# 18.4: Criteria for Spontaneous Change: The Second Law of Thermodynamics

Learning Objectives

• State and explain the second and third laws of thermodynamics
• Calculate entropy changes for phase transitions and chemical reactions under standard conditions

In the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy of the systemS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:

$ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr} \label{$$\PageIndex{1}$$}$

To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:

1. The objects are at different temperatures, and heat flows from the hotter to the cooler object. This is always observed to occur spontaneously. Designating the hotter object as the system and invoking the definition of entropy yields the following:
$ΔS_\ce{sys}=\dfrac{−q_\ce{rev}}{T_\ce{sys}}\hspace{20px}\ce{and}\hspace{20px}ΔS_\ce{surr}=\dfrac{q_\ce{rev}}{T_\ce{surr}} \label{$$\PageIndex{2}$$}$
The arithmetic signs of qrev denote the loss of heat by the system and the gain of heat by the surroundings. Since Tsys > Tsurr in this scenario, the magnitude of the entropy change for the surroundings will be greater than that for the system, and so the sum of ΔSsys and ΔSsurr will yield a positive value for ΔSuniv. This process involves an increase in the entropy of the universe.
2. The objects are at different temperatures, and heat flows from the cooler to the hotter object. This is never observed to occur spontaneously. Again designating the hotter object as the system and invoking the definition of entropy yields the following:
$ΔS_\ce{sys}=\dfrac{q_\ce{rev}}{T_\ce{sys}}\hspace{20px}\ce{and}\hspace{20px}ΔS_\ce{surr}=\dfrac{−q_\ce{rev}}{T_\ce{surr}} \label{$$\PageIndex{3}$$}$
The arithmetic signs of qrev denote the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than that for the system, but in this case, the signs of the heat changes will yield a negative value for ΔSuniv. This process involves a decrease in the entropy of the universe.
3. The temperature difference between the objects is infinitesimally small, TsysTsurr, and so the heat flow is thermodynamically reversible. See the previous section’s discussion). In this case, the system and surroundings experience entropy changes that are equal in magnitude and therefore sum to yield a value of zero for ΔSuniv. This process involves no change in the entropy of the universe.

These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table $$\PageIndex{1}$$.

 ΔSuniv > 0 spontaneous ΔSu niv < 0 nonspontaneous (spontaneous in opposite direction) ΔSuniv = 0 reversible (system is at equilibrium)

Definition: The Second Law of Thermodynamics

All spontaneous changes cause an increase in the entropy of the universe.

For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, qsurr is a good approximation of qrev, and the second law may be stated as the following:

$ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \label{$$\PageIndex{4}$$}$

We may use this equation to predict the spontaneity of a process as illustrated in Example $$\PageIndex{1}$$.

Example $$\PageIndex{1}$$: Will Ice Spontaneously Melt?

The entropy change for the process

$\ce{H2O}(s)⟶\ce{H2O}(l) onumber$

is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?

Solution

We can assess the spontaneity of the process by calculating the entropy change of the universe. If ΔSuniv is positive, then the process is spontaneous. At both temperatures, ΔSsys = 22.1 J/K and qsurr = −6.00 kJ.

At −10.00 °C (263.15 K), the following is true:

\begin{align} ΔS_\ce{univ}&=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} onumber\\ &=\mathrm{22.1\: J/K+\dfrac{−6.00×10^3\:J}{263.15\: K}=−0.7\:J/K} onumber \end{align} onumber

Suniv < 0, so melting is nonspontaneous (not spontaneous) at −10.0 °C.

At 10.00 °C (283.15 K), the following is true:

$ΔS_\ce{univ}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} onumber$

$\mathrm{=22.1\:J/K+\dfrac{−6.00×10^3\:J}{283.15\: K}=+0.9\: J/K} onumber$

Suniv > 0, so melting is spontaneous at 10.00 °C.

Exercise $$\PageIndex{1}$$

Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of Suniv?

Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.

