Chapter 18: Entropy, Free Energy, and Equilibrium

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Chapter 18: Entropy, Free Energy, and Equilibrium Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm.

1 Chapter 18: Entropy, Free Energy, and Equilibrium Spontaneous Physical and Chemical 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 The expansion of a gas in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous spontaneous nonspontaneous 1

2 Does a decrease in enthalpy mean a reaction proceeds spontaneously? Spontaneous reactions CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)!h 0 = kj H + (aq) + OH - (aq) H 2 O (l)!h 0 = kj H 2 O (s) H 2 O (l)!h 0 = 6.01 kj NH 4 NO 3 (s) H 2 O NH 4 + (aq) + NO 3 - (aq)!h 0 = 25 kj Entropy (S) is a measure of the randomness or disorder of a system. order S disorder S!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 Spontaneous Processes for which!h = 0 Mixing of gases. Mixing of very similar liquids. Racemization reactions. There must be something else that matters - 2

3 Entropy S, a state function. Entropy measures the randomness (or, disorder) of a system. More random systems are more favorable than less random systems. S sol < S liq << S gas S sol < S liq << S gas 3

4 Problem about S Values Which has the greater value of S? CO 2 2( (s) or CO 2 2(g) (g)? SO 2 (g) or SO 3 (g)? 2NO 2 (g) or N 2 O 4 (g)? H 2 O(g) at 100 C or H 2 O(g) at 200 C?!S for Reactions Creation of a gas makes!s more positive (or, less negative): CaCO 3 (s) " CaO(s) + CO 2 (g) N 2 O 4 (g) " 2NO 2 (g) H 2 O(l) " H 2 O(g)!S for Reactions Creation of a solid, or!n gas < 0, makes!s more negative (or, less positive): Ag + (aq) + Cl - (aq) " AgCl(s) NH 3 (g) + HCl(g) " NH 4 Cl(s) N 2 (g) + 3H 2 (g) " 2NH 3 (g) H 2 O(l) " H 2 O(s) 4

5 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) 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: Equilibrium process:!s univ =!S sys +!S surr > 0!S univ =!S sys +!S surr = 0 Second Law of Thermodynamics Spontaneous process/reaction: Q < K eq Process or reaction should go forward. (but some go very slowly) Some of the reactants should be converted into products. 5

6 Second Law of Thermodynamics, con t. Criteria:!S univ > 0 easier to understand but harder to calculate!g sys < 0 (usually written!g < 0) harder to understand but easier to calculate Third Law of Thermodynamics Why do tables contain S values rather than!s (or!s f ) values? Third Law: The entropy of a perfect crystal is 0 at T=0 K, so, values of S (in addition to values of!s) can be measured (not possible for the functions H, E, or G) 6

7 Free Energy (G) The Gibbs free energy function combines energy and entropy: G = H - TS or,!g =!H - T!S if T is constant Note that units of H and S are different! Free Energy (G), con t.!g =!H - T!S if T is constant!h < 0 is favorable (exothermic reaction)!s > 0 is favorable (-!S < 0 is favorable)!h and T!S have the same units!g < 0 is favorable Problems Involving Hess s Law Hess s law says that enthalpy is a state function. Nothing more. It applies to all state functions. Given the following data calculate!g for Cr(s) +3/2O 2 (g) " CrO 3 (s) 2Cr(s) +3/2O 2 (g) " Cr 2 O 3 (s) CrO(s) +O 2 (g) " CrO 3 (s) 2CrO(s) +1/2O 2 (g) " Cr 2 O 3 (s)!g = -100 kj!g = -50 kj!g = -250 kj 7

8 Relationship of G to K eq!g =!H -T!S T constant; all pressures 1 atm (1 bar)!h and!s for the reaction can be worked out from the values in tables K eq = exp (-!G /RT) [or,!g = -RTlnK eq ] Am I here in CHE 107? 1. Yes 2. Yes, in mind only 3. Yes, in body only 4. Yes, I think 0% 0% 0% 0%!Yes!Yes,!in!mind!only!Yes,!in!body!only!Yes,!I!think Relationship of G to K eq!g =!H -T!S K eq = exp (-!G /RT);!G = -RTlnK eq If!G = 0, then K eq = 1 If!G < 0, then K eq > 1 The more negative!g the larger K eq If!G > 0, then K eq < 1 (but never < 0) 8

9 When Do Reactants Give Products? The equilibrium constant must be favorable. The values of K c and K p (or, K eq ) are determined by the size of the energy e change (!G ) for the reaction. The reaction must have a reasonable rate. H 2 (g) and O 2 (g) can be stored together indefinitely. Diamonds do not turn into graphite (even though they should). Problem What is!g at 298 K for the reaction H - 2 O(l) = H + (aq) + OH (aq)? kj kj kj 4. Haven t a clue What is!g at 298 K for the reaction H 2 O(l) = H + (aq) + OH - (aq)? 0% 0% 0% 0%!79.9!kJ!0.00!kJ!"56.6!kJ!Haven t!a!clue 9

