chapter 19 principles of reactivity: entropy and free energy

John C. Kotz • State University of New York, College at Oneonta

John C. KotzPaul M. TreichelJohn Townsend

http://academic.cengage.com/kotz

Chapter 19Principles of Reactivity: Entropy and Free Energy





Important – Read Before Using Slides in Class

Instructor: This PowerPoint presentation contains photos and figures from the text, as well as selected animations and videos. For animations and videos to run properly, we recommend that you run this PowerPoint presentation from the PowerLecture disc inserted in your computer. Also, for the mathematical symbols to display properly, you must install the supplied font called “Symb_chm,” supplied as a cross-platform TrueType font in the “Font_for_Lectures” folder in the "Media" folder on this disc.

If you prefer to customize the presentation or run it without the PowerLecture disc inserted, the animations and videos will only run properly if you also copy the associated animation and video files for each chapter onto your computer. Follow these steps:

1. Go to the disc drive directory containing the PowerLecture disc, and then to the “Media” folder, and then to the

“PowerPoint_Lectures” folder. 2. In the “PowerPoint_Lectures” folder, copy the entire chapter

folder to your computer. Chapter folders are named “chapter1”, “chapter2”, etc. Each chapter folder contains the PowerPoint Lecture file as well as the animation and video files.

For assistance with installing the fonts or copying the animations and video files, please visit our Technical Support at http://academic.cengage.com/support or call (800) 423-0563. Thank you.

http://academic.cengage.com/support

3333

© 2009 Brooks/Cole - Cengage

Entropy and Free Entropy and Free EnergyEnergy

Entropy and Free Entropy and Free EnergyEnergy

How to predict if a How to predict if a reaction can occur, reaction can occur, given enough time?given enough time?

THERMODYNAMICSTHERMODYNAMICS

How to predict if How to predict if a reaction can a reaction can occur at a occur at a reasonable rate?reasonable rate?

KINETICSKINETICS

PLAY MOVIE

PLAY MOVIE

4444


ThermodynamicsThermodynamics• If the state of a chemical system is such that a If the state of a chemical system is such that a

rearrangement of its atoms and molecules would rearrangement of its atoms and molecules would

decrease the energy of the system--- decrease the energy of the system---

• ANDAND the K is greater than 1, the K is greater than 1,

• then this is a then this is a product-favoredproduct-favored system.system.

• Most product-favored reactions are Most product-favored reactions are

exothermicexothermic

—but this is not the only criterion—but this is not the only criterion

5555


ThermodynamicsThermodynamics• Both product- and reactant-favored reactions can Both product- and reactant-favored reactions can

proceed to equilibrium in a proceed to equilibrium in a spontaneousspontaneous process.process.

AgCl(s) AgCl(s) ee Ag Ag++(aq) + Cl(aq) + Cl––(aq) (aq) K = 1.8 x 10K = 1.8 x 10-10-10

Reaction is not product-favored, but it moves Reaction is not product-favored, but it moves spontaneously toward equilibrium.spontaneously toward equilibrium.

• Spontaneous does not imply anything about time Spontaneous does not imply anything about time for reaction to occur.for reaction to occur.

6666


Thermodynamics and Thermodynamics and KineticsKinetics

Thermodynamics and Thermodynamics and KineticsKinetics

Diamond is Diamond is thermodynamically thermodynamically favoredfavored to convert to to convert to graphite, but graphite, but not not kinetically favoredkinetically favored..

Paper burns — a Paper burns — a product-favoredproduct-favored reaction. reaction. Also Also kinetically favoredkinetically favored once once reaction is begun.reaction is begun.

PLAY MOVIE

7777


Spontaneous ReactionsSpontaneous Reactions

In general, spontaneous In general, spontaneous reactions are reactions are exothermicexothermic..

FeFe22OO33(s) + 2 Al(s) (s) + 2 Al(s) ff

2 Fe(s) + Al2 Fe(s) + Al22OO33(s)(s)

∆∆rrH = - 848 kJH = - 848 kJ

8888


Spontaneous ReactionsSpontaneous ReactionsBut many spontaneous reactions or But many spontaneous reactions or

processes are endothermic or even processes are endothermic or even have ∆H = 0.have ∆H = 0.

