Chapter 11 Spontaneous Change and Equilibrium 11-1 Enthalpy and Spontaneous Change 11-2 Entropy 11-3 Absolute Entropies and Chemical Reactions 11-4 The Second Law of Thermodynamics 11-5 The Gibbs Function 11-6 The Gibbs Function and Chemical Reactions 11-7 The Gibbs Function and the Equilibrium Constant 11-8 The Temperature Dependence of Equilibrium Constants OFP Chapter 11 1
This entire Chapter deals with a concept called Entropy and the Gibbs function Both are new state properties Entropy = S which has the units JK -1 mol -1 OFP Chapter 11 2
Consider Energy, it obvious that processes occur spontaneously to produce a state of lower energy Gas E Liquid Solid But, a chunk of ice at Room Temperature, spontaneously melts, forming a state of higher Energy Apparently more than energy is involved in determining the direction of spontaneous change OFP Chapter 11 3
This additional factor is the tendency of a system to assume the most random molecular arrangement possible Systems become disordered, more random Natural processes are favored which result in Decreased Energy (favored) Increased Entropy (favored) These two factors can oppose each other. Which one wins out? OFP Chapter 11 4
Recall from Chapter 10 lecture notes Third Law of Thermodynamics S = 0 at 0 K At absolute zero the Entropy term contributes nothing to the direction of spontaneous change The most stable state has the lowest energy A temperature increases, molecular motion increases and the tendency to disorder becomes more important At sufficiently high temperatures the Entropy factor becomes large enough to overcome even an unfavorable energy change OFP Chapter 11 5
For H 2 O (s) H 2 O (l) Above T m, the Entropy is dominant so spontaneous melting takes place Below T m, the energy decrease is dominant so spontaneous freezing take s place Temperature is a critical factor >T m S H2O is dominant <T m E H2O is dominant OFP Chapter 11 6
Disorder and Entropy Entropy is a quantitative measure of the number of microstates available to the molecules in a system Entropy is the degree of randomness or disorder in a system OFP Chapter 11 7
For phase transitions, at temperature T under equilibrium conditions Melting (solid to liquid) Fusion (liquid to solid) Vaporization (liquid to gas) Condensation (gas to liquid) Define S as the absolute molar Entropy, which is the absolute entropy of 1 mol of a substance in standard state Appendix D gives standard molar entropy values, S in units JK -1 mol -1 OFP Chapter 11 8
Entropies of Reaction S r = S products S reactants S r is the sum of products minus the sum of the reactants For a general reaction a A + b B c C + d D OFP Chapter 11 9
Exercise 11-3 (a) Use Data from Appendix D to calculate S r at 298.15 K for the reaction 2H 2 S(g) + 3O 2 (g) 2SO 2 (g) + 2H 2 O(g) (b) Calculate S of the system when 26.71 g of H 2 S(g) reacts with excess O 2 (g) to give SO 2 (g) and H 2 O(g) and no other products at 298.15K OFP Chapter 11 10
Exercise 11-3 (a) Use Data from Appendix D to calculate S r at 298.15 K for the reaction 2H 2 S(g) + 3O 2 (g) 2SO 2 (g) + 2H 2 O(g) Solution Notice that this is minus, which is consistent with 5 to 4 decrease in the amount of gas OFP Chapter 11 11
Exercise 11-3 2H 2 S(g) + 3O 2 (g) 2SO 2 (g) +2H 2 O(g) (b) Calculate S of the system when 26.71 g of H 2 S(g) reacts with excess O 2 (g) to give SO 2 (g) and H 2 O(g) and no other products at 298.15K Solution: OFP Chapter 11 12
Chapter 11 Spontaneous Change and Equilibrium Second Law of Thermodynamics In a real spontaneous process the Entropy of the universe (meaning the system plus its surroundings) must increase. if S universe = 0, then everything is in equilibrium The 2 nd Law of Thermodynamics profoundly affects how we look at nature and processes OFP Chapter 11 13
