Chapter 4: Physical Transformation of Pure Substances Major Topics: Review: Phase, Phase Transition Dependence of G on T, P Phase Diagrams Gibb’s Phase Rule Chemical Potential Dependence on T, P Clapyeron Equation Clausius-Clapyeron Equation Phases Phase: state of matter that is uniform throughout in chemical composition and physical state # of phases, P Examples of P=1: • Pure gas • Gas mixture • Sodium chloride solution • Ice Example of P=3: Examples of P=2: • 퐶푎퐶푂 푠 → 퐶푎푂 푠 + 퐶푂 푔 • Ice & water mixture 3 2 • Oil & water Must distinguish between Components and Constituents Phase Transition Phase transition: spontaneous conversion of one phase to another Transition temperature, Tm: T at which the 2 phases are in equilibrium for a given P Chemical Potential For a pure substance, 휇 = 퐺푚 For a multi-component system, i i Chemical potential, 휇, is the potential for moving material. In this chapter, from one phase to another (phase change) In Ch. 6, from one substance to another (reaction) Criterion of phase equilibrium: At equilibrium the chemical potential is the same throughout the sample. Phase Boundaries Phase boundary: pressure and temperature line where two phases can coexist. Phase transition of a stable phase occurs when P,T cross the phase boundaries. a b Triple point: pressure and temperature where all three phases can coexist: vapor liq solid Vapor Pressure - pressure of a vapor in equilibrium with its condensed phase Vapor pressure increases with temperature. Boiling point: temperature at which the vapor pressure becomes equal to the external pressure. Gibbs Phase Rule Phase rule: F = C – P + 2 F = degrees of freedom, i,e,, the number of parameters that can be varied independently while preserving the number of phases in equilibrium. P = the number of phases C = the number of components, where components are chemically independent constituents of a system. Number of components: Ethanol water mixture: two constituents & two components + - Aqueous NaCl solution: three constituents (Na , Cl , H2O); only 2 components because Na+ and Cl- can’t vary independently For a pure substance, C=1 If P=1, then F = 1 - 1+ 2 = 2. Thus, 2 variables, typically T and P, may be changed independently without changing the number of phases. If P = 2, then F = 1, and only 1 variable can be changed independently. Example: 1-component phase diagram F = C – P + 2 Which row has the correct number of phases present at each point? a b c d A 1 2 3 4 B 1 3 3 2 C 2 2 2 3 D 1 2 2 3 Phase Diagram of H2O Phase Diagram of CO2 Phase Diagram of He Temperature dependence of chemical potential 휕휇 휕퐺 = 푚 = −푆 휕푇 휕푇 푚 푃 푃 • Temperature dependence of chemical potential is determined by the entropy. • The phase with larger entropy shows steeper change of chemical potential vs. temperature: At the temperature where the relative value of chemical potential changes, a phase 푆푚 푔 > 푆푚 ℓ > 푆푚 푠 transition occurs. Change in Chemical Potential with Pressure dG SdT VdP (basic equation) Change in with P at constant T: 휕휇 휕퐺푚 = = 푉푚 휕푃 푇 휕푃 푇 (In practice, the linear lines depicted are o curved, since Gm = Gm + ln P ) In most cases 푉푚 ℓ > 푉푚 푠 , so an increase in P will increase melting Temp. Water is different! Melting point Melting point increases at higher decreases at higher pressure. (typical pressure. (e.g. H2O) material) Vapor pressure of a liquid under pressure Effect of external pressure on vapor pressure of a liquid is given by: 푝 = 푝∗e푉m(l)∆푃Τ푅푇 Where p is the observed vapor pressure, ∆P is the applied pressure, Vm is the molar volume, and p* is the V.P. in absence of applied press. Density of water is 0.997 g/cm-3 at 298 K. Molar volume = 18.1 cm-3/mol Vapor pressure is increased by 0.73% When pressure is increased by 10 bar. (Brief Illustration 4B.2) Slope of phase boundary (Clapeyron equation) On a phase boundary, two phases are in equilibrium: 푑퐺 = 푑μ = −푆푑푇 + 푉푑푃 a (P,T) b (P,T) For infinitesimal changes in P and T: phase a da Sa dT V a dP S b dT V b dP db p (V b V a )dP (S b Sa )dT phase