1. Chemical Thermodynamics AP Chapter 19

2. The First Law Of Thermodynamics • Energy can not be created nor destroyed, only transferred between a system and the surroundings. • The energy in the universe is constant. • Energy is conserved!

3. Spontaneous Processes • Spontaneous processes occur without outside intervention. • Most reactions are spontaneous in one direction and non-spontaneous in the other direction. • This spontaneity is related to the thermodynamic path the system takes from the initial state to the final state.

4. Irreversible Reactions

5. Spontaneity can depend on the temperature.

6. Reversible vs. Irreversible • Reversible process – both the system and the surroundings can be restored to their original states by exactly reversing the process. • Irreversible processes – cannot return to the original process without a permanent change in the surroundings. • Any spontaneous process is irreversible. (It’s not spontaneous in the other direction.) • An iron nail can react with H2O and O2 to form Fe2O3, but the reverse does not happen.

7. Spontaneous Processes • Spontaneous processes may be fast or slow. • Many forms of combustion are fast. • Conversion of a diamond to graphite is slow.

8. Thermodynamics • Thermodynamics can tell us the direction and extent of a reaction, but NOT the speed of a reaction.

9. Entropy (S) • A measurement of the randomness or disorder of a system.

10. The Second Law of Thermodynamics • In any spontaneous process, there is always an increase in the entropy of the universe. • For a given change to be spontaneous, ΔSuniv must be positive.

11. The Second Law of Thermodynamics • The change in entropy of the universe, ΔSuniv = ΔSsystem + ΔSsurroundings • In a reversible process, ΔSuniv = 0. • In an irreversible, (spontaneous) process, ΔSuniv > 0. • Entropy values: J/K

12. Entropy’s Driving Force • The driving force for a spontaneous process is an increase in the entropy of the universe. • Entropy is a thermodynamic function describing the number of arrangements that are available to a system. • Nature proceeds towards the states that have the highest probability of existing.

13. Molecular Motion • Molecules can undergo 3 types of motion: • 1. translational motion • 2. vibrational motion • 3. rotational motion • A particular combination of motions and locations of the atoms in a system at a particular instant is called a microstate.

14. Positional Entropy • The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved. • Entropy generally increases when • liquids or solutions are formed from solids • gases are formed from either solids or liquids • or the number of molecules of gas increases during a chemical reaction.

16. Third Law of Thermodynamics • The entropy (S) of a perfect crystalline solid at 0 K is zero.

18. Constant Temp and Pressure • In reactions involving gaseous molecules, the change in a positional entropy is dominated by the relative numbers of molecules of gas reactants and products • 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g) 9 molecules 10 molecules ΔS increases

19. Entropy and Temperature • Entropy changes in the surroundings are primarily determined by heat flow. • Exothermic reactions in a system at constant temperature increase the entropy of the surroundings. • Endothermic reactions in a system at constant temperature decrease the entropy of the surroundings. • The transfer of a given quantity of energy as heat either to or from the surroundings has a greater impact at lower temperatures.

20. Calculating Entropy Change in a Reaction • ΔS°reaction = ΣnS°products - ΣnS°reactants • Entropy is an extensive property (a function of the number of moles) • Generally, the more complex the molecule, the higher the standard entropy value.

23. Gibbs Free Energy • Gibbs Free Energy (G) is a thermodynamic state function that combines enthalpy and entropy. • For a process at constant temperature & pressure, the sign of ΔG relates to the spontaneity of the process. • ΔG = negative, the process is spontaneous • ΔG = positive, the process is not spontaneous, but the reverse process is spontaneous. • At equilibrium, the process is reversible and ΔG is zero.

24. Free Energy • Free energy is the amount of useful work that can be obtained from a process at constant temperature and pressure.

25. Potential energy and Free energy – an analogy between the gravitational-PE change in a boulder rolling down a hill (position a) and the free energy change in a spontaneous reaction (b). The equilibrium position in (a) is given by the minimum gravitational PE available to the system. The equilibrium position in (b) is given by the minimum free energy available to the system.

26. Potential Energy and Free Energy • In any spontaneous process at constant temperature and pressure, the free energy always decreases.

27. Free Energy and Equilibrium

28. Gibbs Free Energy (G) • Calculating Free Energy Change (G) (constant temperature and pressure) • ΔG = ΔH – TΔS • ΔG is the change in free energy measure in kJ/mol) • Δ H is the change in enthalpy (kJ/mol) • ΔS is the change in entropy (J/mol K) • T is Kelvin Temperature

30. Summary of Important Thermodynamic Quantities

31. Free Energy and the Equilibrium Constant • ΔG = ΔG° + RT ln Q • R is the ideal gas constant, 8.314 J/mol-K • T is the absolute temperature • Q is the reaction quotient that corresponds to the reaction mixture of interest. • Under standard conditions the concentrations of all the reactants and products are equal to 1. Under standard conditions Q = 1 and therefore, ln Q = 0. • Therefore, ΔG = ΔG° under standard conditions.

32. The standard free energy for any reaction is related to the equilibrium constant. • At equilibrium, ΔG = 0 and Q = K, the equilibrium constant. • At equilibrium: ΔG = -RT ln K

33. Example • The standard free energy change for the following reaction at 25°C is -118.4 kJ/mol: • KClO3(s) → KCl(s) + 3/2 O2(g) • Calculate Kp for the reaction at 25°C and the equilibrium pressure of O2 gas. • ΔG° = -RT ln K • -118.4 kJ/mol = -(8.314 J/mol K)(298 K)(1kJ/1000 J) lnKp • Ln Kp = 47.8 • Kp = e47.8 = 5.68 x 1020 • Kp = p3/2 O2 • PO2 = Kp2/3 = (5.68 x 1020 )2/3 = 6.9 x 1013 atm

34. Standard Free Energies of Formation • ΔG° for any process can be calculated from tabulations of standard free energies of formation – see Tables in Appendix C.