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Chapter 17

Chapter 17. Free Energy and Thermodynamics. 1 st Law of Thermodynamics. Chapter 6: Energy is conserved. Energy cannot be created or destroyed, rather it is transferred from one place to another. Energy can be transferred in two ways. __________ or __________

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Chapter 17

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  1. Chapter 17 Free Energy andThermodynamics

  2. 1st Law of Thermodynamics • Chapter 6: Energy is conserved. • Energy cannot be created or destroyed, rather it is transferred from one place to another. • Energy can be transferred in two ways. • __________ or __________ • Internal energy of a system = KE + PE • Enthalpy – the first thermodynamic quantity • H = E + PDV

  3. Spontaneity • The first law does not tell us the extent to which a reaction will happen (or not). • Reactions are said to be spontaneous in one direction (and non-spontaneous in the opposite direction). • Ex) Releasing a ball – which direction will it go – up or down?

  4. Spontaneity • Which reaction is spontaneous? • H2O(l) H2(g) + ½ O2(g) • H2(g) + ½ O2(g)  H2O(l)

  5. Spontaneity • Which reaction is spontaneous? • 2Fe(s) + 3/2O2(g)  Fe2O3(s) • Fe2O3(s)  2Fe(s) + 3/2O2(g)

  6. Spontaneity • Often, it is dependent on the temperature. • H2O(s) H2O(l) • Spontaneous above or below 0oC?

  7. Spontaneity • Scientists first thought that the criteria for spontaneity was based solely on whether a reaction was exothermic or endothermic. • However, ice melting at room temperature is endothermic as is the dissolution of some salts like NH4Cl. • Clearly, a second criteria is needed to predict spontaneity.

  8. Carnot Cycle • Sadi Carnot theorized about an ideal steam engine – one that worked at 100% efficiency. • No heat energy would be lost – all energy is converted to work.

  9. Two Types of Processes • Reversible – a system is changed in such a method that BOTH the system and surroundings can be returned to their former states by EXACTLY reversing the change. • Irreversible – is one that cannot be reversed without altering the system or surroundings permanently.

  10. Reversible Processes • Phase changes at their melting or boiling point temperature are always reversible. • Equilibrium reactions when they reach a steady state are reversible.

  11. Irreversible Processes • In (a), we have a gas occupying the right half of the container. • In (b), the partition is removed and the gas spontaneously expands to fill the container. • In (c), the system is restored by compressing the gas with a piston. But, this requires work done by the surroundings changing it permanently!!!

  12. Entropy • A second quantity in thermodynamics. • A measurement of the randomness of a system. • Also a state function just like internal energy (E) and enthalpy (H). • Thus, DS = Sfinal – Sinitial • For any reversible process, DS = qrev / T • Units for DS = J/K mol • LEP #1

  13. Entropy • When a system undergoes a change, both the system and the surroundings are affected. • DSuniverse = DSsystem + DSsurroundings • When DSuniverse > 0 J/K, process is spontaneous. • When DSuniverse < 0 J/K, process is non-spontaneous. • When DSuniverse = 0 J/K, process is reversible. • 2nd Law of Thermodynamics = the entropy of the universe always increase for a spontaneous process. • LEP #2

  14. Molecular Interpretation • Molecules can undergo three basic types of motions. • Translational • Vibrational • Rotational

  15. Molecular Interpretation • As any gas is heated, its average KE increases – KE is proportional to temperature. • This additional KE can be split up among the three types of motion. • Ludwig Boltzmann – decided to look at entropy from a statistical viewpoint. • S = k ln(W) • k = Boltzmann’s constant (1.38 x 10-23 J/K) • W = number of possible microstates

  16. Molecular Interpretation • Number of microstates depends on the relevant numbers of particles and the positions they can occupy.

  17. Molecular Interpretation • The number of microstates is akin to playing cards.

  18. Molecular Interpretation • An increase in the entropy means that the randomness (or disorder) of the system has increased. • Or – an increase in the number of microstates. • More or less microstates if we have four molecules of gas rather than two? • More or less microstates if we have two decks of cards rather than one?

  19. Entropy and Life But, Mom – cleaning my room violates the 2nd law of thermodynamics! • Human beings (and all life forms) are highly ordered. • Does this violate the 2nd Law of Thermodynamics?

