Chapter 19 Chemical Thermodynamics Lecture Presentation Chapter 19 Chemical Thermodynamics James F. Kirby Quinnipiac University Hamden, CT © 2015 Pearson Education, Inc.

19.1 Spontaneous Processes Spontaneous Physical and Chemical Processes Thermodynamics: study of energy conversions and spontaneity of processes spontaneous Spontaneous Physical and Chemical Processes At 1 atm, water freezes below 0 oC and ice melts above 0 oC Heat flows from a hotter object to a colder object Iron exposed to oxygen and water forms rust A gas expands in an evacuated bulb nonspontaneous

19.1 Spontaneous Processes Does a decrease in enthalpy mean a reaction proceeds spontaneously? Spontaneous reactions CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = −890.4 kJ/mol H+ (aq) + OH− (aq) H2O (l) DH0 = −56.2 kJ/mol H2O (s) H2O (l) DH0 = 6.01 kJ/mol NH4NO3 (s) NH4+(aq) + NO3− (aq) DH0 = 25 kJ/mol H2O Enthaply alone cannot determine spontaneity of a process.

19.2 Entropy and the Second Law of Thermodynamics Entropy (S) is a measure of the randomness or disorder of a system. order S disorder S DS = Sf – Si Sf > Si DS > 0 Processes that lead to an increase in entropy (DS > 0):

19.2 Entropy and the Second Law of Thermodynamics For a substance in different phases: Ssolid < Sliquid << Sgas H2O (s) H2O (l) DS > 0 For substances in the same phase: He(g) < Ne(g) < Ar(g) Molar mass Molecular complexity F2(g) < Cl2(g) < Br2(g) He(g) < H2(g) isoelectronic Ar(g) < F2(g) isoelectronic NO(g) < NO2(g) < N2O4(g)

Ne(g) SO2(g) Na(s) NaCl(s) H2(g) EXERCISE #1 Arrange the following substances in order of increasing standard molar entropy at 25oC: Ne(g) SO2(g) Na(s) NaCl(s) H2(g) 20.2 g/mol 64.1 g/mol 23.0 g/mol 58.5 g/mol 2.0 g/mol Order: 4 5 1 2 3 Ssolid < Sliquid << Sgas Molar mass Molecular complexity

19.2 Entropy and the Second Law of Thermodynamics First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0

19.3 Entropy and the Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero. The entropy of a substance increases with temperature. EXERCISE #2 Which of the following obeys the third law of thermodynamics? a) KBr, solid, amorphous, 0 K b) KBr, solid, perfect crystal, 25 K c) KBr, solid, perfect crystal, 0 K

19.4 Entropy Changes in Chemical Reactions Standard entropy of reaction (∆S0) The entropy change for a reaction carried out at 1 atm and 25oC. aA + bB cC + dD ∆S0 = [cS0(C) + dS0(D)] – [aS0(A) + bS0(B)] ∆S0 = SnS0(products) − SmS0(reactants) Hess’s law S0 is absolute entropy of a substance at 25oC and 1 atm When gases are produced (or consumed): Dn > 0 DS0 > 0 Dn < 0 DS0 < 0 Dn = 0 DS0 may be positive or negative BUT DS0 will be a small number.

EXERCISE #3 Calculate ΔS0 for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) S0 (J/K·mol) Na(s) 51 H2O(l) 70 NaOH(aq) 50 H2(g) 131 x 2 x 2 x 2 ΔS0 = [2S0(NaOH, aq) + S0(H2, g)] – [2S0(Na, s) + 2S0(H2O, l)] = [2(50) + 131] – [2(51) + 2(70)] = –11 J/K

Enthalpy Changes in Chemical Reactions Standard enthalpy of reaction ( ∆𝐇 𝐫𝐱𝐧 𝟎 ) The enthalpy change for a reaction when it occurs under standard-state conditions. aA + bB cC + dD ∆H0 = [c ∆H f 0 (C) + d ∆H f 0 (D)] – [a ∆H f 0 (A) + b ∆H f 0 (B)] ∆H0 = Sn ∆H f 0 (products) − Sm ∆H f 0 (reactants) Hess’s law ∆𝑯 𝐟 𝟎 is standard enthalpy of formation The enthalpy change when 1 mole of a compound is formed from its elements in their standard states. ∆H f 0 of any element in its stable form is zero.

