Thermodynamics Part II: Gibbs Free Energy & Equilibrium

1. ThermodynamicsPart II:Gibbs Free Energy & Equilibrium Jespersen Chap. 19 Sec 7 thru 10 Dr. C. Yau Spring 2014 1

2. When ΔG is negative  spontaneous rxn When ΔG is positive  nonspontaneous rxn When ΔG is zero ???? neither spontaneous nor nonspontaneous What does that mean?? The system is at equilibrium! REMEMBER! At equilibrium, ΔG = 0. 2

3. Dynamic Equilibrium Consider ice at 0oC… ice liquid water Both are existing at the same time… indefinitely. No work can be done at equilibrium

4. ΔG = ΔH – T ΔS At equilibrium, 0 = ΔH – T ΔS soΔH = TΔS Rearrange to get 2 useful equations: ΔH is relatively easy to determine, ΔS is not. The 1st equation allows us to determine ΔS from ΔH and T.

5. At equilibrium, ΔH = TΔS Rearranged eqn is If we have ΔH and ΔS, we can calculate T…. but what exactly is T? T is the temperature at equilibrium. For A (s) A (l) What is the significance of T at equilibrium? It is the mp of the substance. For A (l) A (g), T is the bp.

6. Example 19.7 p. 895 For the phase change Br2(l) Br2(g) ΔHo = +31.0 kJ mol-1 ΔSo = 92.9 J mol-1 K-1 Assuming that ΔH and ΔS are nearly temp independent, calculate the approximate Celsius temp at which Br2(l) will be in equilibrium with Br2(g) at 1 atm (i.e. the normal boiling point of Br2.) Ans. 61oC (experimental bp = 58.8oC) But 31.0/ 92.9 = 0.334. Why is it not 61oC? Do Pract Exer 17, 18 on p. 895

7. Using ΔG to predict equilibrium shift ΔG = ΔGo + RT ln Q (Q = reaction quotient) Example 19.10 p. 900 2 NO2 (g) N2O4 (g) ΔG298o= - 5.40 kJ/mol N2O4 In a rxn mixture, the partial pressure of NO2 is 0.25 atm and the partial pressure of N2O4 is 0.60 atm. In which direction must this reaction proceed to reach equilibrium? R = 8.314 J mol-1 K-1 Ans. ΔG=+ 0.20 kJ/mol, rxn will proceed to left. How do we know that?...ΔG=+ means to the left?? (Some of N2O4 will break up to form NO2.) Do Pract Exer 23, 24 p. 902

8. Relating ΔG to K (equilibrium constant) Remember that at equilibrium… ΔG = 0 and Q = K ΔG = ΔGo + RT ln Q becomes 0 = ΔGo + RT ln K Therefore ΔGo= - RT ln K Know this!

9. Example 19.11 p. 902 (Calculate ΔGo from K?) The brownish haze often associated with air pollution is caused by nitrogen dioxide, NO2, a red-brown gas. Nitric oxide NO, is formed in auto engines and some of it escapes into the air where it is oxidized to NO2 by oxygen. 2 NO (g) + O2(g) 2NO2(g) The value of Kp for the rxn is 1.7x1012 at 25.00oC. What is ΔGo for the rxn in joules per mole? In kilojoules per mole? Ans. -69.80 kJ/mol

10. Example 19.12 p. 903 (Calculate K from ΔGo ) Sulfur dioxide, which is sometimes present in polluted air, reacts with oxygen when it passes over the catalyst in automobile catalytic converters. the product is the very acidic oxide SO3. 2SO2(g) + O2(g) 2SO3(g) For this rxn, ΔGo = - 1.40x102 kJ mol-1 at 25oC. What is the value of Kp? Ans. 3x1024 Do Pract Exer 25, 26 p. 904

11. SUMMARY How many equations can you think of that allow us to calculate ΔG? ΔG =  ΔGf(products) –  ΔGf (reactants) ΔG = ΔH - T ΔS (Calc. ΔH from ΔHf  and ΔS from S) ΔG = 0 (at equilibrium) ΔG = ΔG + RTlnQ (predict how reach equil.) ΔG = - RTlnK (at equilibrium) Only eqn provided at exams

12. Sec. 19.10 Bond Energies We already covered this in CHEM 121.