Free Energy Functions

Helmholtz Free Energy (Constant V and T)

In an isolated system the criterion dS > 0 indicates that a process is spontaneous. In general, we must consider dS_sys for the system and dS_surr for surroundings. Since we can think of the entire universe as an isolated system dS_total > 0. The entropy tends to increase for the universe as a whole. If we decompose dS_total into the entropy change for the system and that for the surroundings we find that we have a criterion for spontaneity for the system that also requires consideration of the entropy change in the surroundings. The free energy functions will allow us to eliminate consideration of the surroundings and to express a criterion for spontaneity solely in terms of parameters that depend on the system.

Starting with the First Law

dU = dw + dq

At constant temperature and volume we have dw = 0 and

dU = dq

Recall that dS ³ dq/T so we have

dU £ TdS

which leads to

dU - TdS £ 0

Since T and V are constant we can write this as

d(U - TS) £ 0

We realize that the quantity in parentheses is a measure of spontaneity of the system that depends on known state functions. We therefore define a new state function

A = U -TS

which means that

dA £ 0.

We call A the Helmholtz free energy. At constant T and V the Helmholtz free energy will decrease until all possible spontaneous processes have occurred. At that point the system will be in equilibrium. The condition for equilibrium is dA = 0.

Expressing the change in the Helmholtz free energy we have

DA = DU – TDS

for an isothermal change from one state to another.

The condition for spontaneous change is that DA is less than zero and the condition for equilibrium is that DA = 0.

We write

DA = DU – TDS £ 0 (at constant T and V)

In a case where DA is greater than zero a process is not spontaneous. It can occur if work is done on the system, however. The Helmholtz free energy has an important physical interpretation. Noting the q_rev = TDS we have

DA = DU – q_rev

According to the first law DU – q_rev = w_rev so

DA = w_rev (reversible, isothermal)

If DA < 0 then it represents the maximum amount of work that can be extracted from the system if the change occurs reversibly and isothermally. Any irreversibility introduced due to friction etc. will result in an amount of work less than DA being extracted.

Gibbs Free Energy (Constant P and T)

Most reactions occur at constant pressure rather than constant volume. To derive the constant pressure free energy function we begin with the first law.

Using the facts that q_rev = TdS and w_rev = -PdV we can state that

dU £ TdS – PdV

which can be written dU - TdS + PdV £ 0. As above the equals sign applies to an equilibrium condition and the < sign means that the process is spontaneous.

or d(U - TS + PV) £ 0 (at constant T and P)

We define a state function G = U + PV – TS = H – TS.

Thus, dG £ 0 (at constant T and P)

The quantity G is called the Gibb's free energy. In a system at constant T and P, the Gibb's energy will decrease as the result of spontaneous processes until the system reaches equilibrium, where dG = 0.

Comparing the Helmholtz and Gibb's free energies we see that A(V,T) and G(P,T) are completely analogous except that one occurs at constant volume and the second at constant pressure.

We can see that

G = A + PV

which is exactly analogous to

H = U + PV

the relationship between enthalpy and internal energy.

For chemical processes we see that

DG = DH – TDS £ 0 (at constant T and P)

is the analog of the equation for DA above.

We can consider various possible scenarios for a chemical process

DH	DS	Description of process
>0	>0	Endothermic, spontaneous for T > DH/DS
<0	<0	Exothermic, spontaneous for T < DH/DS
<0	>0	Exothermic, spontaneous for all T
>0	>0	Never spontaneous

For a phase transition DG = 0 of the two phases are in equilibrium.

For example, for water liquid and vapor are in equilibrium at 373.15 K (at 1 atm of pressure). We can write

where we have expressed G as a molar free energy. From the definition of free energy we have

The magnitude of the molar enthalpy of vaporization is 40.7 kJ/mol and that of the entropy is 108.9 J/mol-K. Thus,

However, if we were to calculate the free energy of vaporization at 363.15 K we would find that it is +1.1 kJ/mol so vaporization is not spontaneous at that temperature. If we consider the free energy of vaporization at 383.15 K it is -1.08 kJ/mol and so the process is spontaneous (DG < 0).

Just as the Helmholtz free energy represents the maximum amount of P-V work that can be extracted from the system, the Gibb's free energy also represents the maximum amount of non P-V work that can be extracted. This statement may seem kind of strange. We have not really discussed non-P-V work (and most books do not either). Non P-V work usually is taken to mean electrical work. Thus, it is the Gibb's free energy that is used in deriving expressions that describe work done in electrochemical cells. To see where this comes from, consider that

G = U - TS + PV or by differentiation

dG = dU - TdS - SdT + PdV + VdP.

From the first law we have

dU = TdS + dw_rev.

Thus,

dG = dw_rev - SdT + PdV + VdP.

If we were to only consider P-V work then we would say that

dw_rev = - PdV.

This leads to one expression of the Gibb's free energy

dG = - SdT + VdP.

Note that at constant T and P dG = 0.

If we allow for dw_rev to represent not only P-V work but also non P-V work such as electrical work, then

dw_rev = - PdV + dw_nonPV.

Substituting this into the expression above for dG we have

dG = dw_nonPV - SdT + VdP.

At constant temperature and pressure this gives

dG = dw_nonPV.

Just as was the case for the maximum P-V, any irreversibility in the system will result in less than the maximum work being extracted.