Spontaneity and Equilibrium

isolated system: 0S

Isothermal process

Aw

dAdw

TSEddw

TSddEdw

TdSdEdw

dwdETdS

dQTdS

dQdQrev

Aw

TSEA

EnergyHelmholz

Maximum work obtained in a process at constant temperature is equal to the decrease in Helmholtz energy of the system.

Calculate the maximum work that can be obtained from the

combustion of 1 mole of methane at 298 K.

Given DHo and DSo of the the combustion of methane.

)(2)(2)(2)(4 22 gggg OHCOOCH

Aw

STRTnHA

STpVHA

STpVHA

STEA

TSEA

max

Transformation at constant temperature and pressure

Gw

dGdw

TSHddw

TdSdHdw

TdSVdppdVdHdwpdV

TdSVdppdVdHdwdw

TdSpVHddw

dwdETdS

dQTdS

dQdQ

Vpnon

Vpnon

Vpnon

Vpnon

Vpnon

VpnonVp

rev

Gw Vpnon

TSHG

EnergyGibbs

Maximum work, over and above pV-work, obtained in a process at constant temperature and pressure is equal to the decrease in the Gibbs energy of the system.

G

Gw Vpnon

0

Special case:No work over and above pV-work

wnon p-V=0

forcedG

mEquilibriuG

eoussponG

0

0

tan0

Calculate the maximum non-pV work that can be obtained from the

combustion of 1 mole of methane at 298 K.

http://www.google.jo/url?sa=i&rct=j&q=thermodynamics%20spontaneity%20entropy%20free%20energy&source=images&cd=&cad=rja&docid=bTtHuQ5gnFu5uM&tbnid=ajnCTHjlkOBBcM:&ved=0CAUQjRw&url=http://lawrencekok.blogspot.com/2012/03/ib-chemistry-on-entropy-gibbs-free.html&ei=R2iDUZP4FcmvO6rPgfAC&psig=AFQjCNGaX7tDm1X3W2MndKzg9APqZ4rjdQ&ust=1367652799910476

Fundamental equations of Thermodynamics

Maxwell Relations

pdVTdSdE

pdVdQdE

dwdQdE

rev

VS

S

p

V

T

VdpTdSdH

VdppdVpdVTdSdH

VdppdVdEdH

pVEH

pSS

V

p

T

pV

ET

S

E

dVV

EdS

S

EdE

VSfE

SV

SV

),(

Vp

HT

S

H

dpp

HdS

S

HdH

pSfH

Sp

Sp

),(

pdVSdTdA

SdTTdSpdVTdSdA

SdTTdSdEdA

TSEA

VTT

p

V

S

VdpSdTdG

SdTTdSVdpTdSdG

SdTTdSdHdG

TSHG

pTT

V

p

S

pV

AS

T

A

dVV

AdT

T

AdA

VTfA

TV

TV

),(

Vp

GS

T

G

dpp

GdT

T

GdG

pTfG

Tp

Tp

),(

1212

12

2

1

2

1

2

1

:/

),(

ppVpGpG

dpVpGpGsolidsliquidsFor

VdpdG

Vdpdpp

GdG

dpp

GdT

T

GdG

pTfG

p

p

p

p

G

G

T

Tp

Transformation at constant temperature

pRTp

p

pRTp

p

pRT

n

G

n

pG

p

pnRTGpG

pp

p

pnRTpGpG

dpp

nRTpGpGgasidealFor

o

oo

o

o

oo

o

p

p

ln

ln

ln

ln

ln

:

