CEq-1Thermodynamics and Equilibrium

One of the most important applications of thermodynamics is chemical equilibrium.

At constant T and P (where most chemistry takes place) we know that the condition for equilibrium is ΔG = 0.

The sign of ΔG dictates whether a process will proceed spontaneously, so if we are not at equilibrium, we can predict which way a reaction will proceed to obtain equilibrium.

We need to apply our thermodynamic knowledge to chemical equilibrium to derive relationships between G and the equilibrium constant for a chemical reaction.

CEq-2Extent of reaction

)()()()( gZvgYvgBvgAv ZYBA +→+

ξAAA vnn −= 0,

ξBBB vnn −= 0,

Start with a gas phase reaction described by a balanced reaction:

Define the extent of reaction, ξ

…

ξZZZ vnn += 0,

ξYYY vnn += 0,

Reactants Products

ξ

varies from 0 to a maximum value dictated by stoichiometry

For example, if nA,0 = vA moles and nB,0 = vB moles, then ξ

varies from…

CEq-3Gibbs energy and ξ

ZZYYBBj

AAjj dndndndndndG μμμμμ +++==∑

ξdvdn AA −= ξdvdn YY =

Recall from the first few Solutions slides…

At constant T and P,

ξdvdn ZZ =ξdvdn BB −=

( ) ξμμμμ dvvvvdG ZzYYBBAA ++−−= At constant T and P

Book defines Δr G as the change in Gibbs energy when the extent of reaction changes by one mole. It has units of energy/mol and only has meaning for a specified balanced chemical reaction. (pg 1051)

CEq-4Δr G relative to standard states

BBAAZzYYr vvvvG μμμμ −−+=Δ

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

o

o

PP

RTPT jjj ln),( μμ

)()()()()( TvTvTvTvTG BBAAZZYYrooooo μμμμ −−+=Δ

( ) ( ) ( ) ( )( )ooooo PPvPPvPPvPPvRTGG BBAAZZYYrr /ln/ln/ln/ln −−++Δ=Δ

Eq

24.13

Let…

Note that P°

is 1 bar in this case and is often not shown in the equation. However, you must remember that there is a reference pressure in the denominator.

( ) ( )( ) ( ) AA

ZY

vA

vA

vZ

vY

PPPPPPPPoo

oo

////ln

CEq-5At equilibrium

QRTTGG rr ln)( +Δ=Δ o

0,

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=ΔPT

rGGξ

eqr QRTTG ln)( −=Δ o

At equilibrium the Gibbs energy is a minimum wrt

any displacement…

)(ln)( TKRTTG Pr −=Δ o

eqv

Bv

A

vZ

vY

P BA

ZY

PPPPTK ⎟⎟

⎠

⎞⎜⎜⎝

⎛=)(

So …

Equilibrium constant only has meaning for a balanced reaction and known reference states

Where…The equilibrium constant

(with the partial pressures at their equilibrium values)

CEq-6From pressures to concentrations…)()()()( gZvgYvgBvgAv ZYBA +→+

BAZY vvvv

cP PRTcKK

−−+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

o

o ( ) ( )( ) ( ) BA

ZY

vB

vA

vZ

vY

cccccccccKoo

oo

////

=

Using the ideal gas law, P = cRT, we can convert KP into concentrations

eqv

Bv

A

vZ

vY

P BA

ZY

PPPPTK ⎟⎟

⎠

⎞⎜⎜⎝

⎛=)(

BAZY

BA

ZYvvvv

vB

vA

vZ

vY

P PRT

ccccK

−−+

⎟⎠⎞

⎜⎝⎛=

o

cRTP =

Add a reference state wrt

concentration…

where

Often select c°

as 1 mole per liter and don’t display it…

but don’t forget about it!

CEq-7Kp from Δf G°

and other thermodynamic terms

)(ln)( TKRTTG Pr −=Δ o

][][][][)( BGvAGvZGvYGvTG fBfAfZfYrooooo Δ−Δ−Δ+Δ=Δ

One way to find Δr G°(T)…

Another way to find Δr G°(T)…

ooo STHG rrr Δ−Δ=Δ

How can you find these?

