chapter 6 : gibbs free energy - university of...
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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
Winter 2013 Chem 254: Introductory Thermodynamics
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 66
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)
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 67
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 68
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 69
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 )
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 70
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 71
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 72
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 73
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 74
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 75
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 76
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 77
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 78
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 79
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 80
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 81
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
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 6 : Gibbs Free Energy 82
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!!)