entropy and free energy how to predict if a reaction can occur, given enough time? thermodynamics...

Entropy and Free EnergyEntropy and Free EnergyEntropy and Free EnergyEntropy and Free EnergyHow 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 a How to predict if a reaction can occur at reaction can occur at a reasonable rate?a reasonable rate?

KINETICSKINETICSCopyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

ThermodynamicsThermodynamicsThermodynamicsThermodynamics• Is the state of a chemical system such that a Is the state of a chemical system 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?

• If yes, system is favored to react — a If yes, system is favored to react — a product-product-favoredfavored system.system.

• Most product-favored reactions are Most product-favored reactions are exothermic.exothermic.

• Often referred to as Often referred to as spontaneousspontaneous reactions.reactions.

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


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

Thermodynamics and KineticsThermodynamics and KineticsThermodynamics and KineticsThermodynamics and Kinetics


Thermodynamics and KineticsThermodynamics and KineticsThermodynamics and KineticsThermodynamics and KineticsDiamond is Diamond is

thermodynamically thermodynamically favored to convert to favored to convert to graphite, but not graphite, but not kinetically favored.kinetically favored.

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


Product-Favored ReactionsProduct-Favored ReactionsIn general, product-In general, product-

favored reactions are favored reactions are exothermicexothermic..

FeFe22OO33(s) + 2 Al(s) (s) + 2 Al(s) ------

> 2 Fe(s) + Al> 2 Fe(s) + Al22OO33(s)(s)

H = - 848 kJH = - 848 kJ


Product-Favored ReactionsProduct-Favored ReactionsBut many spontaneous reactions or But many spontaneous reactions or

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

NHNH44NONO33(s) + heat ---> NH(s) + heat ---> NH44NONO33(aq)(aq)


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

product-favored processes is product-favored processes is that the final state is more that the final state is more DISORDEREDDISORDERED or or RANDOMRANDOM than the original.than the original.

Spontaneity is related to an Spontaneity is related to an increase in randomness.increase in randomness.

The thermodynamic property The thermodynamic property related to randomness is related to randomness is ENTROPY, SENTROPY, S..

Reaction of K Reaction of K with waterwith water


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


How probable is it that reactant How probable is it that reactant molecules will react? molecules will react?

PROBABILITYPROBABILITY suggests that a suggests that a product-favored reaction will product-favored reaction will result in the result in the dispersal of energy dispersal of energy or of matter or of matter or both.or both.

Directionality of ReactionsDirectionality of Reactions


Probability suggests that a product-Probability suggests that a product-favored reaction will result in the favored reaction will result in the dispersal of energy or of matter or dispersal of energy or of matter or both.both.

Matter DispersalMatter Dispersal



Probability suggests that a product-Probability suggests that a product-favored reaction will result in the favored reaction will result in the dispersal of energy or of matter or dispersal of energy or of matter or both.both.

Energy DispersalEnergy Dispersal



Directionality of Directionality of Reactions —Reactions —

Energy DispersalEnergy DispersalExothermic 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 product-favored.product-favored.


S (gases) > S (liquids) > S (solids)S (gases) > S (liquids) > S (solids)

SSoo (J/K•mol) (J/K•mol)

HH22O(liq)O(liq) 69.9169.91

HH22O(gas)O(gas) 188.8 188.8


HH22O(liq)O(liq) 69.9169.91

HH22O(gas)O(gas) 188.8 188.8

Entropy, SEntropy, SEntropy, SEntropy, S


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.



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


CHCH44 248.2248.2

CC22HH66 336.1 336.1

CC33HH88 419.4419.4


CHCH44 248.2248.2

CC22HH66 336.1 336.1

CC33HH88 419.4419.4



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



Entropy usually increases when a Entropy usually increases when a pure liquid or solid dissolves in a pure liquid or solid dissolves in a solvent.solvent.



