Copyright © 2012, The McGraw-Hill Compaies, Inc. Permission required for reproduction or display. Chemistry Third Edition Julia Burdge Lecture PowerPoints Chapter 18 Entropy, Free Energy, and Equilibrium

CHAPTER 18 Entropy, Free Energy, and Equilibrium Spontaneous Processes 18.2Entropy 18.3Entropy Changes in a System 18.4Entropy Changes in the Universe 18.5Predicting Spontaneity 18.6Free Energy and Chemical Equilibrium 18.7Thermodynamics in Living Systems

Topics 18.1Spontaneous Processes 3

18.1Spontaneous Processes 4 A process that does occur under a specific set of conditions is called a spontaneous process. One that does not occur under a specific set of conditions is called nonspontaneous.

18.1Spontaneous Processes 5

18.1Spontaneous Processes 6 These exothermic reactions are spontaneous at room temperature: This exothermic reaction is not spontaneous at temperatures above 0°C:

18.1Spontaneous Processes 7 So, while exothermicity favors spontaneity, it cannot be the sole factor that determines spontaneity.

Topics 18.2Entropy 8 A Qualitative Description of Entropy A Quantitative Definition of Entropy

18.2Entropy A Qualitative Description of Entropy 9 Qualitatively, the entropy (S) of a system is a measure of how spread out or how dispersed the system’s energy is. The simplest interpretation of this is how spread out a system’s energy is in space. In other words, for a given system, the greater the volume it occupies, the greater its entropy. Just as spontaneity is favored by a process being exothermic, spontaneity is also favored by an increase in the system’s entropy.

18.2Entropy A Quantitative Definition of Entropy 10 where k is the Boltzmann constant (1.38×10 –23 J/K) and W is the number of energetically equivalent different ways the molecules in a system can be arranged.

18.2Entropy A Quantitative Definition of Entropy 11

18.2Entropy A Quantitative Definition of Entropy = 4 possible arrangements 4 2 = 16 possible arrangements

18.2Entropy A Quantitative Definition of Entropy 13 There are three different states possible for this system. 1.One molecule on each side (eight possible arrangements) 2.Both molecules on the left (four possible arrangements) 3.Both molecules on the right (four possible arrangements) The most probable state is the one with the largest number of possible arrangements. In this case, the most probable state is the one with one molecule on each side of the container.

Topics 18.3Entropy Changes in a System 14 Calculating  S sys Standard Entropy, S° Qualitatively Predicting the Sign of  S sys

18.4Entropy Changes in the Universe Calculating  S sys 15

18.4Entropy Changes in the Universe Calculating  S sys 16

SAMPLE PROBLEM Determine the change in entropy for 1.0 mole of an ideal gas originally confined to one-half of a 5.0-L container when the gas is allowed to expand to fill the entire container at constant temperature. Setup R = J/K · mol, n = 1.0 mole, V final = 5.0 L, and V initial = 2.5 L.

SAMPLE PROBLEM Solution

18.3Entropy Changes in a System Standard Entropy, S° 19 It is possible to determine the absolute value of the entropy of a substance, S; something we cannot do with either energy or enthalpy. Standard entropy is the absolute entropy of a substance at 1 atm.

18.3Entropy Changes in a System Standard Entropy, S° 20

18.3Entropy Changes in a System Standard Entropy, S° 21 For a given substance, the standard entropy is greater in the liquid phase than in the solid phase. [Compare the standard entropies of Na(s) and Na(l).] For a given substance, the standard entropy is greater in the gas phase than in the liquid phase. For two monatomic species, the one with the larger molar mass has the greater standard entropy.

18.3Entropy Changes in a System Standard Entropy, S° 22 For two substances in the same phase, and with similar molar masses, the substance with the more complex molecular structure has the greater standard entropy.

18.3Entropy Changes in a System 23

18.3Entropy Changes in a System Standard Entropy, S° 24 In cases where an element exists in two or more allotropic forms, the form in which the atoms are more mobile has the greater entropy.

