1. Chemical Thermodynamics Chapter 19 (except 19.7!)

2. THERMODYNAMICS The BIG question!!! How do we know whether a reaction will happen or not?

3. THERMODYNAMICS Sample Problem 1 Which of the following processes are spontaneous (happen naturally)? (a) a ball rolling up a hill (b) the freezing of water at -3 o C and 1 atm pressure (c) the freezing of water at 3 o C and 1 atm pessure (d) the decomposition of carbon dioxide into carbon and oxygen at 25 o C and 1 atm pressure (e) the dissolving of NaCl in water (a) not spontaneous (b) spontaneous (c) not spontaneous (d) not spontaneous (e) spontaneous Processes that occur spontaneously in one direction are NOT spontaneous in the opposite direction.

4. THERMODYNAMICS The two driving forces for reactions are (1) the release of energy (gain of stability) (2) increasing entropy In nature, reactions move toward lowest energy and highest entropy

5. THERMODYNAMICS Sample Problem 2 At 25 o C, the ΔH value for the reaction 2 H 2 (g) + O 2 (g)  2 H 2 O(l) is - 572 kJ Can you conclude from the sign of ΔH that the reaction is spontaneous? NO! Many exothermic reactions are spontaneous, but not all!

6. THERMODYNAMICS Entropy is a thermodynamic state function represented by S. Entropy is a measure of the number of ways energy is distributed in a system. The more ways energy is distributed in a system, the more random or the more disordered the system is said to be. The greater the entropy of a system, the greater is its randomness or disorder. Entropies are usually positive for both elements and compounds. The units of entropy are J/moleK

8. THERMODYNAMICS We usually look at the change in entropy during a process: ΔS = S final - S initial (state function) ΔS > 0 (is positive) if the system changes to a state of greater disorder. ΔS < 0 (is negative) if the system changes to a state of greater order

9. THERMODYNAMICS 2nd Law of Thermodynamics: In any spontaneous process, the entropy of the universe always increases. ΔS universe = ΔS system + ΔS surroundings > 0 A process that involves an increase in order (ΔS system < 0) CAN be spontaneous provided it causes an even greater positive entropy changes in the surroundings. An isolated system (no exchange of energy between the system and surroundings) always increases its entropy when it undergoes a spontaneous change. For such a change, ΔS surroundings = 0

10. THERMODYNAMICS Standard entropies (S°) have been determined for many substances, and these values are based on the 3rd law of thermodynamics. 3rd Law of Thermodynamics The entropy of a perfect, pure crystalline substance at 0 K is zero.

11. THERMODYNAMICS Qualitative "Rules" About Entropy: 1)Entropy increases as one goes from a solid to a liquid, or even more dramatically, a liquid to a gas 2)Entropy increases as the number of moles of gas increases 3) The Entropy of any material increases with increasing temperature 4) Entropy increases if a solid or liquid is dissolved in a solvent. 5) Entropy is higher for weakly bonded compounds than for compounds with very strong covalent bonds 6) Entropy increases as the number of particles in a system increases 7) Entropy increases as the number of electrons increases: Entropy increases as the mass, # of atoms, # of heavier atoms, etc. of a molecule increases

12. THERMODYNAMICS Entropy changes dramatically at a phase change.

13. THERMODYNAMICS Sample Problem 3 For each of the following pairs, which has the greater entropy: (a) CO 2 (s) or CO 2 (g) (b) NH 3 (l) or NH 3 (g) (c) a crystal of pure Mg at 0 K or a crystal at 200 K (a) CO 2 (g)

14. THERMODYNAMICS Sample Problem 3 For each of the following pairs, which has the greater entropy: (a) CO 2 (s) or CO 2 (g) (b) NH 3 (l) or NH 3 (g) (c) a crystal of pure Mg at 0 K or a crystal at 200 K (a) CO 2 (g) (b) NH 3 (g)

15. THERMODYNAMICS Sample Problem 3 For each of the following pairs, which has the greater entropy: (a) CO 2 (s) or CO 2 (g) (b) NH 3 (l) or NH 3 (g) (c) a crystal of pure Mg at 0 K or a crystal at 200 K (a) CO 2 (g) (b) NH 3 (g) (c) crystal at 200 K

