Gibbs Energy and Chemical Potentials

Calculation of TunivS two system parametersrS rH

Define system parameters that determine if a given process will be spontaneous?

2

Distinguish between a reversible and an irreversible transformation.

3

revsurr dqdST

Tdq

Sd irrevsurr

From the first law

4

dwdqdU

dwdUdSTsurr

Our definition of work can be extended to include other types of work.

• Electrical work.

• Surface expansion.

• Stress-strain work.

dw=-Pext dV+dwa

where dwa includes all other types of work

5

For a general system

6

0ext e surrdU P dV dw T dS

In an isolated system where dq=0; dw=0; dU=0

dS 0 Now allow the system to make thermal

contact with the surroundings. For an isentropic process (dS = 0)

dU 0

7

For a systems where the temperature is constant and equal to Tsurr

8

0 dwTdSdU

Define the Helmholtz energy A

A(T,V) =U – TS Note that for an isothermal process

dA dw

A w For an isochoric, isothermal process

A 0

9

The Helmholtz energy is a function of the temperature and volume

10

dVVA

dTTA

dATV

PVA

T

STA

V

For an ideal gas undergoing an isothermal volume change

11

1

2ln

2

1

2

1

VV

nRT

PdVdVVA

AV

V

V

V T

For an isothermal, isobaric transformation

12

0ext adU TdS P dV dw

0adU TdS PdV dw

Define the Gibbs energy G

G(T,P) =U – TS+PV Note that for an isothermal process

dG dwa

G wa

For an isothermal, isobaric process

G 0

13

The Gibbs energy is a function of temperature and pressure

14

P T

G GdG dT dP

T P

P

GS

T T

GV

P

For an ideal gas undergoing an isothermal pressure change

15

1

2ln

2

1

2

1

PP

nRT

VdPdPPG

GV

V

P

P T

Define the chemical potential = G/n

16

1

212 lnPP

RTn

GGnG

1

212 ln

PP

RT

For P1 = P = 1 bar, we define the standard state chemical potential

°= (T, 1bar)

17

barP

RTbarTT 1

ln) 1 ,()(

Solids and liquids are essentially incompressible

18

PVPPV

VdPGGGP

P

)( 12

12

2

1

Under isochoric conditions

19

dTSdTTA

AT

T

T

T V

2

1

2

1

SdTdTTA

dAV

Consider the calculation of Helmholtz energy changes at various temperatures

20

dTSTATAT

T

2

1

12

1212 TTSTATA

Under isobaric conditions

21

dTSdTTG

GT

T

T

T P

2

1

2

1

SdTdTTG

dGP

The Gibbs energy changes can be calculated at various temperatures

22

dTSTGTGT

T

2

1

12

1212 TTSTGTG

The Gibbs-Helmholtz relationship

23

2 TH

TT

G

P

2 TH

TT

G

P

Differentiating the chemical potential with temperature

24

JmP

J ST ,

PPRTSTS

JmJm ln,,

For a reversible process

dU = TdS – PdV The Fundamental Equation of

Thermodynamics!! Internal energy is a function of entropy

and volume

25

The total differential

26

dVVU

dSSU

dUSV

PVU

S

TSU

V

The systems is described by • Mechanical properties (P,V)

• Three thermodynamic properties (S, T, U)

• Three convenience variables (H, A, G)

27

The Maxwell relationships are simply consequences of the properties of exact differentials • The equality of mixed partials

28

VS SP

VT

Obtain relationships between the internal energy and the enthalpy

29

PVS

TVU

TT

The Thermodynamic Equation of State!!

A simple relationship between (H/P)T and other parameters.

30

TT PS

TVPH

VdpSdTdG

For a system at fixed composition

...2

,,2

1

,,1

dnnG

dnnG

VdpSdTdG

j

j

nPT

nPT

If the composition of the system varies

KnPTJJ n

G

,,

Using the chemical potential definition

J

JJ dnVdpSdTdG

JO

JJ pRTT ln

Chemical potential is an intensive property For an ideal gas

Note - J (T) is the Standard State Chemical Potential

of substance J

pxRTT JO

JJ ln

The chemical potential of any gas in a mixture is related to its mole fraction in the mixture

Jpure

JJ xRTTp ln,

JJJ

J

OJJJmix

xnRT

nG

ln

In a non-reacting mixture, the chemical potentials are calculated as above.

