# 16.4 Free energy

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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
• Explain how temperature affects the spontaneity of some processes
• Relate standard free energy changes to equilibrium constants

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:

$G=H-TS$

Free energy is a state function, and at constant temperature and pressure, the standard free energy change (Δ G °)    may be expressed as the following:

$\text{Δ}G=\text{Δ}H-T\text{Δ}S$

(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:

$\text{Δ}{S}_{\text{univ}}=\text{Δ}S+\phantom{\rule{0.2em}{0ex}}\frac{{q}_{\text{surr}}}{T}$

The first law requires that q surr = − q sys , and at constant pressure q sys = Δ H , and so this expression may be rewritten as the following:

$\text{Δ}{S}_{\text{univ}}=\text{Δ}S-\phantom{\rule{0.1em}{0ex}}\frac{\text{Δ}H}{T}$

Δ H is the enthalpy change of the system . Multiplying both sides of this equation by − T , and rearranging yields the following:

$\text{−}T\text{Δ}{S}_{\text{univ}}=\text{Δ}H-T\text{Δ}S$

Comparing this equation to the previous one for free energy change shows the following relation:

$\text{Δ}G=\text{−}T\text{Δ}{S}_{\text{univ}}$

The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, Δ S univ . [link] 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
Δ S univ >0 Δ G <0 spontaneous
Δ S univ <0 Δ G >0 nonspontaneous
Δ S univ = 0 Δ G = 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 that have undergone some change. 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 [link] .

$\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}$

## Evaluation of δ G ° change from δ H ° and δ S °

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 Δ G ° say about the spontaneity of this process?

## Solution

The process of interest is the following:

${\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)$

The standard change in free energy may be calculated using the following equation:

$\text{Δ}{G}_{298}^{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}$

From Appendix G , here is the data:

Substance $\text{Δ}{H}_{\text{f}}^{°}\text{(kJ/mol)}$ ${S}_{298}^{°}\text{(J/K·mol)}$
H 2 O( l ) −286.83 70.0
H 2 O( g ) −241.82 188.8

Combining at 298 K:

$\begin{array}{l}\text{Δ}H\text{°}=\text{Δ}{H}_{298}^{°}=\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\\ =\left[\text{−241.82 kJ}-\left(\text{−285.83}\right)\right]\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}=\text{44.01 kJ/mol}\end{array}$
$\begin{array}{c}\text{Δ}S\text{°}=\text{Δ}{S}_{298}^{°}={S}_{298}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-{S}_{298}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\\ =188.8\phantom{\rule{0.2em}{0ex}}\text{J/mol·K}-70.0\phantom{\rule{0.2em}{0ex}}\text{J/K}=118.8\phantom{\rule{0.2em}{0ex}}\text{J/mol·K}\end{array}$
$\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}$

Converting everything into kJ and combining at 298 K:

$\begin{array}{c}\text{Δ}{G}_{298}^{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\\ \\ =\text{44.01 kJ/mol}-\left(\text{298 K}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}118.8\phantom{\rule{0.2em}{0ex}}\text{J/mol·K}\right)\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\frac{\text{1 kJ}}{\text{1000 J}}\end{array}$
$\text{44.01 kJ/mol}-\text{35.4 kJ/mol}=\text{8.6 kJ/mol}$

At 298 K (25 °C) $\text{Δ}{G}_{298}^{°}>0,$ 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 Δ G ° say about the spontaneity of this process?

${\text{C}}_{2}{\text{H}}_{6}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+{\text{C}}_{2}{\text{H}}_{4}\left(g\right)$

## Answer:

$\text{Δ}{G}_{298}^{°}=\text{102.0 kJ/mol};$ the reaction is nonspontaneous ( not spontaneous) at 25 °C.

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