NHNH44NONO33(s) + heat (s) + heat ff NH NH44NONO33(aq)(aq)∆∆H = 0H = 0

PLAY MOVIE

9999


Entropy, SEntropy, SOne property common to spontaneous One property common to spontaneous

processes is that the energy of the processes is that the energy of the final state is more final state is more disperseddispersed..

In a spontaneous process energy goes In a spontaneous process energy goes from being more concentrated to from being more concentrated to being more dispersed.being more dispersed.

The thermodynamic property related to The thermodynamic property related to energy dispersal is energy dispersal is ENTROPY, S.ENTROPY, S.

2nd Law of Thermo — a 2nd Law of Thermo — a spontaneous process results in spontaneous process results in an increase in the entropy of an increase in the entropy of the universe. the universe.

Reaction of K Reaction of K with waterwith water

10101010


Probability suggests that a spontaneous Probability suggests that a spontaneous reaction will result in the dispersal of reaction will result in the dispersal of energy.energy.

Energy DispersalEnergy Dispersal

Directionality of Directionality of ReactionsReactions

PLAY MOVIE

11111111



Energy DispersalEnergy Dispersal

Exothermic reactions involve a release of Exothermic reactions involve a release of stored chemical potential energy to the stored chemical potential energy to the surroundings. surroundings.

The stored potential energy starts out in a few The stored potential energy starts out in a few molecules but is finally dispersed over a molecules but is finally dispersed over a great many molecules. great many molecules.

The final state—with energy dispersed—is The final state—with energy dispersed—is more probable and makes a reaction more probable and makes a reaction spontaneous.spontaneous.

12121212


Energy DispersalEnergy DispersalTo begin, particle 1 has 2 packets of energy and 2-4 have none To begin, particle 1 has 2 packets of energy and 2-4 have none ((upper leftupper left). With time it is more probable energy is dispersed ). With time it is more probable energy is dispersed over two particles. Each of these ways to distribute energy is over two particles. Each of these ways to distribute energy is called a called a microstatemicrostate. .

BeginBeginBeginBegin

See Figure 19.4See Figure 19.4

Energy Energy over 2 over 2 particlesparticles

13131313


Matter & energy dispersalMatter & energy dispersal


As the size of the As the size of the container increases, container increases, the number of the number of microstates accessible microstates accessible to the system to the system increases, and the increases, and the density of states density of states increases. Entropy increases. Entropy increases.increases.

PLAY MOVIE

14141414


• The entropy of liquid water is greater than the entropy of solid water (ice) at 0˚ C.

• Energy is more dispersed in liquid water than in solid water.

15151515


S (solids) < S (liquids) < S (gases)S (solids) < S (liquids) < S (gases)

SSoo (J/K•mol) (J/K•mol)

HH22O(liq)O(liq) 69.9569.95

HH22O(gas)O(gas) 188.8 188.8


HH22O(liq)O(liq) 69.9569.95

HH22O(gas)O(gas) 188.8 188.8

Entropy, SEntropy, S

Energy dispersalEnergy dispersal

PLAY MOVIE

16161616


Entropy and States of Entropy and States of MatterMatter

S˚(BrS˚(Br22 liq) < S˚(Br liq) < S˚(Br22 gas) gas) S˚(HS˚(H22O sol) < S˚(HO sol) < S˚(H22O liq)O liq)

17171717


Entropy of a substance increases Entropy of a substance increases with temperature.with temperature.

Molecular motions Molecular motions of heptane, Cof heptane, C77HH1616

Molecular motions of Molecular motions of heptane at different temps.heptane at different temps.


PLAY MOVIEPLAY MOVIE

18181818


Increase in molecular Increase in molecular complexity generally complexity generally leads to increase in S.leads to increase in S.


PLAY MOVIE

19191919


Entropies of ionic solids depend on Entropies of ionic solids depend on coulombic attractions.coulombic attractions.


MgOMgO 26.926.9

NaFNaF 51.551.5


MgOMgO 26.926.9

NaFNaF 51.551.5


MgMg2+2+ & O & O2-2- NaNa++ & F & F--

PLAY MOVIE

20202020


Liquids or solids dissolve in a solvent in a Liquids or solids dissolve in a solvent in a spontaneous process owing to the increase spontaneous process owing to the increase in entropy. Matter (and energy) are more in entropy. Matter (and energy) are more dispersed. dispersed.

