Chemistry Third Edition Julia Burdge Lecture Power Points

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

Thermodynamics Energy conversion Conversion of the chemical energy of fuels into heat energy of combustion, and this heat is in turn converted to mechanical or electrical work.

Aim f the this chapter - To understand laws of thermodynamics - How to use thermodynamic quantities to determine whether or a process is spontaneous (or energy efficient) or not?

CHAPTER 18 Entropy, Free Energy, and Equilibrium 18. 1 18. 2 18. 3 18. 4 18. 5 18. 6 18. 7 Spontaneous Processes Entropy Changes in a System Entropy Changes in the Universe Predicting Spontaneity Free Energy and Chemical Equilibrium Thermodynamics in Living Systems (Omitted) 4

18. 1 Spontaneous Processes Topics Spontaneous Processes 5

18. 1 Spontaneous Processes 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. 6

18. 1 Spontaneous Processes 7

18. 1 Spontaneous Processes These exothermic reactions are spontaneous at room temperature: 8

• Spontaneous Processes H 2(g) + 1/2 O 2(g) H 2 O(l) = -58. 5 k. J/mole This exothermic reaction is not spontaneous at temperatures above 0°C: - Enthalpy change - Entropy change

18. 1 Spontaneous Processes So, while exothermicity favors spontaneity, it cannot be the sole factor that determines spontaneity. The other factor which tells about spontaneity is Entropy (S). 10

What is Entropy? • Entropy (S): is a thermodynamic quantity that measures the disorder of a system. In general, a greater disorder means a greater entropy, and a greater order means a smaller entropy.

Melting of ice is explained in terms of entropy change Ordered Disordered H 2 O(S) H 2 O(l) = 6. 01 k. J/mole

18. 3 Entropy Changes in a System Topics Qualitatively Predicting the Sign of Ssys Determination of Standard Entropy, S° 13

Qualitatively Predicting the Sign of Ssys (Sometimes it’s useful just to know the sign of S°rxn).

We must ask few questions while predicting the sign of Entropy… If the volume is increasing and the system is becoming more disordered? Is the system more spreading? Is the system has more probability? If the answer is yes, the entropy of system is increasing. An increase in Entropy favors the reaction to occur!

18. 3 Entropy Changes in a System Qualitatively Predicting the Sign of Ssys 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 16

Volume change Increase in volume increases disorder, and hence entropy

Entropy Change in a System • One can predict qualitatively whether the entropy of a system increases (ΔS > 0) or decreases (ΔS < 0). (a) Melting: ΔSsys > 0 solid liquid Particles are confined to fixed positions. They are highly ordered. Became less ordered compared to the solid state. Ssolid < Sliquid

Entropy Change in a System (b) Vaporization: ΔSsys > 0 liquid gas Sliquid << Sgas Particles in the gaseous phase are highly disordered as compared to solid and liquid phases.

In general… S(s) < S(l) << S(g)

Entropy Change in a System (c) Dissolving: ΔSsys > 0 pure substance When the highly ordered particles of the pure substance and the ordered solvent molecules are mixed, the order in both are disrupted, and hence entropy increases. solution 18. 3

Entropy Change in a System Translational motion (d) Heating: Heating always increases Vibrational motion both the energy and the Rotational motion entropy of a system. ΔSsys > 0 Slower temp. < Shigher temp.

Entropy Change in a System (e) Chemical reactions: A reaction that results in an increase in the number of moles of gas always increases the entropy of the system. (g) (g) ΔSsys > 0 Sfewer moles < Smore moles

Smaller (f) Molar mass: Ssmaller molar mass < Slarger molar mass Bigger

18. 3 Molecular complexity (F 2 vs. O 3) Translational motion Vibrational motion Rotational motion 25

Predicting the Entropy Change of Dissolution Process (ΔSsoln) of Ions

18. 3 I. Solvation of nonionic compounds such as sugar. The dissolving process causes dispersal of the molecules into a larger volume - resulting in an increase in entropy 27

Entropy increases due to the dispersion of ionic species throughout the solution Entropy decreases due to the less mobility of water molecules