CEq-8KP (T) for a given reaction

)()()( 235 gClgPClgPCl +→

eqPCl

ClPClP P

PPTK ⎟

⎟⎠

⎞⎜⎜⎝

⎛=

5

23)(

totjj PxP =

PKeq

eqP 2

2

1 ξξ−

=

Initially: 1 mol 0 mol 0 mol

Use to find Peq

CEq-9But KP is only a function of T!

PKeq

eqP 2

2

1 ξξ−

= Looks like a function of T and P!

But we know KP depends only on T (see CEq-5)…

Since KP is a constant at a fixed temperature, if you have a change in P, there must be a concomitant change in ξeq to maintain the same KP .

Remember what this principle is called?!

CEq-10Le Châtelier’s

Principle

Figure 26.1

Following a change in the conditions that displaces equilibrium,

a reaction will adjust to new equilibrium state (e.g., changing pressure).

)()()( 235 gClgPClgPCl +→

PKeq

eqP 2

2

1 ξξ−

=

If I raise P, what must happen to ξeq ?

CEq-11G(ξ) is a minimum at equilibrium

ξξξ

ξξξξξξ

++

+−

−+Δ+Δ−=12ln2

11ln)1(2)1()(

242RTRTGGG NOfONf

oo

Figure 26.2

ξeq = 0.1892 mol

So how do I find G(ξ) at equilibrium?

)(2)( 242 gNOgON →

CEq-12Reaction quotient, QP

Prr QRTTGG ln)( +Δ=Δ o

⎟⎟⎠

⎞⎜⎜⎝

⎛=

BA

ZY

vB

vA

vZ

vY

P PPPPTQ )(

PPr QRTKRTG lnln +−=Δ

)(ln)( TKRTTG Pr −=Δ o

eqv

Bv

A

vZ

vY

P BA

ZY

PPPPTK ⎟⎟

⎠

⎞⎜⎜⎝

⎛=)(

In general:

At equilibrium:

Arbitrary pressures

Equilibrium pressures

Together…

CEq-13QP tells which way things will go…

p

Pr K

QRTG ln=Δ ⎟⎟⎠

⎞⎜⎜⎝

⎛=

BA

ZY

vB

vA

vZ

vY

P PPPPQ

At equilibrium, Δr G = 0 and QP = KP .

If QP < KP , then QP must increase to proceed toward equilibrium.

If QP > KP , then QP must decrease to proceed toward equilibrium.

)()()()( gZvgYvgBvgAv ZYBA +→+

eqv

Bv

A

vZ

vY

P BA

ZY

PPPPK ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

CEq-14How does K depend on T?

2

/TH

TTG

P

oo Δ−=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂

Δ∂

)(ln)( TKRTTG Pr −=Δ o

We can apply the Gibbs-Helmholtz

equation

And substitute

van’t

Hoff Equation2

)(lnRT

HdT

TKd rPoΔ

=

Endothermic: Δr H°

> 0 KP (T) with T. Exothermic: Δr H° < 0 KP (T) with T.

CEq-15Integrate van’t

Hoff with constant Δr H

2

)(lnRT

HdT

TKd rPoΔ

= ∫Δ

= 2

12

1

2 )()()(ln

T

Tr

P

P dTRT

THTKTK o

⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ−=

121

2 11)()(ln

TTRH

TKTK r

P

Po

)()()()( 222 gOHgCOgCOgH +→+

RHr

oΔ−=slope

Integrate

900 °C

600 °C

Endothermic or Exothermic Reaction?

Figure 26.3

CEq-16Integrate van’t

Hoff without constant Δr H

∫Δ

= 2

12

1

2 )()()(ln

T

Tr

P

P dTRT

THTKTK o

)()(21)(

23

322 gNHgNgH →+

If we can’t consider Δr H(T) constant over the temperature range…

The relationship is no longer linear…

∫+Δ=Δ 2

1

)()()( 12

T

T Prr dTTCTHTH ooo

We know how to calculate temperature dependence of ΔH. Recall:

Where Cp (T) is often reported as a series of T from an experimental fit. (See Ex 26-7)

Endothermic or Exothermic Reaction?

Figure 26.4

CEq-17Equilibrium and Partition Functions)()()()( gZvgYvgBvgAv ZYBA +→+

0=−−+=Δ BBAAZZYYr vvvvG μμμμ

),,(),,(),,(),,(),,,,( TVNQTVNQTVNQTVNQTVNNNNQ ZYBAZYBA =

!),(

!),(

!),(

!),(),,,,,(

Z

NZ

Y

NY

B

NB

A

NA

ZYBA NTVq

NTVq

NTVq

NTVqTVNNNNQ

ZYBA

=

A

A

TVNAA N

TVqRTN

QRTj

),(lnln

,,

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−=μ

General reaction:

At equilibrium:

Recall Partition functions…

For each constituent…

Using Stirling’s

approx for NA

!