Summarize a few Concepts 1 st Law of Thermodynamics In any process, the total energy of the universe remains unchanged: energy is conserved A process and its reverse are equally allowed E forward = - E reverse (conservation of energy) 2 nd Law of Thermodynamics S, the entropy of a universe, increases in only one of the two directions of a reaction Processes that decrease S are impossible. Or improbable beyond conception OFP Chapter 11 14
Gibbs Function How are Enthalpy and Entropy related? G has several names 1. Gibbs function 2. Gibbs free energy 3. Free Enthalpy For the change in the Gibbs Energy of system, at constant Temperature and Pressure Notice that if a process is exothermic, H sys is negative, and S surroundings is positive OFP Chapter 11 15
Recall the 2 nd Law S universe > 0 (spontaneous process) S universe = S system + S surroundings > 0 From before S surroundings = - H sys /T Substituting S universe = S system - H sys /T > 0 Multiplying both sides by T T S system - H sys > 0 Or equivalently (by multiplying by -1) H sys -T S system = G system >0 (spontaneous process) OFP Chapter 11 16
From Earlier S univ > 0 S univ = 0 S univ < 0 Spontaneous Equilibrium Non-spontaneous OFP Chapter 11 17
Typical example using Gibbs Free energy Benzene, C 6 H 6, boils at 80.1 C. H vap = 30.8 kj a) Calculate S vap for 1 mole of benzene B) at 60 C and pressure = 1 atm does benzene boil? OFP Chapter 11 18
Typical example using Gibbs Free energy Benzene, C 6 H 6, boils at 80.1 C. H vap = 30.8 kj a) Calculate S vap for 1 mole of benzene B) at 60 C and pressure = 1 atm does benzene boil? OFP Chapter 11 19
All most all liquids have about the same S vap molar entropy of vaporization Called Trouton s rule S vap = 88 ± 5 JK -1 mol -1 In the last benzene boiling point problem, we calculated 87.2 JK -1 mol -1. This fits Trouton s rule. OFP Chapter 11 20
The Gibbs Function and Chemical Reactions G = H - T S G f is the standard molar Gibbs function of formation Because G is a State Property, for a general reaction a A + b B c C + d D OFP Chapter 11 21
Example 11-7 Calculate G for the following reaction at 298.15K. Use Appendix D for additional information needed. 3NO(g) N 2 O(g) + NO 2 (g) OFP Chapter 11 22
Effects of Temperature on G For temperatures other than 298K or 25C G = H - T S Typically H and S are constant over a broad range As Temperature increases G becomes more positive, i.e., less negative Temperature (C) 0 0 100 200 300-50 -100-150 Gr Hr Sr -200 Temperature (C) OFP Chapter 11 23
For temperatures other than 298K or 25C G = H - T S OFP Chapter 11 24
The Gibbs Function and the Equilibrium Constant What about non-standard states, other than 1 atm or a conc. [X] = 1 mol/l? aa + bb cd + dd OFP Chapter 11 25
G = G + RT ln Q Where Q is the reaction quotient a A + b B c C + d D If The amount of products are too high relative to the amounts of reactants present, and the reaction shifts in reverse (to the left) to achieve equilibrium If Q = K equilibrium If compare The amounts of reactants are too high relative to the amounts of products present, and the reaction proceeds in the forward direction (to the right) toward equilibrium Q K P = P c C a A P = P c C a A P P d D b B P P d D b B any conditions equilibrium OFP Chapter 11 26
G = G + RT ln Q Where Q is the reaction quotient a A + b B c C + d D If Q<K the rxn shifts towards the product side If Q=K equilibrium If Q>K the rxn shifts toward the reactant side At Equilibrium conditions NOTE: we can now calculate equilibrium constants (K) for reactions from standard G f functions of formation OFP Chapter 11 27
