1. Chapter 19 Chemical 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:  G =  H – T  S Sign of  G  G < 0 Reaction is spontaneous, reaction will proceed in the forward direction  G < 0 Reaction is not spontaneous, however the reaction going in the reverse direction is spontaneous  G = 0The 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 a)H 2 + I 2 2HI K c = 51 b)N 2 + O 2 2NOK c = 1 x 10 -30 Which reaction is spontaneous in the forward direction? Which reaction has a negative  G 0 ?

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  E system = -  E surroundings 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 S 0 is the entropy of a pure substance relative to its entropy at absolute zero

11. S: Entropy Generalizations 1.Entropies of gases >>> entropy of liquids > entropies of solids 2.Entropies of complex molecules are > than entropies of simpler molecules 3.Entropies of ionic solids become smaller as the attractions between ions become stronger

12. 4.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  S is positive or negative for the following processes CaCO 3 (s) --- > CaO(s) + CO 2 (g) 2CO(g) + O 2 --- > 2CO 2 (g) Ag + (aq) + Cl - (aq) --- > AgCl(s)

14. The “naught” Notation and Standard States S 0, H 0, G 0 Solid:Pure Liquid:Pure Gas: 1 atm Solution: 1 M Temperature: 25 o C

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

16. Free Elements (continued)  H f 0 of a free element in its standard state is always equal to zero  G f 0 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  S > 0

18. Calculating Entropy Changes for a reaction  S 0 reaction =  n S 0 products –  n S 0 reactants where S 0 is the absolute entropy of each compound  S 0 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  S 0 for the Haber Process N 2 (g) + 3H 2 (g) --- > 2NH 3 (g)  S 0 reaction = 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.  H f 0 =  H f 0 (compound) –  n H f 0 (elements) =  H f 0 (compound) – 0 Look up  H f 0 in a Table  S f 0 = S 0 (compound) –  n S 0 (elements) Look up S 0 (compound) and S 0 (element) for each element in a Table and substract

21.  S 0 reaction vs.  S f 0  S 0 is the entropy of the reaction  the entropy of formation  S 0 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 S 0 values for the elements or compounds involved in the reaction and calculate  S 0 using  S 0 =  n S 0 products –  n S 0 reactants

22.  S f 0 is the entropy change for the reaction which forms the compound from its elements in their standard states Look up all the S 0 values and calculate  S f 0 using  S f 0 = S 0 (compound) –  n S 0 (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  S 0 reaction = S 0 [CH 3 OH(l)] – S 0 [CO(g)] -2S 0 [H 2 (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) + 2H 2 (g) + ½ O 2 (g) --- > CH 3 OH (l)  S f 0 = S 0 [CH 3 OH (l)] – S 0 [C(s)] – 2S 0 [H 2 (g)] - ½ S 0 [O 2 (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 NOT Kilojoules / 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  G is a quantity which tell us –Whether a reaction is spontaneous or not –Relates enthalpy and entropy –Units: kilojoules / mol

28. Calculating  G  G reaction =  H reaction – T  S reaction  G reaction =  n  G f products –  n  G f reactants  G reaction =  G 0 reaction + RT ln Q

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

30. Standard Free Energy  G 0  G 0 reaction 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  G 0 and K  G 0 < 0 Reaction is spontaneous as written, product favored at equilibrium K > 1  G 0 = 0 Very rare, at equilibrium [C] c [D] d = [A] a [B] b  G 0 > 0 Not spontaneous, reactant favored at equilibrium K < 1

32.  G 0 and K At equilibrium  G reaction = 0 Therefore since  G reaction =  G 0 reaction + RT ln Q At equilibrium 0 =  G 0 reaction + RT ln K  G 0 reaction = - RT ln K And K = e -  G 0 / RT