1. Thermodynamics Part 5 - Spontaneity

2. Thermodynamics Thermodynamics = the study of energy changes that accompany physical and chemical changes. Enthalpy (H): the total energy “stored” within a substance Enthalpy Change (ΔH): a comparison of the total enthalpies of the product & reactants. ΔH = H products - H reactants

3. Exothermic vs. Endothermic Exothermic reactions/changes: release energy in the form of heat; have negative ΔH values. H 2 O (g)  H 2 O (l) ΔH = -2870 kJ Endothermic reactions/changes: absorb energy in the form of heat; have positive ΔH values. H 2 O (l)  H 2 O (g) ΔH = +2870 kJ

4. Reaction Pathways Changes that involve a decrease in enthalpy are favored! EndothermicExothermic time EaEa EaEa P P RR

5. Entropy Entropy (S): the measure of the degree of disorder in a system; in nature, things tend to increase in entropy, or disorder. ΔS = S products – S reactants All physical & chemical changes involve a change in entropy, or ΔS. (Remember that a high entropy is favorable)

6. Entropy

10. Driving Forces in Reactions Enthalpy and entropy are DRIVING FORCES for spontaneous reactions (rxns that happen at normal conditions) It is the interplay of these 2 driving forces that determines whether or not a physical or chemical change will actually happen.

11. Free Energy Free Energy (G): relates enthalpy and entropy in a way that indicates which predominates; the quantity of energy that is available or stored to do work or cause change.

12. Free Energy ΔG = ΔH – TΔS Where: ΔG = change in free energy (kJ) ΔH = change in enthalpy (kJ) T = absolute temp (K) ΔS = change in entropy (kJ/K)

13. Free Energy ΔG: positive (+) value means change is NOT spontaneous ΔG: negative (-) value means change IS spontaneous

14. Relating Enthalpy and Entropy to Spontaneity ExampleΔHΔHΔSΔSSpontaneity 2K + 2H 2 O  2KOH + H 2 -+always spon. H 2 O(g)  H 2 O(l) --spon. @ lower temp. H 2 O(s)  H 2 O(l) ++spon. @ higher temp. 16CO 2 +18H 2 O  2C 8 H 18 +25O 2 +-never spon.

15. Example #1 For the decomposition of O 3 (g) to O 2 (g): 2O 3 (g)  3O 2 (g) ΔH = -285.4 kJ/mol ΔS = 137.55 J/mol·K @25 °C a) Calculate ΔG for the reaction. ΔG = (-285.4 kJ/mol ) – (298 K )(0.1375 5KJ/mol·K ) ΔG = -326.4 kJ

16. Example #1 For the decomposition of O 3 (g) to O 2 (g): 2O 3 (g)  3O 2 (g) ΔH = -285.4 kJ/mol ΔS = 137.55 J/mol·K @25 °C b) Is the reaction spontaneous? YES

17. Example #1 For the decomposition of O 3 (g) to O 2 (g): 2O 3 (g)  3O 2 (g) ΔH = -285.4 kJ/mol ΔS = 137.55 J/mol·K @25 °C c) Is ΔH or ΔS (or both) favorable for the reaction? Both ΔS and ΔH are favorable (both are driving forces)

18. Example #2 What is the minimum temperature (in °C) necessary for the following reaction to occur spontaneously? Fe 2 O 3 (s) + 3CO(g)  2Fe(s) + 3CO 2 (g) ΔH = +144.5 kJ/mol; ΔS = +24.3 J/K·mol (Hint: assume ΔG = -0.100 kJ/mol) ΔG = ΔH – TΔS -0.100 = (144.5) – (T)(0.0243) T ≈ 5950 K T = 5677 °C