Thermodynamics and Equilibrium
4.4 FREE ENERGY
The spontaneity of a process depends upon both the entropy and the enthalpy change of the system, but the effects of these two thermodynamic properties are combined into one, the entropy of the universe, in the second law. To separate the effects of ΔH and ΔS explicitly and to make the predictive power of the second law more manageable, we require an expression for ΔSuniv in terms of system variables. We begin by dividing the thermodynamic universe into a system and its surroundings:
ΔSuniv = ΔSsur + ΔS
Entropy changes in the surroundings result from changes in the thermal surroundings, which are caused by the transfer of heat between the surroundings and the system. At constant pressure and temperature, the heat flowing from the system into the surroundings is -ΔH. Setting qrev = -ΔH in Equation 4.6, we obtain
ΔSsur = - ΔH T
Thus, exothermic reactions are favored because they add heat to the surroundings, which increases the entropy of the surroundings. Substitution of this expression for ΔSsur into the expression for ΔSuniv yields
ΔSuniv = - ΔH T + ΔS
Multiplying the above equation by -T, we obtain
-TΔSuniv = ΔH - TΔS
The right side of the above expression depends only on system quantities, so it is also a property of the system. This new thermodynamic property is known as the Gibb’s free energy and given the symbol ΔG.*
ΔG = ΔH - TΔS = -TΔSuniv Eq. 4.8
According to the second law, a spontaneous process is one in which ΔSuniv > 0 or -TΔSuniv < 0. Consequently, a spontaneous process at constant temperature and pressure is one in which ΔG < 0; i.e., reactions proceed in the forward direction (→) when ΔG < 0, in the reverse direction (←) when ΔG > 0, and they are at equilibrium (U) when ΔG = 0.
The second law can also be stated as follows:
ΔG < 0 for all spontaneous processes at constant temperature and pressure; i.e., the sign of ΔG indicates the spontaneous direction.
Equation 4.8 shows that the free energy results from the interplay of two energy terms:
ΔH and TΔS. ΔH is the potential energy difference between the reactants and the products.
It arises from differences in bond energies and intermolecular interactions. ΔH < 0 when the energy of the reactants is greater than that of the products (Figure 4.7a). The decrease in enthalpy results in the release of energy, which decreases the free energy of the system.
However, reactions in which the products are at higher energy (Figure 4.7b) require energy, which increases the free energy of the system.
TΔS is the free energy change resulting from entropy differences in the reactants and products. If TΔS > 0 (Figure 4.8a), then the entropy of the products is greater than that of the reactants and the reaction releases TΔS joules of energy, which decreases the free energy. When TΔS is negative (Figure 4.8b), the reaction produces a system that has less entropy, which requires an input of -TΔS joules and increases the free energy.
ΔG is called the ‘free energy’ because it is the energy that is free to do work at constant temperature and pressure.
-wmax = -ΔG Eq. 4.9
* The Gibb’s free energy, ΔG, is the free energy only for processes carried out at constant T and P. However, these are the only conditions considered in this text, so ΔG will be referred to as simply the “free energy.”
Figure 4.7 Enthalpy change Figure 4.8 Entropy change Systems seek to lower their
potential energy, so ΔH < 0 is favorable.
Systems seek to increase their entropy, so ΔS > 0 is favorable.
-wmax is the maximum amount of work* that can be done by a system during a process at constant T and P. If ΔG is negative, free energy is released and the process can do work, but, if it is positive, ΔG joules must be supplied to force the process uphill in free energy.
For example, consider a process in which ΔH = -100 J and TΔS = -90 J. The ΔH term is favorable as 100 J of energy are given off by the change in bonds and other interactions.
However, the TΔS term is unfavorable because ΔS < 0. At the temperature of the process, 90 J of energy is required to decrease the entropy, and it must come from the ΔH term if the process is to be spontaneous. Thus, ΔG = -100 - (-90) = -10 J. ΔG < 0, so the process is spontaneous, but only 10 J of energy are free to do work.
Most of our calculations will be for the standard state, so we apply Equation 4.8 to the standard state to obtain the standard free energy of reaction
ΔGo = ΔHo - TΔSo Eq. 4.10
ΔGo is the free energy of a reaction when all reactants and products are in their standard state, so its sign indicates the spontaneous direction under these specific set of conditions.
Consider the reaction A(g) → B(g). ΔGo is the value of ΔG when both A and B are in their standard states, which is a partial pressure of 1 atm for each gas. If ΔGo < 0, the spontaneous direction is A → B. If A is consumed and B is formed when they are at equal pressures, then B will be present in the greater amount when the reaction is complete; i.e., the reaction is extensive because there is more product than reactant at completion. If ΔGo
> 0, the spontaneous process is A ← B when they are at equal pressures, so more A than B will be present at completion and the reaction is not extensive. We conclude that
The sign of ΔGo indicates the side of the reaction that is favored at equilibrium. If ΔGo > 0, the reactants are favored, but if ΔGo < 0, the products are favored.
Thus, ΔG and ΔGo have very different meanings, but they are often confused. For example, ΔGovap = +8.6 kJ.mol-1 for water at 25 oC, which is often misinterpreted to mean that the evaporation of water is not spontaneous at 25 oC. However, ΔGo > 0 means that the water cannot evaporate spontaneously at standard conditions. Consistent with ΔGo > 0, water vapor would spontaneously condense if its pressure were 1 atm at 25 oC. The fact that ΔGo > 0 simply means that the equilibrium pressure of water vapor is less than 1 atm.
In fact, the equilibrium pressure of water vapor at 25 oC is only 0.031 atm.
* It is the maximum amount of work because some of the energy change in real processes is lost, usually through heat.
PRACTICE EXAMPLE 4.2
Determine the standard entropies of formation of the following:
a) SO3(g)
formation reaction (Hint: The standard state of S is the state that has a zero heat of formation):
ΔSof = ___________ J.mol-1.K-1 b) NH4Cl(g)
formation reaction:
ΔSof = ___________ J.mol-1.K-1 c) O2(g)
formation reaction:
ΔSof = ___________ J.mol-1.K-1