Chemical Equilibrium
Recall our favorite hypothetical chemical reaction,
aA + bB  cC + dD.
(1)
We can get rG o from the free energies of formation, fG o,
rGo  cf GCo  d f GDo  af GAo  b f GBo . (2)
The quantity rGo tells us whether or not the reaction "wants to go." Recall, if rGo < 0
the reaction will proceed spontaneously as written and if rGo > the reaction will not
proceed spontaneously as written (but the reverse reaction will proceed spontaneously).
rGo can also be written in terms of chemical potentials of the components.
rG o = cCo + dDo - aAo - bBo.
(3)
In these equations all the components are in their standard states.
Usually rG o  0 because chemical reactions are not equilibrium processes.
Suppose we don’t want the components to be in their standard states. Then we don’t
want to use the standard state chemical potentials so we must have an expression for the
chemical potentials when they are not necessarily in their standard states. We will write
the general form of the chemical potential of component i in terms of the activity, ai
because we do not want to restrict our discussion to gases or ideal solutions, etc. That is,
we write,
i  io  RT ln ai .
(4)
Then we replace the activity, ai, by whatever is appropriate for the particular
circumstance. That is, we write ai = pi/1 atm for an ideal gas, or ai = fi/1atm for a real
gas, etc. We can still write rG as a sum and difference of the chemical potentials. That
is,
rG = cC + dD - aA - bB
= c(Co + RTlnaC) + d(Do + RTlnaD)  a(Co + RTlnaA)  b(Bo + RTlnaB)
= cCo + dDo  aAo  bBo + RT(clnaC + dlnaD – alnaA – blnaB) (5a)
rG = rGo + RT(lnaCc + lnaDd – lnaAa – lnaBb)
(5b)
or
 rG   rG o  RT ln
aCc aDd
.
aAa aBb
(6)
(There is an interesting problem in going from Equation 5a to Equation 5b which is not
usually discussed. In Equation 5a the coefficients of the balanced chemical equation, a,
b, etc. have units of moles, so that, for example, cC is moles times Joules per mole
which leaves just units of Joules. However, the a, b, etc. in Equation 5b must be unitless.
That means that we wrote, for example, c mol = 1 mole  c unitless. When we bring the
unitless value of c inside the logarithm we left the 1 mole out to multiply the gas
constant, R, so that 1 molRT has units of Joules. Most people just ignore this, but if
you track the units carefully at the end of this page you will see that it is necessary to
keep the 1 mol out of the logarithm for the units to make sense.)
The quantity inside the logarithm has the form of an equilibrium constant, but it does not
necessarily have the value of the equilibrium constant. We will give this product and
quotient of activities (with their powers) the symbol Q. Then,
 rG   rG o  RT ln Q .
(7)
The values of the activities appearing in Q are whatever we choose them to be and we
can make any choice we wish. The equation then gives us the rG for our choice of the
activities. If we choose the reactants and products to be in their standard states then all
the activities are 1 and we get rGo back again. If we choose some different values for
the various activities then we get rG for that choice.
Let’s now take a mixture of reactants and products, held at constant p and T, and ask
what would be the condition for equilibrium. We know that the criterion for equilibrium
at constant p and T is
dGT,p  0,
(8)
or
dGT,p =
i dni
 0. (9)
Let the reaction go by an amount dn ( dn > 0). That is,
dnC
dnD
dnA
dnB
=
=
=
=
Then
c dn
d dn
 a dn
 b dn.
dGT,p = cCdn + dD dn - aA dn - bB dn
(10)
= (cC + dD - aA - bB )dn
dGT,p = rG dn  0.
(11)
Since dn > 0 by construction we conclude that rG  0.
Away from equilibrium the  sign holds so rG < 0. This tells us something we already
knew. If the reaction is to proceed in the direction which makes dn > 0 then rG must be
negative so that the Gibbs free energy goes down. At equilibrium the = sign holds so
rG = 0. We conclude that
rG o + RT ln Qeq = 0,
(12)
when the reaction mixture is at equilibrium.
But at equilibrium Q eq = Ka , where Ka is the thermodynamic equilibrium constant (that
is, the equilibrium constant written in terms of activities rather than concentrations or
pressures).
Then
0 = r G o + RTlnKa
r G o =  RTlnKa
(13)
(14)
Drop the subscript “r” for a while.
G o =  RTlnKa
(15)
or
Ka  e
e
e