22

22

22

1

1

212

12

2

1

Chemical potential m

121

1

2

2

121

1

2

2

121

1

2

2

2

/

/

2

22

22

11

11

11

/

./

/

11/

/11/

2

1

22

11

TTH

T

TG

T

TG

TTH

T

TG

T

TG

TTH

T

TG

T

TG

dTT

HTGd

eqHelmholzGibbsdTT

HTGd

T

H

T

S

T

H

T

S

T

TG

TTSH

T

S

TG

T

S

T

TG

T

TG

T

G

TT

TG

T

T

TG

TG

p

p

ppp

The value of DGfo of Fe(g)

is 370 kJ/mol at 298 K. If

DHfo of Fe(g) is 416 kJ/mol

(assumed to be constant

in the range 250-400 K),

calculate DGfo of Fe(g) at

400 K.

Transformation at constant pressure

G dependence on n

H2O

H2O

nMwtmnG

Mwtn

m

n

G

GG

propertyextensivenondependsE

EE

cmE

21

21

2

22112211

2121

nMwtnMwtmnnG

mmmGGG

Given a system consisting of two substances:

ii

i nG

iiipT

iii

nTnp

nTnp

npTnpTnTnp

i

dndG

dndpp

GdT

T

GdG

dndndpp

GdT

T

GdG

dnn

Gdn

n

Gdp

p

GdT

T

GdG

nnnpTfG

ii

ii

iiii

,

,,

2211

,,

2

,,21

,,1,,

22

...

...

),...,,,,(

21

j

npTjji

n

G

,,

If there is no change in composition:

0

0

,

pT

i

dG

dn

System at constant T and p

a

dn1

Each subsystem is a mixture of substances.

b

Chemical Equilibrium

11 substsubst

1111111

11

11

dndndndG

dndG

dndG

dGdGdG

tot

tot

mequilibriudGif

eousspondGif

tot

tot

0

tan0

11

11

Equilibrium is established if chemical potential of all substances in the system is equal in all parts of the system.

Matter flows from the part of system of higher chemical potential to that of lower chemical potential.

pTpT

xRTpTpT

xRTpTpT

xRTpRTT

xRTpRTT

pxRTTpRTT

pTpRTTpRTT

oHpure

mixtureH

io

ipurei

Ho

HpuremixtureH

HoH

mixtureHH

HoH

mixtureHH

HoHH

oH

mixtureHH

oHpure

oHH

oH

pureHH

,,

ln,,

ln,,

lnln

lnln

lnln

,lnln

22

222

2222

2222

222222

222222

a

Pure H2

b

N2 + H2

Pd membrane

Equilibrium never reached

constant T & p

pp

DG and DS of mixing of gases

iiimix

if

oof

oof

mixturemixturemixturei

iif

oopurei

iii

ifmix

xnRTxRTnxRTnG

xRTnxRTnGG

xRTnxRTnpTnpTnG

xRTpTnxRTpTnG

pTnpTnnG

pTnpTnnG

GGG

lnlnln

lnln

lnln,,

ln,ln,

,,

,,

2211

2211

22112211

222111

2211

2211

.0

0ln1

lnlnln

lnlnln

2211

22

11

2211

spontalwaysG

xx

xxRTnxRTxnxRTxnG

n

nx

n

nx

xnRTxRTnxRTnG

mix

ii

iiitottottotmix

tottot

iiimix

mix

p

mix

pp

ST

GS

T

GS

T

G

0ln

ln

ln

iiitotmix

iiitot

iiitot

p

mix

xxRnS

xxRndT

xxRTnd

T

G

0.0 0.2 0.4 0.6 0.8 1.00

2

4

6

Sm

ix (J/

mol.K

)

x1

0.0 0.2 0.4 0.6 0.8 1.0

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

x1

Gm

ix/n

RT

STHG

TdSdHdG

dTconstT

SdTTdSdHdG

TSHG

0.

0

mixmixmix

mixmixmix

STGH

STHG

mix

T

mix

TT

Vp

G

Vp

GV

p

G

Chemical reactions

CH4(g) +2O2(g) → CO2(g) + 2H2O(g)

i

oii

o

iii

CHOCOOH

ii

iim

CHmOmCOmOHm

Rpif

G

G

G

n

GG

GGGGG

GGGGG

4222

4222

22

22

,

,,,,

Heat of Formation

Formation reaction:

