chemia fizyczna technologia chemiczna ii rok wykład 1 · •2 gibbs free energy of a chemical...

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Physical Chemistry GTM/08 1

A quote of the week

(or camel of the week):

A life spent making mistakes is not only more honorable, but more useful

than a life spent doing nothing.

George Bernard Shaw

2

Reaction progress

For reaction: A = M (numerous cases are known, e.g., isomerization)

Number of reaction progress, ξ, is defined as:

i

idnd

Example: if nA,0= 3, nB,0= 0 and =0.6 then nA=3–1.8=1.2 and nB=1.8

ξ 0 0.2 0.4 0.6 0.8 1.0

nA 3 2.4 1.8 1.2 0.6 0

nB 0 0.6 1.2 1.8 2.4 3.0

Where ni is number of moles of ith component (negative for reactants, po-

sitive for products) and νi is stoichiometric coefficient at ith component.

Physical Chemistry GTM/08

3

Gibbs Free Energy of a

Chemical Reaction (1) For reaction: A = M (numerous cases are known, e.g., isomerization)

If infinitesimal change in composition of the reaction mixture occurs, or

the reaction progresses by d, then dnA= –d, and dnM= d, then:

dG = AdnA + MdnM = –Ad + Md = (M – A)d

Reaction Gibbs free energy

may be defined as: AM

TP

r

GG

,


•2


Chemical Reaction (2)

yet its actual changes are shown

by the red line. This is caused by

the contribution of the Gibbs free

energy of mixing the gaseous

components.

Integration of the last equation from ξi to ξf at standard pressure gives:

00000 1)()()( rifAMifAMifr GGGGGG

with ΔG changing linearly from

to (black line) 0A 0

M


Chem. Fiz. TCH II/06 5


Chemical Reaction(3)

G of reaction as a derivative

6

Gibbs Free Energy of

Mixing (1)

)/ln(

)/ln(

00

00

PPRTn

PPRTn

nnG

MM

AA

MMAAi

Initial state: PA = PM = P

mixing

Final state: PA+ PM = P

PA PM

T = const.

)/ln(

)/ln(

00

00

PPRTn

PPRTnG

MMM

AAAf

)lnln(

)/ln()/ln(

MMAA

MMAA

ifmix

xxxxnRT

PPRTnPPRTn

GGG

n = nA + nB

PA= xAP; PM= xMP


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7

Gibbs Free Energy of

Mixing (2)

-0.8

-0.4

0

0.4

0.8

0 0.25 0.5 0.75 1

molar fraction of A, yA

Thermodynamic functions of mixing

TΔSmix/nR

ΔGmix/nRT

ΔHmix ΔGmix TΔSmix

0)lnln(,,

MMAAmiesz

nnP

miesz xxxxnRST

G

MA


Chem. Fiz. TCH II/06 8

Chemical Equilibrium

For reaction: A = M, if A and M behave like perfect gas,

where Q is one of the possible forms (the simplest) of the

reaction quotient.

00 /ln PPRT AAA and 00 /ln PPRT MMM

QRTGP

PRTG

PP

PPRTG

rA

Mr

A

MAMAMr

lnln

/

/ln

00

0

000

Physical Chemistry EPM/04 9

Gibbs Free Energy of a Chemical

Reaction (4)


•4

10


Reaction(5)


11


Reaction(6)


12

Chemical equilibrium (2)

QRTGG ln0

For reaction: aA + bB + ... = mM + nN + ...

reaction quotient Q is defined as:

b

B

a

A

n

N

m

M

aa

aaQ

where: a means activity of the reactant or product indicated.


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13


b

B

a

A

n

N

m

M

aa

aaQ

Activity:

• for liquids and solids in pure form a=1

• for gases ai=Pi/P0 or better fi/P

0 where f is fugacity

• for solutions ai=fi·ci, where c is concentration and f is activity

coefficient. For perfect solutions f=1. For real solutions f→1

when c→0. Standard state for compounds in solutions is when

a=1.


14


At equilibrium:

QRTGG ln0

0G KRTG ln0

b

eqB

a

eqA

n

eqN

m

eqM

eqaa

aaQK

K is known as the reaction equilibrium constant. It is reaction

specific and depends on temperature only. (Guldberg–Waage law).


15


At standard conditions:

QRTGG ln0

0ln ;1 QQ 0GG

One can say, that the reaction quotient term in the

top equation represents the deviation in Gibbs free

energy from the standard state. Standard Gibbs

free energy is reaction specific, while the second

term depends on composition of the reaction

mixture.


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16

Conclusions from the

Guldberg-Waage law

If, in the reaction mixture at equilibrium, one increases activities (concentrations, pressures) of the reactants, the reaction will shift to the right (toward products), reducing the denominator and increasing the numerator to maintain the constancy of K;

If one increases activities of the products of the reaction at equi-librium, the reaction will shift to the left.

