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Chapter 17 Spontaneity, Entropy, and Free Energy.

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1 Chapter 17 Spontaneity, Entropy, and Free Energy

2 Chapter 17 Table of Contents Copyright © Cengage Learning. All rights reserved Spontaneous Processes and Entropy 17.2 Entropy and the Second Law of Thermodynamics 17.3 The Effect of Temperature on Spontaneity 17.4Free Energy 17.5Entropy Changes in Chemical Reactions 17.6Free Energy and Chemical Reactions 17.7The Dependence of Free Energy on Pressure 17.8Free Energy and Equilibrium 17.9Free Energy and Work

3 Section 17.1 Spontaneous Processes and Entropy Return to TOC Product favored ; reactant favored (spontaneous rxn) (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) + 2H 2 O (l) 2NaOH (aq) + H 2(g)

4 Section 17.1 Spontaneous Processes and Entropy Return to TOC Copyright © Cengage Learning. All rights reserved 4

5 Section 17.1 Spontaneous Processes and Entropy Return to TOC (3) why product - favored ?  H < 0  energy  ? (exothermic)  How about H 2 O (s) → H 2 O (l)  H > 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

6 Section 17.1 Spontaneous Processes and Entropy Return to TOC Copyright © Cengage Learning. All rights reserved 6 The Microstates That Give a Particular Arrangement (State)

7 Section 17.1 Spontaneous Processes and Entropy Return to TOC (5) S  0H, E ? ( S = 0 for a perfect crystal at 0 K) (6)  S = S f - S i (7) generalization : a) S (g) >> S (l) > S (s) b) S : more complex molecules > simpler (ex) : CH 3 CH 2 CH 3 > CH 3 CH 3 > CH 4 c) S : ionic solid (weaker force) > ionic solid (stronger force) (ex) : NaF (s) > MgO (s) d) liq and solid dissolve in a solvent  S > 0 gas dissolve in a solvent  S < 0

8 Section 17.1 Spontaneous Processes and Entropy Return to TOC Copyright © Cengage Learning. All rights reserved 8 Concept Check Predict the sign of  S for each of the following, and explain: a)The evaporation of alcohol b)The freezing of water c)Compressing an ideal gas at constant temperature d)Heating an ideal gas at constant pressure e)Dissolving NaCl in water +––+++––++

9 Section 17.2 Atomic MassesEntropy and the Second Law of Thermodynamics Return to TOC In any spontaneous process, there is always an increase in the entropy of the universe  S univ. > 0 for spontaneous rxn  S univ. =  S sys. +  S surr. System: rxn R P  S univ. > 0rxn →  S univ < 0rxn ←  S univ = 0rxn at equilibrium

10 Section 17.3 The MoleThe Effect of Temperature on Spontaneity Return to TOC ex) H 2 O (l) ─→ H 2 O (g) (1) System (l) → (g)18 ml (1 mol) → 31 L  S sys. > 0 (2) Surrounding a) sign ( + or - ) depend on “heat flow” exothermic process   S surr > 0 endothermic process   S surr < 0

11 Section 17.3 The MoleThe Effect of Temperature on Spontaneity Return to TOC b) magnitude depend on the temp. T  100 o C for rxn H 2 O (l) ─→ H 2 O (g) (3)  S univ =  S sys +  S surr const. T & P q p = |  H |

12 Section 17.3 The MoleThe Effect of Temperature on Spontaneity Return to TOC Copyright © Cengage Learning. All rights reserved 12 ΔS surr The sign of ΔS surr depends on the direction of the heat flow. The magnitude of ΔS surr depends on the temperature.

13 Section 17.4 Free Energy Return to TOC G is a thermodynamic function that gives information about the spontaneity of the system. G = H - TS(for system)  G =  H - T  S sys (const. T & P)  G 0 R ← Pnonspontaneous  G = 0 R  Pequilibrium

14 Section 17.4 Free Energy Return to TOC  H o < 0  S o > 0  H o > 0  S o > 0  H o < 0  S o < 0  H o > 0  S o < 0 at All temp.at High temp. spontaneous at Low temp.No Way !! Table  G =  H - T  S sys

15 Section 17.4 Free Energy Return to TOC Copyright © Cengage Learning. All rights reserved 15 Concept Check A liquid is vaporized at its boiling point. Predict the signs of: w q  H  S  S surr  G Explain your answers. –+++–0–+++–0

16 Section 17.4 Free Energy Return to TOC Copyright © Cengage Learning. All rights reserved 16 Exercise The value of  H vaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C. Calculate  S,  S surr, and  G for the vaporization of one mole of this substance at 72.5°C and 1 atm.  S = 132 J/K·mol  S surr = -132 J/K·mol  G = 0 kJ/mol

17 Section 17.5 Entropy Changes in Chemical Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 17 Concept Check Gas A 2 reacts with gas B 2 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  S surr  S  S univ Explain. – + 0 +