10 What Determines K eq? Reactions that are very exothermic almost always have K eq values well above 1. eq BUT, many reactions that have!h = 0 and some reactions that have!h > 0 have K eq values > 1. How can this be? Contributions to!g!h < 0!H > 0!S < 0 K eq large at low T, but smaller at high T!S > 0 K eq very large (but decreases with T) K eq very small (but increases with T) K eq small at low T, but larger at high T # H Rxn " # H Rxn " # H Rxn " 10

11 Examples!H <0,!S >0: C 8 H 18 (l) O 2 (g) " 8CO 2 (g) + 9H 2 O(g)!H >0,!S <0: 3 O 2 (g) " 2O 3 (g) All molecules fall apart if T is high!h >0,!S >0: enough N 2 O 4 (g) " 2NO 2 (g)!h <0,!S <0: N 2 (g) + 3 H 2 (g) " 2NH 3 (g) How Does K eq Change with T when!h and!s Have the Same Sign? Simplest estimate: assume!h and!s are constant. If!S > 0 then!g and K eq become more favorable as T rises. If!S < 0 then!g and K eq become less favorable as T rises. For the reaction N 2 O 4 (g) = 2NO 2 (g) Problem Reasonable assumption in many cases, but T is an estimate.!h is 57.2 kj/mol;!s is175.9 J/(mol-K) If!H and!s are assumed to be constant with T, at what T does the reaction become spontaneous?!g become zero? 11

g and K eq become less favorable as T rises. Contributions to!g!h < 0!H > 0!

12 How Does K eq Change with T when!h and!s Have the Same Sign? More complete answer: If!H > 0 then!g and K eq become more favorable as T rises. The effect is largest if K eq is near 1. If!H < 0 then!g and K eq become less favorable as T rises. Contributions to!g!h < 0!H > 0!S < 0 K eq large at low T, but smaller at high T!S > 0 K eq very large (but decreases with T) K eq very small (but increases with T) K eq small at low T, but larger at high T CaCO 3 (s) " CaO(s) + CO 2 (g) 12

13 Caution! A negative value of!g does not guarantee that reactants will be converted completely into products! Reactants are converted completely into products only if!g is very negative.!g and the Conversion of Reactants to Products What is the magnitude of K eq? (If!G is only slightly negative, then K eq will not be much larger than 1). How big is the difference between!g and!g? (It is!g, not!g, that determines whether the reaction goes forward). Relationship between!g and K eq at 298 K!G (kj/mol) K eq Errors in!g are magnified when K eq is E+17 calculated E

G = -RTlnK eq + RT ln Q = RTln(Q/K eq ) Problem about!g K p is 4.

14 Relationship between!g and!g!g : All gases have P = 1 atm (or, 1 bar)!g: Gas pressures can have any value. This is the important difference!!g =!G + RT ln Q!G = -RTlnK eq + RT ln Q = RTln(Q/K eq ) Problem about!g K p is 4.40 at 2000 K for the reaction H 2 (g) + CO 2 (g) = H 2 O(g) + CO(g) What is!g? What is!g if the partial pressures are H atm CO atm H 2 O 0.66 atm CO 1.20 atm 14

15 How % Conversion of Reactants to Products Changes with!g G % conversion One More Thing about!g (anticipates Chpt. 19)!G is the energy that can be extracted as work when the reaction is run. (Important application is batteries). at least theoretically - never quite get as much energy out in practice Phase Transitions When two phases are in equilibrium then they have the same free energy (G).!G = 0 for the phase change.!g = 0 =!H - T!S, so if know two of values!h tr,!s tr, and T tr, can calculate the third. 15

16 Problems Given the following values of!h f and S at 298 K estimate the boiling point of water at 1 atm. H 2 O(l) kj/mol 69.9 J/(mol-K) H 2 O(g) kj/mol J/(mol-K) Then, what is the equilibrium vapor pressure of water at 298 K? At 288 K? At 308 K? Problem What is K p for the reaction N 2 O 4 (g) $ 2NO 2 (g) if!h f is 9.16 kj/mol for N 2 O 4 (g), and kj/mol for NO 2 (g) and if S is J/(mol-K) for N 2 O 4 (g), and J/(mol-K) for NO 2 (g)? Important Relationships!G =!H -T!S (and!g =!H - T!S)!G = -RT lnk eq!g =!G + RT lnq!g = 0 at equilibrium (not!g )!G < 0 for a spontaneous reaction K eq increases with T if!h > 0 16

17 Relationship between!g and!g, con t.!g =!G + RT ln Q;!G = RTln(Q/K eq ) If Q < K eq, then!g < 0 and the reaction should go forward (is spontaneous). If Q = K eq, then!g = 0 and the reaction is at equilibrium. If Q > K eq, then!g > 0 and the reaction should go backward (is not spontaneous). When Can We Expect the Reaction to Be Spontaneous? When!H is negative and!s is positive When Q << K all reactions give at least a few molecules of products When!S is positive and the temperature is high When!H is negative and the temperature is low Important Relationships S sol < S liq << S gas!s > 0 for a reaction if!n gas > 0 For molecules l in the same phase, S increases with the number of atoms!h < 0 for an exothermic reaction!h > 0 for an endothermic reaction 17

Chapter 19. Entropy, Free Energy, and Equilibrium

Chapter 19. Entropy, Free Energy, and Equilibrium Chapter 19 Entropy, Free Energy, and Equilibrium Spontaneous Physical and Chemical 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