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Chemical Thermodynamics 2013/2014 10th Lecture: Phase Equilibria – Pure Substances Valentim M B Nunes, UD de Engenharia Introduction In previous lectures we studied the equilibrium in chemical reactions. Let us now study the equilibrium in phase transitions or Phase Equilibria. Our goal is to understand the thermodynamics of phase transitions and coexistence for single component systems. Fusion, vaporization, sublimation… Phase: An homogenous portion of a given system (for instance, one gas or a mixture of gases, mixture of miscible liquids, etc..) 2 Gibbs Phase Rule Before examining the one component systems, we will derive the Gibbs Phase Rule. This rule allow us to calculate the number of intensive variables necessary to determinate the thermodynamic state of a system with any number of phases and components. What is a component of the system? Substance whose concentration varies independently. For instance, for the reaction: CaCO3(s) = CaO(s) + CO2(g) we have 2 components, since one of them its not independent! So the number of components of a system is: C SR Where S is the number of substances and R is the number of chemical equilibriums. 3 Gibbs Phase Rule Consider a system with several phases and components: Thermal equilibrium: Tα= Tβ = Tγ = … α Mechanical equilibrium: pα = pβ = pγ = … β Chemical equilibrium: μ1α = μ1β = μ1γ = … γ μ2α = μ2β = μ2γ = … . . …… If we have C components and F phases, then the maximum number of variables, g, it will be: g = FC+2. But in each phase we have: x i 1 i We can subtract one variable for each phase! Also μ1α = μ1β = μ1γ = …, then we can subtract F-1 variables for each component. 4 Gibbs Phase Rule The total number of variables its then: g FC 2 F C ( F 1) FC 2 F FC C g F C2 This is the final expression for the Gibbs Phase Rule, a very important result when we are studying phase equlibria. 5 Examples Systems with one component, C=1: F g Observations 1 2 We have to define temperature and pressure 2 1 For instance liquid/vapor equilibrium: if we set the temperature, pressure is already known. 3 0 There is only one condition that satisfies this condition: the triple point Several components: CaCO3(s) = CaO(s) + CO2(g) In this case, F = 3, C = 2, then g = 1. We have to define only temperature. 6 One Component Systems Phase transitions are accompanied by a change of entropy and also of enthalpy. For a given temperature, T, and pressure, p, the more stable phase is the one with lower chemical potential, that is, the lower molar Gibbs energy: phase2 phase1 The chemical potential controls phase transitions and phase equlibria. At equilibrium μ must be equal throughout the system. When multiple phases are present μ must be same in all phases. 7 How does μ depends on T? We know that dG = -SdT + Vdp or per mole dμ = -SmdT + Vmdp , and Sm T p The 3rd Law tell us that Sm>0, so as a consequence: 0 T We also know that Sm,gas > Sm,liq > Sm,sol then: gas S m, gas T liq S m,liq T sol T S m, sol This means that the negative slope is steepest in gas phase, less steep in liquid and even less in the solid phase. 8 How does μ depends on T? Figure shows the change of chemical potential with temperature for a pure substance: μ μsol = μliq μliq = μgas Tfus Tb T At the temperatures Tfus and Tb two phases coexist in equilibrium. 9 Clapeyron,s Equation As we saw, the equilibrium condition is μα (T,p) = μβ (T,p). Let us suppose that temperature shifts from T to T+dT and pressure to p+dp. Then the equilibrium condition it will be μα (T,p) + d μα = μβ (T,p) + d μβ . As a consequence d μα = d μβ Since the chemical potential is the Gibbs energy per mole we can write: S dT V dp S dT V dp S Rearranging S dT V V dp dp S dT V This is the Clapeyron’s Equation. From this equation we can calculate phase diagrams. 