Learning Objectives
- Define Gibbs free energy, and describe its relation to spontaneity
- Calculate free energy change for a process using free energies of formation for its reactants and products
- Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products
One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that we must determine the entropy change for the system and the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs. This new property is called the Gibbs free energy change (G) (or simply the free energy), and it is defined in terms of a system’s enthalpy and entropy as the following:
Free energy is a state function, and at constant temperature and pressure, the standard free energy change (ΔG°) may be expressed as the following:
(For simplicity’s sake, the subscript “sys” will be omitted henceforth.)
We can understand the relationship between this system property and the spontaneity of a process by recalling the previously derived second law expression:
The first law requires that qsurr = −qsys, and at constant pressure qsys = ΔH, and so this expression may be rewritten as the following:
ΔH is the enthalpy change of the system. Multiplying both sides of this equation by −T, and rearranging yields the following:
Comparing this equation to the previous one for free energy change shows the following relation:
The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, ΔSuniv. Table 12.3 summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.
| Relation between Process Spontaneity and Signs of Thermodynamic Properties | ||
|---|---|---|
| ΔSuniv > 0 | ΔG < 0 | spontaneous |
| ΔSuniv < 0 | ΔG > 0 | nonspontaneous |
| ΔSuniv = 0 | ΔG = 0 | reversible (at equilibrium) |
Calculating Free Energy Change
Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes according to the following relation as demonstrated in Example 12.7.
Example 12.7
Evaluation of ΔG° from ΔH° and ΔS°
Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for ΔG° say about the spontaneity of this process?Solution
The process of interest is the following:The standard change in free energy may be calculated using the following equation:
From Appendix G, here is the data:
| Substance | ||
|---|---|---|
| H2O(l) | −286.83 | 70.0 |
| H2O(g) | −241.82 | 188.8 |
Combining at 298 K:
Converting everything into kJ and combining at 298 K:
At 298 K (25 °C) and so boiling is nonspontaneous (not spontaneous).
Check Your Learning
Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of this process?Answer:
the reaction is nonspontaneous (not spontaneous) at 25 °C.
The standard free energy change for a reaction may also be calculated from standard free energy of formation values of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpy of formation, is by definition zero for elemental substances under standard state conditions. The approach used to calculate for a reaction from values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction
the standard free energy change at room temperature may be calculated as
Example 12.8
Calculation of
Consider the decomposition of yellow mercury(II) oxide.Calculate the standard free energy change at room temperature, using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?
Solution
The required data are available in Appendix G and are shown here.| Compound | |||
|---|---|---|---|
| HgO (s, yellow) | −58.43 | −90.46 | 71.13 |
| Hg(l) | 0 | 0 | 75.9 |
| O2(g) | 0 | 0 | 205.2 |
(a) Using free energies of formation:
(b) Using enthalpies and entropies of formation:
Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature.
Check Your Learning
Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?Answer:
141.5 kJ/mol, nonspontaneous