G & phase changes: Example: H 2 O(l ) H 2 O(g) Physical Equilibrium G & phase changes. Generalization

Transcription

1 Physical Equilibrium G & phase changes CHEM 102 T. Hughbanks G & phase changes: Example: H 2 O(l ) H 2 O(g) For P = 1 atm; T < 373 K (100 C) G > 0 (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 At the boiling point, the liquid and vapor 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 : H 2 O(l ) H 2 O(g) more From Appendix 2A: H 2 O(g) H 2 O(l ) H f (kj/mol) S (J/mol K) For the vaporization process at 298 K: H 298 = H f (gas) - H f (liq.) = ( 285.8) = kj S 298 = S (gas) S (liq.) = = J/K = kj/k G 298 = H 298 (298 K) S 298 = +8.5 kj G : H 2 O(l ) H 2 O(g) more We saw that at 298 K, G > 0 ( H 298 / S 298 ) = (44.0/0.119) K = 370 K this is close to 373 K, the boiling point! Why? rearrange: H 298 = (370 K) S 298 H (370 K) S 298 = 0 recall: H 373 -(373 K) S 373 = G 373 = 0 (why?) if H 373 H 298 and S 373 S 298 then G 373 = 0 H (373 K) S 298 Generalization H and S often don t change much for a process or reaction from H 298 and S 298 Often, can estimate for G at other temps.: G(T) H 298 -T S 298 Ex: Phase transitions for NH 3 The molar enthalpy and entropy of melting and vaporization for ammonia are given below. Estimate the melting and boiling points of ammonia. NH 3 (s) NH 3 (l) NH 3 (l) NH 3 (g) H 0 : 5.65 kj/mol S 0 : 28.9 J/mol K kj/mol 97.5 J/mol K

2 Example: Allotropes of Tin Solid tin exists in two common forms (allotropes), white and gray tin. The standard molar enthalpy and molar free energy for the transformation between these two allotropes are given below Sn white Sn gray H = kj/mol; G = kj/mol (a) Under standard conditions (298 K, 1 atm), which allotrope is thermodynamically stable? Example - continued Sn white Sn gray H = kj/mol; G = kj/mol (b) The standard molar entropy, S, of gray tin is J mol -1 K -1. What is the standard molar entropy of white tin? (c) The melting point of tin is 505 K. For temperatures below 505 K, give the temperature ranges (if any) over which white and gray tin are the most stable allotropes. Phase Diagrams P-T Phase diagram for H 2 O 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 CO 2 note: normal boiling point is not defined for CO 2. not to scale

3 Phase diagram for NH 3 The triple point for ammonia occurs at K, 0.60 atm. Draw an approximate phase (P-T) diagram for ammonia; label the gas, liquid and solid regions. Include a dashed line on your drawing corresponding to a pressure of 1 atm. Be sure that the points you have obtained from part (a) and the triple point appear at the proper places on your drawing. (T m = 195.5, T b = K) Allotropes of Sulfur (Example) At a pressure of 1 atm, orthorhombic sulfur is stable below K, monoclinic sulfur is stable above K, and these two forms are stand at equilibrium at K (the transition temp.). Consider the process: S orthorhombic S monoclinic Example - continued (a) At K, H = 402 J/mol for this process. What is S for this process at K? (b) What must be the sign of G 298 for this process? Estimate the value of G 298. (c) Bonus! Monoclinic sulfur is a little less dense than orthorhombic sulfur: V cm 3 /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 The Phase Rule Critical Properties Gas-Liquid Equilibrium

4 Recall: Henry s Law Solubility (s) of a gas in a liquid is proportional the the partial pressure of the gas: s = k H P Henry s constant, k H, depends on the strength of solvent-solute intermolecular forces and temperature. E.g., k H for O 2 in water is mol L -1 atm -1 at 20 C. Is the number of O 2 molecules per liter greater in water or in air? Boiling, Vapor Pressures The boiling point of a substance is temperature at which the equilibrium vapor pressure of the gas over the liquid reaches 1.0 atm. Water here is made to boil at 55 C by reducing the pressure below 1.0 atm. H 2 O(l ) H 2 O(g ) Recall: H 2 O Phase Diagram H 2 O(l ) H 2 O(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 eq = (at equilibrium) 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, H 2 O(l ), has an activity of 1. G, G, and Equilibrium Text gives this equation for any equilibrium: G = G + RT ln Q p. 876 (eqn. 19.6) G is the standard molar (P = 1 bar, n = 1 mol) free energy and Q is the reaction quotient. When a system is at equilibrium Q is numerically equal to K eq, the equilibrium constant. This expression can be applied the physical and chemical equilibria, as long as we are careful to use the correct expression the equilibrium constant. G, G, and Equilibrium At equilibrium, the quantities of reactants and products do not change and G = 0 and becomes equal to K eq. (In phase equilibria, reactants and products are just the two phases in question.) Solving the free energy equation, 0 = G + RT ln K eq lnk eq = - G /RT which gives us this very important expression relating the value of the equilibrium constant to G found on page 876 (eqn. 19.7). We need to spell out what forms Q and K eq take!

5 Phase Equilibria For molecules or ions dissolved in solution, equilibria involve concentrations. For gases, equilibria involve pressures. When a solid or liquid is in equilibrium with a gas, the concentration of the condensed phase = 1.0. Also, when a solid is in equilibrium with dissolved species, the concentration of the solid = 1.0. Gas-Liquid and Gas-Solid Equilibria H 2 O(l ) H 2 O(g ) K P = P H2 O CO 2 (s ) CO 2 (g) K P = P CO2 When gases are nearly ideal (usually OK), activity is numerically equal to pressure (in bar). There is no concentration of H 2 O(l ) or CO 2 (s ) in these expressions. As long as at least some H 2 O(l ) or CO 2 (s) are present, the equilibrium vapor pressures of the gases will not not depend on how much of the condensed phases there are. Vapor Pressure, Quantitative Relationship H 2 O(l ) H 2 O(g) Now we can get the relationship between pressure and the boiling temperature. K = P H2O = exp{- G vap /RT} ln P H2O - G vap/rt Temperature Dependence of K eq ln K eq = G/RT and G = H T S ln K eq ( H /RT) + ( S /R) For vapor pressure equilibrium, K eq = P and the relevant enthalpy and entropy terms are H vap and S vap. ln (P 2 /P 1 ) ( H vap /R)(1/T 2 1/T 1 ) The Clausius-Clapeyron equation 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 The normal boiling point of iodomethane, CH 3 I, is C and its vapor pressure at 0 C is 140 torr. Calculate (a) standard enthalpy of vaporization ( H vap ) of iodomethane (b) standard entropy of vaporization ( S vap ) of iodomethane (c) the vapor pressure of CH 3 I at 25 C Problem CaCO 3 (s) CaO(s) + CO 2 (g) limestone lime Given: H = kj; at 25 C, K P = Give an estimate for K P at 800 C.

6 Solubility Equilibrium (Chap. 18) The process is in equilibrium when rate of ions (or molecules) leaving the solid = rate returning to the solid. That is, when dissolution rate = precipitation rate Solubility Product, K sp A x X y (s) x X y+ (aq) + y X x- (aq) (eg., BaCl 2 (s) Ba 2+ (aq) + 2 Cl - (aq)) What is the form for the equilibrium constant? K eq = activities of products activities of reactants (at equilibrium) Write the equilibrium constant expression.