Spontaneity, Entropy, and Free Energy Chapter 17 Spontaneity, Entropy, and Free Energy

17.1 Spontaneous Processes and Entropy 17.2 Entropy and the Second Law of Thermodynamics 17.3 The Effect of Temperature on Spontaneity 17.4 Free Energy 17.5 Entropy Changes in Chemical Reactions 17.6 Free Energy and Chemical Reactions 17.7 The Dependence of Free Energy on Pressure 17.8 Free Energy and Equilibrium 17.9 Free Energy and Work Copyright © Cengage Learning. All rights reserved

ex. 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) (1) spontaneous process : it occurs without outside intervention. (2) spontaneous rxn & speed  thermodynamics vs kinetics the science of E transfer predict occur study of the rates of rxn why somes are fast or slow? vs ex. 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) Product favored ; reactant favored (spontaneous rxn)

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(3) why product - favored ? DH < 0  energy  ? (exothermic) How about H2O(s) → H2O(l) DH > 0 at 0℃ (4) something else !  Entropy ( S ) S = the driving force for a spontaneous process is an increase in the entropy of the universe. = a measure of randomness or disorder of a system the great the randomness, the greater S. = the number of arrangements  probability the higher probability, the higher S

The Microstates That Give a Particular Arrangement (State) Copyright © Cengage Learning. All rights reserved

( S = 0 for a perfect crystal at 0 K) (6) DS = Sf - Si (5) S  0 H , E ? ( S = 0 for a perfect crystal at 0 K) (6) DS = Sf - Si (7) generalization : a) S(g) >> S(l) > S(s) b) S : more complex molecules > simpler (ex) : CH3 CH2 CH3 > CH3 CH3 > CH4 c) S : ionic solid (weaker force) > ionic solid (stronger force) (ex) : NaF(s) > MgO(s) d) liq and solid dissolve in a solvent DS > 0 gas dissolve in a solvent DS < 0

Predict the sign of S for each of the following, and explain: Concept Check Predict the sign of S for each of the following, and explain: The evaporation of alcohol The freezing of water Compressing an ideal gas at constant temperature Heating an ideal gas at constant pressure Dissolving NaCl in water + – a) + (a liquid is turning into a gas) b) - (more order in a solid than a liquid) c) - (the volume of the container is decreasing) d) + (the volume of the container is increasing) e) + (there is less order as the salt dissociates and spreads throughout the water) Copyright © Cengage Learning. All rights reserved

In any spontaneous process, there is always an increase in the entropy of the universe DSuniv. > 0 for spontaneous rxn DSuniv. = DSsys. + DSsurr. System: rxn R P DSuniv. > 0 rxn → DSuniv < 0 rxn ← DSuniv = 0 rxn at equilibrium

a) sign ( ＋ or －) depend on “heat flow” ex) H2O(l) ─→ H2O(g) (1) System (l) → (g) 18 ml (1 mol) → 31 L DSsys. > 0 (2) Surrounding a) sign ( ＋ or －) depend on “heat flow” exothermic process  DSsurr > 0 endothermic process  DSsurr < 0

b) magnitude depend on the temp. T  100 oC for rxn H2O(l) ─→ H2O(g) (3) DSuniv = DSsys + DSsurr const. T & P qp = | DH |

The sign of ΔSsurr depends on the direction of the heat flow. The magnitude of ΔSsurr depends on the temperature. Copyright © Cengage Learning. All rights reserved

DG < 0 R → P spontaneou DG > 0 R ← P nonspontaneous G is a thermodynamic function that gives information about the spontaneity of the system. G = H - TS (for system) DG = DH - TDSsys (const. T & P) DG < 0 R → P spontaneou DG > 0 R ← P nonspontaneous DG = 0 R  P equilibrium

DG = DH - TDSsys DHo < 0 DSo > 0 DHo > 0 DSo < 0 at All temp. at High temp. spontaneous at Low temp. No Way !! Table. 17.5

A liquid is vaporized at its boiling point. Predict the signs of: w q Concept Check A liquid is vaporized at its boiling point. Predict the signs of: w q H S Ssurr G Explain your answers. – + As a liquid goes to vapor, it does work on the surroundings (expansion occurs). Heat is required for this process. Thus, w = negative; q = H = positive. S = positive (a gas is more disordered than a liquid), and Ssurr = negative (heat comes from the surroundings to the system); G = 0 because the system is at its boiling point and therefore at equilibrium. Copyright © Cengage Learning. All rights reserved

Exercise The value of Hvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C. Calculate S, Ssurr, and G for the vaporization of one mole of this substance at 72.5°C and 1 atm. S = 132 J/K·mol Ssurr = -132 J/K·mol G = 0 kJ/mol S = 132 J/K·mol Ssurr = -132 J/k·mol G = 0 kJ/mol Copyright © Cengage Learning. All rights reserved

