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Slide 1 of 44 Chapter 19: Spontaneous Change: Entropy and Free Energy.

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Presentation on theme: "Slide 1 of 44 Chapter 19: Spontaneous Change: Entropy and Free Energy."— Presentation transcript:

1 Slide 1 of 44 Chapter 19: Spontaneous Change: Entropy and Free Energy

2 Slide 2 of 44 Contents 19-1Spontaneity: The Meaning of Spontaneous Change 19-2The Concept of Entropy 19-3Evaluating Entropy and Entropy Changes 19-4Criteria for Spontaneous Change: The Second Law of Thermodynamics 19-5Standard Free Energy Change, ΔG° 19-6Free Energy Change and Equilibrium 19-7ΔG° and K eq as Functions of Temperature

3 Slide 3 of Spontaneity: The Meaning of Spontaneous Change

4 Slide 4 of 44 Spontaneous Process  A process that occurs in a system left to itself.  Once started, no external actions is necessary to make the process continue.  A non-spontaneous process will not occur without external action continuously applied. 4 Fe(s) + 3 O 2 (g) → 2 Fe 2 O 3 (s) H 2 O(s) H 2 O(l)

5 Slide 5 of 44 Spontaneous Process  Potential energy decreases.  For chemical systems the internal energy U is equivalent to potential energy.  Berthelot and Thomsen 1870’s.  Spontaneous change occurs in the direction in which the enthalpy of a system decreases.  Mainly true but there are exceptions.

6 Slide 6 of The Concept of Entropy  Entropy, S.  The greater the number of configurations of the microscopic particles among the energy levels in a particular system, the greater the entropy of the system. ΔS > 0 spontaneous ΔU = ΔH = 0

7 Slide 7 of 44 Entropy Change ΔS = q rev T For changes occurring at constant temperature

8 Slide 8 of Evaluating Entropy and Entropy Changes  Phase transitions.  Exchange of heat can be carried out reversibly. ΔS = ΔHΔH T tr H 2 O(s, 1 atm) H 2 O(l, 1 atm) ΔH fus  = 6.02 kJ at K ΔS fus = ΔH fus T tr ° = 6.02 kJ mol K = 2.20  kJ mol -1 K -1

9 Slide 9 of 44 Trouton’s Rule ΔS = ΔH vap T bp  87 kJ mol -1 K -1

10 Slide 10 of 44 Absolute Entropies  Third law of thermodynamics.  The entropy of a pure perfect crystal at 0 K is zero.  Standard molar entropy.  Tabulated in Appendix D. ΔS = [  p S°(products) -  r S°(reactants)]

11 Slide 11 of 44 Entropy as a Function of Temperature

12 Slide 12 of 44 Vibrational Energy and Entropy

13 Slide 13 of 44  Worked Examples Follow

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18 Slide 18 of 44  CRS Questions Follow:

19 Slide 19 of 44 Which of the following processes would you expect to result in a greater positive change in entropy? None of these processes involve positive entropy changes.

20 Slide 20 of 44 Which of the following processes would you expect to result in a greater positive change in entropy? None of these processes involve positive entropy changes.

21 Slide 21 of 44 Which of the following statements is incorrect? 1.The absolute entropy for gaseous C 2 H 6 is lower than that for gaseous C 3 H 8 at the same temperature. 5.  fus S (solid to liquid) for H 2 O is greater than that for C 3 H 8. 4.The absolute entropy for gaseous C 3 H 8 decreases when the temperature is increased. 2.The entropy of formation for gaseous C 3 H 8 is negative. 3.The entropy of formation of gaseous C 3 H 8 is more negative than the entropy of formation for C 2 H 6.

22 Slide 22 of 44 Which of the following statements is incorrect? 1.The absolute entropy for gaseous C 2 H 6 is lower than that for gaseous C 3 H 8 at the same temperature. 5.  fus S (solid to liquid) for H 2 O is greater than that for C 3 H 8. 4.The absolute entropy for gaseous C 3 H 8 decreases when the temperature is increased. 2.The entropy of formation for gaseous C 3 H 8 is negative. 3.The entropy of formation of gaseous C 3 H 8 is more negative than the entropy of formation for C 2 H 6.

23 Slide 23 of 44 The entropy change for the decomposition of ozone forming diatomic oxygen, 1.is positive because two moles of gas are forming three moles of gas. 2.is close to zero because there are the same number of atoms on each side of the equation. 3.is negative because energy is released as the reaction proceeds. 4.is close to zero because both ozone and oxygen are in the gas phase. 5.None of the above answers makes sense.

24 Slide 24 of 44 The entropy change for the decomposition of ozone forming diatomic oxygen, 1.is positive because two moles of gas are forming three moles of gas. 2.is close to zero because there are the same number of atoms on each side of the equation. 3.is negative because energy is released as the reaction proceeds. 4.is close to zero because both ozone and oxygen are in the gas phase. 5.None of the above answers makes sense.

25 Slide 25 of 44 Which of the following reactions, occurring when propane is burned, would you expect to be the most entropically favored (most positive change in entropy)?

26 Slide 26 of 44 Which of the following reactions, occurring when propane is burned, would you expect to be the most entropically favored (most positive change in entropy)?

27 Slide 27 of 44 Which of the following compounds would obey Trouton’s rule most closely?

28 Slide 28 of 44 Which of the following compounds would obey Trouton’s rule most closely?


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