Chapter 19 Chemical Thermodynamics

1. Chapter 19Chemical Thermodynamics Entropy, Enthalpy, and Free Energy

2. Thermodynamic Quantities • H : Enthalpy, heat energy • E : Total energy, q + w • S : Entropy, disorder • G : Free energy, measure of spontaneity, energy available to do work

3. Relationship between thermodynamic quantities: DG = DH – TDS • Sign of DG DG < 0 Reaction is spontaneous, reaction will proceed in the forward direction DG < 0 Reaction is not spontaneous, however the reaction going in the reverse direction is spontaneous DG = 0 The reaction is at equilibrium, nothing will happen

4. The Meaning of “Spontaneous” • Spontaneous reaction: • Product favored • Given sufficient time, a combination of reactants will be converted to products K >> 1 • Non spontaneous reaction: • Reactant favored • Given sufficient time, nothing will happen K << 1

5. Problems • H2 + I2 < ----- > 2HI Kc = 51 • N2 + O2 < ----- > 2NO Kc = 1 x 10-30 • Which reaction is spontaneous in the forward direction? • Which reaction has a negative DG0?

6. What Drives a Reaction to Occur? • Driving Forces: • The tendencies of concentrated energy and matter to disperse • Enthalpy: • Exothermic reactions disperse energy • Energy flows from hot area to cold area • Entropy: • The tendency of concentrated matter to disperse

7. Will a process be spontaneous? • If energy and matter are both dispersed in a reaction, it is definitely spontaneous • If only energy or matter is dispersed, then the relative effects of enthalpy and entropy determine spontaneity • If neither matter nor energy is dispersed, then that process will not be spontaneous and reactants will remain, no matter how long we wait

8. The Three Laws of Thermodynamics First Law: The total energy of the universe is a constant DEsystem = -DEsurroundings and E = q+w Second Law: The total entropy of the universe is always increasing Third Law: The entropy of a pure perfectly formed crystalline substance at absolute zero is zero.

9. Entropy • Boltzman’s Expression for Entropy: A quantitative measure of matter dispersal or disorder Ludwig Boltzman (1844 – 1906) S = k log W S = entropy k = constant W = number of ways atoms or molecules can be arranged

10. Entropy • Entropy of a substance is determined by calorimetry S = q / T • Absolute entropy of a substance at any temperature can be determined • Absolute entropy at 0 K is 0. • Absolute entropy S0 is the entropy of a pure substance relative to its entropy at absolute zero

11. S: Entropy Generalizations • Entropies of gases >>> entropy of liquids > entropies of solids • Entropies of complex molecules are > than entropies of simpler molecules • Entropies of ionic solids become smaller as the attractions between ions become stronger

12. Entropy usually increases when a pure liquid or solid dissolves in a solvent 5. Entropies of liquids comprised of molecules with similar structures are smaller when hydrogen bonding is possible 6. Entropy increases when a dissolved gas escapes from a solution

13. Problem • Predict whether DS is positive or negative for the following processes CaCO3(s) --- > CaO(s) + CO2(g) 2CO(g) + O2 --- > 2CO2(g) Ag+(aq) + Cl-(aq) --- > AgCl(s)

14. The “naught” Notation and Standard States S0, H0, G0 Solid: Pure Liquid: Pure Gas: 1 atm Solution: 1 M Temperature: 25oC

15. Free Elements • The absolute entropy S0 of a free element in its standard state is NOT zero • S0 refers to the entropy increase in warming a pure substance from absolute zero where its entropy is zero, to 25oC S0 = S0(at 25oC) – S0(at –273oC) = S0 – 0 = S0

16. Free Elements (continued) DHf0 of a free element in its standard state is always equal to zero DGf0 of a free element in its standard state is always equal to zero

17. Second Law of Thermodynamics • The total entropy of the universe is constantly increasing • Whenever anything happens, matter, energy or both become more dispersed or disordered • For any spontaneous process DS > 0

18. Calculating Entropy Changes for a reaction DS0reaction= S n S0products– S n S0reactants where S0 is the absolute entropy of each compound DS0 is the entropy change that occurs when reactants in their standard states (pure, 1atm or 1M) are converted completely to products in their standard states.

19. Calculate DS0 for the Haber Process N2(g) + 3H2(g) --- > 2NH3(g) DS0reaction = 2mol (195.2 J/mol K) -1 mol (191.5 J/mol K) -3 (130.6 J/mol K) = -198.4 J/mol K

20. Enthalpies and Entropies of Formation The enthalpy or entropy change associated with forming a compound from its free elements. DHf0 = DHf0(compound)– S n Hf0(elements) = DHf0(compound)– 0 Look up DHf0in a Table DSf0 = S0(compound)– S n S0(elements) Look up S0(compound) and S0(element) for each element in a Table and substract

21. DS0reaction vs. DSf0 DS0 is the entropy of the reaction NOT the entropy of formation DS0 is the weighted sum of all the absolute entropies of the product minus the weighted sum of all the absolute entropies of the reactants Look up all the S0 values for the elements or compounds involved in the reaction and calculate DS0 using DS0 = S n S0products– S n S0reactants

22. DSf0 is the entropy change for the reaction which forms the compound from its elements in their standard states Look up all the S0 values and calculate DSf0 using DSf0 = S0(compound)– S n S0(elements)

23. Problem • Calculate the entropy change for the following reaction and calculate DSf0 for CH3OH (l) • CO(g) + 2H2(g) --- > CH3OH (l)

24. Solution DS0reaction = S0 [CH3OH(l)] – S0[CO(g)] -2S0 [H2(g)] = 1mol 126.8 J/mol K – 1 mol 197.6 J/mol K -2 mol 130.7 J/mol K = -332.2 J/K

25. Formation Reaction C(s) + 2H2(g) + ½ O2(g) --- > CH3OH (l) DSf0 = S0 [CH3OH (l)] – S0[C(s)] – 2S0[H2(g)] - ½ S0 [O2(g)] = 1 mol (126.8 J/ mol K) – 5.69 J/ mol K - 2 mol (130.58 J/mol K) – ½ (205 J/mol K) = -242.6 J/ mol K

26. Notes: • Absolute entropies are always positive • The units for S are Joules / mol K NOTKilojoules / mol K • The units for G and H are in kilojoules / mol

27. Gibbs Free Energy • A measure of the amount of energy involved in a reaction which is available (free) to do work DG is a quantity which tell us • Whether a reaction is spontaneous or not • Relates enthalpy and entropy • Units: kilojoules / mol

28. Calculating DG DGreaction = DHreaction – TDSreaction DGreaction = S nDGf products – S nDGf reactants DGreaction = DG0reaction + RT ln Q

29. The sign of DG DG < 0 Reaction is spontaneous as written DG = 0 Reaction is at equilibrium DG > 0 Reverse reaction is spontaneous Work must be done to make the reaction occur

30. Standard Free Energy DG0 DG0reaction is the free energy change when a mixture of only reactants all in their standard states is completely converted to a mixture of all products in their standard states.

31. The sign of DG0 and K DG0 < 0 Reaction is spontaneous as written, product favored at equilibrium K > 1 DG0 = 0 Very rare, at equilibrium [C]c[D]d = [A]a[B]b DG0 > 0 Not spontaneous, reactant favored at equilibrium K < 1

32. DG0 and K At equilibrium DGreaction = 0 Therefore since DGreaction = DG0reaction + RT ln Q At equilibrium 0 = DG0reaction + RT ln K DG0reaction = - RT ln K And K = e- DG0/ RT