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Lecture 9 : D G, Q, and K The Meaning of D G

Lecture 9 : D G, Q, and K The Meaning of D G. Reading: Zumdahl 10.10, 10.11, 10.12 Outline Relating D G to Q Relating D G to K A descriptive example of D G. The Second Law. The Second Law: there is always an increase in the entropy of the universe .

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Lecture 9 : D G, Q, and K The Meaning of D G

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  1. Lecture 9 : DG, Q, and KThe Meaning of DG • Reading: Zumdahl 10.10, 10.11, 10.12 • Outline • Relating DG to Q • Relating DG to K • A descriptive example of DG

  2. The Second Law • The Second Law: there is always an increase in the entropy of the universe. • From our definitions of system and surroundings: DSuniverse = DSsystem + DSsurroundings

  3. Defining DG • Recall, the second law of thermodynamics: DSuniv = DStotal = DSsystem + DSsurr • Also recall: DSsurr = -DHsys/T • Then, DStotal = DSsystem -DHsys/T -TDStotal = -TDSsystem+DHsys

  4. Defining DG (cont.) • We then define: DG = -TDStotal • Then: DG = -TDSsys+DHsys • Or: DG = DH - TDS DG = The Gibbs Free Energy w/ Pconst

  5. Relating DG to Q • Recall from Lecture 6: DS = R ln (Wfinal/Winitial) • For the expansion of a gas Wfinal Volume

  6. Relating DG to Q (cont.) • Given this relationship DS = R ln (Vfinal/Vinitial)

  7. Relating DG to Q (cont.) • This equation tells us what the change in entropy will be for a change in concentration away from standard state. Entropy change for process occurring under standard conditions Additional term for change in concentration. (1 atm, 298 K) (P ≠ 1 atm)

  8. Relating DG to Q (cont.) • How does this relate to DG?

  9. Relating DG to Q (cont.) • Generalizing to a multicomponent reaction: • Where

  10. Relating DG to Q (cont.)

  11. An Example • Determine DGrxn at 298 K for: C2H4(g) + H2O(l) C2H5OH(l) where PC2H4 = 0.5 atm (others at standard state) DG°rxn = -6 kJ/mol (from Lecture 9)

  12. An Example (cont.) C2H4(g) + H2O(l) C2H5OH(l) DGrxn = -6 kJ/mol + (8.314 J/mol.K)(298K)ln(2) = -4.3 kJ/mol

  13. DG and K • The Reaction Quotient (Q) corresponds to a situation where the concentrations of reactants and products are not those at equilibrium. • At equilibrium, we have K. • What is the relationship between DG and K?

  14. DG and K (cont.) • At equilibrium, DGrxn = 0 0 K 0 = DG°rxn +RTln(K) DG°rxn = -RTln(K)

  15. DG and K (cont.) • Let’s look at the interaction between DG° and K DG°rxn = -RTln(K) If DG° < 0 then K > 1 Products are favored over reactants

  16. DG and K (cont.) • Let’s look at the interaction between DG° and K DG°rxn = -RTln(K) If DG° = 0 then K = 1 Products and reactants are equally favored

  17. DG and K (cont.) • Let’s look at the interaction between DG° and K DG°rxn = -RTln(K) If DG° > 0 then K < 1 Reactants are favored over products

  18. Temperature Dependence of K • We now have two definitions for DG° DG°rxn = -RTln(K) = DH° - TDS° • Rearranging: y = m x + b • Plot of ln(K) vs 1/T is a straight line

  19. T Dependence of K (cont.) • If we measure K as a function of T, we can determine DH° by determining the slope of the line slope intercept

  20. T Dependence of K (cont.) • Once we know the T dependence of K, we can predict K at another temperature: - the van’t Hoff equation.

  21. An Example • For the following reaction: CO(g) + 2H2(g) CH3OH(l) DG° = -29 kJ/mol What is K at 340 K? • First, what is Keq when T = 298 K? DG°rxn = -RTln(K) = -29 kJ/mol ln(K298) = (-29 kJ/mol) = 11.7 -(8.314 J/mol.K)(298K) K298 = 1.2 x 105

  22. An Example (cont.) • Next, to use the van’t Hoff Eq., we need DH° CO(g) + 2H2(g) CH3OH(l) DHf°(CO(g)) = -110.5 kJ/mol DHf°(H2(g)) = 0 DHf°(CH3OH(l)) = -239 kJ/mol DH°rxn=SDH°f (products) - SDH° f (reactants) = DH°f(CH3OH(l)) - DH°f(CO(g)) = -239 kJ - (-110.5 kJ) = -128.5 kJ

  23. An Example (cont.) • With DH°, we’re ready for the van’t Hoff Eq. Why is K reduced? Reaction is Exothermic. K340 = 2.0 x 103 Increase T, Shift Eq. To React. Keq will then decrease

  24. An Example • For the following reaction at 298 K: HBrO(aq) + H2O(l)BrO-(aq) + H3O+(aq) w/ Ka = 2.3 x 10-9 What is DG°rxn? DG°rxn = -RTln(K) = -RTln(2.3 x 10-9) = 49.3 kJ/mol

  25. An Example (cont.) • What is DGrxn when pH = 5, [BrO-] = 0.1 M, and [HBrO] = 0.2 M ? HBrO(aq) + H2O(l) BrO-(aq) + H3O+(aq)

  26. An Example (cont.) • Then: = 49.3 kJ/mol + (8.314 J/mol.K)(298 K)ln(5 x 10-6) = 19.1 kJ/mol DGrxn < DG°rxn “shifting” reaction towards products

  27. A Descriptive Example • Consider the following gas-phase equilibrium: ClOO (g) OClO (g) • Now, DGrxn = DG°rxn +RTln(Q) = -RTln(K) + RTln(Q) = RTln(Q/K)

  28. A Descriptive Example (cont.) • From inspection, DGrxn = RTln(Q/K) If Q/K < 1 DGrxn < 0 OClO (g) ClOO (g) If Q/K > 1 DGrxn > 0 OClO (g) ClOO (g) If Q/K = 1 DGrxn = 0 equilibrium OClO (g) ClOO (g)

  29. A Descriptive Example (cont.) • K = 0.005 DG°rxn = -RTln(K) = 13.1 kJ/mol • Let’s start the rxn off with 5 atm of OClO and 0.005 atm of ClOO. DGrxn = RTln(Q/K) Q = PClOO/POClO = .001 = RTln(.001/.005) = RTln(.2) = -1.6RT < 0 DGrxn < 0, reaction is proceeding towards products

  30. A Descriptive Example (cont.) • Let’s start the rxn off with 5 atm. of ClOO and 0.5 atm of OClO. DGrxn = RTln(Q/K) Q = PClOO/POClO = 10 = RTln(10/0.005) = RTln(2000) >> 0 DGrxn > 0, reaction is proceeding towards reactants

  31. A Descriptive Example (cont.) • Where is equilibrium? Back to Chem 142 OClO (g) ClOO (g) Pinit (atm) 5 0.005 Pequil. (atm) 5 - x 0.005 + x x = 0.02 PClOO = 0.025 atm POClO = 5 - .02 = 4.98 atm

  32. A Descriptive Example (cont.) DGrxn = RTln(Q/K) = RTln(1) = 0 equilibrium. K Pictorally: “Amount” of OClO Present

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