Chapter 16 Spontaneity, Entropy, and Free Energy Why do some things happen and others don’t?

Chapter 16Spontaneity, Entropy,

and Free Energy

Why do some things happen and others

don’t?

Chapter 16

Table of Contents

Copyright © Cengage Learning. All rights reserved 2

16.1Spontaneous Processes and Entropy

16.2 Entropy and the Second Law of Thermodynamics

16.3 The Effect of Temperature on Spontaneity

16.4 Free Energy

16.5 Entropy Changes in Chemical Reactions

16.6 Free Energy and Chemical Reactions

16.7 The Dependence of Free Energy on Pressure

16.8 Free Energy and Equilibrium

16.9 Free Energy and Work

Section 16.1

Spontaneous Processes and Entropy

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Objectives

• To compare kinetics with thermodynamics• To define spontaneity• To answer why some things happen and others

don’t.• To define Entropy• To assign positive or negative sign depending

on the process • To understand why a diverging diamond

interchange is safer than other designs

Section 16.1


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Thermodynamics vs. Kinetics

• Domain of Kinetics Rate of a reaction

depends on the pathway from reactants to products.

• Thermodynamics tells us whether a reaction is spontaneous based only on the properties of reactants and products.

Section 16.1


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• Thermodynamics lets us predict whether a process will occur but gives no information about the amount of time required for the process.

• A spontaneous process is one that occurs without outside intervention.

• Examples??

Section 16.1


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Section 16.1


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Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant.

What should happen to the gas when you open the valve?

Section 16.1


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The Expansion of An Ideal Gas Into an Evacuated Bulb

What is the driving force for the process? Why does it happen?

Section 16.1


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Entropy

• The driving force for a spontaneous process is an increase in the entropyentropy (S) of the universe.

• Entropy (S) is a measure of molecular randomness or disorderdisorder.

Section 16.1


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Entropy

• Thermodynamic function that describes the number of arrangements that are available to a system existing in a given state.

• Nature spontaneously proceeds toward the states that have the highesthighest probabilitiesprobabilities of existing.

• Is a deck of cards usually ordered or disordered?• There are so many ways for a deck of cards to be

disordered, but• Only one way for it to be ordered…

Section 16.1


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The Microstates That Give a Particular Arrangement (State)

Section 16.1


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Positional Entropy

• A gas expands into a vacuum because the expanded state has the highesthighest positionalpositional probabilityprobability of states available to the system.

• Therefore: Ssolid < Sliquid << Sgas

• Why do solutions form?• Why do ice cubes get smaller, even though they

are in the freezer?• If a process moves in a direction of greater

disorder or higher entropy, the sign is positive.

Section 16.1


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Concept Check

Predict the sign of S for each of the following, and explain:

a) The evaporation of alcohol

b) The freezing of water

c) Compressing an ideal gas at constant temperature

d) Heating an ideal gas at constant pressure

e) Dissolving NaCl in water

+

–

–

+

+

Section 16.1


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Objectives Review• To compare kinetics with thermodynamics• To define spontaneity• To answer why some things happen and others

don’t.• To define Entropy• To assign positive or negative sign depending

on the process • Why is a diverging diamond safer? #12• Work Session Page 782 # 1, 19, 23 (think IGL

and volume), 25, 33, 35

Section 16.2

Atomic MassesEntropy and the Second Law of Thermodynamics

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Objectives

• To define the Second Law of Thermodynamics• Wait, what was the first? I’m constantly

forgetting it…

Section 16.2


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Second Law of Thermodynamics

• In any spontaneous process there is always an increase in the entropy of the universe.

• What is the sign of a spontaneous process?• The entropy of the universe is increasing.• The total energyenergy of the universe is constant, but the

entropyentropy is increasing.

Suniverse = ΔSsystem + ΔSsurroundings

Section 16.2


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Second Law of Thermodynamics

Suniverse = ΔSsystem + ΔSsurroundings

• How can life exist if it requires cells to take small molecules to make bigger ones?

