Entropy, Free Energy, and Equilibrium

Entropy, Free Energy, and Equilibrium
Chemistry Third Edition Julia Burdge Lecture PowerPoints Chapter 18 Entropy, Free Energy, and Equilibrium Copyright © 2012, The McGraw-Hill Compaies, Inc. Permission required for reproduction or display. 1

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?

18 Entropy, Free Energy, and Equilibrium 18.1 Spontaneous Processes
18.2 Entropy 18.3 Entropy Changes in a System 18.4 Entropy Changes in the Universe 18.5 Predicting Spontaneity 18.6 Free Energy and Chemical Equilibrium 18.7 Thermodynamics in Living Systems (Omitted) Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4

Spontaneous Processes
18.1 Spontaneous Processes Spontaneous Processes Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 5

18.1 Spontaneous Processes 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. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6

18.1 Spontaneous Processes Spontaneous Processes These exothermic reactions are spontaneous at room temperature: Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 8

H2(g) + H2O(l) 1/2O2(g) = kJ/mole This exothermic reaction is not spontaneous at temperatures above 0°C: Enthalpy change Entropy change

18.1 Spontaneous Processes 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). Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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 H2O(S) H2O(l) = 6.01 kJ/mole

Entropy Changes in a System
18.3 Entropy Changes in a System Qualitatively Predicting the Sign of Ssys Determination of Standard Entropy, S° Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 13

(Sometimes it’s useful just to know the sign of S°rxn).
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!

Qualitatively Predicting the Sign of Ssys
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 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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

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

In general… S(s) < S(l) << S(g)

18.3 (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

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

(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) (g) ΔSsys > 0 Sfewer moles < Smore moles

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

18.3 Standard Entropy, S° Molecular complexity (F2 vs. O3)
Translational motion Vibrational motion Rotational motion Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 25

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

I. Solvation of nonionic compounds such as sugar.
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

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 Al3+ 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:
decomposition of CaCO3(s) to give CaO(s) and CO2(g) heating bromine vapor from 45°C to 80°C condensation of water vapor on a cold surface reaction of NH3(g) and HCl(g) to give NH4Cl(s) dissolution of sugar in water. > 0 > 0 < 0 < 0 > 0 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 31

C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O (g)
Determine the sign of S for the system C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O (g) > 0 O3(g) O2(g) < 0 2O3(g) 3O2(g) > 0 C (s) + 2 Cl2 (g) CCl4 (g) > 0 2F(g) F2(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. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 36

Entropy Changes in the Universe
18.4 Entropy Changes in the Universe Calculating Ssys R = J/K · mol, n = mole, V = Volume. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 37

R = 8.314 J/K · mol, n = 1.0 mole, Vfinal = 5.0 L, and
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 = J/K · mol, n = 1.0 mole, Vfinal = 5.0 L, and Vinitial = 2.5 L. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 38

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) = Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 40

S°[H2(g)] = 131.0 J/K · mol S°[NH3(g)] = 193.0 J/K · mol
18.2 From the standard entropy values given below, calculate the standard entropy changes for the following reactions at 25°C: Setup S°[CaCO3(s)] = 92.9 J/K · mol S°[CaO(s)] = 39.8 J/K · mol S°[CO2(g)] = J/K · mol S°[N2(g)] = J/K · mol S°[H2(g)] = J/K · mol S°[NH3(g)] = J/K · mol S°[Cl2(g)] = J/K · mol S°[HCl(g)] = J/K · mol Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 41

- In reality, N2 and H2 react and form NH3 (exothermic & spontaneous reaction, though slow)
- Entropy is not the sole factor - We need to look beyond

The Second Law of Thermodynamics
18.4 Entropy Changes in the Universe The Second Law of Thermodynamics Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 46

18.4 Entropy Changes in the Universe The Second Law of Thermodynamics The second law of thermodynamics says that for a process to be spontaneous as written (in the forward direction), Suniv 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. ΔSuniv = ΔSsys + ΔSsurr > 0 “Spontaneous” ΔSuniv = ΔSsys + ΔSsurr = 0 “Equilibrium” Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 47

18.4 Entropy Changes in the Universe 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

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 ①  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.