Criteria for Spontaneity in a Chemical Reaction Spontaneous Processes Equilibrium Processes Nonspontaneous Processes Suniv > 0 Suniv = 0 Suniv < 0 Conditions All conditions Gf < 0 Gf = 0 Gf > 0 Constant P and T Q < K Q = K Q > K Constant P and T OFP Chapter 11 28
Example 11-9 The G r for the following reaction at 298.15K was obtained in example 11-7. Now, calculate the equilibrium constant for this reaction at 25C. 3NO(g) N 2 O(g) + NO 2 (g) Strategy Use - G = RT ln K G = - 104.18 kj from example 11-7 OFP Chapter 11 29
Example 11-9 3NO(g) N 2 O(g) + NO 2 (g) Solution Use Rearrange - G = RT ln K G ln K = RT Use G r = - 104.18 kj from Ex. 11-7 OFP Chapter 11 30
The Temperature Dependence of Equilibrium Constants Where does this come from? Recall G = H - T S Divide by RT, then multiply by -1 G H S = RT RT R OFP Chapter 11 31
Notice that this is y= mx + b the equation for a straight line A plot of y= mx + b or ln K vs. 1/T OFP Chapter 11 32
If we have two different Temperatures and K s (equilibrium constants) Now given H and T at one temperature, we can calculate K at another temperature, assuming that H and S are constant over the temperature range OFP Chapter 11 33
Exercise 11-11 The reaction 2 Al 3 Cl 9 (g) 3 Al 2 Cl 6 (g) Has an equilibrium constant of 8.8X103 at 443K and a H r = 39.8 kjmol -1 at 443K. Estimate the equilibrium constant at a temperature of 600K. OFP Chapter 11 34
The Variation of Vapor Pressure with Temperature Used for equilibrium between pure liquids and their vapors. P 2 and P 1 = vapor pressure at different temperatures At the boiling point of a substance at 1 atm this simplifies to OFP Chapter 11 35
Chapter 11 Summary Gas Gas E Liquid S Liquid Solid Solid Natural processes are favored which result in Decreased Energy (favored) Increased Entropy (favored) Entropy is a quantitative measure of the number of microstates available to the molecules in a system Entropy is the degree of randomness or disorder in a system The Entropy of all substances is positive E.g., S = S l S s > 0 OFP Chapter 11 36
Chapter 11 Summary Gas Gas E Liquid S Liquid Solid Solid Natural processes are favored which result in Decreased Energy (favored) Increased Entropy (favored) Entropy is a quantitative measure of the number of microstates available to the molecules in a system Entropy is the degree of randomness or disorder in a system The Entropy of all substances is positive E.g., S = S l S s > 0 OFP Chapter 11 37
Chapter 11 Summary Entropies of Reaction S r = S products S reactants For a general reaction a A + b B c C + d D S o = c S o (C) + d S o (D) a S o (A) b S o (B) S univ > 0 S univ = 0 S univ < 0 Spontaneous Equilibrium Non-spontaneous OFP Chapter 11 38
G a G G = H - T S G f is the standard molar Gibbs function of formation Because G is a State Property, for a general reaction a A + b B c C + d D o f = c G o f (A) o f (C) + b G d G o f (B) For a change at constant temperature and pressure o f (D) G sys < 0 G sys = 0 G sys > 0 Spontaneous Equilibrium Non-spontaneous, but the reverse is spontaneous OFP Chapter 11 39
aa + bb cd + dd G = G + RT ln Q Where Q is the reaction quotient ln K = ln K G RT G = RT = H RT + S R K ln K 2 1 = H R van't Hoff 1 1 T1 T2 Equation o o K2 P2 Hvap 1 1 ln = ln = K1 P1 R T1 T2 Clausius - Clapeyron Equation OFP Chapter 11 40
Chapter 11 Spontaneous Change and Equilibrium Examples / Exercises 11-2, 11-3, 11-4, 11-5, 11-6 11-7, 11-8, 11-9, 11-10, 11-11, 11-12 Problems 13, 15, 23, 29, 31, 37, 39, 43, 49, 53, 57, 63, 69 OFP Chapter 11 41