1. V. Determining Spontaneity It would be easier if we could determine spontaneity by just considering changes in the system. We derive an equation from the relationships we have learned and define a new quantity – the Gibbs free energy.

2. V. Gibbs Free Energy Equation

3. V. Gibbs Free Energy This equation calculates the change in the Gibbs free energy. Gibbs free energy (G) is formally defined as G = H – TS. From the derivation, ΔG = -TΔS univ.  Since ΔS univ is the criterion for spontaneity, ΔG can be used as a criterion for spontaneity.

4. V. Using ΔG The change in free energy can be calculated w/ respect to the system, and the result can be used to determine spontaneity.  If ΔG is negative, the process is spontaneous.  If ΔG is positive, the process is nonspontaneous.

5. V. Possible Combos in ΔG There are 3 factors that determine the outcome of the sign on ΔG. Sometimes ΔH and ΔS work together; sometimes they don’t.

6. V. Sample Problem The reaction C 2 H 4(g) + H 2(g)  C 2 H 6(g) has ΔH = -137.5 kJ and ΔS = -120.5 J/K. Calculate ΔG at 25 °C and determine whether the reaction is spontaneous at this temperature. Does ΔG become more negative or more positive as temperature increases?

7. V. Calculating ΔS° rxn Recall that the “not” symbol means standard conditions – 25 °C and 1 atm.  Standard state would be the state the substance exists as at these conditions.  For a solution, [ ] must be exactly 1 M. The standard entropy change for a reaction (ΔS° rxn ) is the change in entropy for a process in which all reactants and products are in their standard states.

8. V. 3 rd Law of Thermodynamics In order to calculate changes in entropy, we need to have a reference point. 3 rd Law of Thermodynamics: the entropy of a perfect crystal at absolute zero (0 K) is zero. Standard molar entropies can thus be tabulated.

9. V. Standard Molar Entropies

10. V. Factors Influencing S° The standard entropy of a substance is basically the amount of energy dispersed into one mole of that substance at 25 °C. Thus, the more “places” the compound can put the energy, the more “entropic” that compound will be.

11. V. S° and Physical State As mentioned before, entropy increases as a substance goes from solid to liquid, to gas.  Liquid H 2 O has S° = 70.0 J/mole. K.  Gaseous H 2 O has S° = 188.8 J/mole. K.

12. V. S° and Molar Mass As the molar mass increases, the entropy increases. The “why” is outside the scope of this course, but it has to do with energy levels.

13. V. S° and Allotropes Recall that allotropes are different forms of the same element. Less constrained allotropes have more entropy than more constrained ones.

14. V. S° and Molecular Complexity The more complex the compound, the more places it can put energy due to its different modes of motion. e.g. Ar (g) vs. NO (g).

15. V. S° and Dissolution Dissolved ionic solids have more entropy than the nondissolved solid. Energy that was concentrated in the crystalline solid becomes dispersed when dissolved in a solution. e.g. KClO 3(s) with S° = 143.1 J/mole. K vs. KClO 3(aq) with S° = 265.7 J/mole. K.

16. V. Calculating ΔS° rxn Calculating changes in standard entropies for a reaction is done via a Hess’s Law type of calculation.

17. V. Rxn Free Energy Changes So, to determine whether or not a process is spontaneous, we can calculate ΔG° rxn. Of course, this is the standard free energy change of a reaction. There are three ways to calculate ΔG° rxn, depending on what information is given.

18. V. Calculating ΔG° rxn with ΔH° rxn and ΔS° rxn The first method involves using the Gibbs free energy equation.  ΔG° rxn = ΔH° rxn - TΔS° rxn To use this method, typically need to do Hess’s Law type calculations for enthalpy and entropy.

19. V. Sample Problem Determine whether the reaction NO (g) + ½ O 2(g)  NO 2(g) is spontaneous at 25 °C. Note that the enthalpies of formation for NO (g) and NO 2(g) are 91.3 kJ/mole and 33.2 kJ/mole, respectively and the standard entropies for NO (g), O 2(g), and NO 2(g) are 210.8 J/mole. K, 205.2 J/mole. K, and 240.1 J/mole. K, respectively.

20. V. Calculating ΔG° rxn Via Hess’s Law standard free energy of formation, ΔG° f : change in free energy when 1 mole of a compound forms from its constituent elements in their standard states. If a table of ΔG° f ’s is available, ΔG° rxn can be calculated using a Hess’s Law type of calculation.

21. V. Sample Problem Calculate ΔG° rxn for the reaction 2CO (g) + 2NO (g)  2CO 2(g) + N 2(g) given that the standard free energies of formation for CO (g), NO (g), and CO 2(g) are -137.2 kJ/mole, 87.6 kJ/mole, and -394.4 kJ/mole, respectively.

22. V. Calculating ΔG° rxn Stepwise Just like thermochemical equations, reactions w/ associated free energies can be manipulated with the same rules applying. 1)If the equation is multiplied by a factor, then ΔG rxn is multiplied by the same factor. 2)If an equation is reversed, then ΔG rxn changes sign. 3)If a series of reactions adds up to an overall reaction, the ΔG rxn for the overall process is the sum of ΔG rxn for each step.

23. V. Sample Problem Using the equations below, find the ΔG° rxn for the reaction N 2 O (g) + NO 2(g)  3NO (g). 2NO (g) + O 2(g)  2NO 2(g) ΔG° rxn = -71.2 kJ N 2(g) + O 2(g)  2NO (g) ΔG° rxn = 175.2 kJ 2N 2 O (g)  2N 2(g) + O 2(g) ΔG° rxn = -207.4 kJ

24. V. Why Is It “Free” Energy? The energy of a reaction available to do work is the free energy. The free energy is the theoretical maximum; usually, it will be less. The only way to get the maximum is by using an infinitesimally slow reaction known as a reversible reaction.

25. V. Nonstandard Free Energies Of course, most of the time, a reaction will not be under standard conditions. We need a way to calculate spontaneity under a variety of conditions.

26. V. Sample Problem The reaction 2H 2 S (g) + SO 2(g)  3S (s, rhombic) + 2H 2 O (g) has a standard free energy change of -102 kJ. Calculate ΔG rxn when the partial pressures of H 2 S, SO 2, and H 2 O are 2.00 atm, 1.50 atm, and 0.0100 atm. Is the reaction more or less spontaneous under these conditions?

27. V. Relationship Between ΔG° rxn and K Recall that K is a measure of how far a reaction goes towards products which is another way of describing spontaneity. Thus, there must be a relationship between free energy and equil. constants. If ΔG rxn 0, reaction is nonspontaneous. If ΔG rxn = 0, reaction is at equilibrium.

28. V. Relationship Between ΔG° rxn and K

29. V. Temp. Dependence of K Previously, we discovered that equilibrium constants depend on temperature. With the relationship between standard free energy and the equilibrium constant, we can see how temperature is affected by temperature.

30. V. Temp. Dependence of K