Concept Check Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The bond energy of AB is much greater than that of either reactant. Predict the signs of: H Ssurr S Suniv Explain. Since the average bond energy of the products is greater than the average bond energies of the reactants, the reaction is exothermic as written. Thus, the sign of H is negative; Ssurr is positive; S is close to zero (cannot tell for sure); and Suniv is positive. – + 0 + Copyright © Cengage Learning. All rights reserved

 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) (1) ex.  N2 (g) + 3H2(g) → 2NH3(g) DS < 0  H2(g) → 2H(g)  4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) (2) enthalpy (H) (at const. P) DH determines a rxn exothermic endothermic

(3) free energy (G) (at const. T & P) spontaneous DG determines a rxn nonspontaneous equilibrium (4) S → absolute values DST1→T2 S = 0 for a perfect crystal at 0 oK 3rd law of thermodynamics (5) standard entropy So (298 K & 1 atm)

(1) DHo ： measured by calorimeter (2) DGo ： standard free energy change 25 ℃ , 1 atm  The more negative the value of DGo , the further a rxn will go to the right.  Can be calculated by DGo = DHo - T DSo ex. 17.9  DGorxn can be calculated by using Hess’s law ex. 17.10

thermodynamically unstable rxn ← when high temp.  DGorxn can be calculated using Hess’s law kinetically stable thermodynamically unstable rxn ← when high temp.

 DGof = standard free energy of formation DGof = 0 for elements ex. 17.11 & 17.12

A stable diatomic molecule spontaneously forms from its atoms. Concept Check A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of: H° S° G° Explain. The reaction is exothermic, more ordered, and spontaneous. Thus, the sign of H is negative; S is negative; and G is negative. – – – Copyright © Cengage Learning. All rights reserved

Free Energy and Pressure G = G° + RT ln(P) or ΔG = ΔG° + RT ln(Q) Copyright © Cengage Learning. All rights reserved

(1) Slarge V > Ssmall V  Slow P > Shigh P G = Go + RT ln(P) Go : the gas at the P = 1 atm G : the gas at the P = P ex. N2(g) + 3H2(g) → 2NH3(g) DG = Gproducts – Greactants = 2GNH3 – (GN2 + 3GH2) where GNH3 = GoNH3 + RT ln(PNH3) GN2 = ? GH2 = ?

(2) The meaning of DG for a chemical rxn. DG = (2GoNH3 – GoN2 – 3GoH2) + RT [2ln(PNH3) – ln(PN2) – 3ln(PH2)] = DGo + RTln[(PNH3)2/ (PN2)(PH2)3] DG = DGo + RTlnQ (2) The meaning of DG for a chemical rxn.

The equilibrium point occurs at the lowest value of free energy available to the reaction system. ΔG = 0 = ΔG° + RT ln(K) ΔG° = –RT ln(K) Copyright © Cengage Learning. All rights reserved

(1) Fig. 17.8 & Fig. 17.9 A(g)  B (g) at equilibrium : Gproducts = Greactants  DG = GP – Gr = 0 & DG = DGo + RTlnQ i.e. DG = DGo + RTlnk = 0 (equilibrium) DGo = –RTlnk  DGo = 0  k = 1  DGo < 0  k > 1 products favor  DGo > 0  k < 1 reactants favor

(2) Temp. dependence DGo = –RTlnK = DHo – TDSo y = mx + b (ln K) vs. (1/T)

DGo = –33.3 kJ at 25oC ex. 17.14: N2 (g) + 3H2(g)  2NH3(g) (a) PNH3 = 1.00, PN2 = 1.47 atm, PH2 = 1.0010-2 atm predict the rxn direction to reach equilibrium? Solution : DG = DGo + RTlnQ

Solution : DG = DGo + RTlnQ DGo = DHo – TDSo ex. 17.15: Calculate k for the follow rxn at 25 oC 4Fe (s) + 3O2(g)  2Fe2O3(s) DHof (kJ/mol) -826 DSof (J/kmol) 27 205 90 Solution : DG = DGo + RTlnQ DGo = DHo – TDSo

ex. Calculate k for k = 3.7  10-6 at T = 900 K k = ? at T = 550 K

Change in Free Energy to Reach Equilibrium Copyright © Cengage Learning. All rights reserved

Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction Copyright © Cengage Learning. All rights reserved

Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy. wmax = ΔG Copyright © Cengage Learning. All rights reserved

All real processes are irreversible. Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy. All real processes are irreversible. First law: You can’t win, you can only break even. Second law: You can’t break even. Copyright © Cengage Learning. All rights reserved