  20. Entropy and Life • You can’t break even! • To recharge a battery with 100 kJ of useful energy will require more than 100 kJ • because of the Second Law of Thermo! • Every energy transition results in a “loss” of energy • Its an “Energy Tax” demanded by nature!

  21. Predicting DS • Solids are rigid and ordered = low entropy • Liquids are confined to a specific volume, but are free to move = more entropy • Gases are free to move anywhere = high entropy • In general, we can predict an increase in the entropy if: • More molecules or particles are produced. • More gases are produced. • Temperature is increased. • Volume is increased. • LEP #3

  22. 3rd Law of Thermodynamics • The entropy of a pure crystalline substance at absolute zero is zero. • As the substance is warmed, its entropy increases. • Large increases are seen for phase changes.

  23. Entropy Change • Absolute entropies, under standard conditions, can be determined for all substances. • Values are found in Appendix C. • Sosolid < S0liquid < Sogas • So increases with molar mass • So increases with more atoms in formula

  24. Entropy Change • The change in entropy for any reaction can be calculated just like DH was in Chapter 5. • DSo = SnSo(products) – SnSo(reactants) • n = coefficients in chemical reaction • LEP #5

  25. Entropy Change • What is DSo for: N2(g) + 3 H2(g) 2 NH3(g) Given So for N2(g)=191.5 J/K mol, H2(g)=130.6 J/k mol, and NH3(g)=192.5 J/K mol? • This value is for the system. • How does the entropy of the surroundings change? • DSsurr. = -DHsys. / T • If DHsys. = -92.38 kJ, then what is DSsurr.? • What is DS universe?

  26. Gibbs Free Energy • DSuniv. = DSsys. + DSsurr. • We have just seen that DSsurr. = -DHsys. / T. • So, DSuniv. = DSsys. + -DHsys. / T. • Multiplying both sides by –T yields: • -TDSuniv. = -TDSsys. + DHsys. • Josiah Gibbs decided to label -TDSuniv. As DG. • DG = DH – TDS. • Signs for DG and their interpretation.

  27. Free Energy and Reactions

  28. Why is energy “Free”? • The change in free energy (DG) represents the maximum amount of energy available to do work. • Consider the reaction: • C(s, graphite) +2 H2(g) → CH4(g) • DH°rxn= −74.6 kJ = exothermic • DS°rxn= −80.8 J/K = unfavorable • DG°rxn= −50.5 kJ = spontaneous • DG° is less than DH° because some of the released heat energy is lost to increase the entropy of the surroundings

  29. Standard Free Energies • Like DHfo, there is also a standard free energy of formation for substances – DGfo. • These can then be used to calculate the DGo for any reaction using the values in Appendix C. • DGo = SnDGfo(products) – SnDGfo(reactants). • LEP #6

  30. Free Energy and Temperature • Some reactions are ALWAYS spontaneous whereas some are ALWAYS non-spontaneous. • Some reactions are DEPENDENT on the temperature

  31. Free Energy and Temperature • When a reaction becomes just spontaneous (or non-spontaneous), the DG = 0. • DG = DH – TDS. • 0 = DH – TDS. • DH = TDS. • T = DH / DS. • Warning – DH is in kJ and DS is in J. • LEP #7

  32. Applying to a Reaction • Once we can predict DS based on looking at the reaction AND knowing our relationship between DH and DS, we can also predict the outcome on DG. • Ex) 2 SO2(g) + O2(g) 2 SO3(g) ; DHo = -196.6 kJ • What would we predict for DS? • What effect does this have on DG? • LEP #8

  33. Free Energy and Equilibrium • The change in free energy along the reaction path is given by the equation: DG = DGo + RT lnQ. • At equilibrium, DG = 0 and Q = K. • 0 = DGo + RT lnK • DGo = -RT lnK • LEP #9, 10, 11

  34. Free Energy and Equilibrium

  35. Equilibrium and Temperature • From Ch 14, we saw that the equilibrium constant is temperature dependent. We can show why with: • Combining these two equations • DG° = DH° − TDS° • DG° = −RTln(K) • It can be shown that • This equation is in the form y = mx + b

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