Gibbs Free Energy and Temperature: The Gibbs-Helmholtz Equation

Frequently we wish to run reactions at temperatures other than 25oC. Since we know that the change in the Gibbs free energy, ΔrG 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 ΔrG o < 0 the reaction is spontaneous and if ΔrG o > 0 the reaction is not spontaneous.)

We also know that there are two components to ΔrG o. That is, ΔrG o = ΔrH orS o, where the ΔrH o term is only weakly dependent on temperature, but the rS 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 ΔrG o on temperature? The simple answer is obtained from the derivative of G with respect to T.

If we apply Equation 1 to ΔrG o = G oproductsG oreactants we get,
                (2a, b, c)
Equation 2c shows that if ΔrS o is positive ΔrG o decreases with temperature, but if ΔrS o is negative ΔrG o increases with temperature. This will tell us whether ΔrG o increases or decreases with increasing temperature.

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 ΔrG o /T and not on ΔrG 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 ΔrG 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,

Or, applying the procedure we used in Equations 2a, b, and c, we can write,
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 ΔG for G and end up with ΔH in the above sequence of equations.

So what is this good for? We will see that it is tremendously useful after we know the relationship between the equilibrium constant and ΔrG o /T , but for now let's use the Gibbs-Helmholtz equation to calculate ΔrG o at a temperature, T2 other than 25oC (which we will call T1.

Set up Equation 3c for integration.



Equation 5 integrates to give,
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 ΔrH o is constant. With this approximation, Equation 5 integrates to,

We will apply Equation 8 to calculate ΔG for the freezing of super-cooled water at − 20oC. The process is,
H2O(l,− 20oC) → H2O(s, − 20oC).
Do we know ΔG for this process at 0oC? Note that freezing water at its normal melting point is a reversible process so that the heat of fusion is a reversible heat at constant temperature. Therefore,
                (10a, b, c)
Equation 8, for our problem, becomes,
Then ΔG is − 440 J which is what we would expect because freezing super-cooled water is a spontaneous (and irreversible) process at − 20oC.

Why ΔG/T ?

We said above that ΔrG 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 ΔrG 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,

Let's divide Equation 11 by T to see what ΔG/T looks like,
Note that ΔH and ΔS in Equation 12 refer to things happening in the system. Rewrite Equation 12 to indicate that this is true,
Now, ΔHsys is heat absorbed by the system. Where did this heat come from? It had to come from the surroundings so.
               (14a, b)
Most likely our process is irreversibly, but that doesn't matter because H is a state function so that ΔH is independent of path. We can always find a reversibly path to change the enthalpy of the surroundings by an amount ΔHsurr. (Our process is at constant temperature and pressure so that ΔH in both cases is a heat at constant pressure.) If we add heat ΔHsurr to the surroundings isothermally and reversibly then the entropy change in the surroundings is
               (15a, b)
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 ΔG/T is more important than simply ΔG in determining how strongly spontaneous a process is. It is −ΔG/T which is related to the entropy change of the universe, not ΔG directly. Equation 17b also shows that the Gibbs free energy manages to include the entropy effects in the surroundings without ever telling us that it is doing so.


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Last updated 5 Nov 04