• Is this a spontaneous process? • Does S increase?• Is this consistent with the 2nd Law of

Thermodynamics?

• For a spontaneous process, how are ΔSsys & ΔSsurr related?

Section 16.2


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Objectives Review

• To define the Second Law of Thermodynamics• 1st Law of Thermodynamics:• The amount of energy in the Universe is constant• Work Session #8

Section 16.3

The Mole The Effect of Temperature on Spontaneity

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Objectives

• To predict spontaneity of a process based on ΔS and ΔH

• To understand the effect of temperature on Spontaneity

Section 16.3


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Which side is favored by Entropy?

• H2O (l) H2O (g) water is the system…

• 18 ml 31 L @1atm, 100C• Positional Probability?

• What is the sign of ΔSsystem?

• What direction is the heat flow? ΔHsystem?

• …what about the surroundings?• Positional Probability?

• What is the sign of ΔSsurr? Heat flow? ΔHsurr?

• What is the overall ΔS of the universe?

• ΔSuniv = ΔSsys + ΔSsurr

Section 16.3


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Concept Check

For the process A(l) A(s), which direction involves an increase in energy randomness(-ΔH)? Positional randomness(+ΔS)? Explain your answer.

As temperature increases/decreases , does energy or positional randomness take precedence? Why?

At what temperature is there a balance between energy randomness and positional randomness?

Section 16.3


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ΔSsurr = - ΔH T

• The sign of ΔSsurr depends on the direction of the heat flow. –ΔH, +ΔSsurr

• The magnitude of ΔSsurr depends on the temperature.

• Why does ice freeze, but only at lower temperatures? Balance between energy randomness and positional randomness.

• Liquid Solid

Section 16.3


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ΔSsurr = - ΔH enthalpy determines the sign T

Section 16.3


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ΔSsurr = - ΔH temperature determines the magnitude T

Heat flow (constant P) = change in enthalpy = ΔH

Section 16.3


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Interplay of Ssys and Ssurr in Determining the Sign of Suniv

ΔSuniv = ΔSsys + ΔSsurr

Section 16.3


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Objectives Review

• To predict spontaneity of a process based on ΔS and ΔH

• To understand the effect of temperature on Spontaneity

Section 16.4

Free Energy

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Objectives

• To understand Gibbs Free Energy (ΔG) in relation to ΔS and ΔH

• To predict spontaneity based on Gibbs Free Energy

• To conceptually and mathematically use Gibbs Equation

Section 16.4

Free Energy

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Free Energy (G)

• ΔG = ΔH – TΔS (at constant T and P)

• Negative ΔG means positive ΔSuniv

• positive ΔSuniv means Spontaneous!

• Negative ΔG means Spontaneous!• ΔG < 0 means spontaneous

Section 16.4

Free Energy

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How to determine spontaneity based on ΔS and ΔH …

For a process to be considered spontaneous, it must be both :

exothermic (favor energy randomness (-ΔH) and

Favor positional randomness (entropy (+ΔS))

ΔG = ΔH – TΔS

If the opposite of the above conditions exist, we can conclude that the process is not spontaneous.

Spontaneous Not Spontaneous Don’t Know Don’t Know

ΔH - + + -

ΔS + - + -

Section 16.4

Free Energy

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Describe the following as

spontaneous/non-spontaneous/cannot tell, and explain.

A reaction that is:

a) Exothermic and becomes more positionally random

Spontaneous

b) Exothermic and becomes less positionally random

Cannot tell

a) Endothermic and becomes more positionally random

Cannot tell

a) Endothermic and becomes less positionally random

Not spontaneous

Explain how temperature affects your answers.