If one reduces activities of the reactants ...

If one reduces activities of the products ...

b

eqB

a

eqA

n

eqN

m

eqM

eqaa

aaQK


17

Le Chatelier principle

If a system, containing a chemical reaction at equilibrium, is subject to an external factor (P, T) disrupting this equilibrium,

the system reacts in a way leading to minimization of this external factor. (perverseness, contrariness, cussedness)

Temperature: increase in temperature shifts the equilibrium of

exothermic reaction to the left, while that of endothermic reactions – to

the right.

Pressure can influence state of equlibrium only in reactions

leading to change in number of moles of gaseous substances.


Physical Chemistry EPM/04 18

Le Chatelier principle (2)

The principle says: in reactions leading to increase in number of moles of

gaseous substances, increase in pressure shifts the equilibrium to the

left, while decrease – to the right.

Influence of pressure upon a dissociation in the gas phase: A 2B

initial state, Pi external change, P2>Pi system response,

P2>Pf>Pi

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19


Chemical Reaction (3)

Exoergic reactions: they produce work that may drive other pro-

cesses or supply the non expansion work when ΔG<0. For exam-

ple, reaction in a galvanic cell.

Endoergic reactions: they consume work using other processes,

delivering non expansion work, when ΔG>0. For example, externally

driven reaction in an electrolyzer.


20


; ; ; ; ddnddnddnddn NNMMBBAA

General case. For reaction: AA + BB + ... = MM + NN + ...

or generally: ddn YY

then: ddndGY

YY

Y

YY

and:

Y

YY

TP

r

GG

,

substituting: YYY aRT ln0 one can get: QRTGG rr ln0

where: Y

YtwYr GG 0

,

0 and: Y

YYaQ


21

Równowaga chemiczna (7)

General case. For reaction: AA + BB + ... = MM + NN + ...

We may notice, that (accounting for the signs of Y):

0,

0,

0,

0,

0,

0BfBAfANfNMfM

YYfYr GGGGGG

and the second part of equation: Y

YYrr aRTGG ln0

Y Y Y Y

YYYYYYYY aRTaRTaRTaRT

lnlnlnln

again, accounting for the signs of Y:

Qaa

aaaaaaa

BA

NM

BANMY

BA

NMBANM

Y

Y


•8

22


W stanie równowagi:

0 rG 0ln0 KRTGr KRTGr ln0

RSRTHRTSTHRTG rrrrr eeeeK//// 00000

The latter equation may also be derived from the

Boltzmann distribution.


23

Influence of Pressure on

Dissociation in a Gas Phase (1)

Let’s assume that A and M behave like a perfect gas. Then:

A 2M n

n

nn

nn

n

nn M

MA

AA 21

21

A M

initially n 0

change to the state of equilibrium

–αn +2αn

at equilibrium, number of moles, nY

n(1–α) 2nα

at equilibrium, molar fraction xY

at equilibrium, partial pressure PY

0

2

0

20

/

)/(

PP

P

PP

PPK

A

M

A

M

1

1

1

2

P

1

1P

1

2

02

2

0

2

0

2

2

2

1

4

11

4

11

14

P

P

P

P

PP

PK


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A 2M

041

1

KP

P

0

0.2

0.4

0.6

0.8

1

0 4 8 12 16

deg

ree o

f d

isso

cia

tio

n,

P/P0

K=10

K=1

K=0.1

pure M

pure A

Influence of Pressure on

Dissociation in a Gas Phase (2)


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25

Influence of Temperature on

the Equilibrium Constant (1)

van’t Hoff equation: Jacobus

Henricus

van’t Hoff

For evaporation we know the C.-C. equation:

we obtain: 2

0ln

RT

H

dT

Kd r

because:

R

H

Td

Kd r

0

)/1(

ln


If we treat boiling like a chemical reaction: X(l) = X(v)

dTRT

L

P

dP

RT

L

dT

Pd bb

22or

ln

or:

00

and 1

P/Prb HLK

26

Influence of Temperature on

the Equilibrium Constant (2) One can integrate van’t Hoff equation under the usual assumptions:

because:


Integrating from K1 to K2 and from T1 to T2 assuming Hr=const in the

temperature range of interes

11

ln121

2

TTR

H

K

K rwe get:

1

2

01

02

1

2

/

/

P

P

PP

PP

K

K

27

Relations Between Different

Equilibrium Constants(1) Equilibrium constants in full meaning are only thermodynamic

equilibrium constants, defined as:

means where;0a

aaaK

YeqYY

means where;0P

fffK

YeqYY

The following equations define also quantities frequently known as

equilibrium constants:

means where;0c

cccK

YeqYcY

means where;

0P

PPPK

YeqYPY

These „constants” should be treated as approximations only (besides

KP for perfect gases, which alone is a model only) and should be rather

named equilibrium quotients of concentrations, pressures, etc.)