The total Gibbs energy of the mixture

0

ln

ln

mixmix

JJJ

mixmix

JJJ

J

OJJJmix

VH

xnRTG

S

xnRT

nG

In an ideal gas mixture

J

JJ A0

Consider a closed system at constant pressure

The system consists of several reacting species governed by

J

JJ dndG

At constant T and P, the Gibbs energy change results from the composition change in the reacting system

JOJJ nn

Suppose we start the reaction with an initial amount of substance J

• nJ0

Allow the reaction to advance by moles - the extent of reaction

Examine the derivative of the Gibbs energy with the reaction extent

JJJ

PT

GG

,

G – the non-standard Gibbs energy change

0,

PT

GG

The equilibrium condition for any chemical reaction or phase change

0

eqJJJG

For the simple reaction

Extent of Reaction,

GA*

GB*

Pure components

0

maxmin

A (g) ⇌ B (g)

Adding in the contribution from mixG.

Extent of Reaction,

GA*

GB*

Pure components

0

Mixing Contribution

maxmin

The Gibbs energy of reaction.

Extent of Reaction,

GA*

GB*

min

Pure components

0rG

max

eq

DDD pRTG ln

For the reaction

aA (g) + bB (g) pP (g) + qQ (g)⇌

AAA pRTG ln

BBB pRTG ln

CCC pRTG ln

bB

aA

dD

cC

rrpp

ppRTGG ln

The Gibbs energy change can be written as follows

The Gibbs energy change for a chemical reaction?

47

fG = Jø = the molar formation

Gibbs energy (chemical potential) of the substance

J

oJJ

JfJr JGG

Define the reaction quotient

c d

C Da b

A B

p pQ

p p

lnr rG G RT Q

At equilibrium, rG = 0

eq

eq

c d

C Deq a b

A B

p pK Q

p p

0 ln

eqrG RT Q

At equilibrium, the non-standard Gibbs energy change is 0.

lnr eqG RT K

0 lnr eqG RT K

molkJelements 0

Examine the following reaction

CO2 (g) – C (s) – ½ O2 (g) = 0 The standard state chemical potentials for the

elements in their stable state of aggregation

J

Jr JG

sCgOgCOGr 22

gCOgCOGf 22

JJGf

Note – since the (elements) = 0 kJ/mol

In general

RTG

K rP

ln

We can write the equilibrium constant as

dT

TGd

RdTKd

r

P

1ln

Differentiating

J

Jr JG

dT

TJ

d

RdTKd J J

P

1ln

2T

HdT

TGd

2TH

dT

TGd

r

r

For a chemical reaction, with a standard Gibbs

energy change, rG

2

lnRT

HdT

Kd rP

The van’t Hoff equation relates the temperature dependence of Kp to the reaction enthalpy change

2

1

2

1

2

ln

ln

2

ln

ln

T

T

r

K

KP

rP

dTRT

HKd

dTRT

HKd

P

P

Assuming the reaction enthalpy change is constant with temperature

121,

2, 11ln

TTRH

K

Kr

P

P

If the enthalpy change for the reaction is know, we can estimate the Kp value at any temperature

reaction

y = -6937.6x - 8.9829-35.00-34.50-34.00-33.50-33.00-32.50-32.00-31.50-31.00-30.50-30.00

0.003 0.0032 0.0034 0.0036 0.0038

1/T (K-1)

ln K

w

H2O (l) ⇌H+ (aq) + OH- (aq)

Revisit the Gibbs energy profile!

Extent of Reaction,

GA*

GB*

0

maxmin

rG = 0

At equilibrium, the Gibbs energy is at a minimum

The second derivative of the Gibbs energy with the extent of reaction,

= G’’ is positive!!

PT

G

,2

2

The change in the extent of reaction with temperature.

T / K

max

min

rHo >0

rHo < 0

The change in the extent of reaction with pressure.

P / bar

max

min

rVo < 0

rVo > 0