## Gibbs Energy and Changes of Gibbs Energy

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:

$G=H−TS \label{2}$

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

$ΔG=ΔH−TΔS \label{3}$

(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:

$ΔS_\ce{univ}=ΔS+\dfrac{q_\ce{surr}}{T} \label{4}$

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

$ΔS_\ce{univ}=ΔS−\dfrac{ΔH}{T} \label{$$\PageIndex{5}$$}$

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

$−TΔS_\ce{univ}=ΔH−TΔS \label{$$\PageIndex{6}$$}$

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

$ΔG=−TΔS_\ce{univ} \label{$$\PageIndex{7}$$}$

The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, $$ΔS_{univ}$$. Table $$\PageIndex{2}$$ expands on Table $$\PageIndex{2}$$ and summarizes the relation between the spontaneity of a process and the arithmetic signs of $$\Delta G$$ and $$\Delta S$$ indicators.

 ΔSuniv > 0 ΔG < 0 spontaneous ΔSuniv < 0 ΔG > 0 nonspontaneous ΔSuniv = 0 ΔG = 0 reversible (at equilibrium)

J. Willard Gibbs (1839–1903)

Born in Connecticut, Josiah Willard Gibbs attended Yale, as did his father, a professor of sacred literature at Yale, who was involved in the Amistad trial. In 1863, Gibbs was awarded the first engineering doctorate granted in the United States. He was appointed professor of mathematical physics at Yale in 1871, the first such professorship in the United States. His series of papers entitled “On the Equilibrium of Heterogeneous Substances” was the foundation of the field of physical chemistry and is considered one of the great achievements of the 19th century. Gibbs, whose work was translated into French by Le Châtelier, lived with his sister and brother-in-law until his death in 1903, shortly before the inauguration of the Nobel Prizes.

Gibbs 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 Example $$\PageIndex{1}$$.

$ΔG°=ΔH°−TΔS° \label{$$\PageIndex{7}$$}$

Example $$\PageIndex{2}$$: Evaluation of ΔG°

Change 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:

$\ce{H2O}(l)⟶\ce{H2O}(g) \label{$$\PageIndex{8}$$} onumber$

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

$ΔG^\circ_{298}=ΔH°−TΔS° \label{$$\PageIndex{9}$$} onumber$

From Appendix G, here is the data:

Substance $$ΔH^\circ_\ce{f}\ce{(kJ/mol)}$$ $$S^\circ_{298}\textrm{(J/K⋅mol)}$$
H2O(l) −286.83 70.0
H2O(g) −241.82 188.8

Combining at 298 K:

$ΔH°=ΔH^\circ_{298}=ΔH^\circ_\ce{f}(\ce{H2O}(g))−ΔH^\circ_\ce{f}(\ce{H2O}(l)) onumber\\ =\mathrm{[−241.82\: kJ−(−285.83)]\:kJ/mol=44.01\: kJ/mol} onumber$
$ΔS°=ΔS^\circ_{298}=S^\circ_{298}(\ce{H2O}(g))−S^\circ_{298}(\ce{H2O}(l)) onumber\\ =\mathrm{188.8\:J/mol⋅K−70.0\:J/K=118.8\:J/mol⋅K} onumber$
$ΔG°=ΔH°−TΔS° onumber$

Converting everything into kJ and combining at 298 K:

$ΔG^\circ_{298}=ΔH°−TΔS° onumber$

$\mathrm{=44.01\: kJ/mol−(298\: K×118.8\:J/mol⋅K)×\dfrac{1\: kJ}{1000\: J}} onumber$

$\mathrm{44.01\: kJ/mol−35.4\: kJ/mol=8.6\: kJ/mol} onumber$

At 298 K (25 °C) $$ΔG^\circ_{298}>0$$, and so boiling is nonspontaneous (not spontaneous).

Exercise $$\PageIndex{2}$$

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?

$\ce{C2H6}(g)⟶\ce{H2}(g)+\ce{C2H4}(g) onumber$

$$ΔG^\circ_{298}=\mathrm{102.0\: kJ/mol}$$; the reaction is nonspontaneous (not spontaneous) at 25 °C.

Free energy changes may also use the standard free energy of formation $$(ΔG^\circ_\ce{f})$$, for each 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 enthalpies of formation, $$(ΔG^\circ_\ce{f})$$ is by definition zero for elemental substances under standard state conditions. The approach to computing the free energy change for a reaction using this approach is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction

$m\ce{A}+n\ce{B}⟶x\ce{C}+y\ce{D},$

the standard free energy change at room temperature may be calculated as

$ΔG^\circ_{298}=ΔG°=∑νΔG^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants})$

$=[xΔG^\circ_\ce{f}(\ce{C})+yΔG^\circ_\ce{f}(\ce{D})]−[mΔG^\circ_\ce{f}(\ce{A})+nΔG^\circ_\ce{f}(\ce{B})].$

Example $$\PageIndex{3}$$: Calculation of $$ΔG^\circ_{298}$$

Consider the decomposition of yellow mercury(II) oxide.