reaction of forming 1 mole of product from the

elements in their stable form at 25ºC and 1

atm.Heat of formation = DH of formation reaction = DFH

Standard heat of formation = DHº of formation reaction = DFHº

DFHº(NO(g)): ½ N2(g)+½ O2(g) → NO(g) DHº

DFHº(CO(g)): Cgraphite(s)+½ O2(g) → CO(g) DHº

DFHº(O(g)): ½ O2(g) → O(g) DHº

DFHº(Cdiamond(s)): Cgraphite(s) → Cdiamond(s) DHº

DFHº(O2(g)): O2(g) → O2(g) DHº=0

DFHº(Cgraphite(s)): Cgraphite(s) → Cgraphite(s) DHº=0

DG of Formation

Gibbs energy of formation = DG of formation reaction = DFG

Standard Gibbs energy of formation = DGº of formation reaction =

DFGº DFGº(NO(g)): ½ N2(g)+½ O2(g) → NO(g) DGº

DFGº(CO(g)): Cgraphite(s)+½ O2(g) → CO(g) DGº

DFGº(O(g)): ½ O2(g) → O(g) DGº

DFGº(Cdiamond(s)): Cgraphite(s) → Cdiamond(s) DGº

DFGº(O2(g)): O2(g) → O2(g) DGº=0

DFGº(Cgraphite(s)): Cgraphite(s) → Cgraphite(s) DGº=0

01,25

atmC

elementsstableG o

oF

oiFi

orct

oiFi

orct GGHH

Applying Hess’s Law

oiF

oi

i

oiFi

i

oii

o

G

GG

01,25

01,25

atmC

elementsstable

atmC

elementsstableG

ooi

ooiF

Chemical reactions

CH4(g) +2O2(g) → CO2(g) + 2H2O(g)

., constpT

As the reaction proceeds: The number of moles of involved substances

changes. G of system will change:

ii

i

ii

i

dndG

nG

ioii nn

ddn ii

xExtent of reactionReaction

advancementDegree of reaction

ii

i

pT

ii

iii

i

d

dG

ddndG

,

As the forward reaction proceeds:

x grows, dx positive, d x > 0

destablishemequilibriudGd

dGif

spontproceedtdoesnrctforwarddGd

dGif

spontproceedsrctforwarddGd

dGif

pT

pT

pT

00

.'00

.00

,

,

,

.spontproceedsrctreversed

even though G of products larger than G of

reactants, the reaction proceeds!!!!!!!!!!

Reason:

DGmix

0,

ii

i

pTd

dG

mequilibriuat

For a mixture:

i

oii

oi

i

oi

oi

ii

oi

oi

iipure

mixpureoii

ii

oi

iitotal

oii

oi

ii

oi

oii

iitotal

ii

itotal

nnnG

GGnnG

nnG

nG

mixpuretotal

mixpuretotal

GGG

GGG

DG more negative if

DGpure is small

DGmix largely

negative

Pure R

Pure P

Gpure

DGmix

Gtotal

Equilibrium constant

DdCcBbAa

po

io

bB

aA

cC

dDo

aA

bB

cC

dD

o

aA

bB

cC

dD

oA

oB

oC

oD

AoAB

oB

CoCD

oD

AoAB

oB

CoCD

oD

ioii

ABCDi

ii

QRTGpRTGG

pp

ppRTGG

ppppRTGG

pRTpRTpRTpRT

TaTbTcTdG

pRTaTapRTbTb

pRTcTcpRTdTdG

pRTTapRTTb

pRTTcpRTTdG

pRTTT

abcdG

i lnln

ln

lnlnlnln

lnlnlnln

lnln

lnln

lnln

lnln

ln

atm

pRTTT

p

pRTTT

ioii

oio

ii

1ln

ln

R, P: Ideal gases

RT

GK

KRTG

KQpp

pp

constpmequilibriuat

pp

ppRTGG

Gmequilibriuat

o

p

po

peqp

eq

bB

aA

cC

dD

i

eq

bB

aA

cC

dDo

ln

0ln

.:

