∆G & phase changes:
Example: H2O(l ) ⇌ H2O(g)
Physical Equilibrium For P = 1 atm; T < 373 K (100 ˚C) ∆G > 0 ∆G & phase changes (we know water doesn’t boil below 100 ˚C) For P = 1 atm; T > 373 K ∆G < 0 For P = 1 atm; T = 373 K ∆G = 0 CHEM 102 At the boiling point, the liquid and vapor T. Hughbanks are in equilibrium. The boiling point is defined as the temp. at which the liquid and vapor phases are at equilibrium at 1 atm.
∆G : H2O(l ) ⇌ H2O(g) — more ∆G : H2O(l ) ⇌ H2O(g) — more
From Appendix 2A: H2O(g) H2O(l ) We saw that at 298 K, ∆G > 0
∆Hf˚ (kJ/mol) -241.8 -285.8 (∆H298/∆S298) = (44.0/0.119) K = 370 K this is close S˚ (J/mol K) 188.8 69.9 to 373 K, the boiling point! Why? For the vaporization process at 298 K: rearrange: ∆H298 = (370 K) ∆S298 ∆H298 = ∆Hf˚(gas) - ∆Hf˚(liq.) = ∆H - (370 K) ∆S = 0 –241.8 – (–285.8) = +44.0 kJ 298 298 recall: ∆H -(373 K)∆S = ∆G = 0 (why?) ∆S298 = S˚(gas) – S˚(liq.) 373 373 373 = 188.8 - 69.9 = 118.9 J/K = 0.1189 kJ/K if ∆H373 ≈ ∆H298 and ∆S373 ≈ ∆S298 ∆G298 = ∆H298 – (298 K)∆S298 = +8.5 kJ then ∆G373 = 0 ≈ ∆H298 - (373 K)∆S298
Generalization Ex: Phase transitions for NH3 ∆H and ∆S often don’t change much for a The molar enthalpy and entropy of melting and vaporization for ammonia are given process or reaction from ∆H298 and ∆S298 below. Estimate the melting and boiling Often, can estimate for ∆G at other temps.: points of ammonia. ∆G(T) ≈ ∆H298 -T∆S298 NH3(s) → NH3(l) NH3(l) ⇌ NH3(g) ∆H0: 5.65 kJ/mol 23.36 kJ/mol ∆S0: 28.9 J/mol K 97.5 J/mol K Example: Allotropes of Tin Example - continued Solid tin exists in two common forms Snwhite → Sngray (allotropes), “white” and “gray” tin. The ∆H˚ = -2.090 kJ/mol; ∆G˚ = 0.130 kJ/mol standard molar enthalpy and molar free energy (b) The standard molar entropy, S˚, of gray tin is 44.14 J for the transformation between these two mol-1 K-1. What is the standard molar entropy of allotropes are given below white tin? (c) The melting point of tin is 505 K. For temperatures Snwhite → Sngray ∆H˚ = -2.090 kJ/mol; ∆G˚ = 0.130 kJ/mol below 505 K, give the temperature ranges (if any) over which white and gray tin are the most stable (a) Under standard conditions (298 K, 1 atm), allotropes. which allotrope is thermodynamically stable?
Phase Diagrams P-T Phase diagram for H2O
All phase changes are determined by ∆G, phases are in equilibrium when ∆G = 0.
P-T phase diagrams tell us what phases are stable at various values of T and P.
A line separating two regions of a phase diagram represents the temperatures and pressures where ∆G = 0 for the change from one phase to another.
not to scale
Free Energies and Phase Diagrams Phase diagram for CO2 note: normal boiling point is not defined
for CO2.
not to scale Allotropes of Sulfur (Example) Phase diagram for NH 3 At a pressure of 1 atm, The triple point for ammonia occurs at 195.4 orthorhombic sulfur is K, 0.60 atm. Draw an approximate phase stable below 368.5 K, (P-T) diagram for ammonia; label the gas, monoclinic sulfur is stable liquid and solid regions. Include a dashed line on your drawing corresponding to a above 368.5 K, and these pressure of 1 atm. Be sure that the points two forms are stand at you have obtained from part (a) and the equilibrium at 368.5 K triple point appear at the proper places on (the transition temp.). your drawing. (Tm = 195.5, Tb = 239.6 K) Consider the process:
Sorthorhombic → Smonoclinic
Example - continued (a) At 368.5 K, ∆H = 402 J/mol for this process. What is ∆S for this process at 368.5 K? The Phase (b) What must be the sign of ∆G˚298 for this process? Estimate the value of ∆G˚298. Rule (c) Bonus! Monoclinic sulfur is a little less dense than orthorhombic sulfur: ∆V ≅ + 0.4 cm3/mol For a P = 100 bar, estimate the transition temp. (Hint: assume ∆E and ∆S don’t depend on pressure.) 1.0 L-bar = 100 J
Critical Properties Gas-Liquid Equilibrium Recall: Henry’s Law Boiling, Vapor Pressures Solubility (s) of a gas in a liquid is proportional The boiling point of a substance is temperature the the partial pressure of the gas: at which the equilibrium vapor pressure of the gas over the liquid reaches 1.0 atm. s = kHP Henry’s constant, k , depends on the strength of H H2O(l ) ⇌ H2O(g ) solvent-solute intermolecular forces and Water here is temperature. made to “boil” at -3 -1 -1 E.g., kH for O2 in water is 1.3 × 10 mol•L •atm 55 ˚C by at 20 ˚C. Is the number of O2 molecules per reducing the liter greater in water or in air? pressure below 1.0 atm.
