Chapter 6 : Gibbs Free Energy - University of...

Winter 2013 Chem 254: Introductory Thermodynamics

64

Chapter 6 : Gibbs Free Energy ...................................................................................................... 64

Definition of G, A ...................................................................................................................... 64

Maxwell Relations ..................................................................................................................... 65

Gibbs Free Energy G(T,P) (Pure substances) ............................................................................ 67

Gibbs Free Energy for Mixtures ................................................................................................ 68

ΔG of ideal mixing ..................................................................................................................... 71

Chemical Equilibrium and Equilibrium Constant Kp .................................................................. 74

Temperature Dependence of KP ............................................................................................... 75

Chemical Equilibrium: Calculating Equilibrium Composition ................................................... 76

Principle of LeChatelier ............................................................................................................. 80

Use and Significance of ΔG, ΔA ................................................................................................. 81

Chapter 6 : Gibbs Free Energy

Definition of G, A

For single substances

dU q w

For reversible process

rev revdU q w

dU TdS PdV (correct relation always)

P is internal pressure of gas.

dU : expressed in term of functions of state , , , , , ,P V T U H G A

Define: H U PV

A U TS Aarbeit (work): Helmholtz free energy

G H TS U PV TS G Gibbs free energy

Differential forms:

dU TdS PdV

dH dU d PV TdS PdV PdV VdP TdS VdP

dA dU d TS TdS PdV TdS SdT SdT PdV

dG dH d TS TdS VdP TdS SdT SdT VdP


Chapter 6 : Gibbs Free Energy 65

y x

F FdF dx dy

x y

For exact differentials

y x yx

F F

y x x y

if true then ( , )F x y exists

V

UT

S

;

S

UP

V

( , )U S V

P

HT

S

;

S

HV

P

( , )H S P

V

AS

T

;

T

AP

V

( , )A T V

P

GS

T

;

T

GV

P

( , )G T P

Maxwell Relations

U S V

T P

V S

H S P

T V

P S

A T V

S P

V T

1

P

V

V T

;

1

T

V

V P

G T P

S V

P T

Derivations using Maxwell:

dU TdS PdV

TdS dU PdV

1 P

dS dU dVT T

1

V T

U U PdS dT dV dV

T T V T

A) 1V

T

C UdS dT P dV

T T V



B) V T

S SdS dT dV

T V

V

V

CS

T T

1

T T

S UP

V T V

using Maxwell relation

T V

S P

V T

V T

P UT P

T V

OR

T V

U PT P

V T

V T

S SdS dT dV

T V

V

V

C PdS dT dV

T T

VCdT dV

T

We have used all of these relations without derivation before.

dH TdS VdP

TdS dH VdP

dH V

dS dPT T

1

P T

H H VdS dT dP dP

T T P T

A) 1P

T

C HdS dT V dP

T T P

B) P T

S SdS dT dP

T P

P

P

CS

T T

1

T T P

S H VV

P T P T

(Maxwell)



T P

H VV T

P T

P

P

C VdS dT dP

T T

Gibbs Free Energy G(T,P) (Pure substances)

G independent of path, a function of state, depends only on ,i iT P and ,f fT P

dG SdT VdP

G H TS U PV TS

V

GS

T

T

GV

P

changes in P and T

Changes in G with P

f f

i i

P P

P PT

GG dP VdP

P

For liquids and solids

f iG V P P V P

For (ideal) gases nRT

VP

lnf

i

PG nRT

P

Changes in G with T

f f

i i

T T

T TT

GG dT SdT

T

But S always depends on T , not convenient, use a trick

2

1 1

P P

G GG

T T T T T

2 2

1 1SG TS G

T T T H G TS

2

1

P

GH

T T T



2

f f

i i

T T

T TP

G HdT

T T T

2

( )ff

ii

TT

TT

G H T

T T

assume ( ) ( )fH T H T (constant H , approximation)

1 1f i

f i f i

G GH

T T T T

1 1f i

f i f i

G GH

T T T T

Pretend we have absolute Gibbs free energy, but in thermodynamics this is

never the case. We always only consider changes in , .G H

Calculate o

r TG , analogous to o

r TH

298.15

o

f G found in tables along with absolute S

298.15

298.15

1 1

298.15 298.15

oo

orr T

r

f

GGH

T T

298.15

o

r H assumed to be constant (small error)

In real life you need to know o

r TG , don’t use this formula, use T dependence of H .

