1. THERMODYNAMICS: ENTROPY, FREE ENERGY, AND EQUILIBRIUM Chapter 17

2. Remember K? Equilibrium constant… K > 1 means the reaction moves toward completion K < 1 means the reaction doesn’t proceed very far before equilibrium What determines the value of K?

3. A. Spontaneous Processes Defined as a process that proceeds on its own without any external intervention The reverse of a spontaneous process is always nonspontaneous Example… you falling down the stairs is spontaneous. Can you fall up the stairs?

4. Spontaneous Processes The spontaneity of a reaction is affected by temperature, pressure, and reaction mixture composition. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) At 300K, K p = 4.4 * 10 5 If partial pressures are 1 atm, Q p = 1 and reaction will proceed forward. But, at 700K, K p = 8.8 * 10 -5, and even with the same partial pressures for individual items, Q p > K p and the reaction is not spontaneous.

5. Spontaneous Processes A spontaneous reaction will always move a reaction mixture towards equilibrium. (reactions naturally tend toward equilibrium…) Nonspontaneous reactions move a reaction mixture away from equilibrium But remember: just because a reaction is spontaneous doesn’t mean that it’s fast! (thermodynamics: where a reaction is headed. This tells us nothing about the kinetics/rate of the reaction!)

6. B. Enthalpy, Entropy, and Spontaneous Processes: A Brief Review Many spontaneous chemical reaction involve the release of heat (exothermic) In these reactions,  H is negative. Where does the heat go? The heat, of course, is gained by the surroundings… However, not all spontaneous reactions are exothermic. Ice melts above O o C, for instance. How does this happen? Indication that enthalpy alone is not an indication of reaction spontaneity.

7. Enthalpy, Entropy, and Spontaneity… The other driving force for reaction spontaneity: entropy Measure of randomness or disorder Systems tend to move spontaneously toward more disorder  S = S final – S inital Positive  S = increase in entropy = favorable Thinking back to our ice… the molecules in a liquid are more disordered than the molecules in a solid. This is why the melting of ice at 0 o C is favorable and spontaneous.

8. Entropy Melting and vaporization increase entropy; freezing and condensation decrease entropy Dissolving NaCl in water also will increase entropy of the NaCl. However, S of the water molecules decreases – they become more ordered around the Na+ and Cl- ions. In total, the S of the system is positive.

9. C. Entropy and Probability Disordered states are preferred because they are more likely to occur – they are more probable. If you have 20 coins in a box, and you shake the box, what’s the likelihood that all 20 will come up heads? Is it more likely that some will be heads and some will be tails? Which is the more random/disordered situation? The “20 heads” or “20 tails” situation only has one possible way it can occur. Therefore, it is less likely to happen. A “5 heads” situation has many more possible configurations.

10. D. Standard Molar Entropies and Standard Entropies of Reaction We know standard molar entropies for many substances. They are calculated and listed for one mole of the pure substance at 1 atm pressure and specified temperature, usually 25 o C. Note that entropies of gaseous substances tend to be larger than entropies for liquids… To calculate a standard entropy of reaction,  S o, subtract S of reactants from S of products.  S o = S o (prod) – S o (react) taking coefficients into account, as we do in the example on the next page. (Entropies are listed as per-mole values)

11. example Calculate the standard entropy of reaction at 25oC for the decomposition of calcium carbonate: CaCO3(s)  CaO(s) + CO2(g)  S o = [S o (CaO) + S o (CO 2 )] - S o (CaCO 3 ) = [(1 mol)(39.7 J/K*mol) + (1 mol)(213.6 J/K*mol)] - (1 mol)(92.9 J/K*mol) = +160.4 J/K

12. E. Entropy and the Second Law of Thermodynamics Overall system goal: minimum enthalpy, maximum entropy But not all reactions in the universe meet these criteria Combination of H and S – determining the free energy change for the reaction – determines whether the reaction will be spontaneous at a given temperature  G =  H – T  S  G 0, nonspontaneous. Whether this energy is consumed (endothermic) or released (exothermic), the total energy of the system and the surroundings remains constant. FIRST LAW OF THERMODYNAMICS: In any process, spontaneous or nonspontaneous, the total energy of a system and its surroundings is constant.

