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CHAPTER 18 Entropy, Free Energy, and Equilibrium

CHAPTER 18 Entropy, Free Energy, and Equilibrium. Review of Thermodynamics The following are general concepts from thermodynamics as applied to chemical systems. 1) First law of thermodynamics: U = q + w w = work ( = - pV for mechanical work for a system at constant pressure)

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CHAPTER 18 Entropy, Free Energy, and Equilibrium

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  1. CHAPTER 18 Entropy, Free Energy, and Equilibrium

  2. Review of Thermodynamics The following are general concepts from thermodynamics as applied to chemical systems. 1) First law of thermodynamics: U = q + w w = work ( = - pV for mechanical work for a system at constant pressure) q = heat U = Uf - Ui = change in internal energy Note that U is a state function, while q and w are not.

  3. Sign convention w > 0 work done on system w < 0 work done by system (volume decreases) (volume increases) q > 0 heat flows into system q < 0 heat flows out of system (endothermic) (exothermic) 2) Enthalpy and heat H (enthalpy) is defined as H = U + pV (and so a state function) At constant volume U = qV At constant pressure H = qp

  4. 3) Calculating the enthalpy change for a chemical reaction Hrxn = [  Hf(products) ] - [  Hf(reactants) ] The above is in terms of the enthalpy of formation. Recall that by definition the formation reaction is the reaction that produces one mole of a single product out of elements in their standard (most stable) state. For example, the formation reaction for C2H5Cl(g) is 2 C(s) + 5/2 H2(g) + 1/2 Cl2(g)  C2H5Cl(g) Hf(C2H5Cl(g)), the enthalpy of formation for C2H5Cl(g), is the value for H when one mole of C2H5Cl(g) is formed by the above reaction.

  5. Spontaneous Processes Consider the following two processes. Why does only one occur?

  6. Entropy To determine which processes consistent with the first law of thermodynamics actually occur we must first define a new thermo-dynamic function, called entropy (S). Entropy measures the amount of “disorder” in a system (in a precise mathematical sense)*. Entropy is a state function. We may talk about the entropy change for the system (Ssyst) and the entropy change for the surroundings (Ssurr). The entropy change for the universe (Suniv) is then given by the expression Suniv = Ssyst + Ssurr We will discuss ways to find values for S for specific types of processes. * In fact, S = if (đq)rev/T

  7. The Second Law of Thermodynamics The importance of entropy lies in the fact that it can be used to decide which processes that are consistent with the first law of thermodynamics will occur, and which will not. To decide, we need a new thermodynamic law, called the second law of thermodynamics Suniv  0 for any allowed process In words, what the above means is the amount of randomness or disorder in the universe (as measured by entropy) must always increase or remain constant.

  8. Comments Concerning the Second Law 1) The units for entropy are J/K (or J/mol.K) 2) There are three types of processes according to the second law Suniv > 0 , spontaneous, will eventually occur Suniv = 0 , process where equilibrium has been established Suniv < 0 , process will not occur 3) The second law of thermodynamics is based on observation. 4) Since Suniv = Ssyst + Ssurr, it is possible for Ssyst or Ssurr to be negative and still have a spontaneous process (though both cannot be negative for the same process). 5) There is a connection between entropy and the way in which states can be arranged on a molecular level, though we will not explore this (this is studied in statistical thermodynamics).

  9. Molar Entropy The standard molar entropy of a substance is the value for the entropy of 1 mole of the substance for standard pressure (1 atm) and some tempera-ture T. We usually choose T = 25. C.

  10. Entropy Change For a Chemical Reaction When a chemical reaction takes place, Ssyst = Srxn. The expression for the change in entropy, for standard conditions, is Srxn = [  S(products) ] - [  S(reactants) ] Note that elements in their standard state must be included in the above calculation. For the entropy change for the surroundings, we may show that Ssurr = qsurr/T = - qsyst/T = - Hrxn/T where Hrxn is the enthalpy change for the chemical reaction.

  11. Example Consider the following chemical reaction N2(g) + 3 H2(g)  2 NH3(g) What are Ssyst, Ssurr, and Suniv for the above reaction when carried out under standard conditions (pgas = 1.0 atm, T = 25.C)? Is the reaction spontaneous for these conditions?