Entropy, SEntropy, SEntropy, SEntropy, S

21212121


Standard Molar EntropiesStandard Molar Entropies

22222222


Entropy Changes for Phase Entropy Changes for Phase ChangesChanges

For a phase change, For a phase change, ∆S = q/T∆S = q/T

where q = heat transferred in where q = heat transferred in phase changephase change

For HFor H22O (liq) O (liq) ff H H22O(g)O(g)

∆∆H = q = +40,700 J/molH = q = +40,700 J/mol

∆S =

q

T =

40,700 J/mol

373.15 K = +109 J/K•mol

∆S =

q

T =

40,700 J/mol

373.15 K = +109 J/K•mol

PLAY MOVIE

23232323


Entropy and TemperatureEntropy and TemperatureEntropy and TemperatureEntropy and Temperature

S increases S increases slightly with Tslightly with T

S increases a S increases a large amount large amount with phase with phase changeschanges

24242424


Consider 2 HConsider 2 H22(g) + O(g) + O22(g) (g) ff 2 H 2 H22O(liq)O(liq)

∆∆SSoo = 2 S = 2 Soo (H (H22O) - [2 SO) - [2 Soo (H (H22) + S) + Soo (O (O22)])]

∆∆SSoo = 2 mol (69.9 J/K = 2 mol (69.9 J/K··mol) - mol) - [2 mol (130.7 J/K[2 mol (130.7 J/K··mol) + mol) +

1 mol (205.3 J/K1 mol (205.3 J/K··mol)]mol)]

∆∆SSoo = -326.9 J/K = -326.9 J/K

Note that there is a Note that there is a decrease in Sdecrease in S because 3 because 3 mol of gas give 2 mol of liquid.mol of gas give 2 mol of liquid.

Calculating ∆S for a Calculating ∆S for a ReactionReaction

∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)

25252525


2nd Law of 2nd Law of ThermodynamicsThermodynamics

A reaction is spontaneous if ∆S for the A reaction is spontaneous if ∆S for the universeuniverse is is positive.positive.

∆∆SSuniverseuniverse = ∆S = ∆Ssystemsystem + ∆S + ∆Ssurroundingssurroundings

∆∆SSuniverseuniverse > 0 for spontaneous process > 0 for spontaneous process

Calculate the entropy created by energy dispersal Calculate the entropy created by energy dispersal in the system and surroundings. in the system and surroundings.

26262626


Dissolving NH4NO3 in water—an entropy driven process.


∆∆SSuniverseuniverse = =

∆∆SSsystemsystem + ∆S + ∆Ssurroundingssurroundings

PLAY MOVIE

PLAY MOVIE

27272727


2 H2 H22(g) + O(g) + O22(g) (g) ff 2 H 2 H22O(liq)O(liq)

∆∆SSoosystemsystem = -326.9 J/K = -326.9 J/K

ΔSosurroundings =

qsurroundingsT

= -ΔHsystem

T ΔSosurroundings =

qsurroundingsT

= -ΔHsystem

T

ΔSosurroundings =

- (-571.7 kJ )(1000 J /kJ )298.15 K

ΔSosurroundings = - (-571.7 kJ )(1000 J /kJ )

298.15 K


∆∆SSoosurroundingssurroundings = +1917 J/K = +1917 J/K

Can calc. that ∆Can calc. that ∆rrHHoo = ∆H = ∆Hoosystemsystem = -571.7 kJ = -571.7 kJ

28282828


2 H2 H22(g) + O(g) + O22(g) (g) ff 2 H 2 H22O(liq)O(liq)

∆∆SSoosystemsystem = -326.9 J/K = -326.9 J/K

∆∆SSoosurroundingssurroundings = +1917 J/K = +1917 J/K

∆∆SSoouniverse universe = +1590. J/K= +1590. J/K

The entropy of the The entropy of the universe is universe is increasing, so increasing, so the reaction is the reaction is product-favored.product-favored.


29292929


Spontaneous or Not?Spontaneous or Not?