10 Phase Diagrams Phase diagrams describe the phase properties as a function of state variables, like pressure and temperature (p,T) Solid – Liquid equilibrium (melting line) dp S fus dT V fus At equilibrium temperature it’s a reversible change, so ΔSfus = ΔHfus/Tfus Since ΔHfus > 0, ΔSfus > 0. On the other hand ΔVfus can be positive (general case) or negative water p S L Typical values are: ΔSfus : 2 ~ 6 cal.K-1.mol-1 ΔVfus : ± 1 ~ 10 cm3.mol-1 Melting line is very steep T 11 Liquid - Gas equilibrium In this case, ΔSvap > 0 and ΔVvap > 0 (all substances), then dp/dT > 0 Near the normal boiling point (p=1atm) for most substances we have: ΔSvap ≈ 20 cal.K-1.mol-1 ΔVvap ≈ 20 000 cm3.mol-1 p S L G Boiling line is less steep T This line intersects the melting line in a point for which there are three phases in equilibrium: the triple point. Accordingly with the phase rule g = 0. 12 Measuring vapor pressures 13 Solid - Gas equilibrium In this case, as for liquid gas equilibrium, ΔSsubl > 0 and ΔVsubl > 0 (all substances), then dp/dT > 0 This curve as a higher slope, near triple point, than the liquid gas curve equilibrium, since, at triple point, ΔHsubl= ΔHfus+ ΔHvap > ΔHvap and ΔVsubl ≈ ΔVvap. p S L G pt Tt T This is the general aspect of a p,T phase diagram. We will examine later some phase diagrams of real substances! 14 Critical point At the critical point the gas-liquid curve stops. Beyond the critical point the liquid and gas phase become indistinguishable, and we have a supercritical fluid (no phase change!). 15 Integration of Clapeyron Equation From Clapeyron’s equation and for the solid liquid equilibrium we have: dp S fus V fus p2 dT dp p1 H fus 1 T V fus T dT 1 T2 Assuming that ΔHfus and ΔVfus are T independent then: p2 p1 H fus V fus T2 ln T1 16 Clausius-Clapeyron Equation For the liquid – vapor equilibrium ΔV its not constant, but we can make some approximations. Ignoring the molar volume of the condensed phase, compared with gas, we have: Vvap Vgas Vliq Vgas Then the Clapeyron equation comes: Assuming an ideal gas, then Vgas dp H vap dT TVgas RT p Substituting in the equation we obtain: dp pH vap dT RT 2 or H vap 1 dp dT 2 p RT 17 Clausius-Clapeyron Equation H vap 1 dT Integration of equation leads to: dp 2 p RT p0 T0 p T Assuming now that ΔHvap is constant in the temperature interval we obtain: p H vap 1 1 ln p0 R T0 T If p0 = 1 atm, the T0 is the normal boiling point, Tb, then: p H vap H vap 1 ln 1 RTb R T This is the Clausius – Clapeyron equation. It allow us, for example, to calculate the boiling point at any pressure. 18 Clausius-Clapeyron Equation Graphically it can be observed that for many substances the Clausius Clapeyron equation is obeyed for relatively large temperature intervals. ΔHvap/RTb Slope =-ΔHvap/R ln p = f(1/T) 19 Example of phase diagram: Water The figure shows the simplified phase diagram for water: Negative slope! Critical point Triple point Water sublimes bellow 0.006 atm (~4.58 mmHg): lyophilization 20 Some considerations about the critical point At critical point we have some “thermal anomalies”. For instance if we rearrange the Clapeyron equation: H vap TVvap dp dT Now, since ρL ρG, ΔVvap 0, then: lim H vap T Tc 0 Above Tc an pc, liquid and gas become indistinguishable, a single fluid phase known as supercritical fluid. These fluids are finding remarkably practical applications. Supercritical water (Tc=374 ºC and pc = 221 bar) readily dissolve organic molecules and inorganic salts are nearly insoluble. 21 Phase diagram: Carbon Dioxide Some remarkable features of the CO2 phase diagram are: Solid liquid line has a positive slope CO2 sublimates bellow 5.1 atm, for instance at atmospheric pressure CO2 is a supercritical fluid just above room temperature! Supercritical CO2 is a reaction solvent. It can replace chlorinated and volatile organic compounds. It is a “dry cleaning” solvent. 22 Phase diagram: Carbon Synthetic diamonds! 23