1. Chapter 16.4-16.6 in our book Spontaneity, Entropy, and Free Energy

2. 2 Section 16.4 Free Energy Copyright © Cengage Learning. All rights reserved Return to TOC Spontaneous Reactions http://college.cengage.com/chemistry/discipline/thinkwell/2999.html

3. 3 Section 16.4 Free Energy Copyright © Cengage Learning. All rights reserved Return to TOC Concept Check ? A liquid is vaporized at its boiling point. Predict the signs of: w q ΔH ΔS ΔS surr ΔG Explain your answers. – + + + – ? 0 w = –PΔV

4. 4 Section 16.5 Entropy Changes in Chemical Reactions Copyright © Cengage Learning. All rights reserved Return to TOC Concept Check ? Gas A 2 reacts with gas B 2 to form gas AB at constant temperature and pressure. The bond energy of AB is much greater than that of either reactant. Predict the signs of: ΔHΔS surr ΔSΔS univ Explain. – + 0 +

5. Temperature Dependence of K So, ln(K) ∝ 1/T

6. Free Energy and Work  The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy  The amount of work obtained is always less than the maximum  Henry Bent's First Two Laws of Thermodynamics  First law: You can't win, you can only break even  Second law: You can't break even

7. 7 Chapter 16 Table of Contents Copyright © Cengage Learning. All rights reserved 16.4Free Energy16.4Free Ener 16.5Entropy Changes in Chemical Reactions16.5Entropy Changes in Chemical Reaction 16.6Free Energy and Chemical Reactions16.6Free Energy and Chemical Reaction 16.7The Dependence of Free Energy on Pressure16.7The Dependence of Free Energy on Pressur 16.8Free Energy and Equilibrium16.8Free Energy and Equilibriu 16.9Free Energy and Work16.9Free Energy and Wor

8. 8 Chapter 16 Table of Contents Copyright © Cengage Learning. All rights reserved CW: Finish Notes 16.3-16.6 CW: 2 Online Quizzes Ch. 16 and ch. 6 (Links on podcast page) Turn in Calorimetry Lab report IN BOX before leaving today. HW: Ch. 16 question packet due Thursday HW: Make sure you completed the 9 question QUIA.com quiz. You may link from podcast page for Tests/Quizzes.

9. 9 Section 16.3 The MoleThe Effect of Temperature on Spontaneity Copyright © Cengage Learning. All rights reserved Return to TOC ΔS surr Heat flow (constant P) = change in enthalpy = ΔH

10. In terms of temperature, how would you describe an object that has an entropy value of 0? 0 K Perfect solid crystal with no motion Only Theoretical It is not possible to reach absolute 0! Entropy of universe is always increasing!

11. 3rd Law of Thermodynamics the entropy of a perfect crystalline substance is zero at absolute zero See appendix in book for values *Based on 0 entropy as a reference point, and calculations involving calculus beyond the scope of this course, data has been tabulated for Standard Molar Entropies ΔSº Pure substances, 1 atm pressure, 298 K

12. Standard Molar Entropies ΔSº 1)Standard molar entropies of elements are not 0 (unlike ΔHº f ). (0 entropy is only theoretical; not really possible) S.M.E of gases > S.M.E of liquids and solids. (gases move faster than liquids) 3)S.M.E. increase with increasing molar mass. (more potential vibrational freedom with more mass) 4)S.M.E. increase as the number of atoms in a formula increase. (same as above)

13. Calculating Entropy Change in a Reaction  Entropy is an extensive property (a function of the number of moles)  Generally, the more complex the molecule, the higher the standard entropy value

14. Calculating Entropy Change in a Reaction Units for Δ S ΔS=J/molK Since we are considering ΔS° J/K are often used because moles are assumed and cancel in the calculations when considering standard states.

15. Calculate the standard entropy change ( Δ Sº) for the following reaction at 298K Al 2 O 3(s) + 3H 2(g) → 2Al (s) + 3H 2 O (g) SubstanceSº(J/mol-K) at 298K Al28.32 Al 2 O 3 51.00 H 2 O (g) 188.8 H 2(g) 130.58

16. Δ S o = [2Sº(Al) + 3Sº(H 2 O)] - [Sº(Al 2 O 3 ) + 3Sº(H 2 )] Δ S o rxn = ∑ n S o (products) - ∑ m S o (reactants) Al 2 O 3(s) + 3H 2(g) → 2Al (s) + 3H 2 O (g) = 180.3 J/K

17. Predict the sign of ΔSº of the following reaction. 2SO 2(g) + O 2(g) → 2SO 3(g) Entropy decreases, - Lets’ Calculate

18. Calculate the standard entropy change ( Δ Sº) for the following reaction at 298K 2SO 2(g) + O 2(g) → 2SO 3(g) SubstanceSº(J/mol-K) at 298K SO 2 (g)248.1 SO 3 (g)256.7 O 2(g) 205.0 ΔSº = -187.8 J K -1

19. Spontaneous reactions result in an increase in entropy in the universe. Reactions that have a large and negative ΔΗ tend to occur spontaneously. Spontaneity depends on enthalpy, entropy, and temperature. Predicting spontaneous reactions

20. Provides a way to predict the spontaneity of a reaction using a combination of enthalpy and entropy of a reaction. Gibbs Free Energy (G)

21. 21 Section 16.4 Free Energy Copyright © Cengage Learning. All rights reserved Return to TOC Free Energy (G) A process (at constant T and P) is spontaneous in the direction in which the free energy decreases.  Negative ΔG means positive ΔS univ.