Remember that –∆H˚Remember that –∆H˚syssys is proportional to ∆S˚ is proportional to ∆S˚surrsurr

An exothermic process has ∆S˚An exothermic process has ∆S˚surrsurr > 0. > 0.

Remember that –∆H˚Remember that –∆H˚syssys is proportional to ∆S˚ is proportional to ∆S˚surrsurr

An exothermic process has ∆S˚An exothermic process has ∆S˚surrsurr > 0. > 0.

30303030


Gibbs Free Energy, GGibbs Free Energy, G

Multiply through by -TMultiply through by -T

-T∆S-T∆Sunivuniv = ∆H = ∆Hsyssys - T∆S - T∆Ssyssys

-T∆S-T∆Sunivuniv = change in Gibbs free energy = change in Gibbs free energy for the system = ∆Gfor the system = ∆Gsystemsystem

Under Under standard conditionsstandard conditions — —

∆∆GGoosyssys = ∆H = ∆Hoo

syssys - T∆S - T∆Soosyssys

ΔSuniv =

−ΔHsys

T + ΔSsys

∆∆SSunivuniv = ∆S = ∆Ssurrsurr + ∆S + ∆Ssyssys

J. Willard GibbsJ. Willard Gibbs1839-19031839-1903

31313131


∆∆GGoo = ∆H = ∆Hoo - T∆S - T∆Soo

Gibbs Gibbs free energyfree energy change = change =

total energy change for system total energy change for system

- energy lost in energy dispersalenergy lost in energy dispersal

If reaction isIf reaction is

•• exothermic (negative ∆Hexothermic (negative ∆Hoo) )

•• and entropy increases (positive ∆Sand entropy increases (positive ∆Soo) )

•• then ∆Gthen ∆Goo must be NEGATIVE must be NEGATIVE

• reaction is spontaneous (and product-favored).

32323232


∆∆GGoo = ∆H = ∆Hoo - T∆S - T∆Soo

Gibbs free energy change = Gibbs free energy change =

total energy change for system total energy change for system

- energy lost in energy dispersalenergy lost in energy dispersalIf reaction isIf reaction is

•• endothermic (positive ∆Hendothermic (positive ∆Hoo))•• and entropy decreases (negative ∆Sand entropy decreases (negative ∆Soo))•• then then ∆G∆Goo must be must be POSITIVEPOSITIVE •• reaction is reaction is not spontaneousnot spontaneous (and is (and is reactant-favored).reactant-favored).

33333333



∆ ∆GGoo = ∆H = ∆Hoo - T∆S - T∆Soo

∆∆HHoo ∆S∆Soo ∆G∆Goo ReactionReaction

exo(–)exo(–) increase(+)increase(+) –– Prod-favoredProd-favored

endo(+)endo(+) decrease(-)decrease(-) ++ React-favoredReact-favored

exo(–)exo(–) decrease(-)decrease(-) ?? T dependentT dependent

endo(+)endo(+) increase(+)increase(+) ?? T dependentT dependent

34343434




Two methods of calculating ∆GTwo methods of calculating ∆Goo

a)a) Determine ∆ Determine ∆rrHHo o and ∆and ∆rrSSoo and use Gibbs equation. and use Gibbs equation.

b) Use tabulated values of b) Use tabulated values of free energies of free energies of

formation, ∆formation, ∆ffGGoo..

∆∆rrGGo o = = ∆ ∆ffGGoo (products) - (products) - ∆ ∆ffGGoo (reactants) (reactants)∆∆rrGGo o = = ∆ ∆ffGGoo (products) - (products) - ∆ ∆ffGGoo (reactants) (reactants)

35353535


Free Energies of Free Energies of FormationFormation

Note that ∆Note that ∆ffG˚ for an element = 0G˚ for an element = 0

36363636


Calculating ∆Calculating ∆rrGGoo

Combustion of acetyleneCombustion of acetylene

CC22HH22(g) + 5/2 O(g) + 5/2 O22(g) (g) ff 2 CO 2 CO22(g) + H(g) + H22O(g)O(g)

Use enthalpies of formation to calculateUse enthalpies of formation to calculate

∆∆rrHHoo = -1238 kJ = -1238 kJ

Use standard molar entropies to calculateUse standard molar entropies to calculate

∆∆rrSSoo = -97.4 J/K or = -97.4 J/K or -0.0974 kJ/K-0.0974 kJ/K

∆∆rrGGoo = -1238 kJ - (298 K)(-0.0974 kJ/K) = -1238 kJ - (298 K)(-0.0974 kJ/K)

= = -1209 kJ-1209 kJ

Reaction is Reaction is product-favoredproduct-favored in spite of negative ∆ in spite of negative ∆rrSSoo. .