$\ce{HgO}(s,\,\ce{yellow})⟶\ce{Hg}(l)+\dfrac{1}{2}\ce{O2}(g) onumber$

Calculate the standard free energy change at room temperature, $$ΔG^\circ_{298}$$, 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 $$ΔG^\circ_\ce{f}\:\mathrm{(kJ/mol)}$$ $$ΔH^\circ_\ce{f}\:\mathrm{(kJ/mol)}$$ $$S^\circ_{298}\:\textrm{(J/K⋅mol)}$$
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:

$ΔG^\circ_{298}=∑νGS^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants}) onumber$

$=\left[1ΔG^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔG^\circ_{298}\ce{O2}(g)\right]−1ΔG^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) onumber$

$\mathrm{=\left[1\:mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)\right]−1\: mol(−58.43\: kJ/mol)=58.43\: kJ/mol} onumber$

(b) Using enthalpies and entropies of formation:

$ΔH^\circ_{298}=∑νΔH^\circ_{298}(\ce{products})−∑νΔH^\circ_{298}(\ce{reactants}) onumber$

$=\left[1ΔH^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔH^\circ_{298}\ce{O2}(g)\right]−1ΔH^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) onumber$

$\mathrm{=[1\: mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)]−1\: mol(−90.46\: kJ/mol)=90.46\: kJ/mol} onumber$

$ΔS^\circ_{298}=∑νΔS^\circ_{298}(\ce{products})−∑νΔS^\circ_{298}(\ce{reactants}) onumber$

$=\left[1ΔS^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔS^\circ_{298}\ce{O2}(g)\right]−1ΔS^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) onumber$

$\mathrm{=\left[1\: mol(75.9\: J/mol\: K)+\dfrac{1}{2}mol(205.2\: J/mol\: K)\right]−1\: mol(71.13\: J/mol\: K)=107.4\: J/mol\: K} onumber$

$ΔG°=ΔH°−TΔS°=\mathrm{90.46\: kJ−298.15\: K×107.4\: J/K⋅mol×\dfrac{1\: kJ}{1000\: J}} onumber$

$ΔG°=\mathrm{(90.46−32.01)\:kJ/mol=58.45\: kJ/mol} onumber$

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.

Exercise $$\PageIndex{3}$$

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?

$\ce{C2H4}(g)⟶\ce{H2}(g)+\ce{C2H2}(g) onumber$

−141.5 kJ/mol, nonspontaneous

## Key Concepts and Summary

The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, Suniv > 0. If ΔSuniv < 0, the process is nonspontaneous, and if ΔSuniv = 0, the system is at equilibrium. Gibbs free energy (G) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A negative value for ΔG indicates a spontaneous process; a positive ΔG indicates a nonspontaneous process; and a ΔG of zero indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible.

## Key Equations

• $$ΔS^\circ=ΔS^\circ_{298}=∑νS^\circ_{298}(\ce{products})−∑νS^\circ_{298}(\ce{reactants})$$
• $$ΔS=\dfrac{q_\ce{rev}}{T}$$
• ΔSuniv = ΔSsys + ΔSsurr
• $$ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T}$$
• ΔG = ΔHTΔS
• ΔG = ΔG° + RT ln Q
• ΔG° = −RT ln K

## Glossary

Gibbs free energy change (G)
thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in G
standard free energy change (ΔG°)
change in free energy for a process occurring under standard conditions (1 bar pressure for gases, 1 M concentration for solutions)
standard free energy of formation $$(ΔG^\circ_\ce{f})$$
change in free energy accompanying the formation of one mole of substance from its elements in their standard states
second law of thermodynamics
entropy of the universe increases for a spontaneous process
standard entropy (S°)
entropy for a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K and denoted $$S^\circ_{298}$$
standard entropy change (ΔS°)
change in entropy for a reaction calculated using the standard entropies, usually at room temperature and denoted $$ΔS^\circ_{298}$$