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Chapter 19. Chemical Thermodynamics.

Chapter 19. Chemical Thermodynamics. Δ G = Δ H – T Δ S. Enthalpy (H) and entropy (S).

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Chapter 19. Chemical Thermodynamics.

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  1. Chapter 19. Chemical Thermodynamics. ΔG = ΔH – TΔS

  2. Enthalpy (H) and entropy (S) In Chapter 5 we discussed the role of enthalpy ( ΔH) in driving reactions. We saw that highly exothermic reactions such as the burning of gasoline proceeded very strongly, while the reverse reaction was highly endothermic, and did not proceed spontaneously at all. (CO2 and H2O do not react spontaneously to form gasoline and oxygen). However, enthalpy is not the only driving force in reactions. There is a second contribution, the entropy. Entropy relates to probability, and the increase in disorder of a system. We can think about probability in terms of a gas moving to occupy two flasks instead of only one:

  3. We connect two flasks as shown at the right by a closed stopcock, one of which is evacuated. When the stopcock is opened, the air will rush into the evacu- ated flask to equalize the pressure. What is the driving force for this? It is in fact probability, or entropy.

  4. 19.1 Spontaneous Processes. A spontaneous process is one that proceeds on its own without any outside assistance. A gas will expand from one flask into a second as a spontaneous process, but the reverse will never occur. Processes that are spontaneous in one direction are non-spontaneous in the opposite direction.

  5. The rusting of nails in air – a spontaneous process: Fresh nails rusted nails spontaneous not spontaneous 2 Fe (s) + 3 O2 (g)  Fe2O3 (s) (‘rust’)

  6. Examples of spontaneous processes water Ice at 25 oC will spontaneously melt: ice spontaneous not spontaneous The reverse will never happen. Water at 25 oC will never freeze.

  7. Examples of spontaneous processes But water at -10 oC will spontaneously freeze: ice spontaneous water not spontaneous

  8. Equilibrium: A mixture of water and ice at 0 oC is at equilibrium. The relative quantity of ice and water remains the same if we don’t add or remove heat. There is no spontaneous process in either direction. If we remove some heat, the temperature remains at 0 oC, but some water turns to ice. ice water (at 0 oC)

  9. Reversible processes: These occur with infinitesmal changes in the system. The melting and freezing of ice at 0 oC is reversible. Thus, if we add some heat, some of the ice melts, and if we remove some heat, some of the water freezes, but the temperature remains at 0 oC. At all other temperatures an irreversible process occurs. Melting and freezing of a substance at its melting point are reversible. Boiling and condensation of a substance at its boiling point are also reversible. Reversible processes are not spontaneous.

  10. Example: When benzene vapor converts to benzene liquid at 80.1 oC (B.Pt. of benzene) and 1 atm is this a spontaneous process? The condensation of a gas to a liquid at the boiling point of a substance is not a spontaneous process. To move the equilibrium in either direction we have to add or subtract energy.

  11. 19.2 Entropy and the second Law of Thermodynamics. Entropy is associated with randomness. The entropy, S, of a system is a state function just like the internal energy. The change in entropy is given by ΔS = Sfinal - Sinitial

  12. Relating entropy to heat transfer and temperature: For a process that occurs at constant temperature such as melting of a solid at its melting point: ΔSsystem = qrev/T To calculate the entropy change of a system we need to know the heat change, which is qrev, and divide it by the temperature in K. The term qrev is the reversible flow of heat in a reversible process such as enthalpy of melting of a solid at its melting point, or of boiling of a liquid at its boiling point.

  13. Example: The molar enthalpy of fusion of Hg is 2.29 kJ/mol. What is the entropy change when 50 g of Hg freezes at the normal melting point of -38.9 oC? (fusion = melting) The important thing to realize here is that qrev = ΔH(fusion) Hg(s) → Hg(l) ΔH = +2.29 kJ/mol So for freezing: Hg(l) → Hg(s) ΔH = -2.29 kJ/mol

  14. ΔH (freezing) Hg(l) Moles Hg = 50.0 g x 1mole = 0.249 mol 200.59 g kJ = 0.249 mole x -2.29 kJ = -0.571 kJ = -571 J 1 mole T = 273.15 – 38.9 = 243.3 K ΔS = qrev/T = -571/243.3 = -2.44 J/K Note that units of entropy are J/K. A positive change in entropy means an increase in entropy of the system.