16. THERMODYNAMICS Sample Problem 4 Predict whether ΔS is > 0 or < 0 for each of the following: (a) sugar + water  sugar dissolved in water (b) Na 2 SO 4 (s)  2 Na + (aq) + SO 4 -2 (aq) (c) 2 H 2 (g) + O 2 (g)  2 H 2 O(g)

17. THERMODYNAMICS Sample Problem 4 Predict whether ΔS is > 0 or < 0 for each of the following: (a) sugar + water  sugar dissolved in water (b) Na 2 SO 4 (s)  2 Na + (aq) + SO 4 -2 (aq) (c) 2 H 2 (g) + O 2 (g)  2 H 2 O(g) (a) > 0 (positive)

18. THERMODYNAMICS Sample Problem 4 Predict whether ΔS is > 0 or < 0 for each of the following: (a) sugar + water  sugar dissolved in water (b) Na 2 SO 4 (s)  2 Na + (aq) + SO 4 -2 (aq) (c) 2 H 2 (g) + O 2 (g)  2 H 2 O(g) (a) > 0 (positive) (b) > 0 (positive)

19. THERMODYNAMICS Sample Problem 4 Predict whether ΔS is > 0 or < 0 for each of the following: (a) sugar + water  sugar dissolved in water (b) Na 2 SO 4 (s)  2 Na + (aq) + SO 4 -2 (aq) (c) 2 H 2 (g) + O 2 (g)  2 H 2 O(g) (a) > 0 (positive) (b) > 0 (positive) (c) < 0 (negative)

20. THERMODYNAMICS Entropy is a state function and can be calculated for a reaction: ΔS° rxn = Σ n S° products - Σ m S° reactants values are tabulated in appendix C in your textbook On AP Equation Sheet!

21. THERMODYNAMICS Sample Problem 5 Calculate ΔS° rxn for the following at 25 o C 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) standard entropy values at 25 o C (from book) SO 2 (g) 248.5 J/K mole O 2 (g) 205.0 J/K mole SO 3 (g)256.2 J/K mole ΔS° rxn = Σ n S° products - Σ m S° reactants = 2 S° SO 3 – (2 S° SO 2 + S° O 2 ) = 2(256.2) – [2(248.5) + 205.0] = - 189.6 J/K negative sign indicates a decrease in entropy (decrease in # of gas molecules also indicates a decrease in entropy)

22. THERMODYNAMICS Exothermic processes (-ΔH) are more likely to be spontaneous than endothermic processes (+ΔH). Processes that involve an increase in entropy (ΔS > 0) are more probable than those showing a decrease in entropy (ΔS < 0).

23. THERMODYNAMICS Enthalpy and entropy are connected together by the thermodynamic state function called free energy (G). The change in free energy for any process or reaction at constant temperature and pressure is ΔG = ΔH - T ΔS if ΔG < 0 (is negative), a reaction spontaneously proceeds in the forward direction if ΔG = 0, a system is at equilibrium if ΔG > 0 (is positive), a reaction is not spontaneous as written, but the reverse direction is spontaneous On AP Equation Sheet!

24. THERMODYNAMICS ΔG is a state function and tabulated values exist for standard free energy of formation for substances - in appendix C ΔG° f for any element in its most stable state is zero. ΔG° rxn = Σ n ΔG° f (products) - Σ m ΔG° f (reactants) On AP Equation Sheet!

25. THERMODYNAMICS Sample Problem 6 Given the ΔG value for the following phase changes at 1 atm, predict whether each change is spontaneous: (a) at 283 K, ΔG = -250 kJ/mole for H 2 O(s)  H 2 O(l) (b) at 273 K, ΔG = 0 kJ/mole for H 2 O(s)  H 2 O(l) (c) at 263 K, ΔG = 210 kJ/mole for H 2 O(s)  H 2 O(l)

26. THERMODYNAMICS Sample Problem 6 Given the ΔG value for the following phase changes at 1 atm, predict whether each change is spontaneous: (a) at 283 K, ΔG = -250 kJ/mole for H 2 O(s)  H 2 O(l) (b) at 273 K, ΔG = 0 kJ/mole for H 2 O(s)  H 2 O(l) (c) at 263 K, ΔG = 210 kJ/mole for H 2 O(s)  H 2 O(l) (a) spontaneous since ΔG < 0