Gibbs Free Energy (G) and Spontaneity For a constant P and T process: G = Gibbs free-energy H = enthalpy T = temperature (in Kelvin) S = entropy G = H – TS For a constant P and T process: DG = DH − TDS DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium.

Standard free-energy of reaction (∆G0) 19.5 Gibbs Free Energy Standard free-energy of reaction (∆G0) The free-energy change for a reaction when it occurs under standard-state conditions. aA + bB cC + dD ∆G0 = [c ∆G f 0 (C) + d ∆G f 0 (D)] – [a ∆G f 0 (A) + b ∆G f 0 (B)] ∆G0 = Sn ∆G f 0 (products) Hess’s law − Sm ∆G f 0 (reactants) ∆𝐆 𝐟 𝟎 is standard free-energy of formation The free-energy change when 1 mole of a compound is formed from its elements in their standard states. ∆G f 0 of any element in its stable form is zero.

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) EXERCISE #4 Given the data below, calculate the standard free-energy change for the following reaction at 25oC. Is the reaction spontaneous? CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ∆𝐺 𝑓 𝑜 (kJ/mol) CO2(g) −394.4 H2O(l) −237.2 CH4(g) −50.8 O2(g) 0 x 2 x 2 ∆G0 = [ ∆G f o (CO2, g) + 2 ∆G f o (H2O, l)] – [ ∆G f o (CH4, g) + 2 ∆G f o (O2, g)] ∆G0 = [−394.4 kJ/mol + 2(−237.2 kJ/mol)] − [−50.8 kJ/mol + 2(0 kJ/mol)] = −818.0 kJ ΔG0 < 0 yes

19.6 Free Energy and Temperature Factors affecting the sign of ∆G in the relationship ∆G = ∆H – T∆S ∆H ∆S ∆G Spontaneity + + ∆G < 0 at high T. Spontaneous in the forward direction at high T. + – ∆G > 0 at all T. Nonspontaneous in the forward direction at all T. – + ∆G < 0 at all T. Spontaneous in the forward direction at all T. – – ∆G < 0 at low T. Spontaneous in the forward direction at low T.

19.6 Free Energy and Temperature Spontaneity of Physical Processes at 1 atm: DG0 = ∆H vap 0 – Tb ∆S vap 0 = 0 EXERCISE #5 The enthalpy of vaporization and the entropy change of ammonia at 1 atm is 23.35 kJ/mol and 9.74 x 101 J/K∙mol, respectively. Calculate the boiling point of ammonia. NH3(l) NH3(g) DG0 = ∆H vap 0 – Tb ∆S vap 0 = 0 ∆H vap 0 = Tb ∆S vap 0 ∆H vap 0 ∆S vap 0 = 23.35 x 10 3 J/mol 97.4 J/K∙mol Tb = = 239.733 K = −33.4oC

19.6 Free Energy and Temperature Spontaneity of Chemical Reactions at 1 atm: DG0 = DH0 – TDS0 < 0 EXERCISE #6 The standard enthalpy and the standard entropy change of the reaction at 1 atm are 177.8 kJ/mol and 160.5 J/K·mol, respectively. CaCO3 (s) CaO (s) + CO2 (g) a) Calculate the standard free energy of the reaction at 25oC. DG0 = DH0 – TDS0 = (177.8 x 103 J/mol) – (298 K) (160.5 J/K·mol) DG0 = 129971 J/mol = 130.0 kJ/mol DG0 > 0 nonspontaneous at 25oC b) At what temperatures will the reaction be spontaneous? ∆H0 ∆S0 = 177.8 x 10 3 J/mol 160.5 J/K∙mol DG0 = DH0 – TDS0 = 0 Tb = = 1108 K = 835oC DG0 = 0 at 835 oC DG0 > 0 at 25 oC DG < 0 when T > 835oC Spontaneous at T > 835oC