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28


Equilibrium Constants(2) Introducing aY = YxY (concentrations as molar fractions):

Y

eqeqxxeqYY

eqYY

eqYeqY QQKKxxK YYYY

where Kx and K or Qeqx and Qeq are related equilibrium quotients.

For gaseous substances one can write:

Y

eqeqPPeqYY

eqYY

eqYeqY QQKKPPK YYYY

also: 0

lim

P

PKK meaning: 1lim0

P

K


29


Equilibrium Constants(3) For reations in gas phase there are more possibilities to express KP:

0P

RTKK cP

because: cRTV

nRTP

where c is in mol/m3

3

00 mol/m0224,0

11

mVc BANM

Y

Y

is about the gas substances only

Similarly, if we consider the Dalton’s law:

0P

PKK xP

Y

Y

nPnP

PKK

0

where: P is the total equilibrium pressure

Y

YnYnK

nY means number of moles of Yth gas at

equilibrium Physical Chemistry GTM/08

KKP KKP

30

Particular Cases of

Equilibrium Constants (1) Ionic product of water:

K 298at 10 14

3

OHOHw aaK

Acidity Ka and basicity constants Kb:

(aq)A(aq)OHO(c)HHA(aq) 32

(aq)OHHA(aq)O(l)H(aq)A 2

(for a conjugate acid-base pair)

HA

OHA

aa

aaK

3

A

OHHA

ba

aaKwba KKK

Concept of pK: pK = –log K

K 298at 14 wba pKpKpK

(aq)OH(aq)OHO(l)2H 32


•11

31

Particular Cases of

Equilibrium Constants (2)

Solubility product:

(aq)yA(aq)xM(s)AM xy

yx

y

A

x

Msp aaK

Relation to solubility s expressed in mol/dm3:

yxyx

sp syxK


32

Particular Cases of


Complex formation constants and stability (instability) constants:

OHLO)M(HLO)M(H 21n2n2

OHLO)M(HLLO)M(H 222n21n2

OHMLLO)LM(H 2n1n2

Formation constants (for consecutive reactions):

LO)M(H

LO)M(H

n2

1n21

K

LO)M(H

LO)M(H

1n2

22n22

K

LO)LM(H

ML

1n2

n

nK

Usually: K1 < K2 < ... < Kn


……

33

Particular Cases of


Stability constants (for all the reactions up to a given level):

11 K

212

n2

22n22

LO)M(H

LO)M(HKK

nn KKK ...

LO)M(H

ML21n

n2

n


OHLO)M(HLO)M(H 21n2n2

OHLO)M(HLLO)M(H 222n21n2

OHMLLO)LM(H 2n1n2

……

Complex formation constants and stability (instability) constants:

•12

34

Decomposition Pressure (1)

Decomposition pressure is a partial pressure of a gaseous product of

decomposition of a given substance, being at equilibrium with this

solid or liquid substance.

If, at given temperature, partial pressure is higher then the

decomposition pressure, then decomposition does not occur (or

occurs in reversed direction), if lower – decomposition does occur.

Example: What pressure will be achieved in a stiff container and

what percentage of the solid substance will decompose, if we place in

this container of 2 dm3 20 g of pure chalk (CaCO3), evacuate the

container (pump the air out of it) and warm it up to 800oC?


35


Solution:

1. From thermodynamic data for the products and reactant of decompo-sition, I find G0

298=130051 J, giving K298=1.6452·10-23. One can observe that at ambient temperature and normal pressure CaCO3 is thermodynamically stable.

2. I calculate (with variable CP) H01073

(169000 J), that does not dif-fer much from H0

298 (177886 J). I assume constant H0=173500 J. 3. From van’t Hoff equation I find K1073= K298·e

50,54706=0.1474 4. For reaction CaCO3(s) = CaO(s) + CO2(g), KP=PCO2/P

0. It gives PCO2=14933Pa and 3,35·10-3 mole of CO2.

5. The same amount of CaCO3 got decomposed. This is 1.67% of its initial amount (0.2 mole).

6. More accurate calculations of G01073 yielded K1073=0.3282,

PCO2=33245 Pa and 3.73% of CaCO3 decomposition. Physical Chemistry GTM/08

36


For reaction:

A(s) + B(g) = AB(s), for which

1

0

P

PK BP

where: PB is decomposition

pressure of compound AB

nA nAB


•13

37


For reactions:

A(s) + B(g) = AB(s)

AB(s) + B(g) = AB2(s)

1

0

11

P

PK BP

where: PB1 is decomposition pressure

of AB, while PB2 – of AB2

nA

nAB

1

0

22

P

PK BP


of B(g)

nAB2

38

Howgh!!!


chemia fizyczna technologia chemiczna ii rok wykład 1 · •2 gibbs free energy of a chemical...

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