22. 22 Section 16.4 Free Energy Copyright © Cengage Learning. All rights reserved Return to TOC Free Energy (G) ΔG = ΔH – TΔS (at constant T and P) all of these are in reference to the system

23. ΔH, ΔS, ΔG and Spontaneity Value of ΔH Value of TΔS Value of ΔG Spontaneity Negative PositiveNegativeSpontaneous Positive NegativePositiveNonspontaneous Negative ???Spontaneous if the absolute value of Δ H is greater than the absolute value of T Δ S (low temperature) Positive ???Spontaneous if the absolute value of T Δ S is greater than the absolute value of Δ H (high temperature) ΔG = ΔH - TΔS ΔG = ΔH - TΔS H is enthalpy, T is Kelvin temperatureH is enthalpy, T is Kelvin temperature

24. 24 Section 16.4 Free Energy Copyright © Cengage Learning. All rights reserved Return to TOC Effect of DH and DS on Spontaneity ΔG = ΔH - TΔS ΔG = ΔH - TΔS H is enthalpy, T is Kelvin temperatureH is enthalpy, T is Kelvin temperature

25. Standard Free Energy Change  ΔG 0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states  ΔG 0 cannot be measured directly  The more negative the value for ΔG 0, the farther to the right the reaction will proceed in order to achieve equilibrium  Equilibrium is the lowest possible free energy position for a reaction

26. 26 Section 16.4 Free Energy Copyright © Cengage Learning. All rights reserved Return to TOC Exercise The value of ΔH vaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C. Calculate ΔS, ΔS surr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm.

27. 27 Section 16.4 Free Energy Copyright © Cengage Learning. All rights reserved Return to TOC Exercise The value of ΔH vaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C. Calculate ΔS surr, ΔS surr = (-45,700 J/mol)/(72.5 +273 K) = -132.3 =-132 J/K·mol

28. 28 Section 16.4 Free Energy Copyright © Cengage Learning. All rights reserved Return to TOC Exercise The value of ΔH vaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C. Calculate ΔS, ΔS surr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm. ΔS = 132 J/K·mol ΔS surr = -132 J/K·mol ΔG = 0 kJ/mol

29. 29 Section 16.4 Free Energy Copyright © Cengage Learning. All rights reserved Return to TOC FOR ME At the boiling point of a substance, the liquid and gaseous forms of the substance are in equilibrium. That means that for the reaction: X_liquid = X_gas the free energy change is zero: 0 = ΔG = ΔH_vap - T_boiling*ΔS_vap where ΔH_vap and ΔS_vap are the enthalpy and entropy of vaporization, respectively, and T_boiling is the boiling temperature (all this assume that a specific pressure has been picked, and is held constant). Rearranging this, we have: ΔH_vap = T_boiling*ΔS_vap ΔH_vap/ΔS_vap = T_boiling The boiling temperature is therefore given by the ratio of the enthalpy to the entropy of vaporization. If the enthalpy change required to vaporize a material increases, so does the boiling temperature. If the entropy change accompanying vaporization increases, the boiling temperature decreases. Note that ΔH_vap is the energy needed to convert a specified amount of the substance from a liquid to a gas at constant pressure. Because it always takes energy to do this (boiling is always an endothermic reaction), ΔH_vap is always positive. T is measured on the thermodynamic scale (in kelvins), and is always positive. Because ΔH_vap is always positive, this means that ΔS_vap is also always positive; the molar entropy of a gas is always larger than the molar entropy of the liquid.

30. For reactions at constant temperature: Δ G 0 = Δ H 0 - T Δ S 0 Calculating Free Energy Method #1

31. Calculating Free Energy: Method #2 An adaptation of Hess's Law: C diamond (s) + O 2 (g)  CO 2 (g) Δ G 0 = -397 kJ C graphite (s) + O 2 (g)  CO 2 (g) Δ G 0 = -394 kJ CO 2 (g)  C graphite (s) + O 2 (g) Δ G 0 = +394 kJ C diamond (s)  C graphite (s) Δ G 0 = C diamond (s) + O 2 (g)  CO 2 (g) Δ G 0 = -397 kJ -3 kJ

32. Calculating Free Energy Method #3 Using standard free energy of formation ( Δ G f 0 ): Δ G f 0 of an element in its standard state is zero

33. The Dependence of Free Energy on Pressure  Enthalpy, H, is not pressure dependent  Entropy, S  entropy depends on volume, so it also depends on pressure – S large volume > S small volume –S low pressure > S high pressure

34. Free Energy and Equilibrium  Equilibrium point occurs at the lowest value of free energy available to the reaction system  At equilibrium, ΔG = 0 and Q = K ΔG0ΔG0 K ΔG 0 = 0K = 1 ΔG 0 < 0K > 1 ΔG 0 > 0K < 1

35. 35 Section 16.5 Entropy Changes in Chemical Reactions Copyright © Cengage Learning. All rights reserved Return to TOC Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero. The entropy of a substance increases with temperature.