CEq-18Put it together0=−−+ BBAAZZYY vvvv μμμμ

A

AA N

TVqRT ),(ln−=μ

BA

ZY

BA

ZY

vB

vA

vZ

vY

vB

vA

vZ

vY

qqqq

NNNN

=

BA

ZY

BA

ZY

vB

vA

vZ

vY

vB

vA

vZ

vY

c ccccK

ρρρρ

==

Insert:

Algebra to get:

The equilibrium constant from molecular partition functions!

From definition of Kc

:

CEq-19A chemical example: All diatomic)(2)()( 22 gHIgIgH →+

( ))/)(/(

/

22

2

VqVqVqK

IH

HIc =

TkhDeT

T

rot

B B

vib

vib

ege

eTh

TMkV

TVq /)2/(1/

2/2/3

20

12),( υ

σπ +

Θ−

Θ−

−Θ⎟⎠⎞

⎜⎝⎛=

( )( )( )

( ) RTDDD

T

TT

HIrot

Irot

Hrot

IH

HIc

IHHI

HIvib

Ivib

Hvib

ee

eeMM

MTK /22/

//

2

2/32

20

200

2222

22 1

11)(

4)( −−

Θ−

Θ−Θ−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−−⎟⎟⎠

⎞⎜⎜⎝

⎛Θ

ΘΘ⎟⎟⎠

⎞⎜⎜⎝

⎛=

Remember from Chapter 18… For an ideal diatomic we have:

K directly from molecular quantum energy levels!

BAZY vvvv

cP PRTcTKTK

−−+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

o

o

)()(

CEq-20Compare with experiment)(2)()( 22 gHIgIgH →+

( )( )( )

( ) RTDDD

T

TT

HIrot

Irot

Hrot

IH

HIc

IHHI

HIvib

Ivib

Hvib

ee

eeMM

MTK /22/

//

2

2/32

20

200

2222

22 1

11)(

4)( −−

Θ−

Θ−Θ−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−−⎟⎟⎠

⎞⎜⎜⎝

⎛Θ

ΘΘ⎟⎟⎠

⎞⎜⎜⎝

⎛=

All parameters are listed in Table 18.2

Figure 26.5 Points = Experimental

Line = Stat Mech

Why the disagreement?

CEq-21Liquids, Solids, and Solutions

o

o

PP

RT jj ln+= μμ

arr QRTGG ln+Δ=Δ o

ar KRTG ln−=Δ o

Treatment of chemical equilibrium not so different than gas phase… we just don’t use pressures in our equations!

CEq-22a for condensed phases (liquids and solids)

jj aRT ln+= oμμ aRTdd ln=μ

dPVd =μ

aRTddPV ln=

∫ ∫ ∫ === 1

1 bar111

11ln

a P

PdP

RTVda

aad

o

so

(see Sol-24)

EX-CEq5

CEq-23a for solutions

))(( −+−+−+−+± == vvvvv mmaa γγ

))(( −+−+−+−+± == vvvvv ccaa γγ

1≈neutralγ

Recall Sol-27 to Sol 29…

On molality

scale

On molarity

scale

What about neutral components?

222211 ±== γcaa 33

2112 4 ±== γcaa44

3113 27 ±== γcaa

Summary of Table 25.3

So…

CEq-24Let’s put this into practice…

1.

What is the pH at equilibrium for a solution 0.100 M CH3

COOH in water?

2.

What is the solubility (i.e., concentration) of Ba2+

and F-

at equilibrium?

)(2)()( 22 aqFaqBasBaF −+ +→ 6107.1 −×=spK

CEq-25Summary

•

We can use the Gibbs energy to evaluate the position of equilibrium for chemical reactions.

•

The Gibbs energy also allows us to understand how chemical systems will respond when displaced from equilibrium (such as changes in pressure and temperature).

•

The equilibrium constant can be obtained directly from the partition function.

•

Thermodynamic equilibrium constants are expressed in terms of activities –

something very important when

discussing reactions involving ionic species.