Section 16.4

Free Energy

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Effect of ΔH and Δ S on SpontaneityΔG = ΔH – TΔS

H S Result

+ spontaneous at all temps

+ + spontaneous at high temps

spontaneous at low temps

+ not spontaneous at any temp

Section 16.4

Free Energy

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Concept Check

A liquid is vaporized at its boiling point. Predict the signs of:

w

q

H

S

Ssurr

G

Explain your answers.

–

+

+

+

–

0

Section 16.4

Free Energy

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Exercise

Predict the spontaneity of: H2O(s) H2O(l)

Given: ΔH = 6.03 X 103 J/mol ΔS = 22.1 J/K*mol

5-Step with Gibbs!

ΔG = ΔH – TΔS

Try at -10 oC, 0 oC, and 10 oC

-10 oC, ΔG = +2.2 X 102 J/mol

0 oC, ΔG = 0 J/mol

10 oC, ΔG = -2.2 X 102 J/mol

Section 16.4

Free Energy

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Exercise

The value of Hvaporization of substance X is 45.7 J/mol, and its normal boiling point is 72.5°C.

Calculate S, Ssurr, and G for the vaporization of one mole of this substance at 72.5°C and 1 atm.

ΔSsurr = - ΔH T

S = 132 J/K·mol

Ssurr = -132 J/K·mol

G = 0 kJ/mol

Section 16.4

Free Energy

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Exercise

Find the normal boiling point of liquid Br2.

Br2(l) Br2(g)

ΔH = 31.0 KJ/mol ΔS = 93.0 J/K*mol

5-Step with Gibbs!

ΔG = ΔH – TΔS

Section 16.4

Free Energy

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Spontaneous Reactions

Section 16.4

Free Energy

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Objectives Review

• To understand Gibbs Free Energy (ΔG) in relation to ΔS and ΔH

• To predict spontaneity based on Gibbs Free Energy

• To conceptually and mathematically use Gibbs Equation

• Work Session: 27, 29, 30, 31

Section 16.5

Entropy Changes in Chemical Reactions

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Objectives

• To predict the sign of ΔS for a given chemical reaction.

• To calculate ΔS°reaction given ΔS° values for products and reactants

• To calculate ΔS°reaction , ΔH°reaction , ΔG°reaction

Section 16.5


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ΔS changes in Chemical Reactions

• Entropy changes in chemical reactions are determined by positional probability.

• This means that the more molecules, the more possible configurations, the more disorder.

• Predict the sign of ΔS for the following reaction:

• H2 2H

• N2(g) + 3H2(g) 2NH3(g)

• 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

• CaCO3(s) CaO(s) + CO2(g)

• 2SOS(g) + O2(g) 2SO3(g)

Section 16.5


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Concept Check

Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The potential energy of AB is much greater than that of either reactant.

Predict the signs of:

H Ssurr Suniv

Explain.

+ - +

Section 16.5


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Third Law of Thermodynamics…there is a 3rd Law??

• So far, we’ve discussed the relative change in entropy, ΔS. Can we assign an absolute S?

• The entropy of a perfect crystal at 0 K is zero.• A perfect crystal represents the lowest possible

entropy.• Or in other words, everything is perfectly ordered, in

its place, just plain ducky….• Therefore, from 0 K, as the temperature increases, so

does the randomness of vibrational motion and, therefore, the entropy of the substance.

Section 16.5


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Third Law of Thermodynamics…there is a 3rd Law??

• Oh, yeah, …• Since a perfect crystal at absolute zero is an

unattainable ideal, we take the Third Law as a standard although we have never actually observed it…….

• You are welcome….

Section 16.5


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Standard Enthalpy Values (H°)

• Represent the increase in enthalpy of formation of a compound.

ΔH°reaction = ΣnpH°products – ΣnrH°reactants

n = moles (balanced coefficient)

H° = standard enthalpy (from a given list in problem # or on pages A20-A21)

See formula on AP reference sheet.

Section 16.5


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Standard Entropy Values (S°)

• Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure.