(we will do it in an exercise)

Gibbs Free Energy for Mixtures

We have various species: # of moles for each: 1 2 3, ,n n n

1

1

, , 1 , ,i i iP n T n T P n n

G G GdG dT dP dn

T P n



2 3

2 3

2 3, , , ,i iT P n n T P n n

G Gdn dn

n n

1 1 2 2 3 3dG SdT VdP dn dn dn

1

i i

i

dG SdT VdP dn

, , j i

i

i T P n n

G

n

i idn : the change in G when adding idn moles of species i , while keep ,P T and

j in n constant. (i =chemical potential of species and has units kJ/mol)

Understand i better for pure substance, molecule ( )A G ,

,A A mG n G (true because G

is a function of ,T P , and ,T P don’t change with the size of system)

,

,

A m A

A T P

GG

n

Pure substance A : molar Gibbs free energy at particular ,T P

What is the physical significance of ?

dG SdT VdP 1 1

1i

dn

i I i II

I i II idG dn dn

I II

i idn dn mass balance

i i I

I II idG dn

0dG Always at constant ,P T

If 0I

idn then 0i i

I II

2 Phases: Matter flows from phase with high chemical potential to phase with low

chemical potential

i i

I II molecules flow from I to II , 0I

idn

Chemical potential strongly depends on concentration ix

At equilibrium i i

I II for any species

Analogy

Volume changes (changing iP between phases) until

pressure is equal (equal )



Chemical potential is equal between different components in a system: Fundamental

principle of chemical equilibrium, at constant ,P T

How to calculate G for mixtures (ideal mixtures)

Ideal mixtures: Ideal gases,

dilute solution (no electrolyte),

components do not interact substantially

Ideal gas, pure compound, change P

( , ) ( , ) lnf

f i

i

PG T P G T P nRT

P

nRT

dG VdP dPP

lnf

i

PG nRT

P

( , )

( , ) ( , ) lno

o

PG T PT P T P RT

n P

Likewise for an ideal gas in a mixture

(Partial )i i totalP P x P

( , ) ( , ) lno

i

i

o

PT P T P RT

P

( , ) lno

i total

o

x PT P RT

P

( , ) ln lno

total

i

o

PT P RT RT x

P

Since ( , ) ( , ) lno

total

o

PT P T P RT

P

( , , ) ( , ) lni total iT P x T P RT x

( , , ) ( , ) lni i iT P x T P RT x

where ( , )i T P is the chemical potential of the pure substance



ΔG of ideal mixing

For pure substances ( , ) , lno

o

PG T P G T P nRT

P

Mixtures i i totalP x P

( , ) , ,i iG T P G T P x

, ln , ln i

i o

o

PG T P nRT x G T P nRT

P

Molar Gibbs free energy = chemical potential

( , , ) , lno

i iT P x T P RT x

( , , ) , lno

i iG T P x G T P nRT x

,o T P , ,oG T P are chemical potentials for pure substances (constant T )

Example I

2 2H H

pure mix at equilibrium

2 , lnH o

pure o

o

PT P RT

P

2 , lnH o i omix o

o

x PT P RT

P

2 2

ln ln

H Hpure mix

o o

P P

P P

2 2H H

pure i mixP P

Total pressure is different in two compartments!

2Hpure

total pureP P 2Hmix Ar

total mixP P P

2H Ar

mixP P



Example II

0mixG

o

mix NeG G ln o

Ne Ne Arn RT x G ln o

Ar Ar Nen RT x G o

ArG

Ne total Nen n x

Ar total Arn n x

ln lnmix total Ne Ne Ar ArG n x RT x x RT x

ln lnmix total Ne Ne Ar ArG n RT x x x x

general expression for mixing

lnmix total i iG n RT x x

0 1ix ; ln 0i ix x

0

lim ln 0i ix

x x

0mixG always

What is mixS ?

i i

i

dG SdT VdP dn

, iP n

GS

T

lnmix i iS nRT x xT

lnmix i iS nR x x 0mixS spontaneous

Calculate mixingS directly:

V change : lnf

i

VS nR

V

at constant T

ln lntotal total

mix Ne Ar

Ne total Ar total

V VS n R n R

x V x V

ln lnmix total Ne Ne Ar ArS n R x x x x



Example

ln 0mix total i i

i

G n RT x x

ln 0mix i i

i

S nR x x

0mix mix mixH G S T (ideal mixes only)

G H TS H G T S

1 1 1 1

ln ln ln 22 2 2 2

G nRT nRT

ln 2totalS n R

What about?

1) ln 2G nRT ?

2) 0G ?

Molecules are indistinguishable in principle.

Gibbs paradox?

0G , we know. Good understanding of this only from Quantum Mechanics



Chemical Equilibrium and Equilibrium Constant Kp

Consider a generic reaction in the gas phase

( ) ( ) ( ) ( )A g B g C g D gn A n B n C n D

r C C D D A A A BG n n n n

, , lno I

I I i o

o

PT P T P RT

P

,o

r i I I

I

G T P

i in for reactants; i in for products

, lno I

r i I o I

I I o

PG T P RT

P

, ln

I

o I

r i I o

I I o

PG T P RT

P

ln

I

o I

r r

I o

PG G T RT

P

o

rG T , pure substance at oP ;

I

= product of all factors, analogous to I

for sums.

o o

r T I f T

I

G G I

i

i

P

I o

PQ

P

Reactions Quotient: change during a reaction, PQ is an instantaneous quantity

Back to initial reaction A B C D

C D A Bn n n n

C D A B

P

o o o o

P P P PQ

P P P P

C D

A B

n n

C D

o o

P n n

A B

o o

P P

P PQ

P P

P P



lno

r r T PG G RT Q

Rules:

- If initially 0rG the reaction will proceed to the right, product ,C D will

increase rG increases until it is 0

- If initially 0rG reaction will proceed to reactants, ,A B increases

At equilibrium

P PQ K equilibrium constant

ln 0o

r T r T PG G RT K

Temperature Dependence of KP

lno

r T

P

GK

RT

2

ln 1 1o o

P r T r T

P P

K G H

T R T T R T

2

ln 1f f

i i

oT T

P r T

T TP

K HdT dT

T R T

2

1ln

ff

i i

oTT r T

P T T

HK dT

R T

assume o

r TH is independent of T

1 1ln ( ) ln ( )

o

r T

P f P i

f i

HK T K T

R T T

Example:

( ) 2 ( ) 2( ) 2( )g l g gCO H O CO H



298.15

298.15ln o

P rRT K G

298.15 2 298.15 2 298.15 2 298.15( ) ( ) ( ) ( )o o o o

f f f fn G CO n G H n G H O n G CO

298.15 298.15

o o

r I fH H

PQ : gases only, liquids and solids has little pressure dependence

2 2CO H

o o

CO

o

P P

P P

P

P

at equilibrium PK

Chemical Equilibrium: Calculating Equilibrium Composition

A B C Dm A m B m C m D

i im (reactants)

i im (products)

m : stoichiometric coefficients (here)

in : actual number of moles for species i

Can we calculate in for each at chemical equilibrium?

Can we calculate ix or

iP ?

How do we do it?

Extent of reaction

o

i i in n o

in being the initial number of moles for molecule i

For any value of you get number of moles in with moles of reaction

Example:

2 2 33 2N H NH

For any mole of 2N that reacts, I gain 2 moles of

3NH and lose 3 moles of 2H

Initial # of moles has to be given, in this example, we start with 2 moles of 2H , 1 mole of

2N and 0.1 moles of 3NH



For all 0in , for all in

max : (reactants) 0in

1 0 1

2 3 0 2

3

max

2

3

min : (products) 0in

0.1 2 0 0.05

min 0.05

2

0.053

bounded

o

i i in n

i idn d

i idG SdT VdP dn

i i i idG dn d at constant ,T P

,

,

i i r T P

iT P

GG

For spontaneous possible process:

,

0T P

GdG d

Always

Initially

If ,

,

0r T P

T P

GG

0d : to the right

i o

in in max min/

2N 1 1 1

2H 3 2 2 3 2

3min

3NH 2 0.1 0.1 2 min 0.05

totaln 3.1 2



If ,

,

0r T P

T P

GG

0d : to the left

If ,

,

0r T P

T P

GG

chemical equilibrium

,

,

r T P i i

iT P

GG

, lno i

i i o

o

PT P RT

P

, lno i

i o

o

x PT P RT

P

, ln lno

i o i

o

PT P RT RT x

P

, lno

i iT P RT x

,r T P i i

i

G

, lno

i i i i

i i

T P RT x

, lnio

r T P i

i

G RT x

, ln io

r T P i

i

G RT x

, lno

r T P xG RT Q

i

x i

i

Q x

; ( )i ix x

i

x i

i

Q x



D

3

2 2

2

3

NH

x

N H

xQ

x x

2 1 3

0.1 2 1 2 3

3.1 2 3.1 2 3.1 2xQ

2 2

3

0.1 2 3.1 2

1 2 3xQ

0xQ as 0.05min

xQ as 2

3min

There is always precisely one point at which ,ln

o

r T P

x

GQ

RT

: chemical equilibrium

ln xQ : mathematically increases between min and max

i

i

total

nx

n

2N

1

3.1 2

2H

2 3

3.1 2

3NH

0.1 2

3.1 2



Principle of LeChatelier

Easy to understand from thermodynamics:

Change in T : 2

ln Pd K H

dT RT

If 0H , PK increases with increasing product, requires heat (lowers T )

If 0H , PK decreases with increasing reactant, requires heat (lowers T )

gasn

x P

o

PK K

P

i

i

P

o

PK

P

i i

ii

P i

o o

x P PK x

P P

i

x iK x

;

i gasn

o o

P P

P P

gasn

x P

o

PK K

P

a) 0gasn more gas on product side xK decreases, reaction to the left

decreases number of gas molecules, reduces P

b) 0gasn , xK increases, products increases

moles of gas decreases, P decreases



Use and Significance of ΔG, ΔA

Clausius 0dq

T

0revq q

T T

revqdS

T

0q TdS

TdS q du w

0du w TdS

Consider a constant T process

TdS d TS TdS SdT

0dU d TS w

0d U TS w A U TS

0dA w

w dA

Same reasoning for G :

expansion non expansiond U TS w w

non expansion

extd U TS P dV w

constraint extP P constant

non expansion 0d U TS PdV w

non expansion 0d U TS PV w U TS PV G

non expansiondG w



Combustion of Octane:

G = -5296 kJ/mol

A = -5285 kJ/mol G A , PV is very small

H = -5471 kJ/mol

Heat engine: heat from burning octane H

1

2

cold

hot heat

T work

T q

1

2w H

Fuel cell: React octane to generate electricity

Max 0.98non PV

rw G H More efficient (known since 1880!!)

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Chapter 6 : Gibbs Free Energy - University of...

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