0ln

0:

mequilibriuGKQif

spontrctforwardGKQif

spontnonrctforwardGKQif

K

QRTG

QRTKRTG

QRTGG

pp

pp

pp

p

p

pp

po

0

.0

.0

ln

lnln

ln

eq

b

o

B

a

o

A

c

o

C

d

o

D

eq

bB

aA

cC

dD

p

pp

pp

pp

pp

pp

ppK

totalii pxp

i

totalxp

bacd

o

totalxp

eq

b

o

total

a

o

total

c

o

total

d

o

total

eq

bB

aA

cC

dD

eq

b

o

totalB

a

o

totalA

c

o

totalC

d

o

totalD

p

pKK

ppKK

pp

pp

pp

pp

xx

xx

ppx

ppx

ppx

ppx

K

Kp relation to Kx

ptotal in atm

Kp relation to Kc

RTcpRTV

npRTnVp ii

iiii

iRTKK

RTcc

cc

RTcRTc

RTcRTc

pp

ppK

cp

bacd

eq

bB

aA

cC

dD

eq

bB

aA

cC

dD

eq

bB

aA

cC

dD

p

10821.01

0821.01

1/1

KKmolatmL

Latmmol

p

Rc

atmpLmolc

p

RTc

c

c

pc

RTcc

p

RTcp

pppp

ppK

o

o

oo

o

o

o

i

oo

oi

o

i

o

ii

eq

bB

aA

cC

dD

p

eq

bB

aA

cC

dD

c cc

ccK c in mol/L

R=0.0821 atmL/mol.K

Consider the reaction N2O4(g) → 2 NO2(g)

DFGº(NO2(g))=51.31 kJ/mol DFGº(N2O4(g))=102.00 kJ/mol

Assume ideal behavior, calculate

242

42

0,1

?%50)4

1)3

)2

25)1

NOmolONmolmolesofnumberinitial

ddissociateONissystemtheinpressurewhatAt

atmatK

K

CatK

x

c

op

no neq Sni xeq pi

N2O4 1 1-x 1+x (1-x)/(1+x) (1-x)/(1+x)*Ptot

NO2 0 2x 2x/(1+x) 2x/(1+x)*Ptot

5.01

deg x

n

nondissociatiofree

oreacted

p

xx

xK p

11

2 2

Temperature dependence of Kp

CTR

HK

T

dT

R

HKd

TTR

HTKTK

TK

TK

T

dT

R

HKd

dTRT

HKd

RT

H

T

H

RdT

TG

d

RdT

Kd

T

G

RRT

GK

o

p

o

p

o

ppp

p

T

T

oTK

TK

p

o

p

oo

o

p

oo

p

p

p

1ln

ln

11lnlnln

ln

ln

11ln

1ln

2

1212

1

2

2

)(

)(

2

22

2

1

2

1

bxay

0 1000 2000 3000 4000 5000

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

.0 endothH

R

H

o

o

500 1000 1500 2000 2500 3000

0

10

20

30

40

50

60

70

1/T

T

ln K

p

Kp

0.0005 0.0010 0.0015

0

2

4

.0 exothH

R

H

o

o

R

SconstIntercept

R

S

RT

H

RT

STH

RT

GK

o

ooooo

p

.

ln

For the reaction N2O4(g) → 2 NO2(g)

DFHº(NO2(g))=51.31 kJ/mol DFHº(N2O4(g))=102.00 kJ/mol

Kp(25oC)=0.78 atm, calculate Kp at 100oC?

For a given reaction, the equilibrium constant is 1.80x103 L/mol at 25oC and 3.45x103 L/mol at 40oC. Assuming DHo to be independent of temperature, calculate DHo and DSo.

Heterogeneous Equilibria

)(2)()(3 gss COCaOCaCO

i

oii

o

iii

G

G

eqCOp

COo

COoi

oi

oi

oi

oiCO

oi

oii

oii

COoii

ioii

iii

pK

pRTGG

pRTCaCOCaOCOG

CaCOCaOpRTCOG

CaOCaO

CaCOCaCO

pRTCOCO

pRTp

CaCOCaOCOG

2

2

2

2

2

ln

ln

ln

ln

ln

32

32

33

22

32

DFGº kJ/mol

DFHº kJ/mol

CaCO3(s) -1128.8 -1206.9

CaO(s) -604.0 -635.1

CO2(g) -394.4 -393.51

Calculate The pressure of CO2 at 25oC and at 827oC?

