Library Guides: Chemistry Textbook: Second Law of Thermodynamics and Gibbs Free Energy

The Second Law of Thermodynamics

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

State and explain the second and third laws of thermodynamics
Define Gibb's free energy and calculate the free energy changes in a chemical reaction
Predict spontaneity of a reaction from change in Gibbs Free Energy at different temperatures

In the quest to identify a property that may reliably predict the spontaneity of a process, a promising candidate has been identified: entropy. Processes that involve an increase in entropy of the system (ΔS > 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:

Δ S_{univ} = Δ S_{sys} + Δ S_{surr}

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:

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:
$Δ S_{sys} = \frac{− q_{rev}}{T_{sys}} and Δ S_{surr} = \frac{q_{rev}}{T_{surr}}$
The magnitudes of −q_rev and q_rev are equal, their opposite arithmetic signs denoting loss of heat by the system and gain of heat by the surroundings. Since T_sys > T_surr in this scenario, the entropy decrease of the system will be less than the entropy increase of the surroundings, and so the entropy of the universe will increase:
$\begin{array}{l} | Δ S_{sys} | < | Δ S_{surr} | \\ Δ S_{univ} = Δ S_{sys} + Δ S_{surr} > 0 \end{array}$
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:
$Δ S_{sys} = \frac{q_{rev}}{T_{sys}} and Δ S_{surr} = \frac{− q_{rev}}{T_{surr}}$
The arithmetic signs of q_rev 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 (that is, the direction of the heat flow) will yield a negative value for ΔS_univ. This process involves a decrease in the entropy of the universe.
The objects are at essentially the same temperature, T_sys ≈ T_surr, and so the magnitudes of the entropy changes are essentially the same for both the system and the surroundings. In this case, the entropy change of the universe is zero, and the system is at equilibrium.
$\begin{array}{l} | Δ S_{sys} | \approx | Δ S_{surr} | \\ Δ S_{univ} = Δ S_{sys} + Δ S_{surr} = 0 \end{array}$

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 12.2.

The Second Law of Thermodynamics
ΔS_univ > 0	spontaneous
ΔS_univ < 0	nonspontaneous (spontaneous in opposite direction)
ΔS_univ = 0	at equilibrium

Table 12.2

EXAMPLE 12.6

Will Ice Spontaneously Melt? The entropy change for the process

H_{2} O (s) ⟶ H_{2} O (l)

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 ΔS_univ is positive, then the process is spontaneous. At both temperatures, ΔS_sys = 22.1 J/K and q_surr = −6.00 kJ.

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

\begin{matrix} Δ S_{univ} & = Δ S_{sys} + Δ S_{surr} = Δ S_{sys} + \frac{q_{surr}}{T} \\ = 22.1 J/K + \frac{−6.00 \times 10^{3} J}{263.15 K} = −0.7 J/K \end{matrix}

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

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

\begin{matrix} Δ S_{univ} = Δ S_{sys} + \frac{q_{surr}}{T} \\ = 22.1 J/K + \frac{−6.00 \times 10^{3} J}{283.15 K} = +0.9 J/K \end{matrix}

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

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

Answer:

Entropy is a state function, so ΔS_freezing = −ΔS_melting = −22.1 J/K and q_surr = +6.00 kJ. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.

Gibbs Free Energy

One of the challenges of using the ΔS_system + ΔS_surr>0 to determine the spontaneity of a process is that it requires measurements of the entropy change for the surroundings (ie. ΔS_surr term) which is often inconvenient to measure. An alternative equation can be derived which is based on only the system's properties as shown below.

Condition for spontaneity:

Δ S_{sys} + Δ S_{surr} > 0

ΔS_surr = q_rev/T. But, 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. As a result, q_surr is a good approximation of q_rev, and the second law may be stated as the following:

Δ S_{sys} + \frac{q_{surr}}{T} > 0

The first law requires that q_surr = −q_sys, and at constant pressure q_sys = ΔH_sys. So, this expression may be rewritten as:

Δ S_{sys} - \frac{Δ H_{sys}}{T} > 0

Multiplying both sides of this equation by −T, and rearranging yields the following:

Δ H_{sys} - T Δ S_{sys} < 0

For simplicity’s sake, the subscript “sys” will be omitted henceforth. Also, we introduce a new thermodynamic property, ΔG to represent ΔH−TΔS. Therefore, In a spontaneous process:

Δ G = Δ H - T Δ S < 0

The free energy change (ΔG= ΔH_sys−TΔS_sys) is therefore a reliable indicator of the spontaneity of a process. It is directly related to the previously identified spontaneity indicator, ΔS_univ as ΔG=−TΔS_univ. 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
ΔS_univ > 0	ΔG < 0	spontaneous
ΔS_univ < 0	ΔG > 0	nonspontaneous
ΔS_univ = 0	ΔG = 0	at equilibrium

Table 12.3

Gibb's Free energy is a thermodynamic property that can be interpreted as the energy available to do work. It is mathematically expressed as:

G = H - T S

Under standard conditions (1atm and 25°C): $G ° = H ° - T S °$

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

Δ G ° = \sum ν Δ G_{f}^{°} (products) - \sum ν Δ G_{f}^{°} (reactants)

where ν represents the stoichiometric coefficients, ΔG_f° represents the standard Gibbs Free Energies of formation.

