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Entropy and Free Energy

Entropy and Free Energy. Chapter 19. Spontaneous Change and Equilibrium. reactions proceed until equilibrium is reached some favor reactants, some favor products spontaneous changes occur without outside intervention doesn’t mean quickly!. 19.1 Heat and Spontaneity.

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Entropy and Free Energy

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  1. Entropy and Free Energy Chapter 19

  2. Spontaneous Change and Equilibrium • reactions proceed until equilibrium is reached • some favor reactants, some favor products • spontaneous changes occur without outside intervention • doesn’t mean quickly!

  3. 19.1 Heat and Spontaneity • many exothermic reactions are product-favored, but heat release does not determine spontaneity • gas fills available space (no heat) • ice melts (endothermic) above 0oC • energy transfer as heat

  4. 19.2 Dispersal of Energy: Entropy • Entropy, S, is the driving force for a spontaneous process; measures disorder or chaos • 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

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

  6. 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).

  7. Calculating Entropy ΔS = qrev/T where qrev is heat absorbed 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.

  8. 19.3 Entropy: Dispersal of Energy • In all spontaneous changes, energy changes from being concentrated to becoming dispersed in a process carried out at a given temperature.

  9. Dispersal of Matter • Gas expansion (dispersion of matter) leads to the dispersal of energy. If O2 and N2 are connected by a valve, they diffuse together leading to an even distribution. The mixture will never separate into separate samples on its own!

  10. Homework • After reading sections 19.1-19.3, you should be able to do the following problems… • P. 887 (1-2,31)

  11. 19.4 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 0K; S = 0.

  12. Entropy • Entropies of gases are much higher than those for liquids, and entropies of liquids are higher than those for solids. • Larger and more complex molecules have higher entropies than smaller and more simple molecules. • Entropy increases as temperature is raised.

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

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

  15. 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 NH4Cl(s) in water: NH4Cl(s)  NH4Cl(aq) • The formation of 2.0 mol of NH3(g) from N2(g) and H2(g): N2(g) + 3H2(g)  2NH3(g)

  16. 19.5 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

  17. Calculating ΔSo • ΔSosys = ΣSoproducts – ΣSoreactants • ΔSsurr = - ΔHosys/T • ΔSouniv = ΔSosys + ΔSosurr

  18. Practice Problem • Classify the following as one of the four rxn types in table 19.2. • CH4(g) + 2O2(g)  H2O(l) + CO2(g) ΔHrxn = -890.6kJ ΔSsys = -242.8J/K • 2Fe2O3(s) + 3C(graphite)  4Fe(s) + 3CO2(g) • ΔHrxn = +467.9kJ ΔSsys = +560.7J/K • C(graphite) + O2(g)  CO2(g) • ΔHrxn = -393.5kJ ΔSsys = +3.1J/K • N2(g) + 3F2(g)  2NF3(g) • ΔHrxn = -264.2kJ ΔSsys = -277.8J/K

  19. Practice Problem • Is the reaction of hydrogen and chlorine to give hydrogen chloride gas predicted to be spontaneous? • H2(g) + Cl2(g)  2HCl(g) • Calculate the values for ΔSsys and ΔSsurr

  20. Practice Problem • 2Fe2O3(s) + 3C(graphite)  4Fe(s) + 3CO2(g) • ΔHrxn = +467.9 kJ and ΔSrxn = +560.7 J/K • Show that it is necessary that this reaction be carried out at high temperatures.

  21. Homework • After reading sections 19.4 and 19.5, you should be able to do the following… • P. 888 (3-14)

  22. 19.6 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

  23. ΔGo 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.

  24. 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 = -RTlnK

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

  26. 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).

  27. Dependence of Spontaneity on Temperature

  28. 19.7 Free Energy and Chemical Reactions • Standard free energy change can be calculated (free energy under standard conditions) ΔGo = ΔHo – TΔSo

  29. 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)

  30. 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 NH3(g) from the elements at standard conditions. N2(g) + 3H2(g)  2NH3(g)

  31. Practice Problem • Calculate the standard free energy change for the oxidation of 1.00 mol of SO2(g) to form SO3(g).

  32. Homework • After reading sections 19.6-19.7, you should be able to do the following… • P. 888 (15-30)

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