ΔS°reaction = ΣnpS°products – ΣnrS°reactants

n = moles (balanced coefficient)

S° = standard entropy (from a given list in problem # or on pages A20-A21)

See formula on AP reference sheet.

Section 16.5


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Exercise

Calculate Δ S° for the following reaction:

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

Given the following information:

S° (J/K·mol)

Na(s) 51

H2O(l) 70

NaOH(aq) 50

H2(g) 131

Δ S°= –11 J/K

Section 16.5


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Exercise


2NiS(s) + 3O2(g) → 2SO2(g) + 2NiO(s)


S° (J/K·mol)

SO2(g) 248

NiO(s) 38

O2(g) 205

NiS(s) 53

Δ S°= –149 J/K

Section 16.5


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Exercise


Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g)


S° (J/K·mol)

Al2O3(s) 51

H2(g) 131

Al(s) 28

H2O(g) 189

Δ S°= 179 J/K

Section 16.6

Free Energy and Chemical Reactions

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Standard Free Energy Change (ΔG°)• The change in free energy that will occur if the

reactants in their standard states are converted to the products in their standard states.

ΔG° = ΔH° – TΔS°

ΔG°reaction = ΣnpG°products – ΣnrG°reactants

ΔH°reaction = ΣnpH°products – ΣnrH°reactants

ΔS°reaction = ΣnpS°products – ΣnrS°reactants

A20-A21 for ΔS°, ΔH°, ΔG° data…

Section 16.6


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Concept Check

A stable diatomic molecule spontaneously forms from its atoms.

Predict the signs of:

H° S° G°

Explain.

– – –

Section 16.6


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Predict the sign of:H° S° G°

Explain

Section 16.5


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Objectives

• To predict the sign of ΔS for a given chemical reaction.

• To calculate ΔS°reaction given ΔS°values for products and reactants

• To calculate ΔS°reaction , ΔH°reaction , ΔG°reaction

• Work Session # 33, 34, 37, 39, 40, 41, 43, 45, 47, 51 (like Hess’s Law!) A20-A21 for ΔS°, ΔH°, ΔG° data…

Section 16.6


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Concept Check

Consider the following system at equilibrium at 25°C.

PCl3(g) + Cl2(g) PCl5(g)

G° = −92.50 kJ

What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCl3 if the temperature is raised? Explain. The ratio will decrease.

Section 16.7

The Dependence of Free Energy on Pressure

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Free Energy and Pressure

G = G° + RT ln(P)

or

ΔG = ΔG° + RT ln(Q)

Section 16.7


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Concept Check

Sketch graphs of:

1. G vs. P

2. H vs. P

3. ln(K) vs. 1/T (for both endothermic and exothermic cases)

Section 16.7


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The Meaning of ΔG for a Chemical Reaction

• A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.

Section 16.8

Free Energy and Equilibrium

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• The equilibrium point occurs at the lowest value of free energy available to the reaction system.

ΔG = 0 = ΔG° + RT ln(K)

ΔG° = –RT ln(K)

Section 16.8


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Change in Free Energy to Reach Equilibrium

Section 16.8


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Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction

Section 16.9

Free Energy and Work

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• Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy.

wmax = ΔG

Section 16.9

Free Energy and Work

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• Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy.

• All real processes are irreversible.• First law: You can’t win, you can only break

even.• Second law: You can’t break even.

Section 16.1


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Concept Check


b) Calculate H, E, q, and w for the process you described above.

All are equal to zero.

Section 16.1


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Concept Check


b) Given your answer to part a, what is the driving force for the process?

Entropy

Section 16.1


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The Microstates That Give a Particular Arrangement (State)

Section 16.4

Free Energy

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Free Energy (G)

• A process (at constant T and P) is spontaneous in the direction in which the free energy decreases. Negative ΔG means positive ΔSuniv.

univ = (at constant and ) G

S T PT

Documents

Chapter 16 Spontaneity, Entropy, and Free Energy Why do some things happen and others don’t?