Chemistry: Atoms First

# 12.4Free Energy

Chemistry: Atoms First12.4 Free Energy

### Learning Objectives

By the end of this section, you will be able to:
• 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:

$G=H−TSG=H−TS$
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Free energy is a state function, and at constant temperature and pressure, the standard free energy change (ΔG°) may be expressed as the following:

$ΔG=ΔH−TΔSΔG=ΔH−TΔS$
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(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:

$ΔSuniv=ΔS+qsurrTΔSuniv=ΔS+qsurrT$
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The first law requires that qsurr = −qsys, and at constant pressure qsys = ΔH, and so this expression may be rewritten as the following:

$ΔSuniv=ΔS−ΔHTΔSuniv=ΔS−ΔHT$
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ΔH is the enthalpy change of the system. Multiplying both sides of this equation by −T, and rearranging yields the following:

$−TΔSuniv=ΔH−TΔS−TΔSuniv=ΔH−TΔS$
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Comparing this equation to the previous one for free energy change shows the following relation:

$ΔG=−TΔSunivΔG=−TΔSuniv$
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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)
Table 12.3

### 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.

$ΔG°=ΔH°−TΔS°ΔG°=ΔH°−TΔS°$
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### 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:
$H2O(l)⟶H2O(g)H2O(l)⟶H2O(g)$
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The standard change in free energy may be calculated using the following equation:

$ΔG298°=ΔH°−TΔS°ΔG298°=ΔH°−TΔS°$
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From Appendix G, here is the data:

Substance $ΔHf°(kJ/mol)ΔHf°(kJ/mol)$ $S298°(J/K·mol)S298°(J/K·mol)$
H2O(l) −286.83 70.0
H2O(g) −241.82 188.8

Combining at 298 K:

$ΔH°=ΔH298°=ΔHf°(H2O(g))−ΔHf°(H2O(l))=[−241.82 kJ−(−286.83)]kJ/mol=45.01 kJ/molΔH°=ΔH298°=ΔHf°(H2O(g))−ΔHf°(H2O(l))=[−241.82 kJ−(−286.83)]kJ/mol=45.01 kJ/mol$
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$ΔS°=ΔS298°=S298°(H2O(g))−S298°(H2O(l))=188.8J/mol·K−70.0J/K=118.8J/mol·KΔS°=ΔS298°=S298°(H2O(g))−S298°(H2O(l))=188.8J/mol·K−70.0J/K=118.8J/mol·K$
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$ΔG°=ΔH°−TΔS°ΔG°=ΔH°−TΔS°$
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Converting everything into kJ and combining at 298 K:

$ΔG298°=ΔH°−TΔS°=45.01 kJ/mol−(298 K×118.8J/mol·K)×1 kJ1000 JΔG298°=ΔH°−TΔS°=45.01 kJ/mol−(298 K×118.8J/mol·K)×1 kJ1000 J$
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$45.01 kJ/mol−35.4 kJ/mol=9.6 kJ/mol45.01 kJ/mol−35.4 kJ/mol=9.6 kJ/mol$
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At 298 K (25 °C) $ΔG298°>0,ΔG298°>0,$ and so boiling is nonspontaneous (not spontaneous).

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?
$C2H6(g)⟶H2(g)+C2H4(g)C2H6(g)⟶H2(g)+C2H4(g)$
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$ΔG298°=102.0 kJ/mol;ΔG298°=102.0 kJ/mol;$ 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 $(ΔGf°)(ΔGf°)$ 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, $ΔGf°ΔGf°$ is by definition zero for elemental substances under standard state conditions. The approach used to calculate $ΔG°ΔG°$ for a reaction from $ΔGf°ΔGf°$ values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction

$mA+nB⟶xC+yD,mA+nB⟶xC+yD,$
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the standard free energy change at room temperature may be calculated as

$ΔG298°=ΔG°=∑νΔG298°(products)−∑νΔG298°(reactants)=[xΔGf°(C)+yΔGf°(D)]−[mΔGf°(A)+nΔGf°(B)].ΔG298°=ΔG°=∑νΔG298°(products)−∑νΔG298°(reactants)=[xΔGf°(C)+yΔGf°(D)]−[mΔGf°(A)+nΔGf°(B)].$
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### Example 12.8

#### Calculation of $ΔG298°ΔG298°$

Consider the decomposition of yellow mercury(II) oxide.
$HgO(s,yellow)⟶Hg(l)+12O2(g)HgO(s,yellow)⟶Hg(l)+12O2(g)$
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Calculate the standard free energy change at room temperature, $ΔG298°,ΔG298°,$ 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 $ΔGf°(kJ/mol)ΔGf°(kJ/mol)$ $ΔHf°(kJ/mol)ΔHf°(kJ/mol)$ $S298°(J/K·mol)S298°(J/K·mol)$
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:

$ΔG298°=∑νGSf°(products)−∑νΔGf°(reactants)ΔG298°=∑νGSf°(products)−∑νΔGf°(reactants)$
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$=[1ΔGf°Hg(l)+12ΔGf°O2(g)]−1ΔGf°HgO(s,yellow)=[1ΔGf°Hg(l)+12ΔGf°O2(g)]−1ΔGf°HgO(s,yellow)$
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$=[1mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−58.43 kJ/mol)=58.43 kJ/mol=[1mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−58.43 kJ/mol)=58.43 kJ/mol$
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(b) Using enthalpies and entropies of formation:

$ΔH298°=∑νΔHf°(products)−∑νΔHf°(reactants)ΔH298°=∑νΔHf°(products)−∑νΔHf°(reactants)$
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$=[1ΔHf°Hg(l)+12ΔHf°O2(g)]−1ΔHf°HgO(s,yellow)=[1ΔHf°Hg(l)+12ΔHf°O2(g)]−1ΔHf°HgO(s,yellow)$
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$=[1 mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−90.46 kJ/mol)=90.46 kJ/mol=[1 mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−90.46 kJ/mol)=90.46 kJ/mol$
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$ΔS298°=∑νΔS298°(products)−∑νΔS298°(reactants)ΔS298°=∑νΔS298°(products)−∑νΔS298°(reactants)$
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$=[1ΔS298°Hg(l)+12ΔS298°O2(g)]−1ΔS298°HgO(s,yellow)=[1ΔS298°Hg(l)+12ΔS298°O2(g)]−1ΔS298°HgO(s,yellow)$
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$=[1 mol(75.9 J/mol K)+12mol(205.2 J/mol K)]−1 mol(71.13 J/mol K)=107.4 J/mol K=[1 mol(75.9 J/mol K)+12mol(205.2 J/mol K)]−1 mol(71.13 J/mol K)=107.4 J/mol K$
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$ΔG°=ΔH°−TΔS°=90.46 kJ−298.15 K×107.4 J/K·mol×1 kJ1000 JΔG°=ΔH°−TΔS°=90.46 kJ−298.15 K×107.4 J/K·mol×1 kJ1000 J$
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$ΔG°=(90.46−32.01)kJ/mol=58.45 kJ/molΔG°=(90.46−32.01)kJ/mol=58.45 kJ/mol$
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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.

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?
$C2H4(g)⟶H2(g)+C2H2(g)C2H4(g)⟶H2(g)+C2H2(g)$
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141.5 kJ/mol, nonspontaneous

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