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Entropy, Free Energy, and Equilibrium Chapter 18 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Presentation on theme: "Entropy, Free Energy, and Equilibrium Chapter 18 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display."— Presentation transcript:

1 Entropy, Free Energy, and Equilibrium Chapter 18 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 The following items will be discussed : 18-1 The three laws of thermodynamic 18-2 Spontaneous Processes 18-3 Entropy 18-4 Second Law of Thermodynamics 18-5 Gibbs Free Energy 18-6 Gibbs Free Energy and Chemical Equilibrium

3 18-1 The three laws of thermodynamic First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. One measure of these changes is the amount of heat given off or absorbed by a system during a constant pressure process, which chemists define as a change in enthalpy. The second law of thermodynamics explains why chemical process tend to favor one direction. The third law is an extension of the second law.

4 18-2 Spontaneous Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0 C and ice melts above 0 0 C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous 18.2 The spontaneous processes occur to decrease the energy of a system

5 spontaneous nonspontaneous 18.2

6 Does a decrease in enthalpy mean a reaction proceeds spontaneously? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H 0 = -890.4 kJ H + (aq) + OH - (aq) H 2 O (l)  H 0 = -56.2 kJ H 2 O (s) H 2 O (l)  H 0 = 6.01 kJ NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq)  H 0 = 25 kJ H2OH2O Spontaneous reactions 18.2 We cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis of energy changes in the system. To make this kind of prediction we need another thermodynamic quantity, which turns out to be entropy

7 18-3 Entropy State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, enthalpy, pressure, volume, temperature, entropy 18.3

8 Entropy (S) is a measure of the randomness or disorder of a system. order increase S increase disorder increase S decrease Like enthalpy, entropy is a state function so,  S = S f - S i If the change from initial to final results in an increase in randomness S f > S i  S > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state S solid < S liquid << S gas H 2 O (s) H 2 O (l)  S > 0 18.3

9 W = 1 W = 4 W = 6 W = number of microstates S = k ln W  S = S f - S i  S = k ln WfWf WiWi W f > W i then  S > 0 W f < W i then  S < 0 Microstates and Entropy 18.3

10 Processes that lead to an increase in entropy (  S > 0) 18.2 Standard entropy is the absolute entropy of a substance at 1 atm and 25 0 C

11 How does the entropy of a system change for each of the following processes? (a) Condensing water vapor Randomness decreases Entropy decreases (  S < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (  S < 0) (c) Heating hydrogen gas from 60 0 C to 80 0 C Randomness increases Entropy increases (  S > 0) (d) Subliming dry ice Randomness increases Entropy increases (  S > 0) 18.3 E) Dissolving NaCl in water Randomness decreases because of the hydration Entropy decreases (  S < 0)

12

13 First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. 18-4 Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.  S univ =  S sys +  S surr > 0 Spontaneous process:  S univ =  S sys +  S surr = 0 Equilibrium process: 18.4  E = q + W = q – P  V Remember that

14 Entropy Changes in the System (  S sys ) aA + bB cC + dD S0S0 rxn dS 0 (D) cS 0 (C) = [+] - bS 0 (B) aS 0 (A) [+] S0S0 rxn n S 0 (products) =  m S 0 (reactants)  - The standard entropy of reaction (  S 0 ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. rxn 18.4 What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = 197.9 J/K mol S 0 (O 2 ) = 205.0 J/K mol S 0 (CO 2 ) = 213.6 J/K mol S0S0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S0S0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K mol

15 Entropy Changes in the System (  S sys ) 18.4 When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes,  S 0 > 0. If the total number of gas molecules diminishes,  S 0 < 0. If there is no net change in the total number of gas molecules, then  S 0 may be positive or negative BUT  S 0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O 2 (g) 2ZnO (s) The total number of gas molecules goes down,  S is negative.

16 Worked Example 18.2

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18 Entropy Changes in the Surroundings (  S surr ) Exothermic Process  S surr > 0 Endothermic Process  S surr < 0 18.4  S surr = -  H surr / T

19 Worked Example 18.5

20 Third Law of Thermodynamics and absolute entropy The entropy of a perfect crystalline substance is zero at the absolute zero of temperature. If the crystal is impure of if it has defects, then its entropy is greater than zero even at 0 0 K 18.3 S = k ln W W = 1 S = 0

21 18-5 Gibbs Free Energy Spontaneous process: Equilibrium process:  S univ =  S sys +  S surr > 0  S univ =  S sys +  S surr = 0 Thus for spontaneous process we can change the equation to - T  S univ =  H sys - T  S sys < 0 Substituting -  H surr / T for  s surr and multiplying both sides of the equation by T gives T  S univ = -  H sys + T  S sys > 0 In order to express the spontaneity of a reaction more directly, we introduce another thermodynamic function called Gibbs free energy G which is a state function: G = H - TS

22 For a constant pressure and temperature process we can write:  G =  H sys -T  S sys Gibbs free energy (G)  G < 0 The reaction is spontaneous in the forward direction.  G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.  G = 0 The reaction is at equilibrium. 18.5

23 aA + bB cC + dD G0G0 rxn d  G 0 (D) f c  G 0 (C) f = [+] - b  G 0 (B) f a  G 0 (A) f [+] G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - The standard free-energy of reaction (  G 0 ) is the free- energy change for a reaction when it occurs under standard- state conditions. rxn Standard free energy of formation (  G 0 ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f  G 0 of any element in its stable form is zero. f

24 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - What is the standard free-energy change for the following reaction at 25 0 C? G0G0 rxn 6  G 0 (H 2 O) f 12  G 0 (CO 2 ) f = [+] - 2  G 0 (C 6 H 6 ) f [] G0G0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0 C?  G 0 = -6405 kJ < 0 spontaneous 18.5

25 Worked Example 18.4

26  G =  H - T  S 18.5

27 CaCO 3 (s) CaO (s) + CO 2 (g)  H 0 = 177.8 kJ  S 0 = 160.5 J/K  G 0 =  H 0 – T  S 0 At 25 0 C,  G 0 = 130.0 kJ  G 0 = 0 at 835 0 C 18.5 Temperature and Spontaneity of Chemical Reactions Equilibrium Pressure of CO 2

28 Gibbs Free Energy and Phase Transitions H 2 O (l) H 2 O (g)  G 0 = 0=  H 0 – T  S 0  S = T HH = 40.79 kJ 373 K = 109 J/K 18.5

29 Efficiency = X 100% T h - T c TcTc Chemistry In Action: The Efficiency of Heat Engines A Simple Heat Engine

30 18-6 Gibbs Free Energy and Chemical Equilibrium  G =  G 0 + RT lnQ R is the gas constant (8.314 J/K mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium  G = 0 Q = K 0 =  G 0 + RT lnK  G 0 =  RT lnK 18.6

31 Worked Example 18.6

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33 18.6 Free Energy Versus Extent of Reaction  G 0 < 0  G 0 > 0

34  G 0 =  RT lnK 18.6

35 Problems 1-2 -3 -4 -5 -7 -8 -9 -10 -12 -13 -14 -15 -16 - 17 -18 -19 -20 -21 -22 -23 -24 -25 -26 -30 - 32 -44 -56 -57 -72 -90 -92 -98


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