Reaction is Reaction is “enthalpy driven”“enthalpy driven”

37373737



Is the dissolution of ammonium nitrate product-Is the dissolution of ammonium nitrate product-favored? favored?

If so, is it enthalpy- or entropy-driven?If so, is it enthalpy- or entropy-driven?

NHNH44NONO33(s) + heat (s) + heat ff NH NH44NONO33(aq)(aq)

PLAY MOVIE

38383838



From tables of thermodynamic data we findFrom tables of thermodynamic data we find

∆∆rrHHoo = +25.7 kJ = +25.7 kJ

∆∆rrSSoo = +108.7 J/K or +0.1087 kJ/K = +108.7 J/K or +0.1087 kJ/K

∆∆rrGGoo = +25.7 kJ - (298 K)(+0.1087 J/K) = +25.7 kJ - (298 K)(+0.1087 J/K)

= -6.7 kJ= -6.7 kJ

Reaction is Reaction is product-favoredproduct-favored in spite of negative in spite of negative ∆∆rrHHoo. .

Reaction is Reaction is “entropy driven”“entropy driven”

NHNH44NONO33(s) + heat (s) + heat ff NH NH44NONO33(aq)(aq)

39393939




Two methods of calculating ∆GTwo methods of calculating ∆Goo

a) Determine ∆a) Determine ∆rrHHoo and ∆ and ∆rrSSoo and use Gibbs equation. and use Gibbs equation.

b) Use tabulated values of b) Use tabulated values of free energies of free energies of

formation, ∆formation, ∆ffGGoo..

∆∆rrGGo o = = ∆ ∆ffGGoo (products) - (products) - ∆ ∆ffGGoo (reactants) (reactants)∆∆rrGGo o = = ∆ ∆ffGGoo (products) - (products) - ∆ ∆ffGGoo (reactants) (reactants)

40404040


Calculating ∆GCalculating ∆Goorxnrxn

Combustion of carbonCombustion of carbon

C(graphite) + OC(graphite) + O22(g) (g) ff CO CO22(g) (g)

∆∆rrGGoo = ∆ = ∆ffGGoo(CO(CO22) - [∆) - [∆ffGGoo(graph) + ∆(graph) + ∆ffGGoo(O(O22)])]

∆∆rrGGoo = -394.4 kJ - [ 0 + 0] = -394.4 kJ - [ 0 + 0]

Note that free energy of formation of an element Note that free energy of formation of an element in its standard state is 0.in its standard state is 0.

∆∆rrGGoo = -394.4 kJ = -394.4 kJ

Reaction is Reaction is product-favoredproduct-favored as expected. as expected.

∆∆rrGGoo = = ∆G ∆Gffoo (products) - (products) - ∆G ∆Gff

oo (reactants) (reactants)∆∆rrGGoo = = ∆G ∆Gffoo (products) - (products) - ∆G ∆Gff

oo (reactants) (reactants)

41414141


Free Energy and Free Energy and TemperatureTemperature

2 Fe2 Fe22OO33(s) + 3 C(s) (s) + 3 C(s) ff 4 Fe(s) + 3 CO 4 Fe(s) + 3 CO22(g)(g)

∆∆rrHHo o = +467.9 kJ= +467.9 kJ ∆∆rrSSoo = +560.3 J/K = +560.3 J/K

∆∆rrGGoo = +300.8 kJ = +300.8 kJ

Reaction is Reaction is reactant-favoredreactant-favored at 298 K at 298 K

At what T does ∆At what T does ∆rrGGoo just change from being just change from being (+) to being (-)? (+) to being (-)?

