Entropy and Free Energy Chapter 18 Learning Objectives

Entropy and Free Energy Chapter 18

Learning Objectives • Students understand • The concept of entropy and its relationship to reaction spontaneity. • The relationship between enthalpy, entropy, and free energy changes for a reaction.

Learning Objectives • Students will be able to • Calculate the change in entropy to determine spontaneity • Use the change in Gibbs free energy to predict spontaneity • Calculate standard free energy change from data • Calculate an equilibrium constant for a process from free energy

Spontaneous Change and Equilibrium • reactions proceed until equilibrium is reached • some favor reactants, some favor products • spontaneous changes occur without outside intervention in the direction that leads to equilibrium • doesn’t mean quickly!

18. 1 Heat and Spontaneity • many exothermic reactions are product-favored, but heat release alone does not determine spontaneity • gas fills available space (energy neutral) • ice melts (endothermic) above 0 o. C • energy transfer as heat

18. 2 Dispersal of Energy: Entropy • Entropy, S, measures disorder or chaos and quantifies the dispersal of energy • positional entropy describes positions in a given state (more positional entropy increases from solid liquid gas) • energy goes from being more concentration to being more dispersed

Second Law of Thermodynamics • In any spontaneous process there is always an increase in the entropy of the universe.

Entropy • Thermal energy is caused by the random motion of particles. • Potential energy is dispersed when it is converted to thermal energy (when energy is transferred as heat).

Calculating Entropy ΔS = qrev/T where qrev is heat transfer under reversible conditions and T is the Kelvin temp at which change occurs • Reversible process: after carrying out a change along a given path, it must be possible to return to the starting point by the same path without altering surroundings.

18. 3 Dispersal of Energy • Boltzmann proposed that the entropy of a system (dispersal of energy at a given temperature) results from the number of microstates (ways to distribute energy) available. • As the number of microstates increases, so does the entropy of the system.

Dispersal of Matter • Gas expansion (dispersion of matter) leads to the dispersal of energy. If O 2 and N 2 are connected by a valve, they diffuse together leading to an even distribution. The mixture will never separate into separate samples on its own!

Homework • After reading sections 18. 1 -18. 3, you should be able to do the following problems… • P. 711 (1 -8)

18. 4 Entropy Measurements and Values • For any substance under certain conditions, a numerical value for entropy can be determined • The Third Law of Thermodynamics states that there is no disorder in a perfect crystal at 0 K; S = 0.

Entropy • Entropy of a substance will increase in going from a solid to a liquid to a gas. • Larger and more complex molecules have higher entropies than smaller and more simple molecules. • Entropy increases as temperature is raised. • Entropy of a gas increases with an increase in volume.

Third Law of Thermodynamics • ΔSsurr = -ΔHsys/T • Sign is positive when exothermic; heat flows to surroundings • Sign is negative when endothermic

Standard Entropy Values, So • Change can also be calculated from tables ΔSorxn = ΣSoproducts - ΣSoreactants

Practice Problem • Calculate the standard entropy changes for the following processes using entropy values in Appendix L. Do the calculated values of ΔSo match predictions? • Dissolving 1. 0 mol of NH 4 Cl(s) in water: NH 4 Cl(s) NH 4 Cl(aq) • The formation of 2. 0 mol of NH 3(g) from N 2(g) and H 2(g): N 2(g) + 3 H 2(g) 2 NH 3(g)

18. 5 Entropy Changes • Change in entropy is equal to the change in the entropy of the system AND the change in entropy of surroundings. ΔSouniv = ΔSosys + ΔSosurr Spontaneous when ΔSuniv > 0, positive Not spontaneous when ΔSuniv < 0, negative

Calculating o ΔS • ΔSosys = ΣSoproducts – ΣSoreactants • ΔSsurr = - ΔHosys/T • ΔSouniv = ΔSosys + ΔSosurr

Practice Problem • Predict whether these reactions are spontaneous – see table 18. 1 p. 694 • CH 4(g) + 2 O 2(g) H 2 O(l) + CO 2(g) ΔHrxn = -890. 6 k. J ΔSsys = -242. 8 J/K • 2 Fe 2 O 3(s) + 3 C(graphite) 4 Fe(s) + 3 CO 2(g) • ΔHrxn = +467. 9 k. J ΔSsys = +560. 7 J/K • C(graphite) + O 2(g) CO 2(g) • ΔHrxn = -393. 5 k. J ΔSsys = +3. 1 J/K • ΔHrxn = -264. 2 k. J ΔSsys = -277. 8 J/K • N 2(g) + 3 F 2(g) 2 NF 3(g)

Practice Problem • Is the reaction of hydrogen and chlorine to give hydrogen chloride gas predicted to be spontaneous? • H 2(g) + Cl 2(g) 2 HCl(g) • Calculate the values for ΔSsys and ΔSsurr

Practice Problem • 2 Fe 2 O 3(s) + 3 C(graphite) 4 Fe(s) + 3 CO 2(g) • ΔHrxn = +467. 9 k. J and ΔSrxn = +560. 7 J/K • Show that it is necessary that this reaction be carried out at high temperatures.

Homework • After reading sections 18. 4 and 18. 5, you should be able to do the following… • P. 711 (9 -14)

18. 6 Free Energy • Free energy, G, describes whether or not a reaction is spontaneous. Gibbs free energy is dependent on change in enthalpy, change in entropy, and temperature of the system. ΔG = ΔH – TΔS where T is Kelvin

o ΔG and Spontaneity • If the value of ΔGrxn is negative, a reaction is spontaneous. • If ΔGrxn = 0, the reaction is at equilibrium. • If the value of ΔGrxn is positive, the reaction is not spontaneous.

Free Energy and Equilibrium • At equilibrium, no net change in concentration of reactants and products occur, so free energy and the equilibrium constant have the following relationship: ΔG = -RTln. K

Free Energy and Equilibrium • When ΔG is negative, K is greater than 1 and the reaction is product favored. K>Q • When ΔG is positive, K is less than 1 and the reaction is reactant favored. K<Q • When ΔG = 0, the reaction is at equilibrium. Q = K

What is “Free” Energy? • The “free” energy is the sum of the energies available from the enthalpy term (dispersal of energy) and the entropy term (dispersal of matter).

Dependence of Spontaneity on Temperature ΔS ΔH ΔG + - Spontaneous at all temps + + Spontaneous at high temps - - Spontaneous at low temps - + Process not spontaneous at any temp

18. 7 Free Energy and Chemical Reactions • Standard free energy change can be calculated (free energy under standard conditions) ΔGo = ΔHo – TΔSo

Standard Free Energy • The standard free energy of formation of a compound is the free energy change when forming one mole of the compound from the component elements. (element in standard state is zero) ΔGrxno = ΣΔGfo(products) – ΣΔGfo(reactants)

Practice Problem • Using values of ΔHfo and So to find ΔHrxno and ΔSrxno respectively, calculate the free energy change, ΔGo, for the formation of 2 mol of NH 3(g) from the elements at standard conditions. N 2(g) + 3 H 2(g) 2 NH 3(g)

Practice Problem • Calculate the standard free energy change for the oxidation of 1. 00 mol of SO 2(g) to form SO 3(g).

Homework • After reading sections 19. 6 -19. 7, you should be able to do the following… • P. 711 b (15 -16, 21 -24, 29 -32)