Entropy Changes for Phase ChangesEntropy Changes for Phase Changes

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

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

For HFor H22O (liq) ---> HO (liq) ---> H22O(g)O(g)

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


Entropy Changes for Phase ChangesEntropy Changes for Phase Changes

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

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

For HFor H22O (liq) ---> HO (liq) ---> H22O(g)O(g)

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

S = qT

= 40, 700 J/mol

373.15 K = + 109 J/K • molS =

qT

= 40, 700 J/mol

373.15 K = + 109 J/K • mol


Consider 2 HConsider 2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 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•mol) - = 2 mol (69.9 J/K•mol) - [2 mol (130.7 J/K•mol) + [2 mol (130.7 J/K•mol) +

1 mol (205.3 J/K•mol)]1 mol (205.3 J/K•mol)]

SSoo = -326.9 J/K = -326.9 J/K

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

Calculating Calculating S for a ReactionS for a Reaction

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


2nd Law of Thermodynamics2nd Law of ThermodynamicsA reaction is spontaneous (product-favored) if ²S A reaction is spontaneous (product-favored) if ²S

for the universe is positive.for the universe is positive.

SSuniverseuniverse = = SSsystemsystem + + SSsurroundingssurroundings

SSuniverseuniverse > 0 for product-favored process > 0 for product-favored process

First calc. entropy created by matter dispersal First calc. entropy created by matter dispersal ((SSsystemsystem))

Next, calc. entropy created by energy dispersal Next, calc. entropy created by energy dispersal ((SSsurroundsurround))


Dissolving NH4NO3 in water—an entropy driven process.

2nd Law of Thermodynamics2nd Law of Thermodynamics


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

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





Sosurroundings =

qsurroundings

T =

-Hsystem

TSo

surroundings = qsurroundings

T =

-Hsystem

T





Can calc. that Can calc. that HHoorxnrxn = = HHoo

systemsystem = -571.7 kJ = -571.7 kJ

Sosurroundings =

qsurroundings

T =

-Hsystem

TSo


T =

-Hsystem

T







Sosurroundings =

qsurroundings

T =

-Hsystem

TSo


T =

-Hsystem

T

Sosurroundings =

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

298.15 KSo

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

298.15 K







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

Sosurroundings =

qsurroundings

T =

-Hsystem

TSo


T =

-Hsystem

T

Sosurroundings =

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

298.15 KSo

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

298.15 K






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

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






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

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


Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G

²S²Sunivuniv = ²S = ²Ssurrsurr + ²S + ²Ssyssys

Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida




Suniv = Hsys

T + Ssys




Multiply through by -TMultiply through by -T

Suniv = Hsys

T + Ssys



SSunivuniv = = S Ssurrsurr + + S Ssyssys


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

Suniv = Hsys

T + Ssys



SSunivuniv = = S Ssurrsurr + + S Ssyssys


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

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

Suniv = Hsys

T + Ssys


Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GSSunivuniv = = SSsurrsurr + + SSsyssys


-T-TSSunivuniv = = HHsyssys - T - TSSsyssys

-T-TSSunivuniv = change in Gibbs free energy for the = change in Gibbs free energy for the system = system = GGsystemsystem

Under standard conditions —Under standard conditions —

GGoo = = HHoo - T - TSSoo

Suniv = Hsys

T + Ssys



GGoo = = HHoo - T - T SSoo

Gibbs free energy change = Gibbs free energy change =

total energy change for system total energy change for system - - energy lost in disordering the systemenergy lost in disordering the system

If reaction is exothermic (If reaction is exothermic (HHoo negative) and negative) and entropy increases (entropy increases (SSoo is +), then is +), then GGoo must be must be negative and reaction product-favored.negative and reaction product-favored.

If reaction is endothermic (If reaction is endothermic (HHoo is +), and is +), and entropy decreases (entropy decreases (SSoo is -), then is -), then GGoo must be must be + and reaction is reactant-favored.+ and reaction is reactant-favored.




HHoo SSoo GGoo 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




Two methods of calculating Two methods of calculating GGoo





a)a) Determine Determine HHoorxnrxn and and SSoo

rxnrxn and use and use

GIbbs equation.GIbbs equation.








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

of formation, of formation, GGffoo..








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

of formation, of formation, GGffoo..

²G²Goorxnrxn = = ²G ²Gff

oo (products) - (products) - ²G ²Gffoo (reactants) (reactants)²G²Goo

rxnrxn = = ²G ²Gffoo (products) - (products) - ²G ²Gff

oo (reactants) (reactants)


Calculating Calculating GGoorxnrxn


Combustion of acetyleneCombustion of acetylene

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

Use enthalpies of formation to calculateUse enthalpies of formation to calculate

HHoorxnrxn = -1238 kJ = -1238 kJ

Use standard molar entropies to calculateUse standard molar entropies to calculate

SSoorxnrxn = -97.4 J/K or -0.0974 kJ/K = -97.4 J/K or -0.0974 kJ/K

GGoorxnrxn = -1238 kJ - (298 K)(-0.0974 J/K) = -1238 kJ - (298 K)(-0.0974 J/K)

= -1209 kJ= -1209 kJ

Reaction is Reaction is product-favoredproduct-favored in spite of negative in spite of negative SSoorxnrxn. .