19.7 Free Energy and the Equilibrium Constant Free Energy and Spontaneity of Chemical Equilibrium DG0 is the standard Gibbs free-energy (kJ/mol) R is the gas constant (8.314 J/K∙mol) DG = DG0 + RT lnQ T is the absolute temperature (K) Q is the reaction quotient DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium.

19.7 Free Energy and the Equilibrium Constant Free Energy and Equilibrium Constant DG = DG0 + RT ln Q DG = 0 Q = K 0 = DG0 + RT ln K DG0 = −RT ln K Relationship between ∆Go and K as predicted by ∆Go = −RT ln K K ln K ∆Go At equilibrium > 1 + − Products favored over reactants. = 1 Products and reactants equally favored. < 1 – + Reactants favored over products.

EXERCISE #7 Consider the decomposition of water 2H2O(l) 2H2(g) + O2(g), and the thermodynamic data below for the reaction at 25oC. ∆H f o (kJ/mol) So(J/K∙mol) H2O(l) −286 70 H2(g) 0 131 O2(g) 0 205 x 2 x 2 x 2 x 2 a) Calculate the standard enthalpy of the reaction. ∆H0 = [ ∆H f o (O2,g) + 2 ∆H f o (H2, g)] – [2 ∆H f o (H2O, l)] ∆H0 = [0 + 2(0)] – [−2(286 kJ/mol)] = 572 kJ b) Calculate the standard entropy of the reaction. ∆S0 = [So(O2, g) + 2So(H2, g)] – [2So(H2O, l)] ∆S0 = [205 J/K∙mol + 2(131 J/K∙mol)] – [2(70 J/K∙mol)] = 327 J/K

EXERCISE #7 Continued c) Calculate the standard Gibbs energy of the reaction. DH0 = 572 kJ DS0 = 327 J/K T = 25oC = 298 K ∆G0 = ∆H0 – T∆S0 = (572 x 103 J) – (298 K) (327 J/K) = 474600 J = 474.6 kJ d) Calculate the equilibrium constant Kp of the reaction. ∆G0 = −RT ln Kp R = 8.314 J/K∙mol ln Kp = ∆G 0 −RT = 474.6 x 10 3 J − 8.314J/K∙mol)(298 K = −191.6 Kp = e−191.6 = 6 x 10−84 e) What does DG0 and Kp tell us about the conditions at equilibrium? ∆G0 >> 0 and Kp << 0 reactants are favored over products f) Assuming DH0 and DS0 are constant, estimate the value of K at 100oC. ln K = ∆G 0 −RT = 474.6 x 10 3 J − 8.314J/K∙mol)(373 K = −153.0 K = e−153.0 = 4 x 10−67

EXERCISE #8 The standard free-energy change for the reaction is 5.40 kJ/mol at 298 K. In a certain experiment, the initial pressures are PNO2 = 0.122 atm and PN2O4 = 0.453 atm. Calculate ΔG for the reaction at these pressures and predict the direction of the net reaction toward equilibrium. N2O4(g) 2NO2(g) (P NO 2 )2 P N 2 O 4 = 0.122 2 0.453 ∆G = ∆G0 + RT lnQp R = 8.314 J/K∙mol Qp = (ln 0.122 2 0.453 ) ∆G = 5.40 x 103 J/mol + (8.314 J/K∙mol) (298 K) ∆G = −3.06 kJ/mol = −3.06 x 103 J/mol ∆G < 0 spontaneous in the forward direction

End of Chapter 19