  12. Consider the following chemical reaction N2(g) + 3 H2(g)  2 NH3(g) What are Ssyst, Ssurr, and Suniv for the above reaction when carried out under standard conditions (pgas = 1.0 atm, T = 25. C)? Is the reaction spontaneous for these conditions? (Appendix 2 for data) Substance H°f(kJ/mol) G°f(kJ/mol) S°(J/mol.K) N2(g) 0.0 0.0 191.6 H2(g) 0.0 0.0 130.7 NH3(g) - 45.9 - 16.4 192.8 Srxn = [  S(products) ] - [  S(reactants) ] = [2 (192.8)] - [(191.6) + 3(130.7)] = - 198.1 J/mol.K Hrxn = [  Hf(products) ] - [  Hf(reactants) ] = [(2 (-45.9)] - [ 0 ] = - 91.8 kJ/mol

  13. Srxn = - 198.1 J/mol.K Hrxn = - 91.8 kJ/mol So Ssyst = Srxn = - 198.1 J/mol.K Ssurr = - Hrxn/T = - (- 91800. J/mol) = 307.8 J/mol.K (298.2 K) Suniv = Ssyst + Ssurr = (- 198.1 J/mol.K) + (307.8 J/mol.K) = 109.7 J/mol.K Note that even though Ssyst < 0, the reaction is still spontaneous for standard conditions.

  14. Entropy For Pure Substances The more random a substance is, the larger the value for entropy. So Sgas >> Sliquid > Ssolid S°(Ca(s)) = 41.6 J/mol.K S°(H2O(l)) = 70.0 J/mol.K S°(Ca(g)) = 154.9 J/mol.K S°(H2O(g)) = 188.8 J/mol.K

  15. Predictions For Entropy Changes For Chemical Reactions Since Sgas >> Sliquid > Ssolid gases are far more disordered than solids and liquids. Based on that, we would predict the following If ngas > 0 then Srxn > 0 If ngas = 0 then Srxn ~ 0 If ngas < 0 then Srxn < 0 Example: Predict the value for Srxn for the following reactions CaCO3(s)  CaO(s) + CO2(g) C(s) + 2 H2(g)  CH4(g) 2 HCl(g) + Br2(g)  2 HBr(g) + Cl2(g)

  16. Example: Predict the value for Srxn for the following reactions CaCO3(s)  CaO(s) + CO2(g) ng = +1, so S > 0 (S = + 160.2 J/mol.K) C(s) + 2 H2(g)  CH4(g) ng = -1, so S < 0 (S = - 80.8 J/mol.K) 2 HCl(g) + Br2(g)  2 HBr(g) + Cl2(g) ng = 0, so S ~ 0 (S = + 1.2 J/mol.K)

  17. Other Trends in Entropy 1) Entropy generally increases as the number of atoms per molecule increases. S°(CH4(g)) = 186.3 J/mol.K S°(C2H6(g)) = 229.2 J/mol.K S°(C3H8(g)) = 270.3 J/mol.K 2) Entropy generally increases when molecular solids or liquids are dissolved in solvents. For ionic compounds, entropy usually (but does not always) also increases. S°(NaCl(s)) = 72.1 J/mol.K S°(NaCl(aq)) = 115.0 J/mol.K

  18. Third Law of Thermodynamics The entropy of a pure substance increases as temperature increases. This is because the degree of randomness in the substance increases. This is particularly true if a phase transition occurs.

  19. Based on low temperature experiments, it appears that the entropy of every pure substance approaches the same value as T  0. K. Third law of thermodynamics: The absolute entropy (S) of a perfect crystal of any pure substance at absolute zero is 0.0 J/mol.K. Because there are standard ways of find the change in entropy for a pure substance as we change the temperature of the substance at constant pressure, the third law of thermodynamics allows us to assign values for entropy for pure substances at any temperature. Standard molar entropy (S) – The value for the entropy of one mole of a pure substance at p = 1.00 atm and a particular temperature. We usually choose T = 25. C = 298. K. Note - The above does not apply to ions in aqueous solution. In that case, the entropy of H+ ion at T = 25. C is defined as 0.0 J/mol.K.

  20. Free Energy - Definition While the second law of thermodynamics is true for all processes, it is difficult to use because it requires calculations on both the system and the surroundings. To help us around this problem we define a new thermodynamic function, called the free energy (G). G = H - TS (a state function) Note that for a process taking place at a particular fixed temperature G = H - (TS) = H - TS

  21. Free Energy and the Second Law For a process taking place at a particular fixed temperature G = H - (TS) = H - TS If we divide both sides of the above equation by –T, we get - (G)/T = - (H)/T + S = Ssurr + Ssyst = Suniv Since the second law requires that Suniv 0, and since we know that T > 0, it follows that the sign for G for a process where pressure and temperature do not change tells us whether or not a process is consistent with the second law of thermodynamics. Since we require Suniv  0, it follows that for processes at a fixed temperature and pressure, G  0.

