Thermodynamics Part II Gibbs Free Energy Equilibrium Jespersen

Thermodynamics Part II: Gibbs Free Energy & Equilibrium Jespersen Chap. 19 Sec 7 thru 10 Dr. C. Yau Spring 2013 1

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

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

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

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. 5

Example 19. 7 p. 895 For the phase change Br 2(l) Br 2(g) ΔHo = +31. 0 k. J 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 Br 2 (l) will be in equilibrium with Br 2 (g) at 1 atm (i. e. the normal boiling point of Br 2. ) Ans. 61 o. C (experimental bp = 58. 8 o. C) But 31. 0/ 92. 9 = 0. 334. Why is it not 61 o. C? Do Pract Exer 17, 18 on p. 895 6

Using ΔG to predict equilibrium shift ΔG = ΔGo + RT ln Q (Q = reaction quotient) Example 19. 10 p. 900 2 NO 2 (g) N 2 O 4 (g) ΔG 298 o= - 5. 40 k. J/mol N 2 O 4 In a rxn mixture, the partial pressure of NO 2 is 0. 25 atm and the partial pressure of N 2 O 4 is 0. 60 atm. In which direction must this reaction proceed to reach equilibrium? R = 8. 314 J mol-1 K-1 Do Pract Exer 23, 24 p. 902 Ans. ΔG=+ 0. 20 k. J/mol, rxn will proceed to left. How do we know that? . . . ΔG=+ means to the left? ? 7 (Some of N 2 O 4 will break up to form NO 2. )

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! 8

Example 19. 11 p. 902 (Calculate ΔGo from K? ) The brownish haze often associated with air pollution is caused by nitrogen dioxide, NO 2, 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 NO 2 by oxygen. 2 NO (g) + O 2 (g) 2 NO 2 (g) The value of Kp for the rxn is 1. 7 x 1012 at 25. 00 o. C. What is ΔGo for the rxn in joules per mole? In kilojoules per mole? Ans. -69. 80 k. J/mol 9

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 SO 3. 2 SO 2(g) + O 2(g) 2 SO 3(g) For this rxn, ΔGo = - 1. 40 x 102 k. J mol-1 at 25 o. C. What is the value of Kp? Ans. 3 x 1024 Do Pract Exer 25, 26 p. 904 10

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 - RTln. Q (predict how reach equil. ) ΔG = - RTln. K (at equilibrium) 11

Sec. 19. 10 Bond Energies We already covered this in CHEM 121. 12