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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.