DGº =130.4 kJ

DHº =178.3 kJ

ln(Kp)=ln(pCO2)=-52.6

pCO2=1.43x10-23 atm

At 1100 K: ln(pCO2)=0.17

pCO2=0.84 atm

Vaporization Equilibria

)()( gl AA )(2)(2 gl OHOH

eqOHp

OHo

loiOHg

oi

ligi

g

g

g

pK

pRTGG

OHpRTOHG

OHOHG

)(2

)(2

)(2

ln

ln )(2)(2

)(2)(2

dTRT

Hpd

dTRT

HKd

ovap

OH

o

p

g 2

2

)(2ln

ln

CTR

Hp

ovap

OH v

1ln

)(2Clausius-Clapeyron Equation

Derive the above relations for the sublimation phase transition!

Mass Action Expression (MAE)

• For reaction: aA + bB cC + dD

Reaction quotient– Numerical value of mass action expression– Equals “Q” at any time, and– Equals “K” only when reaction is known to be at

equilibrium

ba

dc

[B][A][D][C]

Q

41

Calculate [X]equilibrium from [X]initial and KC

Ex. 4 H2(g) + I2(g) 2HI(g) at 425 °C

KC = 55.64

• If one mole each of H2 and I2 are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI?

• Step 1. Write Equilibrium Law

64.55]][[

][

22

2

IHHI

Kc

42

Ex. 4 Step 2. Concentration Table

Conc (M) H2(g) + I2(g) 2HI (g)

Initial 2.00 2.00 0.000ChangeEquil’m

• Initial [H2] = [I2] = 1.00 mol/0.500L =2.00M

• Amt of H2 consumed = Amt of I2 consumed = x

• Amt of HI formed = 2x

– x +2x– x

+2x2.00 – x 2.00 – x

2

22

)00.2(

)2()00.2)(00.2(

)2(64.55

x

xxx

x

43

Ex. 4 Step 3. Solve for x• Both sides are squared so we can take

square root of both sides to simplify

2

2

)00.2(

)2(64.55

x

xK

)00.2(2

459.7x

x

xx 2)00.2(459.7

xx 2459.7918.14

58.1459.9918.14 x

x459.9918.14

44

Ex. 4 Step 4. Equilibrium Concentrations

Conc (M) H2(g) + I2(g) 2HI (g)

Initial 2.00 2.00 0.00ChangeEquil’m

• [H2]equil = [I2]equil = 2.00 – 1.58 = 0.42 M

• [HI]equil = 2x = 2(1.58) = 3.16

– 1.58 +3.16– 1.58

+3.160.42 0.42

45

Calculate [X]equilibrium from [X]initial and KC

Ex. 5 H2(g) + I2(g) 2HI(g) at 425 °C

KC = 55.64

• If one mole each of H2, I2 and HI are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI?

• Now have product as well as reactants initially• Step 1. Write Equilibrium Law

64.55]][[

][

22

2

IHHI

Kc

46

Calculate [X]equilibrium from [X]initial and KC

Ex. 6 CH3CO2H(aq) + C2H5OH(aq) CH3CO2C2H5(aq) +

acetic acid ethanol ethyl acetate H2O(l)

KC = 0.11

• An aqueous solution of ethanol and acetic acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?

47

Calculating KC Given Initial Concentrations and One Final Concentration

Ex. 2a

H2(g) + I2(g) 2HI(g) @ 450 °C• Initially H2 and I2 concentrations are 0.200

mol each in 2.00L (= 0.100M); no HI is present

• At equilibrium, HI concentration is 0.160 M• Calculate KC

• To do this we need to know 3 sets of concentrations: initial, change and equilibrium