Recall our favorite hypothetical chemical reaction,

We can get ΔaA +bB →cC +dD. (1)

. (2)The quantity Δ

Δ_{r}*G*^{o} can also
be written in terms of chemical potentials of the components.

ΔIn these equations all the components are in their standard states._{r}G^{o}=cμ_{C}^{o}+dμ_{D}^{o}−aμ_{A}^{o}−bμ_{B}^{o}. (3)

Usually Δ_{r}*G*^{ 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, *a*_{i}
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,

Δso that,_{r}G=cμ_{C}+dμ_{D}−aμ_{A}−bμ_{B}=

c(μ_{C}^{o}+RTlna_{C}) +d(μ_{D}^{o}+RTlna_{D}) −a(μ_{C}^{o}+RTlna_{A}) −b(μ_{B}^{o}+RTlna_{B})=

cμ_{C}^{o}+dμ_{D}^{o}−aμ_{A}^{o}−bμ_{B}^{o}+RT(clna_{C}+dlna_{D}−alna_{A}−blna_{B}), (5a)

Δor_{r}G= Δ_{r}G^{o}+ RT(lna_{C}+ ln^{c}a_{D}− ln^{d}a_{A}− ln^{a}a_{B}), (5b)^{b}

. . (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,

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

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

dorG≤ 0, (8)_{T,p}

dLet the reaction go by an amountG= &Sigma_{T,p}μ_{i}dn≤ 0. (9)_{i}

Thendn_{C}=c dndn_{D}=d dndn_{A}= −a dndn_{B}= −b dn.

SincedG=_{T,p}cμ_{C}dn+dμ_{D}dn−aμ_{A}dn−bμ_{B}dn(10)

= (cμ_{C}+dμ_{D}−aμ_{A}−bμ_{B})dn

dG= Δ_{T,p}_{r}Gdn≤0. (11)

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

Δwhen the reaction mixture is at equilibrium._{r}G^{o}+RTlnQ_{eq}= 0, (12)

But at equilibrium *Q* _{eq} = *K*_{a} , where
*K*_{a} is the thermodynamic equilibrium constant (that is,
the equilibrium constant written in terms of activities rather than concentrations
or pressures).

Then

0 = ΔDrop the subscript "r" for a while._{r}G^{ o}+RTlnK_{a}, (13)Δ

_{r}G^{ o}= −RTlnK_{a}. (14)

ΔorG^{ o}= −RTlnK_{a}(15)

(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 Δ

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

The ideal gas equilibrium constant then becomes,a=_{i}p/1 atm. (18)_{i}

(19)where we have collected all the "1 atm" together. (Note that both

(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,

where the γf=_{i}p, (21)_{i}γ_{ i}

(22)You could write a "pure pressure" equilibrium constant,

(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

ΔorG^{ o}= −RTlnK_{a}(29)

(30)then,

(31)or

(32)Integrate

(33)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 Δ(34)

(35)

Example 1

First, let's find the equilibrium constant at 25^{o}C for the
formation of liquid water from hydrogen and oxygen. Consider the reaction,

2 H(We have picked a particularly simple example because Δ_{2}(g) + O_{2}(g) → 2 H_{2}O(l).

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 100^{o}C. (We have
picked 100^{o}C in order to avoid the complication of the vaporization
of water above 100^{o}C. The easiest way to get around this problem
would be to use the table values for the reaction which produces H_{2}O(*g*)
at 25^{o}C.)

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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