Frequently we wish to run reactions at temperatures other than 25^{o}C.
Since we know that the change in the Gibbs free energy, Δ_{r}*G*^{
o}, between products and reactants tells us whether or not the reaction will
run spontaneously we will need this quantity at the new temperature.

(Reminder: If Δ_{r}*G*^{
o} < 0 the reaction is spontaneous and if Δ_{r}*G*^{
o} > 0 the reaction is not spontaneous.)

We also know that there are two components to Δ_{r}*G*^{
o}. That is, Δ_{r}*G*^{
o} = Δ_{r}*H*^{ o}
− *TΔ*_{r}*S*^{
o}, where the Δ_{r}*H*^{
o} term is only weakly dependent on temperature, but the *TΔ*_{r}*S*^{
o} term is strongly dependent on temperature due to the presence of
the *T* in the term.

So the question becomes, how do these things balance out? What is the
dependence of Δ_{r}*G*^{ o}
on temperature? The simple answer is obtained from the derivative of *G*
with respect to *T*.

(1)If we apply Equation 1 to Δ

(2a, b, c)Equation 2c shows that if Δ

However, if we want to know whether or not a reaction is favored by an increase or decrease in temperature we really need to be looking at the equilibrium constant. If the equilibrium constant increases with temperature the reaction becomes more favored, but if the equilibrium constant decreases with temperature the reaction becomes less favored.

We will show later that the equilibrium constant for a chemical reaction
depends on Δ_{r}*G*^{ o}
/*T* and not on Δ_{r}*G*^{
o} all by itself. (There is a good reason for this which we will discuss
below.) The Gibbs-Helmholtz equation addresses the question as to how Δ_{r}*G*^{
o} /*T* changes with temperature. The Gibbs-Helmholtz equation
can be presented in two different, but equivalent forms. If we are just
worrying about *G* itself the two forms look like,

(3a)and

(3b)Or, applying the procedure we used in Equations 2a, b, and c, we can write,

(3c)and

(3d)We won't derive any of these equations because the derivation is not particularly instructive. We will just show that the version of Equation 3a is true. (On the way you will also see why Equation 3b is true.)

Start with the left-hand side of Equation 3a and show that it is equivalent to the definition of Gibbs free energy:

We could easily substitute ΔSo what is this good for? We will see that it is tremendously useful
after we know the relationship between the equilibrium constant and Δ_{r}*G*^{
o} /*T* , but for now let's use the Gibbs-Helmholtz equation
to calculate Δ_{r}*G*^{ o}
at a temperature, *T*_{2} other than 25^{o}C (which
we will call *T*_{1}.

Set up Equation 3c for integration.

(4)Equation 5 integrates to give,(5)

(6)To carry out the integration on the right-hand side of Equation 6 we would need to know how the heat of reaction changes with temperature. This information can be obtained if we know the heat capacities of the reactants and products as functions of temperature. However, usually the heat of reaction varies slowly with temperature so that it is a good approximation to regard the heat of reaction as constant.

Notice that Equation 5 is trivial to integrate if we make the approximation
that Δ_{r}*H*^{ o} is constant.
With this approximation, Equation 5 integrates to,

(7)or,

(8)We will apply Equation 8 to calculate Δ

HDo we know Δ_{2}O(l,− 20^{o}C) → H_{2}O(s, − 20^{o}C).

(9)Then

(10a, b, c)Equation 8, for our problem, becomes, Then Δ

**Why Δ G/T ?**

We said above that Δ_{r}*G*^{
o} tells us whether or not a reaction wants to go, but that the equilibrium
constant, the ultimate arbiter of how strongly the reaction wants to go,
depends on Δ_{r}*G*^{ o}
/*T* . We will prove this statement later, but for now it might be
interesting to see why that might be. Consider any process at constant
temperature and pressure. We know that if *T* and *p* are constant
then,

(11)Let's divide Equation 11 by

(12)Note that Δ

(13)Now, Δ

(14a, b)Most likely our process is irreversibly, but that doesn't matter because

(15a, b)or

(16)Applying the result of Equation 16 to Equation 13 we find that,

(17a, b)Equation 17b tells us several things. It tells us that, in the final analysis, the ultimate driving force in nature is entropy, that is, the drive toward disorder. The system plus the surroundings is a closed isolated system so that the only spontaneous processes allowed are those which increase the entropy. Secondly, it explains why Δ

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