18.4 Entropy Changes in the Universe Calculating Ssurr ΔSuniv = ΔSsys + ΔSsurr > 0 “Spontaneous” ΔSuniv = ΔSsys + ΔSsurr < 0 “Non-spontaneous” ΔSuniv = ΔSsys + ΔSsurr = 0 “Equilibrium” Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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

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

18.4 Determine if each of the following is a spontaneous process, a nonspontaneous process, or an equilibrium process at the specified temperature: Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 54

ΔSuniv = ΔSsys + ΔSsurr > 0
18.4 Solution ΔSuniv = ΔSsys + ΔSsurr > 0 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 55

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. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 58

Predicting Spontaneity
18.5 Predicting Spontaneity Gibbs Free-Energy Change, G Standard Free-Energy Changes, G° Using G and G° to Solve Problems ΔSuniv = ΔSsys + ΔSsurr > 0 “Spontaneous” Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 59

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 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 62

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.

For a reaction in which H = 199.5 kJ/mol and
18.5 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. ΔG = ΔH - T ΔS At equilibrium, ΔG = 0 T = ΔH/ΔS Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 65

18.5 Predicting Spontaneity 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 CaO and CO2. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 67

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

18.5 Predicting Spontaneity Calculate the change in entropy for a phase change 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). Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 69

At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.
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 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. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 70

18.7 ΔG = ΔH - T ΔS At equilibrium, ΔG = 0 ΔS = ΔH/ T Setup
The melting point of benzene is = K and the boiling point is = K. ΔG = ΔH - T ΔS At equilibrium, ΔG = 0 ΔS = ΔH/ T Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 71

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 1 atm [ O2(g) , CO2(g) , CH4(g) ] Liquids Pure liquid [ H2O (l) , C2H5OH (l) ] Solids Pure solid [ Na(s) , Mg(s)] Elements The most stable form at 1 atm and 25C Solutions 1 M concentration

18.5 Predicting Spontaneity 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. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 73

Calculate ΔG for the below reaction?
Given below, 2NO2(g) NO(g) + O2(g) 2H2(g) + O2(g) H2O(g) ΔG = 20 kJ ΔG = -5 kJ Calculate ΔG for the below reaction? NO(g) H2(g) + 3/2O2(g) NO2(g) + 2H2O Answer -ΔG = -15 kJ

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

More practical aspect of Gibb’s free energy
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 1 atm [ O2(g) , CO2(g) , CH4(g) ] Liquids Pure liquid [ H2O (l) , C2H5OH (l) ] Solids Pure solid [ Na(s) , Mg(s)] Elements The most stable form at 1 atm and 25C Solutions 1 M concentration Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 80

Free Energy and Chemical Equilibrium
18.6 Free Energy and Chemical Equilibrium Relationship Between G and G° Relationship Between G° and K Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 81

18.6 Free Energy and Chemical Equilibrium Relationship Between G and G° Exercise: Consider the reaction: H2(g) + Cl2(g) HCl(g) How does the value of ΔG change when the pressures of the H2, Cl2 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 ‒ kJ/mol) PRODUCT FAVORED Calculate ΔG. ΔG° = [ 2 ( ‒ kJ/mol)] ‒ [0 + 0] = ‒ kJ/mol Calculate Q . Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

ΔG = ΔG + RT lnQ ΔG = ‒ J/mol + (8.314 J/K·mol)(298 K) ln(0.80) ΔG = ‒ kJ/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 H2 and Cl2 until QP = KP . [HCl]2 [H2][Cl2] H2(g) + Cl2(g)  2HCl(g)

Relationship between ΔG and K
18.6 Relationship between ΔG and K At equilibrium: Q = K and ΔG = 0 Thus, at equilibrium: ΔG = ΔG + RT lnQ 0 = ΔG + RT lnK The larger the value of K, the more negative ΔG becomes, and the reaction proceeds more towards the product side. ΔG = ‒ RT lnK K = Ka, Kb, Kp, Kc, Ksp, and so froth Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

18.9 Calculate the equilibrium constant, KP , for the following reaction at 25°C: Setup ΔG = kJ/mol Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 86

is 1.6 × 10–10. Calculate G° for the process. Setup
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 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 88

18.4 Entropy Changes in the Universe The Third Law of Thermodynamics According to the third law of thermodynamics, the entropy of a perfect crystalline substance is zero at absolute zero. NaCl Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 89

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