1. Chemical Thermodynamics BLB 11 th Chapter 19

2. Chemical Reactions 1. How fast will the reaction occur? Ch. 14 2. How far toward completion will the reaction proceed? Ch. 15 3. Will the reaction occur, i.e. is it spontaneous? Ch. 5, 19

4. Review terms energy heat work pathway state function system surroundings

5. Review terms exothermic endothermic enthalpy enthalpy change standard state std. enthalpy of formation 1 st Law of Thermodynamics

6. 19.1 Spontaneous Processes Spontaneous – proceeds on its own without any outside assistance  product-favored  not necessarily fast Nonspontaneous – requires outside intervention  reactant-favored  not necessarily slow Spontaneity is temperature-dependent.

7. Spontaneity and Energy Examples of spontaneous systems:  Brick falling  Ball rolling downhill  Hot objects cooling  Combustion reactions Are all spontaneous processes accompanied by a loss of heat, that is, exothermic?

8. Reversible & Irreversible Systems Reversible – a change in a system for which the system can restored by exactly reversing the change – a system at equilibrium ex. melting ice at 0°C Irrerversible – a process that cannot be reversed to restore the system and surroundings to their original states – a spontaneous process ex. melting ice at 25°C See p. 806 (last paragraph of section)

10. 19.2 Entropy and the 2 nd Law of Thermodynamics Entropy, S – measure of randomness  State function  Temperature-dependent A random (or dispersed) system is favored due to probability. “Entropy Is Simple – If We Avoid the Briar Patches” Frank Lambert, Occidental College, ret. http://www.entropysimple.com/content.htm

11. Entropy Change ΔS = S final − S initial (a state function) (isothermal) as for phase changes. ΔS > 0 is favorable

12. Calculating ΔS for Changes of State

13. Problem 24

14. System & Surroundings Dividing the universe: 1.System – dispersal of matter by reaction: reactants → products 2.Surroundings – dispersal of energy as heat

15. 2 nd Law of Thermodynamics The total entropy of the universe increases for any spontaneous process.  ΔS univ > 0  ΔS univ = ΔS sys + ΔS surr For a reversible process: ΔS univ = 0. For an irreversible process:  Net entropy increase ►spontaneous  Net entropy decrease ► nonspontaneous

16. 19.3 The Molecular Interpretation of Entropy Molecules have degrees of freedom based upon their motion  Translational  Vibrational  Rotational Motion of water (Figure 19.6, p. 810) Motion of water Lowering the temperature decreases the entropy.

17. Boltzman & Microstates S = k ln W (W = # of microstates) If # microstates ↑, then entropy ↑. Increasing volume, temperature, # of molecules increases the # of microstates.

18. Examples of systems that have increased entropy Entropy increases for: Changes of state: solid → liquid → gas (T) Expansion of a gas (V) Dissolution: solid → solution (V) Production of more moles in a chemical reaction (# of particles) Ionic solids: lower ionic charge S° (J/mol·K) Na 2 CO 3 136 MgCO 3 66

19. Changes of State H 2 O stateS° (J/mol·K) l69.91 g188.83

20. Dissolution

21. Expansion of a Gas

22. 2 NO(g) + O 2 (g) → 2 NO 2 (g)

23. Problem 20 TNT (trinitrotoluene) Detonation 4 C 3 H 5 N 3 O 9 (l) → 6 N 2 (g) + 12 CO 2 (g) + 10 H 2 O(g) + O 2 (g) a)Spontaneous? b)Sign of q? c)Can the sign of w be determined? d)Can the sign of ΔE be determined?

24. 3 rd Law of Thermodynamics The entropy, S, of a pure crystalline substance at absolute zero (0 K) is zero.

26. 19.4 Entropy Changes in Chemical Reactions Standard molar entropy values, S° (J/mol·K): increase in value as temperature increases from 0 K have been determined for common substances (Appendix C, pp. 1112-1114) increase with molar mass increase with # of atoms in molecule

29. Calculating ΔS° sys ΔS° sys = ∑nS°(products) - ∑mS°(reactants) (where n and m are coefficients in the chemical equation)

30. Problem 50

31. Problem

32. Entropy Changes in the Surroundings Heat flow affects surroundings. As T increases, ΔH becomes less important. As T decreases, ΔH becomes more important.

33. Calculating ΔS° univ ΔS univ = ΔS sys + ΔS surr by obtaining ΔS sys and ΔS surr If ΔS univ > 0, the reaction is spontaneous. But there is a better way – one in which only the system is involved.

34. 19.5 Gibbs Free Energy The spontaneity of a reaction involves both enthalpy (energy) and entropy (matter). Gibbs Free Energy, ΔG makes use of ΔH sys and ΔS sys to predict spontaneity. ΔG sys represents the total energy change for a system. G = H – TS or ΔG = ΔH – TΔS or, under standard conditions: ΔG° = ΔH° – TΔS°

36. Gibbs Free Energy If:  ΔG < 0, forward reaction is spontaneous  ΔG = 0, reaction is at equilibrium  ΔG > 0, forward reaction is nonspontaneous In any spontaneous process at constant temperature and pressure, the free energy always decreases. ΔG is a state function. ΔG f ° of elements in their standard state is zero.

38. Calculating ΔG° sys ΔG° sys = ΔH° sys − TΔS° sys or ΔG° sys = ∑nΔG° f (products) - ∑mΔG° f (reactants) (where n and m are coefficients in the chemical equation)

39. Problem 56

40. 19.6 Free Energy and Temperature ΔHΔHΔSΔS−TΔS−TΔSΔGΔGReaction −+−always − spontaneous at all T K > 1 +−+always + nonspontaneous at all T K < 1 −−+ − @ low T + @high T spontaneous at low T nonspontaneous at high T ++− + @ low T − @high T nonspontaneous at low T spontaneous at high T

41. Driving force of a reaction For a reaction where ΔG < 0: Enthalpy-driven – if ΔH < 0 and ΔS < 0; at low temp. Entropy-driven – if ΔH > 0 and ΔS > 0; at high temp. “cross-over point” is where ΔG = 0

42. Problem 66

43. 19.7 Free Energy and K If conditions are non-standard: ΔG = ΔG° + RT lnQR = 8.3145 J/mol·K If at equilibrium: ΔG = ΔG° + RT lnQ = 0 ΔG° = −RT lnK

44. Problem 78

45. Problem