18 Section 17.6 Free Energy and Chemical Reactions Return to TOC (1) ex.  N 2 (g) + 3H 2 (g) → 2NH 3 (g)  S < 0  H 2 (g) → 2H (g)  4NH 3 (g) + 5O 2 (g) → 4NO (g) + 6H 2 O (g) (2) enthalpy (H) (at const. P)  H determines a rxnexothermic endothermic

19 Section 17.6 Free Energy and Chemical Reactions Return to TOC (3) free energy (G) (at const. T & P) spontaneous  G determines a rxn nonspontaneous equilibrium (4) S → absolute values  S T 1 →T 2 S = 0 for a perfect crystal at 0 o K 3 rd law of thermodynamics (5) standard entropy S o (298 K & 1 atm)

20 Section 17.6 Free Energy and Chemical Reactions Return to TOC (1)  H o : measured by calorimeter (2)  G o : standard free energy change 25 ℃, 1 atm  The more negative the value of  G o, the further a rxn will go to the right.  Can be calculated by  G o =  H o - T  S o ex   G o rxn can be calculated by using Hess’s law ex

21 Section 17.6 Free Energy and Chemical Reactions Return to TOC   G o rxn can be calculated using Hess’s law kinetically stable thermodynamically unstable rxn ← when high temp.

22 Section 17.6 Free Energy and Chemical Reactions Return to TOC   G o f = standard free energy of formation  G o f = 0 for elements ex & 17.12

23 Section 17.6 Free Energy and Chemical Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 23 Concept Check A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of:  H°  S°  G° Explain. – – –

24 Section 17.7 The Dependence of Free Energy on Pressure Return to TOC Copyright © Cengage Learning. All rights reserved 24 Free Energy and Pressure G = G° + RT ln(P) or ΔG = ΔG° + RT ln(Q)

25 Section 17.7 The Dependence of Free Energy on Pressure Return to TOC (1)S large V > S small V  S low P > S high P G = G o + RT ln(P) G o : the gas at the P = 1 atm G : the gas at the P = P ex. N 2 (g) + 3H 2 (g) → 2NH 3 (g)  G =  G products –  G reactants = 2G NH 3 – (G N 2 + 3G H 2 ) where G NH 3 = G o NH 3 + RT ln(P NH 3 ) G N 2 = ? G H 2 = ?

26 Section 17.7 The Dependence of Free Energy on Pressure Return to TOC  G = (2G o NH 3 – G o N 2 – 3G o H 2 ) + RT [2ln(P NH 3 ) – ln(P N 2 ) – 3ln(P H 2 )] =  G o + RTln[(P NH 3 ) 2 / (P N 2 )(P H 2 ) 3 ]  G =  G o + RTlnQ (2) The meaning of  G for a chemical rxn.

27 Section 17.8 Free Energy and Equilibrium Return to TOC Copyright © Cengage Learning. All rights reserved 27 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)

28 Section 17.8 Free Energy and Equilibrium Return to TOC (1) Fig & Fig A (g)  B (g) at equilibrium : G products = G reactants   G = G P – G r = 0 &  G =  G o + RTlnQ i.e.  G =  G o + RTlnk = 0 (equilibrium)  G o = –RTlnk   G o = 0  k = 1   G o 1products favor   G o > 0  k < 1reactants favor

29 Section 17.8 Free Energy and Equilibrium Return to TOC (2) Temp. dependence  G o = –RTlnK =  H o – T  S o y = mx + b (ln K) vs. (1/T)

30 Section 17.8 Free Energy and Equilibrium Return to TOC ex : N 2 (g) + 3H 2 (g)  2NH 3 (g)  G o = –33.3 kJ at 25 o C (a) P NH 3 = 1.00, P N 2 = 1.47 atm, P H 2 = 1.00  atm predict the rxn direction to reach equilibrium? Solution :  G =  G o + RTlnQ

31 Section 17.8 Free Energy and Equilibrium Return to TOC ex : Calculate k for the follow rxn at 25 o C 4Fe (s) + 3O 2 (g)  2Fe 2 O 3 (s)  H o f (kJ/mol)  S o f (J/kmol) Solution :  G =  G o + RTlnQ  G o =  H o – T  S o

32 Section 17.8 Free Energy and Equilibrium Return to TOC ex. Calculate k for k = 3.7  at T = 900 K k = ?at T = 550 K

33 Section 17.8 Free Energy and Equilibrium Return to TOC Copyright © Cengage Learning. All rights reserved 33 Change in Free Energy to Reach Equilibrium

34 Section 17.8 Free Energy and Equilibrium Return to TOC Copyright © Cengage Learning. All rights reserved 34 Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction

35 Section 17.9 Free Energy and Work Return to TOC Copyright © Cengage Learning. All rights reserved 35 Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy. w max = ΔG

36 Section 17.9 Free Energy and Work Return to TOC Copyright © Cengage Learning. All rights reserved 36 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.


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