When ∆When ∆rrGGoo = 0 = ∆ = 0 = ∆rrHHo o - T∆- T∆rrSSoo

T = ΔrHΔrS

= 467.9 kJ

0.5603 kJ /K = 835.1 K

T = ΔrHΔrS

= 467.9 kJ

0.5603 kJ /K = 835.1 K

42424242


FACT: ∆FACT: ∆rrGGoo is the change in free energy is the change in free energy

when pure reactants convert COMPLETELY when pure reactants convert COMPLETELY

to pure products.to pure products.

FACT: Product-favored systems have FACT: Product-favored systems have

KKeqeq > 1. > 1.

Therefore, both ∆Therefore, both ∆rrG˚ and KG˚ and Keqeq are related to are related to

reaction favorability.reaction favorability.

Thermodynamics and KThermodynamics and Keqeq

43434343


KKeqeq is related to reaction favorability and so is related to reaction favorability and so to ∆to ∆rrGGoo..

The larger the value of K the more negative The larger the value of K the more negative the value of ∆the value of ∆rrGGoo

∆∆rrGGoo = - RT lnK = - RT lnKwhere R = 8.31 J/K•molwhere R = 8.31 J/K•mol


44444444


Calculate K for the reactionCalculate K for the reaction

NN22OO44 ff 2 NO 2 NO22 ∆∆rrGGoo = +4.8 kJ = +4.8 kJ

∆∆rrGGoo = +4800 J = - (8.31 J/K)(298 K) ln K = +4800 J = - (8.31 J/K)(298 K) ln K

∆∆rrGGoo = - RT lnK = - RT lnK

ln K = -

4800 J

(8.31 J/K)(298K) = -1.94


K = 0.14K = 0.14When ∆When ∆rrGGoo > 0, then K < 1 > 0, then K < 1

45454545


∆∆G, ∆G˚, and KG, ∆G˚, and Keqeq

• ∆∆G is change in free energy at non-G is change in free energy at non-standard conditions.standard conditions.

• ∆∆G is related to ∆G˚G is related to ∆G˚

• ∆∆G = ∆G˚ + RT ln Q G = ∆G˚ + RT ln Q where Q = reaction quotientwhere Q = reaction quotient

• When Q < K or Q > K, reaction is When Q < K or Q > K, reaction is spontaneous.spontaneous.

• When Q = K reaction is at equilibriumWhen Q = K reaction is at equilibrium

• When ∆G = 0 reaction is at When ∆G = 0 reaction is at equilibriumequilibrium

• Therefore, ∆G˚ = - RT ln KTherefore, ∆G˚ = - RT ln K

46464646



Product Favored, ∆G˚ negative, K > 1Product Favored, ∆G˚ negative, K > 1

See Active Figure 19.13See Active Figure 19.13

47474747


• Product-favoredProduct-favored

• 2 NO2 NO22 ee N N22OO44

• ∆∆rrGGoo = – 4.8 kJ = – 4.8 kJ

• State with both reactants State with both reactants and products present is and products present is more stable than more stable than complete conversion.complete conversion.

• K > 1, more products K > 1, more products than reactants.than reactants.


PLAY MOVIE

48484848



Reactant Favored, ∆G˚ positive, K < 1Reactant Favored, ∆G˚ positive, K < 1

See Active Figure 19.13See Active Figure 19.13

49494949


• Reactant-favoredReactant-favored

• NN22OO44 ee 2 NO 2 NO22 ∆∆rrGGoo = +4.8 kJ = +4.8 kJ

• State with both State with both reactants and reactants and products present is products present is more stable than more stable than complete conversion.complete conversion.

• K < 1, more reactants K < 1, more reactants than productsthan products


PLAY MOVIE

50505050


KKeqeq is related to reaction favorability. is related to reaction favorability.

When ∆When ∆rrGGoo < 0, reaction moves < 0, reaction moves

energetically “downhill”energetically “downhill”

∆∆rrGGoo is the change in free energy when is the change in free energy when

reactants convert COMPLETELY to reactants convert COMPLETELY to

products.products.


51515151


A SummaryA Summary

The relation of ∆The relation of ∆rrG, ∆G, ∆rrG˚, Q, K, reaction G˚, Q, K, reaction

spontaneity, and product- or reactant favorability.spontaneity, and product- or reactant favorability.

chapter 19 principles of reactivity: entropy and free energy

Documents