Recall: H O Phase Diagram 2 H2O(l ) ⇌ H2O(g) at equilibrium What is the relationship between pressure and the “boiling temperature’? We have a process at equilibrium when boiling occurs An “equilibrium constant” exists that varies as a function of temperature: activities of products K = (at equilibrium) eq activities of reactants activities of products Q = (reaction quotient, not at equilibrium) activities of reactants
not to scale In the case of boiling, the “reactant”, H2O(l ), has an activity of 1.
∆G, ∆G˚, and Equilibrium ∆G, ∆G˚, and Equilibrium . Text gives this equation for any equilibrium: At equilibrium, the quantities of “reactants” and ∆G = ∆G˚ + RT ln Q p. 876 (eqn. 19.6) “products” do not change and ∆G = 0 and becomes equal to Keq. (In phase equilibria, “reactants” and ∆G˚ is the standard molar (P = 1 bar, n = 1 mol) “products” are just the two phases in question.) free energy and “Q” is the reaction quotient. Solving the free energy equation, When a system is at equilibrium Q is 0 = ∆G˚ + RT ln Keq ⇒ lnKeq = -∆G˚/RT numerically equal to Keq, the equilibrium constant. which gives us this very important expression relating the value of the equilibrium constant to ∆G˚ . This expression can be applied the physical and found on page 876 (eqn. 19.7). chemical equilibria, as long as we are careful to We need to spell out what forms Q and K take! use the correct expression the “equilibrium eq constant. Phase Equilibria Gas-Liquid and Gas-Solid Equilibria
For molecules or ions dissolved in solution, H2O(l ) ⇌ H2O(g ) CO2(s ) ⇌ CO2(g) equilibria involve concentrations. For gases, KP = P KP = P equilibria involve pressures. H2O CO2 When gases are nearly ideal (usually OK), When a solid or liquid is in equilibrium with a activity is numerically equal to pressure (in bar). gas, the “concentration” of the condensed There is no “concentration” of H O(l ) or CO (s ) phase = 1.0. Also, when a solid is in 2 2 in these expressions. As long as at least some equilibrium with dissolved species, the H O( ) or CO ( ) are present, the equilibrium “concentration” of the solid = 1.0. 2 l 2 s vapor pressures of the gases will not not depend on how much of the condensed phases there are.
Temperature Dependence of K Vapor Pressure, Quantitative Relationship eq ln Keq = –∆G/RT and ∆G˚ = ∆H˚ – T∆S˚ ⇒ H2O(l ) ⇌ H2O(g) ln Keq ≈ – (∆H˚/RT) + (∆S˚/R) Now we can get the relationship between pressure For vapor pressure equilibrium, Keq = P and the relevant and the “boiling temperature’. enthalpy and entropy terms are ∆H˚vap and ∆S˚vap. P = ⇒ ln (P /P ) ≈ – (∆H˚ /R)(1/T – 1/T ) K = H2O exp{-∆Gvap/RT} 2 1 vap 2 1 ∆G˚ /RT The Clausius-Clapeyron equation ln PH2O ≈ - vap Assume, as before, that ∆H and ∆S don’t change much as a function of temp. (∆H ≈ ∆H˚; ∆S ≈ ∆S˚), then a plot of P vs. 1/T gives ∆H˚vap and ∆S˚vap.
Problem Problem CaCO3(s) ⇌ CaO(s) + CO2(g) The normal boiling point of iodomethane, CH3I, is 42.43 ˚C and its vapor pressure at 0 ˚C is 140 limestone lime torr. Calculate
(a) standard enthalpy of vaporization (∆H˚vap) of iodomethane Given: ∆H˚ = + 178 kJ; at 25 ˚C, K = 1.39 × 10-23 (b) standard entropy of vaporization (∆S˚vap) of P iodomethane Give an estimate for KP at 800 ˚C.
(c) the vapor pressure of CH3I at 25 ˚C Solubility Equilibrium (Chap. 18) Solubility Product, Ksp The process is in equilibrium when ⇌ y+ x- rate of ions (or AxXy(s) x X (aq) + y X (aq) molecules) leaving 2+ - the solid = rate (eg., BaCl2(s) ⇌ Ba (aq) + 2 Cl (aq)) returning to the solid. What is the form for the equilibrium constant? That is, when dissolution rate = activities of products K = (at equilibrium) precipitation rate eq activities of reactants Write the equilibrium constant expression.