For example, when aluminum chloride dissolves in water, each Al 3+ ion becomes associated with six water molecules. Usually, metals bearing more charges decrease the entropy (due to decrease in entropy of water)

Important Trends • • Look at the physical state (S of gas >> liquids > solids). If the physical state is same, look for number of moles If number of moles are same, look for molar mass (higher molar mass – higher entropy) If phase, moles & molar mass are same, look for molecules with more complex structure (complicated structure will be more disordered)

Exercise For each process, determine the sign of S for the system: (a) decomposition of Ca. CO 3(s) to give Ca. O(s) and CO 2(g) > 0 (b) heating bromine vapor from 45°C to 80°C > 0 (c) condensation of water vapor on a cold surface < 0 (d) reaction of NH 3(g) and HCl(g) to give NH 4 Cl(s) < 0 (e) dissolution of sugar in water. > 0 31

Determine the sign of S for the system C 3 H 8(g) + 5 O 2(g) ==> 3 CO 2(g) + 4 H 2 O (g) > 0 O 3(g) O 2(g) < 0 2 O 3(g) 3 O 2(g) > 0 C (s) + 2 Cl 2 (g) 2 CCl 4 (g) > 0 2 F(g) F 2(g) < 0

What we learned in last class… • Spontaneous process • Entropy • How to determine the sign of entropy (physical states, number moles, molar mass, structure complexity)

What we hope to learn today… • Determination of Standard Entropy (S°) • The second law of Thermodynamics

Determination of Standard Entropy, S°

18. 3 Entropy Changes in a System Standard Entropy, S° 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. 36

18. 4 Entropy Changes in the Universe Calculating Ssys R = 8. 314 J/K · mol, n = mole, V = Volume. 37

18. 1 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 = 8. 314 J/K · mol, n = 1. 0 mole, Vfinal = 5. 0 L, and Vinitial = 2. 5 L. 38

18. 1 Solution 39

18. 3 Entropy Changes in a System Standard Entropy, S° Pay attention to the state of materials, e. g. , Br(l) = 152. 3 Br(g) = 245. 13 40

SAMPLE PROBLEM 18. 2 From the standard entropy values given below, calculate the standard entropy changes for the following reactions at 25°C: Setup S°[Ca. CO 3(s)] = 92. 9 J/K · mol S°[CO 2(g)] = 213. 6 J/K · mol S°[H 2(g)] = 131. 0 J/K · mol S°[Cl 2(g)] = 223. 0 J/K · mol S°[Ca. O(s)] = 39. 8 J/K · mol S°[N 2(g)] = 191. 5 J/K · mol S°[NH 3(g)] = 193. 0 J/K · mol S°[HCl(g)] = 187. 0 J/K · mol 41

SAMPLE PROBLEM 18. 2 Solution 42

SAMPLE PROBLEM 18. 2 Solution 43

SAMPLE PROBLEM 18. 2 Solution 44

SAMPLE PROBLEM - In reality, N 2 and H 2 react and form NH 3 (exothermic & spontaneous reaction, though slow) - Entropy is not the sole factor - We need to look beyond

18. 4 Entropy Changes in the Universe Topics The Second Law of Thermodynamics 46

18. 4 Entropy Changes in the Universe The Second Law of Thermodynamics Topics The second law of thermodynamics says that for a process to be spontaneous as written (in the forward direction), Suniv must be positive. ΔSuniv = ΔSsys + ΔSsurr > 0 “Spontaneous” 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. ΔSuniv = ΔSsys + ΔSsurr = 0 “Equilibrium” 47

18. 4 Entropy Changes in the Universe Topics § The universe is made up of two parts: the system and the surroundings System Reactants and products Surroundings Reaction container, room, and everything else ΔSuniv = ΔSsys + ΔSsurr 48

Entropy Change in the Surroundings (ΔSsurr) • Change in entropy of the surroundings is directly proportional to the enthalpy of the system. At constant temperature: – an exothermic process corresponds to an increase of the entropy of the surroundings. (ΔSsurr > 0) – an endothermic process corresponds to a decrease of the entropy of the surroundings. (ΔSsurr < 0) ΔSsurr α ‒ ΔHsys �Just think about it. Any system naturally tends to lower its energy (lose energy). This is in consistent with the second law of thermodynamics which states that the entropy of the universe is always positive. ①