G o
RT

H o T S o
RT

H
RT
o
e
S
R
(16a, b, c)
o
.
These equations give us some new insight on the "driving forces" for chemical reactions.
First, they remind us that there are two "drives" in nature: there is the drive toward
stability, which shows up here in the Ho term, and the drive toward disorder which
shows up in So. Since the Ho/T term will vary much more rapidly with temperature
than the So term we can make one or the other dominate by changing the temperature.
At high temperatures we can reduce the contribution of Ho and make So dominate and
at low temperatures we can force the Ho term to dominate.
The thermodynamic equilibrium constant for our hypothetical reaction is
aCc aDd
Ka  a b ,
aA aB
(17)
where all the activities have their equilibrium values. For ideal gas reactions we can
write the equilibrium constant in terms of the partial pressures of the gases since for ideal
gases
ai = pi/1 atm.
(18)
The ideal gas equilibrium constant then becomes,
pCc pDd
K a  a b (1atm)abcd ,
pA pB
(19)
where we have collected all the "1 atm" together. (Note that both Q and Ka must be
unitless. It is not unusual for people to omit the "1 atm" term and write,
pCc pDd
Ka  a b ,
pA pB
(20)
but we should not forget that there is an implied 1 atm dividing each pressure.
For nonideal gas reactions we replace the pressure by the fugacity and write the fugacity
as,
fi = pii ,
(21)
where the i , called the "activity coefficient," contains all the nonideality information.
The thermodynamic equilibrium constant for nonideal gases is then (omitting all the "1
atm" terms,
pCc pDd  Cc  Dd
Ka  a b a b .
pA pB  A B
(22)
You could write a "pure pressure" equilibrium constant, Kp, as,
pCc pDd
Ka
 Aa  Bb
K p  a b  c d  Ka c d .
pA pB  C D
 C D
(23)
 Aa  Bb
For a solute in solutions we will write
i  io  RT ln mi
(24)
for ideal solutions, and
i  io  RT ln mi i
(25)
for nonideal solutions, so that the thermodynamic equilibrium constant would be,
mCc mDd  Cc  Dd
Ka  a b a b .
mA mB  A B
(26)
(Note that we have omitted writing an implied 1 molal in the denominator inside the
logarithm above.)
Equilibrium constants at other temperatures
Recall the Gibbs-Helmholtz equation,
 G
 T
 1
 
 T


  H .

p
(27)
This equation also works for chemical reactions with their components in their standard
states, so
 G o
 T

  1
 T
but


o
  H ,

p
(28)
G o = - RTlnKa
(29)
G o
  R ln K a ,
T
(30)
or
then,


  ( R ln K a ) 
o

  H ,
1




T
p
(31)
or


  ln K a 
H o

.

1 
R
 

 T p
(32)
Integrate
H o 1
d
R
T
T1
T2
T2
 d ln K a   
T1

H
R
ln K aT2  ln K aT1
K aT2  K aT1 e

o T2
(33)
1
dT ,
T1
K aT2
H o  1 1 



ln


R  T2 T1 
K aT1
H o  1 1 
  
R  T2 T1 
.
(34)
(35)
Notice that this last equation has the same form as the integrated Clausius-Clapeyron
equation. Notice also that this equation states in mathematical terms something that we
know from qualitative arguments, namely Le Chatelier's principle. That is, if a reaction
is endothermic an increase in temperature favors products and if a reaction is exothermic
an increase in temperature favors reactants. This is easy to see because for an increase in
temperature the temperature part is negative so that a positive H o causes the original
equilibrium constant to be multiplied by a number larger than 1.
Example 1
First, let's find the equilibrium constant at 25oC for the formation of liquid water from
hydrogen and oxygen. Consider the reaction,
2 H2(g) + O2(g)  2 H2O(l).
(We have picked a particularly simple example because rG o for this reaction is just
twice the Gibbs free energy of formation of liquid water, namely 2 mol  ( 237.13
kJ/mol) =  474.26 kJ.
From Equation 16a we write
Ka  e
e

rG o
RT

474.26 kJ1000J / kJ
1 mol8.3145 J/Kmol298.15 K
(36a, b, c, d)
 e191.3
 1.22  1083.
(For an explanation of the "1 mol" in the denominator of the exponent of Equation 36b,
see the parenthetical remark after Equation 6 above.) This equilibrium constant is quite a
large number and indicates that the reaction goes essentially entirely to completion.
Example 2
Let's now see how this equilibrium constant changes with temperature. We will calculate
the equilibrium constant at 100oC. (We have picked 100oC in order to avoid the
complication of the vaporization of water above 100oC. The easiest way to get around
this problem would be to use the table values for the reaction which produces H2O(g) at
25oC.)
From Equation 35 we write,
KT2  KT1 e

H o  1 1 
  
R  T2 T1 
. (35b)
The heat of reaction we obtain from the tables as 2mol(285.83kJ/mol) = 571.66kJ.
KT2  KT1 e

571.66kJ1000J/kJ 
1
1




1mol8.3145J/Kmol  373.15K 298.15K 
 KT1 e46.349
 1.22  10  7.425  10
83
,
(36a, b, c, d)
21
 9.06  1062.
We can understand this result in different ways. First, we know that the reaction is
strongly exothermic so that Le Chatelier's principle tells us that an increase in
temperature favors reactants. Second, we can tell qualitatively that the entropy change
for the reaction is large and negative so that the entropy change favors reactants.
Equation 16c then tells us that the effect of entropy does not change very much with
temperature, but the contribution of the heat of the reaction decreases with increasing
temperature.
WRS