18.3Entropy Changes in a System Standard Entropy, S° 25

SAMPLE PROBLEM From the standard entropy values in Appendix 2, calculate the standard entropy changes for the following reactions at 25°C: Setup S°[CaCO 3 (s)] = 92.9 J/K · molS°[CaO(s)] = 39.8 J/K · mol S°[CO 2 (g)] = J/K · molS°[N 2 (g)] = J/K · mol S°[H 2 (g)] = J/K · molS°[NH 3 (g)] = J/K · mol S°[Cl 2 (g)] = J/K · molS°[HCl(g)] = J/K · mol

SAMPLE PROBLEM Solution

SAMPLE PROBLEM Solution

SAMPLE PROBLEM Solution

18.3Entropy Changes in a System Qualitatively Predicting the Sign of  S sys 30 Sometimes it’s useful just to know the sign of  S° rxn. Several processes that lead to an increase in entropy are Melting Vaporization or sublimation Temperature increase Reaction resulting in a greater number of gas molecules

18.3Entropy Changes in a System Qualitatively Predicting the Sign of  S sys 31

18.3Entropy Changes in a System Qualitatively Predicting the Sign of  S sys 32

SAMPLE PROBLEM For each process, determine the sign of  S for the system: (a)decomposition of CaCO 3 (s) to give CaO(s) and CO 2 (g) (b)heating bromine vapor from 45°C to 80°C (c)condensation of water vapor on a cold surface (d)reaction of NH 3 (g) and HCl(g) to give NH 4 Cl(s) (e)dissolution of sugar in water.

SAMPLE PROBLEM (a)decomposition of CaCO 3 (s) to give CaO(s) and CO 2 (g) (b)heating bromine vapor from 45°C to 80°C (c)condensation of water vapor on a cold surface (d)reaction of NH 3 (g) and HCl(g) to give NH 4 Cl(s) (e)dissolution of sugar in water. Setup Increases in entropy generally accompany solid-to-liquid, liquid-to-gas, and solid-to-gas transitions; the dissolving of one substance in another; a temperature increase; and reactions that increase the net number of moles of gas.

SAMPLE PROBLEM (a)decomposition of CaCO 3 (s) to give CaO(s) and CO 2 (g) (b)heating bromine vapor from 45°C to 80°C (c)condensation of water vapor on a cold surface (d)reaction of NH 3 (g) and HCl(g) to give NH 4 Cl(s) (e)dissolution of sugar in water. Solution (a) positive, (b) positive, (c) negative, (d) negative, and (e) positive.

Topics 18.4Entropy Changes in the Universe 36 Calculating  S surr The Second Law of Thermodynamics The Third Law of Thermodynamics

18.4Entropy Changes in the Universe Calculating  S surr 37

18.4Entropy Changes in the Universe The Second Law of Thermodynamics 38 The second law of thermodynamics says that for a process to be spontaneous as written (in the for- ward direction),  S univ must be positive. An equilibrium process is one that does not occur spontaneously in either the net forward or net reverse direction but can be made to occur by the addition or removal of energy to a system at equilibrium.

SAMPLE PROBLEM Determine if each of the following is a spontaneous process, a nonspontaneous process, or an equilibrium process at the specified temperature:

SAMPLE PROBLEM Solution

SAMPLE PROBLEM Solution

SAMPLE PROBLEM Solution

SAMPLE PROBLEM Solution  S univ is zero; therefore, the reaction is an equilibrium process at 98°C. In fact, this is the melting point of sodium.

18.4Entropy Changes in the Universe The Third Law of Thermodynamics 44 According to the third law of thermodynamics, the entropy of a perfect crystalline substance is zero at absolute zero.

18.4Entropy Changes in the Universe The Third Law of Thermodynamics 45

Topics 18.5Predicting Spontaneity 46 Gibbs Free-Energy Change,  G Standard Free-Energy Changes,  G° Using  G and  G° to Solve Problems

18.5Predicting Spontaneity Gibbs Free-Energy Change,  G 47

18.5Predicting Spontaneity Gibbs Free-Energy Change,  G 48

18.5Predicting Spontaneity Gibbs Free-Energy Change,  G 49  G < 0 The reaction is spontaneous in the forward direction (and nonspontaneous in the reverse direction).  G > 0 The reaction is nonspontaneous in the forward direction (and spontaneous in the reverse direction).  G = 0 The system is at equilibrium.