  22. Free Energy and the Second Law For a process where temperature and pressure are constant, we may say the following: G < 0 process is spontaneous G = 0 system is at equilibrium G > 0 process is not spontaneous (will not occur) Notice this means we need only do a calculation for the system, but only if the process occurs at constant pressure and temperature.

  23. Finding Gsyst For a Chemical Reaction For a chemical reaction Gsyst = Grxn is given by the expression Grxn = [  Gf(products) ] - [  Gf(reactants) ] Note Gf = 0 for an element in its standard state.

  24. Example: What are Hrxn, Srxn, and Grxn for the following process 4 Fe3O4(s) + O2(g)  6 Fe2O3(s) Is the reaction spontaneous for standard conditions? As before, we need a table of thermodynamic data top do this problem. Such data are given in Appendix 2 of the textbook. Substance H°f(kJ/mol) G°f(kJ/mol) S°(J/mol.K) Fe3O4(s) - 1118.4 - 1015.4 146.4 O2(g) 0.0 0.0 205.2 Fe2O3(s) - 824.2 - 742.2 87.4

  25. Example: What are Hrxn, Srxn, and Grxn for the following process 4 Fe3O4(s) + O2(g)  6 Fe2O3(s) Hrxn = [6 Hf(Fe2O3(s))] - [4 Hf(Fe3O4(s))] = [6 (- 824.2 kJ/mol)] - [4 (- 1118.4 kJ/mol)] = - 471.6. kJ/mol Srxn = [6 S(Fe2O3(s))] - [4 S(Fe3O4(s)) + S(O2(g))] = [6 (87.4 J/mol.K)] - [4 (146.4 J/mol.K) + (205.2 J/mol.K)] = - 266.4 J/mol.K Grxn = [6 Gf(Fe2O3(s))] - [4 Gf(Fe3O4(s))] = [6 (- 742.2 kJ/mol)] - [4 (- 1015.4 kJ/mol)] = - 391.6 kJ/mol As a check, Grxn = Hrxn - TSrxn = - 471.6 kJ/mol - (298.2 K) (- 0.2664 kJ/mol.K) = - 392.2 kJ/mol (process is spontaneous)

  26. Finding When Reactions Are Spontaneous Since G = H - TS, it follows that at a particular temperature G =  H - T S To a first approximation the values for Hrxn and Srxn (but not the value for Grxn) are independent of temperature. We can use this to find the approximate value for temperature at which equilibrium occurs for standard conditions of concentration. Grxn = 0 = Hrxn - Teq Srxn Teq  Hrxn /Srxn Example: At what temperature will H2O(l) and H2O(g) exist at equilibrium? (Note this will correspond to the normal boiling point of water).

  27. Example: At what temperature will H2O(l) and H2O(g) exist at equilibrium? (Note this will correspond to the normal boiling point of water). H2O(l)  H2O(g) Hrxn = + 44.0 kJ/mol Srxn = 118.8 J/mol.K Teq  Hrxn /Srxn = (44000. J/mol)/(118.8 J/mol.K) = 370.4 K Since Grxn = Hrxn - T Srxn it follows that if T > 370.4 K, Grxn < 0, and boiling is spontaneous. Of course, the true value for the normal boiling point for water is 373.2 K. The small difference between this and the value calculated above is due to the assumption that Hrxn and Srxn are independent of temperature, which is only approximately correct.

  28. Deciding When Reactions Will Be Spontaneous Recall that Grxn = Hrxn - T Srxn Grxn < 0 for a spontaneous process There are four possible combinations of Hrxn and Srxn. Hrxn Srxn behavior positive positive spontaneous if T > Teq positive negative reaction is never spontaneous negative positive reaction is always spontaneous negative negative spontaneous if T < Teq

  29. Thermodynamic Equilibrium Constant and Reaction Quotient We have previously discussed KC (equilibrium constant in terms of concentration) and Kp (equilibrium constant in terms of pressure), and the corresponding terms QC and Qp. The thermodynamic equilibrium constant (K) and thermo-dynamic reaction quotient (Q) are defined in terms of the following standard states: gases - in terms of a standard pressure of 1.00 atm solids, liquids, solvents - do not appear solutes - in terms of a standard concentration of 1.00 mol/L For reactions involving only gases K = Kp, while for reactions involving only solutes K = KC.