b dP S Clapeyron equation: trs dT trsV T Thus, the expected slope of a phase boundary can be obtained from ΔS and ΔV of the phase transition. Solid - Liquid phase boundary For a solid-liquid phase transition, i.e., melting: H S fus fus T Thus, dP SH fus fus dTfus V T fus V Because ∆fusV is typically very small and usually positive, the slope of the P-T phase boundary is expected to be large. When ∆fusV is negative (e.g., water), the slope of the P-T phase boundary is large, but negative. Liquid/Vapor phase boundary S H trsH dP vap vap trsS and T dT vapV TvapV For the case of a gas-liquid transition: ___ V( g ) V ( ) so Vvap ( g ) V ( g ) Thus, dP H H vap vap (for an ideal gas) dT TV (g) T(RT / P) This rearranges to: Clausius-Clapeyron equation dPln H dPln vap H vap dT RT 2 d(1/ T ) R Problems with Clausius-Clapyeron • Many approaches to problems to find vapH, Tb, Pvapor dln P H v 1 1. Assume Constant Enthalpy, i.e., ∆H ≠ f(T) d T R a. Given P vs. T data. Plot lnP vs. 1/T; fit points to a straight line, and get H from slope. b. Given 2 (P,T) points. Use definite integration: pT 22vap H 1 P vap H 11 H dln P dT ln 2 vap 11 pT2 PP21exp 11RT PRTT1 2 1 RTT21 What information would you need for a more accurate calculation? C-C continued c. Given a Formula for ln p (T) Example: The vapor pressure of benzene in the range 10 - 30 ºC is: 1780 logPtorr 7.960 TK Calculate Hvap: p2 H T2 1 d ln P vap dT R T 2 p1 T1 H 1780 ln P A v 1 2.3037.960 R T TK H v 2.3031780 H 1780 2.303 8.31451 34.08 kJ R v mole C-C given H(T) 2. For accurate work account, must for T dependence of H, i.e., ∆H = f(T). a. With a formula for HT), can integrate the C-C Eqn: Example: Find the boiling temperature for N2 at P = 100 Torr. (The B.P. at P=1 atm is 77.33K) given: 1 vHJ mole 8070 32.07TK 8070 32.07TK Substituting in CC eqn: dln P dT RT2 970.6 3.857 Simplifying: dln P dT dT T2 T P2 1 1 T2 Integrating: ln 970 .6 3.857 ln P1 T1 T2 T1 100 1 1 T Putting in Nos: ln 970 .6 3.857 ln 2 760 77.33 T2 77.33 970 .6 Simplify: 2.028 12.551 3.857 ln T2 16.771 T2 970.6 Rearrange: 0 31.350 3.857lnT2 T2 Can use Excel solver to get: T2=63.2 K C-C Finding H(T) b. If know vapH at one T, can find a general formula for vapH(T), assuming constant heat capacity: vH Hgas Hliq H H H Since: v gas liq C v P T P T P T P then vCP CP,gas CP,liq Therefore, the vapH(T), relative to vapH(Tr) is: T vH(T) vH(Tr ) vCPdT Tr Then use this formula in the C-C equation. -1 -1 Example: Estimate the Pv of CCl4 at 400 K w/ Cp(gas) = 83.30 J-K mole , -1 -1 -1 Cp(l) = 131.74 J-K mole , vapH = 30.0 kJ-mole , and Tb = 349.9 K. To find a formula for the T-dependence of vapH, use Cp’s, Tr, and rH: T 3 vH(T) 30.010 83.3131.74dT 349.9 3 vH 30.010 48.44T 349.9 30.0103 48.44T 349.9 dln P dT RT2 6.65103dT 5.83 dln P dT T2 T P2 3 1 1 T2 ln 6.6510 5.83ln P1 T1 T2 T1 Substituting P1 = 1 atm, T1 = 349.9 K, and T2 = 400 K yields lnP2=1.600, which gives P2 = 4.95 atm, the vapor pressure of CCl4 at 400K Recap 휕휇 휕휇 = −푆푚 = 푉푚 dln P H 휕푇 휕푃 v 푃 푇 1 d T R Clapeyron equation: phase boundary dP S H dT V TV Clausius-Clapeyron equation: vapor phase boundary d ln P H vap dT RT 2 Constant enthalpy: (P, T) data; (P, T) formula H(T): given formula; find formula from Cp Solid-Vapor Boundary Vapor pressure of solids is called sublimation pressure e.g., I2 gas above solid I2 ; CO2 solid sublimes directly to gas at atmospheric pressure Transition from solid to vapor is called sublimation. Transition is governed by Hsub Hsub = Hvap + Hfus because enthalpy is a state function. Only difference in calculations: replace Hvap by Hsub. Because Hsub > Hvap The slope of the P-T boundary for solid- vapor equilibria is greater than that for liquid-vapor equilibria. Equilibria of Condensed Phases Solid - Liquid or Solid - Solid equilibria. From the Clapeyron Equation: dP H V dT or dP dT T V H T To find the dependence of M.P. (Tf) on pressure, integrate: * fus H T PP ln * fusVT f V Vliq Vsolid V M g mole g mL First-Order Transitions • Many different types of phase transitions, e.g.