18.5Predicting Spontaneity Gibbs Free-Energy Change,  G 50

SAMPLE PROBLEM According to Table 18.4, a reaction will be spontaneous only at high temperatures if both  H and  S are positive. For a reaction in which  H = kJ/mol and  S = 476 J/K · mol, determine the temperature (in °C) above which the reaction is spontaneous. Setup

SAMPLE PROBLEM Solution

18.5Predicting Spontaneity Standard Free-Energy Changes,  G° 53 The standard free energy of reaction (  G° rxn ) is the free- energy change for a reaction when it occurs under standard- state conditions—that is, when reactants in their standard states are converted to products in their standard states.

18.5Predicting Spontaneity Standard Free-Energy Changes,  G° 54

18.5Predicting Spontaneity Standard Free-Energy Changes,  G° 55  G f ° is the standard free energy of formation of a compound—that is, the free-energy change that occurs when 1 mole of the com- pound is synthesized from its constituent elements, each in its standard state.

18.5Predicting Spontaneity Standard Free-Energy Changes,  G° 56

SAMPLE PROBLEM Calculate the standard free-energy changes for the following reactions at 25°C: Setup

SAMPLE PROBLEM Solution

SAMPLE PROBLEM Solution

18.5Predicting Spontaneity Using  G and  G° to Solve Problems 60

18.5Predicting Spontaneity Using  G and  G° to Solve Problems 61 Because  G° is a large positive number, the reaction does not favor product formation at 25°C (298 K).

18.5Predicting Spontaneity Using  G and  G° to Solve Problems 62 At T > 835°C,  G° becomes negative, indicating that the reaction would then favor the formation of CaO and CO 2.

18.5Predicting Spontaneity Using  G and  G° to Solve Problems 63 At the temperature at which a phase change occurs (i.e., the melting point or boiling point of a substance), the system is at equilibrium (  G = 0).

SAMPLE PROBLEM The molar heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid-to-liquid and liquid-to-vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C. Setup The melting point of benzene is = K and the boiling point is = K.

SAMPLE PROBLEM Solution

Topics 18.6Free Energy and Chemical Equilibrium 66 Relationship Between  G and  G° Relationship Between  G° and K

18.6Free Energy and Chemical Equilibrium Relationship Between  G and  G° 67

SAMPLE PROBLEM The equilibrium constant, KP, for the reaction is at 298 K, which corresponds to a standard free- energy change of 5.4 kJ/mol. In a certain experiment, the initial pressures are P[N 2 O 4 ] = atm and P[NO 2 ] = atm. Calculate  G for the reaction at these pressures, and predict the direction in which the reaction will proceed spontaneously to establish equilibrium.

SAMPLE PROBLEM Setup Solution Because  G is negative, the reaction proceeds spontaneously from left to right to reach equilibrium.

18.6Free Energy and Chemical Equilibrium Relationship Between  G° and K 70

18.6Free Energy and Chemical Equilibrium Relationship Between  G° and K 71

18.6Free Energy and Chemical Equilibrium Relationship Between  G° and K 72

SAMPLE PROBLEM Using data from Appendix 2, calculate the equilibrium constant, K P, for the following reaction at 25°C: Setup

SAMPLE PROBLEM Solution

SAMPLE PROBLEM The equilibrium constant, K sp, for the dissolution of silver chloride in water at 25°C, is 1.6 × 10 –10. Calculate  G° for the process. Setup

SAMPLE PROBLEM Solution

Topics 18.7Thermodynamics in Living Systems 77 Thermodynamics in Living Systems

18.7Thermodynamics in Living Systems 78 Coupled reactions:

18.7Thermodynamics in Living Systems 79 Coupled reactions play a crucial role in our survival.

18.7Thermodynamics in Living Systems 80

18.7Thermodynamics in Living Systems 81

18.7Thermodynamics in Living Systems 82