  15. The Second Law of Thermodynamics: The total entropy of the universe increases in any spontaneous process. The sum of the entropy change of the system and the surroundings is always positive. Reversible: ΔSuniverse = ΔSsystem + ΔSsurround = 0 Irreversible: ΔSuniverse = ΔSsystem +ΔSsurround > 0

  16. Example: What happens to the entropy of the universe when a lesser spotted thrush begins to do the makarena? Answer: It increases, ΔSuniverse > 0 For any spontaneous process, and that’s any process where anything actually happens, the entropy of the universe increases. You may get questions like the above, and if you bear in mind that (except for a reversible process) when anything happens, the entropy of the universe increases.

  17. 19.3 The Molecular Interpretation of Entropy. Molecules can undergo translational, rotational, and vibrational motion. The number of ways a system can be arranged is the number of microstates it has. The entropy of a system is given by S = k ln W where W is the number of possible microstates, and k is Boltsmann’s constant = 1.38 x 10-23 J/K. Ludwig Boltzmann (1844-1906) Boltzmann worked out the nature of entropy.

  18. Probability and gas molecules in interconnected flasks: Consider a pair of flasks connected by a stop-cock. The probability that all the molecules will be in one flask is (½)n, where n is the number of molecules. one molecule, probability = (1/2)1 = 1/2 two molecules, probability = (1/2)2 = 1/4 four molecules, probability = (1/2)4 = 1/16

  19. The relationship between probability and entropy: So if n = 6.02 x 1023, we have a very small probability of finding all the molecules in one flask! Entropy relates to probability. The gas molecules in the apparatus at right are moving randomly, but the probability aspect keeps them evenly distributed between the two flasks. 6.02 x 1023 molecules, probability = (1/2)6.02 x 10 23 There is thus a chance that all the air molecules in this room would rush into one corner, and we would all die, but it is so small that it would never happen.

  20. The standard molar entropy of a substance (Sº): The entropies of substances may be measured experimentally. The entropy of one mole of a substance in its standard state is known as the standard molar entropy (Sº). The units of Sº are J/mol.K. Some values of standard molar entropies are given overleaf. Note that So values increase solid < liquid < gas, e.g. C6H6 (s) < C6H6 (l) < C6H6 (g); increase with the size of molecules, e.g. HF < HCl < HBr < HI or CH4 < C2H6 < C3H8; and with being a solid or liquid dissolved in solution in another liquid e.g. CH3COOH (l) < CH3COOH (aq).

  21. Standard molar entropies of selected substances at 298 K: (So, J/mol-K): Gases: H2(g) 130.6 N2(g) 191.5 O2(g) 205.0 C6H6(g) 269.2 Liquids: H2O(l) 69.9 C6H6(l) 172.8 Solids: Li(s) 29.1 Fe(s) 27.23 FeCl3(s) 142.3 NaCl(s) 72.3 C(diamond) 2.43 C(graphite) 5.69

  22. Making Qualitative predictions about ΔS. Entropy increases with temperature, since the particles vibrate and move about more rapidly, and so there is greater disorder. It also increases with increasing volume, as with a an expanding gas, and with the number of independ-ently moving particles. Thus, entropy increases solid < liquid < gas. Figure shows increase in entropy of a substance with increasing temp. as we pass solidliquidgas.

  23. Of all phase states, gases have the highest entropy Ssolid < Sliquid < Sgas gas solid liquid In a gas, molecules are more randomly distributed

  24. Larger Molecules generally have a larger entropy Ssmall < Smedium < Slarge Larger molecules have more internal motion

  25. } size of molecules increases } Sgas > Sliquid dissolving a gas in a liquid is accompanied by a lowering of the entropy } dissolving a liquid in another liquid is accompanied by an increase in entropy }

  26. Often, dissolving a solid or liquid will increase the entropy solution liquid dissolves lower entropy more disordered arrangement = higher entropy solid

  27. Dissolving a gas in a liquid decreases the entropy gas dissolves overall more disordered arrangement: higher entropy solution of gas in liquid, lower S

  28. Entropy increases with increasing temperature along the series solid < liquid < gas

  29. Note: S increases with increase in number of moles of gas in a reaction. What is the sign of DS for the following reactions? FeCl2 (s) + H2 (g) → Fe (s) + 2 HCl (g) DS = + solid gas solid gas 1 mol 2 mol Ba(OH)2 (s) → BaO (s) + H2O (g) DS = + solid solid gas 2 SO2 (g) + O2 (g) → 2 SO3 (g) DS = - gas gas gas 2 mol 1 mol 2 mol DS = - Ag+ (aq) + Cl-(aq) → AgCl (s) in solution insoluble

  30. For each of the following pairs of substances, which substance has a higher molar entropy at 25oC ? HCl (l) HCl (s) Li (s) Cs (s) C2H2 (g) C2H6 (g) Pb2+ (aq) Pb (s) O2 (g) O2 (aq) HCl (l) HBr (l) N2 (l) N2 (g) CH3OH (l) CH3OH (aq)

  31. 19.4 Entropy Changes in Chemical Reactions. This parallels the calculation of standard molar enthalpies of reaction. ΔSo = ΣnSo(products) - ΣmSo(reactants) Note that So is not zero for elements. Recall that So is a state function.

