Chemical Equilibrium

Recall our favorite hypothetical chemical reaction,

aA + bB → cC + dD.                (1)
We can get ΔrG o from the free energies of formation, ΔfG o,
.           (2)
The quantity ΔrGo tells us whether or not the reaction "wants to go." Recall, if ΔrGo < 0 the reaction will proceed spontaneously as written and if ΔrGo > the reaction will not proceed spontaneously as written (but the reverse reaction will proceed spontaneously).

ΔrGo can also be written in terms of chemical potentials of the components.

ΔrG o = Co + DoAoBo.                (3)
In these equations all the components are in their standard states.

Usually ΔrG 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,

. (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 ΔrG as a sum and difference of the chemical potentials. That is,
ΔrG = C + DAB

= cCo + RTlnaC) + dDo + RTlnaD) − aCo + RTlnaA) − bBo + RTlnaB)

= Co + DoAoBo + RT(clnaC + dlnaDalnaAblnaB),           (5a)

so that,
ΔrG = ΔrGo + RT(lnaCc + lnaDd − lnaAa − lnaBb),           (5b)
or
.              . (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 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,

.                (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 ΔrG 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 ΔrGo back again. If we choose some different values for the various activities then we get ΔrG 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 = &Sigma μi dni ≤ 0.                (9)
Let the reaction go by an amount dn ( dn > 0). That is,
dnC = c dn
dnD = d dn
dnA = − a dn
dnB = − b dn.
Then
dGT,p = Cdn + D dnA dnB dn                (10)
= (C + DAB )dn

dGT,p = ΔrG dn ≤0.                (11)

Since dn > 0 by construction we conclude that ΔrG ≤ 0.

Away from equilibrium the < sign holds so ΔrG < 0. This tells us something we already knew. If the reaction is to proceed in the direction which makes dn > 0 then ΔrG must be negative so that the Gibbs free energy goes down. At equilibrium the = sign holds so ΔrG = 0. We conclude that

ΔrG 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 = Δr G o + RT lnKa ,               (13)

Δr G o = − RT lnKa .               (14)

Drop the subscript "r" for a while.
ΔG o = − RT lnKa                (15)
or
(16a, b, c)
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 ΔHo term, and the drive toward disorder which shows up in ΔSo. Since the ΔHo/T term will vary much more rapidly with temperature than the ΔSo term we can make one or the other dominate by changing the temperature. At high temperatures we can reduce the contribution of ΔHo and make ΔSo dominate and at low temperatures we can force the ΔHo term to dominate.

The thermodynamic equilibrium constant for our hypothetical reaction is

(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,
(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,
(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 = piγ i ,                (21)
where the γi , called the "activity coefficient," contains all the nonideality information. The thermodynamic equilibrium constant for nonideal gases is then (omitting all the "1 atm" terms),
(22)
You could write a "pure pressure" equilibrium constant, Kp, as,
(23)
For a solute in solutions we will write
(24)
for ideal solutions, and
(25)
for nonideal solutions, so that the thermodynamic equilibrium constant would be,
(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,

(27)
This equation also works for chemical reactions with their components in their standard states, so
(28)
but
ΔG o = − RTlnKa                (29)
or
(30)
then,
(31)
or
(32)
Integrate
(33)

(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 ΔH 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 ΔrG 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

(36a, b, c, d)
(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,

(35b)
The heat of reaction we obtain from the tables as 2mol× (− 285.83kJ/mol) = − 571.66kJ.
(36a, b, c, d)
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

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