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




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?





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

HHoorxnrxn = +25.7 kJ = +25.7 kJ

SSoorxnrxn = +108.7 J/K or +0.1087 kJ/K = +108.7 J/K or +0.1087 kJ/K

GGoorxnrxn = +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 HHoorxnrxn. .

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





Combustion of carbonCombustion of carbon

C(graphite) + OC(graphite) + O22(g) --> CO(g) --> CO22(g) (g)

GGoorxnrxn = = GGff

oo(CO(CO22) - [) - [GGffoo(graph) + (graph) + GGff

oo(O(O22)])]

GGoorxnrxn = -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.

GGoorxnrxn = -394.4 kJ = -394.4 kJ

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

GGoorxnrxn = = ²G ²Gff

oo (products) - (products) - GGffoo (reactants) (reactants)GGoo

rxnrxn = = ²G ²Gffoo (products) - (products) - GGff

oo (reactants) (reactants)


Free Energy and TemperatureFree Energy and Temperature

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

HHoorxnrxn = +467.9 kJ = +467.9 kJ SSoo

rxnrxn = +560.3 J/K = +560.3 J/K

GGoorxnrxn = +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 GGoorxnrxn just change from being just change from being

(+) to being (-)? (+) to being (-)?

When When GGoorxnrxn = 0 = = 0 = HHoo

rxnrxn - T - TSSoorxnrxn


Free Energy and TemperatureFree Energy and Temperature

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

HHoorxnrxn = +467.9 kJ = +467.9 kJ SSoo

rxnrxn = +560.3 J/K = +560.3 J/K

GGoorxnrxn = +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 GGoorxnrxn just change from being just change from being

(+) to being (-)? (+) to being (-)?

When When GGoorxnrxn = 0 = = 0 = HHoo

rxnrxn - T - TSSoorxnrxn

T = HrxnSrxn

= 467.9 kJ

0.5603 kJ/K = 835.1 K


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

When When GGoorxnrxn < 0, reaction moves energetically < 0, reaction moves energetically

“downhill”“downhill”

²G²Goorxnrxn is the change in free energy as reactants is the change in free energy as reactants

convert completely to products.convert completely to products.

But systems often reach a state of equilibrium But systems often reach a state of equilibrium in which reactants have not converted in which reactants have not converted completely to products.completely to products.

In this case In this case GGrxnrxn is < is < GGoorxnrxn , so state with both , so state with both

reactants and products present is more reactants and products present is more stable than complete conversion.stable than complete conversion.

Thermodynamics and KThermodynamics and Keqeq


Product-favored Product-favored reaction. reaction.

2 NO2 NO22 ---> N ---> N22OO44

GGoorxnrxn = -4.8 kJ = -4.8 kJ

Here Here GGrxnrxn is less than is less than GGoo

rxnrxn , so the state , so the state with both reactants with both reactants and products and products present is more present is more stable than complete stable than complete conversion.conversion.



Reactant-favored Reactant-favored reaction. reaction.

NN22OO44 --->2 NO --->2 NO22 GGoo

rxnrxn = +4.8 kJ = +4.8 kJ

Here Here GGoorxnrxn is greater is greater

than than GGrxnrxn , so the , so the 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.



KKeqeq is related to reaction favorability and so is related to reaction favorability and so to to GGoo

rxnrxn..

The larger the value of The larger the value of GGoorxnrxn the larger the the larger the

value of K.value of K.

GGoorxnrxn = - RT lnK = - RT lnK

where R = 8.31 J/K•molwhere R = 8.31 J/K•mol



Calculate K for the reactionCalculate K for the reaction

NN22OO44 --->2 NO --->2 NO22 GGoorxnrxn = +4.8 kJ = +4.8 kJ






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








lnK = -4800 J

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






K = 0.14K = 0.14


lnK = -4800 J

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






K = 0.14K = 0.14

When When GGoorxnrxn > 0, then K < 1 > 0, then K < 1


lnK = -4800 J

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


entropy and free energy how to predict if a reaction can occur, given enough time? thermodynamics...

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