r f f products reactants = 3C+ D 2A B Q = activitiesof products activities of reactants Q = a j v j j For example, for the reaction above: a a Q = a a 3 C 2 A D B

23 The description of equilibrium At equilibrium: v K a j = j j equilibrium RT ln K = Δ G r Example: Find degree of dissociation of water vapour at 23K and 1 bar if standard Gibbs energy for decomposition is 118 kj/mol 1 H 2Og ( ) H2( g) + O2( g) 2 3 Δ G 118*1 1 3 ln K = = K = 2.8*1 RT 8.3*23 K = ph p 2 O p 2 12 p = HO (1 )(2 + ) =.25 2

24 The description of equilibrium RTln K = ΔG r r r r K = e = e e Δ G RT Δ H / RT Δ S / R Increase with reaction entropy decrease with reaction enthalpy Boltzmann distribution interpretation: p = N e E / kt i i Ei / kt e i Due to higher density of energy levels (i.e. higher S), B is dominant at equilibrium

25 The description of equilibrium Relation between equilibrium constants K aa aa γ γ γ γ bb bb = C D = C D C D = At low concentration: A B A B A B K K b KK Using biological standard state If a biological reaction involves H+ ions, we have to take into account that standard biological condition is at γ ph = log a + = 7 NADH ( aq) + H + ( aq) NAD + ( aq) + H ( g) Δ =Δ + = rg rg 7ln1 RT 3 = 21.8 kj / mol kj / K mol 31K = 19.7 kj / mol b H 2

The response of equilibria to the conditions Equilibria i

changes RTln K = ΔG r Pressure dependence: K K = P T

2 Bg ( ) K= p p A Pressure increase by injecting inert

26 The response of equilibria to the conditions Equilibria i will respond to temperature, t pressure and concentration changes RTln K = ΔG r Pressure dependence: K K = P T Depends on standard (standard pressure) Δ G r 2 pb Ag ( ) 2 Bg ( ) K= p p A Pressure increase by injecting inert gas: no change as partial pressures of reactants and products stay the same. Pressure increase by compression: system will adjust partial pressures so the constant stays the same.

27 The response of equilibria to the conditions Le Chatelier principle: A system at equilibrium, when subjected to disturbance responds in a way that tends to minimize the effect of disturbance Extent of dissociation, : 2 pb Ag ( ) 2 Bg ( ) K= p p (1 )n 2 n Mole fractions at equilibrium: (1 ) n 1 2 κa = = κb = (1 ) n+ 2 n pb B p 4 p K = 2 p = κ κ p = 1 A 1 = 1+ 4p Kp A ½ A

28 The response of equilibria to the conditions Temperature response Equilibrium will shift in endothermic direction if temperature is increased and in exothermic direction if temperature is lowered. Van t Hoff equation: RTln K = ΔG r Gibbs-Helmholtz equation dln K 1 d( ΔrG / T) d( ΔrG / T) ΔrH = = 2 dt R dt dt T d ln K Δ r H dln K Δ H = = dt RT d(1/ T ) R r 2 i.e. for exothermic reaction: dln K Δ rh < < dt So, we can predict the equilibrium constant at another temperature: Δ 1 1 ln 2 ln H r K K1 = R T2 T1

29 The response of equilibria to the conditions Boltzmann distribution interpretation

30 The response of equilibria to the conditions Noncalorimetric measuring reaction enthalpy d ln K Δ r = d(1/ T) R r H slope: Δ H r R

31 The response of equilibria to the conditions Value of K at different temperatures d ln K Δ = d(/ (1/ T ) R r H 1/ T2 1 Δ rh 1 1 ln K2 ln K1 = rh d(1/ T) R Δ = R T 1/ T 2 T1 1

32 Equilibria and ph Dissociation of water (autoprotolysis) 2 HOl + 2 ( ) HO 3 ( aq) + OH ( aq ) a + a HO 3 OH 14 Kw = = a a 1 at298k 2 + = HO 3 OH a HO 2 Ionic dissociation constant of water For pure water: a HO 3 = a = 1 + OH 7 ph = log a H 3 O + at low concentration equal to molarity

33 The response of equilibria to ph Arrhenius acid: increases concentration of H 3 O + in solution Arrhenius base: increases concentration of OH - in solution Can be done via donation of OH - or removing of H +. + NaOH ( aq) Na ( aq) + OH ( aq) + NH ( aq ) + HOl ( ) NH ( aq ) + OH ( aq ) Acidity constant, K a: + HA( aq) + H O( l) H O ( aq) + A ( aq) K = 2 3 Conjugate base a a HO 3 a a + HA A pk a = log K a Basicity constant, K b: Conjugate acid a a + HB OH B( aq) + H 2O( l) HB ( aq) + OH ( aq) Kb = a + B pk b = log K b

34 Example: dissociation of formic acid Example: pk of formic acid is 3.77 at 298K. What is ph of.1m solution? What would happen if it were strong acid? HCOOH + H O() l HCOO + H O K a HO HCOO = = [ HCOOH ] x x = x ± ± = 2 b b 4ac 2a x = Answer: [ HO + ] = ph = What would be dissociation degree at ph=4 and ph=1?

35 Class problems: Last lecture Atkins 6.9b: sketch the phase diagram of the system NH 3 /N 2 H 4 given that t the two substances do not form a compound and NH 3 freezes at -78C, N 2H 4 freezes at +2C, eutectic formed with mole fraction of N 2 H 4.7 and melts at -8C. Atkins 6.1b Describe the diagram and what is observed when a and b are cooled down

36 Class problems: Atkins 7.2b: Molecular bromine is 24% dissociated at 16K and 1 bar. Calculate K, r G at 16K and predict K at 2 C, given r H =+112kJ/mol over the temperature range Atkins 7.4b: In the gas phase reaction A+B=C+2D it was found that when 2mol A, 1mol B and 3 mol D were mixed and allowed to come to equilibrium at 25C, the mixture contained.79mol of C at 1 bar. Calculate mol fraction of every species at equilibrium, K x, K and r G.

Brief reminder of the previous lecture

Brief reminder of the previous lecture partial molar quantities: contribution of each component to the properties of mixtures V j V = G µ = j n j n j pt,, n pt,, n dg = Vdp SdT + µ dn + µ dn +... A A B