  30. Example: Give the expression for K (thermodynamic equilibrium constant) for the following reactions 2 H2(g) + O2(g)  2 H2O(g) 2 H3O+(aq) + CO32-(aq)  CO2(g) + 3 H2O(l )

  31. Example: Give the expression for K (thermodynamic equilibrium constant) for the following reactions 2 H2(g) + O2(g)  2 H2O(g) K = (pH2O)2 (pH2)2 (pO2) 2 H3O+(aq) + CO32-(aq)  CO2(g) + 3 H2O(l ) K = (pCO2) [H3O+]2 [CO32-]

  32. Free Energy Change for non-Standard Conditions For a chemical reaction where all the reactants and products are present at standard concentration (1.00 atm pressure for gases, 1.000 M for solutes, and ideal behavior) the free energy change is Grxn. What about the case where standard concentrations are not present? For a chemical reaction taking place at constant temperature and pressure, but any conditions of concentration, the following relationship applies Grxn = Grxn + RT ln Q Grxn - free energy change for the actual concentrations present Grxn - free energy change for standard concentrations Q - thermodynamic reaction quotient

  33. Example: Consider the reaction 2 H2(g) + O2(g)  2 H2O(l) What is Grxn for a) standard conditions (pH2 = pO2 = 1.00 atm), and b) for pH2 = 0.0100 atm, pO2 = 0.0050 atm.

  34. Example: Consider the reaction 2 H2(g) + O2(g)  2 H2O() Q = 1/{(pH2)2 (pO2)} What is Grxn for a) standard conditions (pH2 = pO2 = 1.00 atm), and b) for pH2 = 0.0100 atm, pO2 = 0.0050 atm. a) is standard conditions, so in that case G = Grxn = 2 Gf(H2O(l)) = 2 (- 237.2 kJ/mol) = - 474.4 kJ/mol b) First find Q Q = 1 = 1 = 2.0 x 106 (pH2)2 (pO2) (0.0100)2 (0.0050) G = Grxn + RT ln Q = - 474.4 kJ/mol + (8.314 x 10-3 kJ/mol.K) (298. K) ln (2.0 x 106) = - 474.4 kJ/mol + 35.9 kJ/mol = - 438.5 kJ/mol

  35. Free Energy and Equilibrium Our general expression for the change in free energy for a chemical reaction takes on a particularly interesting form when we are at equilibrium. In general G = Grxn + RT ln Q At equilibrium, G = 0. , and Q = K. Substituting, we get 0 = Grxn + RT ln K Grxn = - RT ln K ln K = - Grxn/RT Notice what this says. Based solely on thermodynamic data we can find the numerical value for the equilibrium constant for a chemical reaction.

  36. Example: Based on thermodynamic data find K for the process NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) K = [NH4+] [OH-] [NH3]

  37. Example: Based on thermodynamic data find K for the process NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) K = [NH4+] [OH-] [NH3] Grxn = [Gf(NH4+(aq)) + Gf(OH-(aq))] - [Gf(NH3(aq)) + Gf(H2O(l))] = [(- 79.4 kJ/mol) + (- 157.3 kJ/mol)] - [(- 26.6 kJ/mol) + (- 237.2 kJ/mol)] = + 27.1 kJ/mol ln K = - Grxn = - 27100. J/mol = - 10.94 RT (8.314 J/mol.K) (298. K) K = e-10.94 = 1.8 x 10-5

  38. End of Chapter 18 “…even though the average disorder in the universe has been increasing ever since the beginning, this increasing disorder is not occurring at every particular place in the universe. You can have local order created in certain locations at the expense of greater disorder elsewhere. Take your refrigerator as an example. You have ice cubes in the freezer that are certainly a manifestation of a great deal of order, but if you go around to the back of the refrigerator there’s a lot of hot air coming out indicating that the ‘cost’ of the order of the ice cubes is paid by the greater amount of disorder somewhere else.” Murray Gell-Mann “[Thermodynamics] is the only physical theory of universal content which, within the framework of the applicability of its basic concepts, I am convinced will never be overthrown.” Albert Einstein

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