EXAMPLE 12.6

Using Standard Free Energies of Formation to Calculate ΔG°consider the decomposition of yellow mercury(II) oxide.

HgO (s, yellow) ⟶ Hg (l) + \frac{1}{2} O_{2} (g)

Calculate the standard free energy change at room temperature, $Δ G °,$ 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	$Δ G_{f}^{°} (kJ/mol)$	$Δ H_{f}^{°} (kJ/mol)$	$S ° (J/K·mol)$
HgO (s, yellow)	−58.43	−90.46	71.13
Hg(l)	0	0	75.9
O₂(g)	0	0	205.2

Table 12.4

(a) Using free energies of formation:

Δ G ° = \sum ν Δ G_{f}^{°} (products) - \sum ν Δ G_{f}^{°} (reactants)

= [1 Δ G_{f}^{°} Hg (l) + \frac{1}{2} Δ G_{f}^{°} O_{2} (g)] - 1 Δ G_{f}^{°} HgO (s, yellow)

= [1 mol (0 kJ/mol) + \frac{1}{2} mol(0 kJ/mol)] - 1 mol(−58.43 kJ/mol) = 58.43 kJ/mol

(b) Using enthalpies and entropies of formation:

Δ H ° = \sum ν Δ H_{f}^{°} (products) - \sum ν Δ H_{f}^{°} (reactants)

= [1 Δ H_{f}^{°} Hg (l) + \frac{1}{2} Δ H_{f}^{°} O_{2} (g)] - 1 Δ H_{f}^{°} HgO (s, yellow)

= [1 mol (0 kJ/mol) + \frac{1}{2} mol (0 kJ/mol)] - 1 mol (−90.46 kJ/mol) = 90.46 kJ/mol

Δ S ° = \sum ν Δ S ° (products) - \sum ν Δ S ° (reactants)

= [1 Δ S ° Hg (l) + \frac{1}{2} Δ S ° O_{2} (g)] - 1 Δ S ° HgO (s, yellow)

= [1 mol (75.9 J/mol K) + \frac{1}{2} mol (205.2 J/mol K)] - 1 mol (71.13 J/mol K) = 107.4 J/mol K

Δ G ° = Δ H ° - T Δ S ° = 90.46 kJ - 298.15 K \times 107.4 J/K·mol \times \frac{1 kJ}{1000 J}

Δ G ° = (90.46 - 32.01) kJ/mol = 58.45 kJ/mol

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?

C_{2} H_{4} (g) ⟶ H_{2} (g) + C_{2} H_{2} (g)

Answer:

(a) 140.8 kJ/mol, nonspontaneous

(b) 141.5 kJ/mol, nonspontaneous

Temperature Dependence of Spontaneity

As was previously demonstrated in this chapter’s section on entropy, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:

Δ G = Δ H - T Δ S

The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:

Both ΔH and ΔS are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is greater than ΔH. If the TΔS term is less than ΔH, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
Both ΔH and ΔS are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔS term’s magnitude is greater than ΔH, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
ΔH is positive and ΔS is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔG will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
ΔH is negative and ΔS is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔG will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.

These four scenarios are summarized in Figure 12.11.

A table with three columns and four rows is shown. The first column has the phrase, “Delta S greater than zero ( increase in entropy ),” in the third row and the phrase, “Delta S less than zero ( decrease in entropy),” in the fourth row. The second and third columns have the phrase, “Summary of the Four Scenarios for Enthalpy and Entropy Changes,” written above them. The second column has, “delta H greater than zero ( endothermic ),” in the second row, “delta G less than zero at high temperature, delta G greater than zero at low temperature, Process is spontaneous at high temperature,” in the third row, and “delta G greater than zero at any temperature, Process is nonspontaneous at any temperature,” in the fourth row. The third column has, “delta H less than zero ( exothermic ),” in the second row, “delta G less than zero at any temperature, Process is spontaneous at any temperature,” in the third row, and “delta G less than zero at low temperature, delta G greater than zero at high temperature, Process is spontaneous at low temperature.” — Figure 12.11 There are four possibilities regarding the signs of enthalpy and entropy changes.

EXAMPLE 12.7

Predicting the Temperature Dependence of Spontaneity The incomplete combustion of carbon is described by the following equation:

2C (s) + O_{2} (g) ⟶ 2CO (g)

How does the spontaneity of this process depend upon temperature?

Solution Combustion processes are exothermic (ΔH < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, ΔS > 0). The reaction is therefore spontaneous (ΔG < 0) at all temperatures.

Check Your Learning Popular chemical hand warmers generate heat by the air-oxidation of iron:

4Fe (s) + {3O}_{2} (g) ⟶ {2Fe}_{2} O_{3} (s)

How does the spontaneity of this process depend upon temperature?

Answer:

ΔH and ΔS are negative; the reaction is spontaneous at low temperatures.

Key Concepts and Summary

The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, S_univ > 0. If ΔS_univ < 0, the process is nonspontaneous, and if ΔS_univ = 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_univ = ΔS_sys + ΔS_surr
$Δ S_{univ} = Δ S_{sys} + Δ S_{surr} = Δ S_{sys} + \frac{q_{surr}}{T}$
ΔG = ΔH − TΔS