27. THERMODYNAMICS Sample Problem 6 Given the ΔG value for the following phase changes at 1 atm, predict whether each change is spontaneous: (a) at 283 K, ΔG = -250 kJ/mole for H 2 O(s)  H 2 O(l) (b) at 273 K, ΔG = 0 kJ/mole for H 2 O(s)  H 2 O(l) (c) at 263 K, ΔG = 210 kJ/mole for H 2 O(s)  H 2 O(l) (a) spontaneous since ΔG < 0 (b) at equilibrium – no net reaction

28. THERMODYNAMICS Sample Problem 6 Given the ΔG value for the following phase changes at 1 atm, predict whether each change is spontaneous: (a) at 283 K, ΔG = -250 kJ/mole for H 2 O(s)  H 2 O(l) (b) at 273 K, ΔG = 0 kJ/mole for H 2 O(s)  H 2 O(l) (c) at 263 K, ΔG = 210 kJ/mole for H 2 O(s)  H 2 O(l) (a) spontaneous since ΔG < 0 (b) at equilibrium – no net reaction (c) non spontaneous since ΔG > 0

29. THERMODYNAMICS Sample Problem 7 Given: ΔG° f for C 6 H 12 O 6 (s) = -907.9 kJ/mole, ΔG° f for CO 2 (g) = -394.6 kJ/mole ΔG° f for H 2 O(l) = -237.2 kJ/mole calculate ΔG° for the oxidation of glucose C 6 H 12 O 6 (s) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O(l) Δ G° rxn = Σ n Δ G°f (products) - Σ m Δ G°f (reactants) = [6(ΔG° f CO 2 ) + 6(ΔG° f H 2 O)] – [ΔG° f C 6 H 12 O 6 + 6(ΔG° f O 2 )] = [6(-394.6) + 6(-237.2)] – [1(-907.9) + 6(0)] = - 2882.9 kJ (spontaneous)

30. THERMODYNAMICS Effect of temperature on Reaction Spontaneity ΔG = ΔH - T ΔS ΔHΔSΔGRxn Characteristics - +always -always spontaneous as written + -always +always nonspontaneous as written reverse rxn is spontaneous - -- at low T spontaneous at low T + at high T nonspontaneous at high T + ++ at low T nonspontaneous at low T - at high T spontaneous at high T

31. THERMODYNAMICS Sample Problem 8 What is the temperature at which sodium chloride reversibly melts (solid and liquid states are in equilibrium)? The enthalpy of melting is 30.3 kJ/mole and the entropy change upon melting is 28.2 J/mole K. Δ G = Δ H - T Δ S at equilibrium Δ G = 0 = Δ H - T Δ S so Δ H = T Δ S and T = Δ H / Δ S 30300 J/mole T = -------------------- = 1070 K (797 o C) 28.2 J/mole K above this temp (1070 K), melting is spontaneous below this temp, melting would be nonspontaneous

32. THERMODYNAMICS Sample Problem 9 At 298 K, Δ G° = -190.5 kJ and Δ H° = -184.6 kJ for the reaction H 2 (g) + Cl 2 (g)  2 HCl(g) calculate the standard entropy Δ S° for the reaction Δ G° = Δ H° - T Δ S° Δ S° = ( Δ H° - Δ G° ) / T - 184.6 kJ – (-190.5 kJ) Δ S° = ------------------------------- =.020 kJ/K = 20 J/K 298 K Typically, S values or  S values are in J/K not kJ/K

33. THERMODYNAMICS Sample problem 10 Given the following data: Substance  H o f (kJ/mol) S o (J/K mol) Fe 2 O 3 (s) -826 90.0 Fe(s) 0 27.0 O 2 (g) 0 205.0 Calculate ΔG o for the reaction: 4 Fe(s) + 3O 2 (g)  2 Fe 2 O 3 (s)  G o =  H o - T  S o  H o =  prod -  react = [2(-826)] – [0 + 0] = - 1650 kJ  S o =  prod -  react = [2(90.0)] – [3(205.0) + 4(27.0)] = -543 J/K Notice that  H is kJ and  S is J/K -543 J/K = -0.543 kJ/K  G o =  H o - T  S o = - 1650 kJ - (298 K)(-0.543 kJ/K) = - 1490 kJ