18. 4 Entropy Changes in the Universe Calculating Ssurr ΔSuniv = ΔSsys + ΔSsurr > 0 ΔSuniv = ΔSsys + ΔSsurr < 0 ΔSuniv = ΔSsys + ΔSsurr = 0 “Spontaneous” “Non-spontaneous” “Equilibrium” 50

What we learned in previous lectures… • Spontaneous process • Entropy • How to determine the sign of entropy (physical states, number moles, molar mass, structure complexity) • Determination of Standard Entropy (S°) • The second law of Thermodynamics

What we hope to learn today… • • The second law of Thermodynamics Gibbs Free-Energy Change, G Standard Free-Energy Changes, G° Using G and G° to Solve Problems

(n. H°(products) – n. H°(reactant) ΔSuniv = ΔSsys + ΔSsurr > 0 “Spontaneous” ΔSuniv = ΔSsys + ΔSsurr < 0 “Non-spontaneous” ΔSuniv = ΔSsys + ΔSsurr = 0 “Equilibrium”

SAMPLE PROBLEM 18. 4 Determine if each of the following is a spontaneous process, a nonspontaneous process, or an equilibrium process at the specified temperature: 54

SAMPLE PROBLEM 18. 4 Solution ΔSuniv = ΔSsys + ΔSsurr > 0 55

SAMPLE PROBLEM 18. 4 Solution 56

SAMPLE PROBLEM 18. 4 Solution 57

SAMPLE PROBLEM 18. 4 Solution Suniv is zero; therefore, the reaction is an equilibrium process at 98°C. In fact, this is the melting point of sodium. 58

18. 5 Predicting Spontaneity Topics Gibbs Free-Energy Change, G Standard Free-Energy Changes, G° Using G and G° to Solve Problems ΔSuniv = ΔSsys + ΔSsurr > 0 “Spontaneous” 59

18. 5 Predicting Spontaneity Gibbs Free-Energy Change, G 60

18. 5 Predicting Spontaneity Gibbs Free-Energy Change, G In terms of system (reactants and products) - Can be used to predict spontaneity of reaction - Can predict the temperature for the reaction to be spontaneous - Can be used to determine the change in Entropy for a phase change 61

18. 5 Predicting Spontaneity Predict spontaneity of reaction 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. 62

18. 5 Predicting Spontaneity Gibbs Free-Energy Change, G 63

Also: It can be used to determine the temperature at which process would be spontaneous or non-spontaneous! - First we determine Temp. at equilibrium - Look at the sign of ΔH and ΔS - Decide Temp.

SAMPLE PROBLEM 18. 5 For a reaction in which H = 199. 5 k. J/mol and S = 476 J/K · mol, determine the temperature (in °C) above which the reaction is spontaneous. ΔG = ΔH - T ΔS At equilibrium, ΔG = 0 T = ΔH/ΔS 65

18. 5 Predicting Spontaneity SAMPLE PROBLEM Predict temperature for spontaneity 66

18. 5 Predicting Spontaneity SAMPLE PROBLEM Predict temperature for spontaneity Think about: If ΔS and ΔH were negative At T > 835°C, G° becomes negative, indicating that the reaction would then favor the formation of Ca. O and CO 2. 67

SAMPLE PROBLEM Also: It can be used to determine the change in Entropy for a phase change!