13. Entropy and the Second Law… The second law better defines spontaneity. SECOND LAW OF THERMODYNAMICS: In any spontaneous process, the total entropy of a system and its surroundings always increases. Given  S total =  S system +  S surroundings If  S total > 0, reaction is spontaneous. If  S total < 0, reaction is nonspontaneous. If  S total = 0, reaction is at equilibrium.

14. Second Law and Spontaneity All reactions proceed spontaneously in the direction that increases the entropy of the system plus surroundings. If your forward reaction has a +  S (spontaneous), your reverse reaction will have a –  S (nonspontaneous) How do you determine  S total ?   S system : calculate using standard molar entropies (last section)  For the surroundings:  S surr = -  H/T (S increases when heat is lost to the surroundings and vice-versa) (If surr are already at high T, addition of more heat will cause a small  S, and vice versa)

15. F. Free Energy Free energy: G = H - TS For a reaction:  If  G<0, the reaction is spontaneous.  If  G = 0, the reaction mixture is at equilibrium.  If  G>0, the reaction is nonspontaneous. If  S is positive and  H is negative, reaction will be spontaneous. Sometimes the factors are in opposition (table 17.2). Consider which is bigger and which will outweigh the other.

16. G. Standard Free Energies of Formation Standard free energy change =  G o Change in free energy when reactants in their standard state are converted to products in their standard states Standard state conditions: Solids, liquids, gases: pure form, 1 atm pressure Solutes: 1 M concentration A specified temperature, usually 25 o C Standard free energy of formation =  G o f For a given substance, the free energy change for formation of one mole of the substance in its standard state from the most stable form of its constituent elements in their standard states. Many substances have a negative  G o f Difficult to synthesize compounds that have a positive  G o f !

17. H. Free-Energy Changes, Composition of Reaction Mixture Reaction mixtures are seldom in standard state concentrations and pressures. How to calculate  G (assuming no access to  H/  S values)? Use the reaction quotient Q  G =  G o + RT ln Q Example: For the formation of ethylene from carbon and hydrogen at 25 o C, P H2 = 100 atm and P C2H4 = 0.10 atm, calculate  G. 2 C(s) + 2 H 2 (g) --> C 2 H 4 (g)  G o = 68.1 kJ

18. Free Energy Changes… example Use the equation  G =  G o + RT ln Q where Q = Q p = P C2H4 /P H2 = 0.10/100 = 1.0 * 10 -5  G = 68.1 kJ/mol + [8.314 * 10 -3 kJ/(K*mol)](298 K) ln 1.0 * 10 -5 = + 39.6 kJ/mol Here,  G > 0, so the reaction would not be spontaneous.

19. J. Free Energy and Chemical Equilibrium Looking again at the equation:  G =  G o + RT lnQ Consider a situation where you have mostly reactants and not much products. Q will be less than 1, and the (RT lnQ) term will be a very large negative number. In this situation, no matter if  G o is positive or negative, RT lnQ will likely outweigh it enough to make  G negative. The same is the case when Q is greater than 1 (mostly products), RT lnQ is a large positive number. To summarize…

20. Free Energy as a Function of Q When the reaction mixture is mostly reactants: Q << 1 RT ln Q << 0  G < 0 When the reaction mixture is mostly products: Q 0

21. Relating Free Energy and Equilibrium Constant Since, at equilibrium, we know that  G = 0, we can also say 0 =  G o + RT ln K (substituting K for Q) or  G o = -RT ln K NOTE: This allows you to calculate the equilibrium constant (K) from the standard free energy change!

22. Example… At 25 o C, K w for the dissociation of water is 1.0 * 10 -14. Calculate  G o for the reaction 2 H 2 O(l) H 3 O + (aq) + OH - (aq).  G o = -RT ln K w = -(8.314 J/K*mol)(298 K)(ln 1.0 * 10 -14 ) = 80 kJ/mol K is very small, so this corresponds to a large positive free energy change.