Chemistry Lecture on Thermodynamics and Gibbs Free Energy

Chem. 31 10/9 Lecture

Announcements I Exam 1 Results Ave = 83.2% Distribution Statistical Calculations Lab Due Wednesday New Homework Assignment (Set 2) Posted Score 100s 90s 80s 70s 60s <60 N 2 36 27 17 7 6

Announcements II Today s Material Chapter 6 Thermodynamics Le Ch telier s Principle Sparingly Soluble Salts Solubility in water Solubility in common ion (if time)

Thermodynamics 1. H, change in enthalpy, is related to heat of reaction - if a reaction produces heat, H < 0 and reaction is exothermic - a reaction that requires heat has H > 0 and is endothermic 2. S, change in entropy, is related to disorder of system - If the final system is more random than initial system, S > 0

Thermodynamics Entropy Entropy A macroscopic analogy to entropy would be to have a box of 50 ping pong balls with half white and half black Even if placed on two separate halves of the box, if the box were shaken to mix the balls, roughly half of each color would be expected in each half leading to a positive S v v initial state final state

Thermodynamics Entropy Examples: (Is S > or < 0?) H2O(l) H2O(g) H2O(s) H2O(l) NaCl(s) 2H2(g) + O2(g) N2(g) + O2(g) S > 0 S > 0 S > 0 S < 0 S > 0 Na+ + Cl- 2H2O(g) 2NO(g)

Thermodynamics G = Change in Gibbs free energy This tells us if a process is spontaneous (expected to happen) or non-spontaneous G < 0 process is spontaneous (favored) G = H - T S (T is absolute temperature) processes that are exothermic ( H < 0) and increase disorder ( S > 0) are favored at all T processes that have H > 0 and S > 0 are favored at high T

Thermodynamics Example Question: The G for the reaction Ca2+ + 2OH- => Ca(OH)2(s) is -52 kJ/mol Determine K at T = 20. C for Ca(OH)2(s) => Ca2+ + 2OH-

Le Chteliers Principle Intuitive Method Addition to one side results in switch to other side Mathematical Method = + ln G G RT Q Q = ln G RT Example: K When Q>K, G>0 (toward reactants) reaction shifts to reactants (more AgCl(s)) AgCl(s) Ag+ + Cl- When Q<K, G<0 (toward products) Addition of Ag+ Example: Q = [Ag+][Cl-] As Ag+ increases, Q>K

Le Chteliers Principle Stress Number 1 Reactant/Products: Addition of reactant: shifts toward product Removal of reactant: shifts toward reactant Addition of product: shifts toward reactant Removal of product: shifts toward product

Le Chteliers Principle Stress Number 1 Example: CaCO3(s) + 2HC2H3O2(aq) Ca(C2H3O2)2(aq) + H2O(l) + CO2(g) 1. Add HC2H3O2(aq) 2. Remove CO2(g) 3. Add Ca(C2H3O2)2(aq) 4. Add CaCO3(s) No effect because (s)

Le Chteliers Principle Stess Number Two: Dilution H + NO = K 2 HNO Side with more moles is favored at lower concentrations 2 2 1 1 + H NO 2 2 = Q Example: HNO2(aq) H+ + NO2- 1 HNO 2 2 If solution is diluted, reaction goes to products 1 = Q K 2 So Q<K, products favored If diluted to 2X the volume:

Le Chteliers Principle Stess Number Two: Dilution Molecular Scale View Concentrated Solution Diluted Solution dissociation allows ions to fill more space H+ NO2- H+ NO2- H+ NO2- H+ NO2- H+ H+ H+ NO2- H+ NO2- H+ NO2- H+ NO2- NO2- NO2-

Le Chteliers Principle Stress Number 3: Temperature If H>0, as T increases, products favored If H<0, as T increases, reactants favored Easiest to remember by considering heat a reactant or product Example: OH- + H+ H2O(l) + heat Increase in T

Some Le Chateliers Principle Examples Looking at the reaction below, that is initially at equilibrium, AgCl(s) Ag+(aq) + Cl-(aq) ( H >0) determine the direction (toward products or reactants) each of the following changes will result in a) increasing the temperature b) addition of water (dilution) c) addition of AgCl(s) d) addition of NaCl

Ch. 6 Solubility Problems Why Solubility is Important Use in gravimetric analysis (predict if precipitation is complete enough) Use in precipitation titrations (in next chapter) Use in separations (e.g. separation of Mg2+ from Ca2+ in tap water for separate analysis) Understand phase in which analytes will exist Problem Overview Dissolution of sparingly soluble salts in water Dissolution of sparingly soluble salts in common ion Precipitation problems (and selective precipitation problems)

Solubility Product Problems - Solubility in Water Example: solubility of Mg(OH)2 in water Solubility defined as mol Mg(OH)2dissolved/L sol n or g Mg(OH)2dissolved/L sol n or other units Use ICE approach: Initial Change Equilibrium Mg(OH)2(s) Mg2+ + 2OH- 0 +x x 0 +2x 2x Note: x = [Mg2+] = solubility

Solubility Product Problems - Solubility of Mg(OH)2 in water Equilibrium Equation: Ksp = [Mg2+][OH-]2 Ksp = 7.1 x 10-12 = x(2x)2 = 4x3 (see Appendix F for Ksp) x = (7.1 x 10-12/4)1/3 = 1.2 x 10-4 M Solubility = 1.2 x 10-4 M = [Mg2+] Conc. [OH-] = 2x = 2.4 x 10-4 M

Solubility Product Problems - Solubility of Mg(OH)2 in Common Ion If we dissolve Mg(OH)2 in a common ion (OH- or Mg2+), from Le Ch telier s principle, we know the solubility will be reduced Example 1) What is the solubility of Mg(OH)2 in a pH = 11.0 buffer? No ICE table needed because, from pH, we know [OH-]eq and buffer means dissolution of Mg(OH)2doesn t affect pH.

Solubility Product Problems - Solubility of Mg(OH)2 at pH 11 cont. [H+] = 10-pH = 10-11 M and [OH-] = Kw/[H+] = 10-3 M Ksp = [Mg2+][OH-]2 Moles Mg(OH)2 dissolved = moles Mg2+ [Mg2+] = Ksp/[OH-]2 = 7.1 x 10-12/(10-3)2 [Mg2+] = 7 x 10-6 M