SAMPLE PROBLEM 18. 7 Calculate the change in entropy for a phase change The molar heats of fusion and vaporization of benzene are 10. 9 and 31. 0 k. J/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. 70

SAMPLE PROBLEM 18. 7 Setup The melting point of benzene is 5. 5 + 273. 15 = 278. 7 K and the boiling point is 80. 1 + 273. 15 = 353. 3 K. ΔG = ΔH - T ΔS At equilibrium, ΔG = 0 ΔS = ΔH/ T 71

SAMPLE PROBLEM Standard Free-Energy Changes, G° • The standard free energy (ΔG rxn) of a system is the change in free energy when reactants in their standard states are converted to products in their standard states. • Standard states are: – Gases – Liquids – Solids – Elements – Solutions 1 atm [ O 2(g) , CH 4(g) ] Pure liquid [ H 2 O (l) , C 2 H 5 OH (l) ] Pure solid [ Na(s) , Mg(s)] The most stable form at 1 atm and 25 C 1 M concentration

18. 5 Predicting Spontaneity SAMPLE PROBLEM Standard Free-Energy Changes, G° Gf ° is the standard free energy of formation of a compound —that is, the free-energy change that occurs when 1 mole of the compound is synthesized from its constituent elements, each in its standard state. 73

SAMPLE PROBLEM 18. 6 Calculate the standard free-energy changes for the following reactions at 25°C: Setup 75

SAMPLE PROBLEM 18. 6 Solution 76

SAMPLE PROBLEM 18. 6 Solution 77

SAMPLE PROBLEM SUMMARY (Gibbs Free Change) Note: will only tell - reactant or products are favored not spontaneity

18. 5 More practical aspect of Gibb’s free energy • The standard free energy (ΔG rxn) of a system is the change in free energy when reactants in their standard states are converted to products in their standard states. • Standard states are: – Gases – Liquids – Solids – Elements – Solutions 1 atm [ O 2(g) , CH 4(g) ] Pure liquid [ H 2 O (l) , C 2 H 5 OH (l) ] Pure solid [ Na(s) , Mg(s)] The most stable form at 1 atm and 25 C 1 M concentration 80

18. 6 Free Energy and Chemical Equilibrium Relationship Between G and G° Relationship Between G° and K 81

18. 6 Free Energy and Chemical Equilibrium Relationship Between G and G° • Exercise: Consider the reaction: H 2(g) + Cl 2(g) 2 HCl(g) How does the value of ΔG change when the pressures of the H 2, Cl 2 and HCl gases are changed to 0. 25 atm, 0. 45 atm and 0. 30 atm, respectively, at 25 C? (ΔG f for HCl(g) is ‒ 95. 27 k. J/mol) PRODUCT FAVORED Calculate ΔG. ΔG° = [ 2 ( ‒ 95. 27 k. J/mol)] ‒ [0 + 0] = ‒ 190. 54 k. J/mol Calculate Q.

ΔG = ΔG + RT ln. Q ΔG = ‒ 190540 J/mol + (8. 314 J/K·mol)(298 K) ln(0. 80) ΔG = ‒ 191. 09 k. J/mol The reaction becomes even more spontaneous at the given non standard states since ΔG is more negative than ΔG. Thus, more HCl will be formed with the given pressures. The reaction will continue producing more HCl and consuming more H 2 and Cl 2 until QP = KP. H 2(g) + Cl 2(g) 2 HCl(g) [HCl]2 [H 2][Cl 2]

18. 6 Relationship between ΔG and K At equilibrium: Q = K and ΔG = 0 Thus, at equilibrium: ΔG = ΔG + RT ln. Q 0 = ΔG + RT ln. K ΔG = ‒ RT ln. K The larger the value of K, the more negative ΔG becomes, and the reaction proceeds more towards the product side. K = Ka, Kb, Kp, Kc, Ksp, and so froth

SAMPLE PROBLEM 18. 9 Calculate the equilibrium constant, KP , for the following reaction at 25°C: ΔG = 474. 4 k. J/mol Setup 86

SAMPLE PROBLEM 18. 9 Solution 87

SAMPLE PROBLEM 18. 10 The equilibrium constant, Ksp, for the dissolution of silver chloride in water at 25°C, is 1. 6 × 10– 10. Calculate G° for the process. Setup 88

18. 4 Entropy Changes in the Universe SAMPLE PROBLEM The Third Law of Thermodynamics According to the third law of thermodynamics, the entropy of a perfect crystalline substance is zero at absolute zero. Na. Cl 89

18. 4 Entropy Changes in the Universe SAMPLE PROBLEM The Third Law of Thermodynamics 90