  32. Standard molar entropies of selected substances at 298 K: (So, J/mol-K): Gases: H2(g) 130.6 N2(g) 191.5 O2(g) 205.0 C6H6(g) 269.2 Liquids: H2O(l) 69.9 C6H6(l) 172.8 Solids: Li(s) 29.1 Fe(s) 27.23 FeCl3(s) 142.3 NaCl(s) 72.3 C(diamond) 2.43 C(graphite) 5.69

  33. Example: Calculate ΔSo for the following reaction: C2H4(g) + H2(g) → C2H6(g) So products = So C2H6(g) = 229.5 J/mol-K So reactants = So C2H4(g) + So H2(g) = 219.4 + 130.58 J/mol-K = 349.98 J/mol-K So(products) – So(reactants) = 229.5 – 349.98 J/mol-K = -120.5 J/mol-K

  34. Entropy Changes in the Surroundings. The change in entropy in the surroundings is caused entirely by heat transferred from the system to the surroundings, and so we have q = ΔH. ΔS(surroundings) = -qsys /T = -(ΔH/T) Thus, if asked to calculate the change in entropy of the surroundings, just use the above equation.

  35. Section 19.5. Gibbs Free Energy. We have seen that both enthalpy changes (ΔH) and entropy changes (ΔS) contribute to the overall energy that drives a reaction. Josiah Willard Gibbs (1839-1903) proposed a new state function, now called G in his honor, known as the free energy. G = H - TS J. Willard Gibbs (1839 – 1903)

  36. Again, we work with changes in G, so we have: ΔG = ΔH - TΔS increase inincrease in entropy ofentropy of surroundingssystem -ΔG is in fact the increase in entropy in the universe.

  37. ΔH is the increase in entropy of the surroundings from ΔS(surroundings) = -q/T. Note: 1) If ΔG is negative, the reaction is spontaneous. 2) If ΔG is zero, the reaction is at equilibrium. 3) If ΔG is positive, the reaction is non- spontaneous , but reverse reaction is spontaneous.

  38. 19.6. Free Energy and temperature. (p. 827) Temperature helps determine whether ΔG will be negative, and the process spontaneous. ΔH ΔS -TΔS ΔG (= ΔH – TΔS) __________________________________________ - + - - (spontaneous) + - + + (non-spontaneous) - - + + or –depends on T + + - + or - depends on T __________________________________________

  39. DG < 0 reaction is spontaneous (“product favored”) “exergonic” DG > 0 reaction is non-spontaneous “endergonic” DG = 0 reaction is at equilibrium

  40. 19.4 Free Energy Changes (ΔG) in Chemical Reactions. This parallels the calculation of standard molar enthalpies of reaction. ΔGo(rxn) = ΣnΔGfo(products) - ΣmΔGfo(reactants) Note that ΔGof is zero for elements. Recall that ΔGof is a state function.

  41. Table of ΔGof , ΔHof, (kJ/mol) and So(J/mol-K) for some substances (p. 1123): Substance ΔHofΔGofSo ________________________________________________________ Al(s) 0 0 28.32 C(s, graphite) 0 0 5.69 C2H6(g) -84.68 -32.89 229.5 O2(g) 0 0 205.0 Ni(s) 0 0 29.9 NiCl2(s) -305.3 -259.0 97.65 Cl2(g) 0 0 222.96

  42. Example: Calculate ΔGo, ΔHo, and ΔSo for the following: (problem 19.54) Ni(s) + Cl2(g) = NiCl2(s) ΔHo = -305.3 – [0 + 0] = -305.3 kJ ΔSo = 97.65 – [29.9 + 222.96] = -155.21 J/K ΔGo = -269.0 – [0 + 0] = -259.0 kJ Alternatively, from ΔG = ΔH – TΔS ΔGo = -305.3 kJ – (-155.21 J x 1 kJ x 298 K) = 1 K 1000 J -259.0 kJ Note units

  43. The reaction of sodium metal with water: 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g) Is the reaction spontaneous? What is the sign of DG? What is the sign of DH? What is the sign of DS?

  44. The reaction of sodium metal with water: 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g) Is the reaction spontaneous? Yes What is the sign of DG? DG = negative What is the sign of DH? DH = negative (exothermic!) What is the sign of DS? DS = positive

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