36. 36 Section 16.5 Entropy Changes in Chemical Reactions Copyright © Cengage Learning. All rights reserved Return to TOC Standard Entropy Values (S°) Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure. ΔS° reaction = Σ n p S° products – Σ n r S° reactants

37. 37 Section 16.5 Entropy Changes in Chemical Reactions Copyright © Cengage Learning. All rights reserved Return to TOC Exercise Calculate DS° for the following reaction: 2Na(s) + 2H 2 O(l) → 2NaOH(aq) + H 2 (g) Given the following information: S° (J/K·mol) Na(s) 51 H 2 O(l) 70 NaOH(aq) 50 H 2 (g) 131 DS°= –11 J/K

38. 38 Section 16.5 Entropy Changes in Chemical Reactions Copyright © Cengage Learning. All rights reserved Return to TOC Basically equilibrium and pressure effects will be covered at a later time We will mainly cover all of chapter 6 and 16.1-16.6 for test. ASSIGNMENTS: Hess’s Quiz graded; see score. QUIA.com HW looked over (9 questions) CW: PROBLEMS CH. 6 AND 16 PRACTICE ONLINE SHOW ME SCORES TURN IN ALL PREVIOUS ASSIGNMENTS HW: SG ch 16 handout” pg. 389-391 #1-32 due Thursday

39. 39 Section 16.5 Entropy Changes in Chemical Reactions Copyright © Cengage Learning. All rights reserved Return to TOC WEDNESDAY - FEB. 13, 2013 We will mainly cover all of chapter 6 and 16.1-16.6 for test - TEST is TUESDAY. Be well prepared. See ch. reviews also. Calorimetry Problems #4-#5 from LAB discussion. ASSIGNMENTS: 2 online quizzes - show me answers QUIA.com HW looked over (9 questions) - not for grade but must complete - only 1/2 of you have done this. CW/HW: Problems handout sheet from ch. 16 #1-32 (also be sure you completed the problem packet from ch. 6 TURN IN ALL PREVIOUS ASSIGNMENTS

40. 40 Section 16.6 Free Energy and Chemical Reactions Copyright © Cengage Learning. All rights reserved Return to TOC Standard Free Energy Change (ΔG°) The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. ΔG° = ΔH° – TΔS° ΔG° reaction = Σ n p G° products – Σ n r G° reactants

41. 41 Section 16.6 Free Energy and Chemical Reactions Copyright © Cengage Learning. All rights reserved Return to TOC Concept Check A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of: ΔH° ΔS°ΔG° Explain. – – –

42. 42 Section 16.6 Free Energy and Chemical Reactions Copyright © Cengage Learning. All rights reserved Return to TOC Concept Check Consider the following system at equilibrium at 25°C. PCl 3 (g) + Cl 2 (g) PCl 5 (g) ΔG° = −92.50 kJ What will happen to the ratio of partial pressure of PCl 5 to partial pressure of PCl 3 if the temperature is raised? Explain. The ratio will decrease.

43. 43 Section 16.7 The Dependence of Free Energy on Pressure Copyright © Cengage Learning. All rights reserved Return to TOC Free Energy and Pressure G = G° + RT ln(P) or ΔG = ΔG° + RT ln(Q) I am not covering section 16.7-16.9 on this test.

44. 44 Section 16.7 The Dependence of Free Energy on Pressure Copyright © Cengage Learning. All rights reserved Return to TOC Concept Check Sketch graphs of: 1. G vs. P 2. H vs. P 3. ln(K) vs. 1/T (for both endothermic and exothermic cases)

45. 45 Section 16.7 The Dependence of Free Energy on Pressure Copyright © Cengage Learning. All rights reserved Return to TOC The Meaning of ΔG for a Chemical Reaction A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.

46. 46 Section 16.8 Free Energy and Equilibrium Copyright © Cengage Learning. All rights reserved Return to TOC The equilibrium point occurs at the lowest value of free energy available to the reaction system. ΔG = 0 = ΔG° + RT ln(K) ΔG° = –RT ln(K)

47. 47 Section 16.8 Free Energy and Equilibrium Copyright © Cengage Learning. All rights reserved Return to TOC Change in Free Energy to Reach Equilibrium

48. 48 Section 16.8 Free Energy and Equilibrium Copyright © Cengage Learning. All rights reserved Return to TOC Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction

49. 49 Section 16.9 Free Energy and Work Copyright © Cengage Learning. All rights reserved Return to TOC Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy. w max = ΔG

50. 50 Section 16.9 Free Energy and Work Copyright © Cengage Learning. All rights reserved Return to TOC Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy. All real processes are irreversible. First law: You can’t win, you can only break even. Second law: You can’t break even.