Chapter 13: Thermodynamics

# 13.3 The Second and Third Laws of Thermodynamics

### Learning Outcomes

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

## The Second Law of Thermodynamics

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 system ($\Delta$S > 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:

$\Delta {S}_{\text{univ}}=\Delta {S}_{\text{sys}}+\Delta {S}_{\text{surr}}$

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:

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:

$\Delta {S}_{\text{sys}}=\dfrac{-{q}_{\text{rev}}}{{T}_{\text{sys}}}\qquad\text{ and }\qquad\Delta {S}_{\text{surr}}=\dfrac{{q}_{\text{rev}}}{{T}_{\text{surr}}}$

The magnitudes of qrev and qrev are equal, their opposite arithmetic signs denoting the loss of heat by the system and the gain of heat by the surroundings. Since Tsys > Tsurr 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:

$\lvert \Delta S_{sys} \rvert < \lvert \Delta S_{surr} \rvert$

$\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} > 0$

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:

$\Delta {S}_{\text{sys}}=\dfrac{{q}_{\text{rev}}}{{T}_{\text{sys}}}$ and $\Delta {S}_{\text{surr}}=\dfrac{-{q}_{\text{rev}}}{{T}_{\text{surr}}}$

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 (that is, the direction of the heat flow) will yield a negative value for ΔSuniv. This process involves a decrease in the entropy of the universe.

3. The objects are at essentially the same temperature, TsysTsurr, 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.

$\lvert \Delta S_{sys} \rvert \approx \lvert \Delta S_{surr} \rvert$
$\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} = 0$

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 13.3.1.

Table 13.3.1. The Second Law of Thermodynamics
$\Delta$Suniv > 0 spontaneous
$\Delta$Suniv < 0 nonspontaneous (spontaneous in opposite direction)
$\Delta$Suniv = 0 reversible (system is at equilibrium)

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:

$\Delta {S}_{\text{univ}}=\Delta {S}_{\text{sys}}+\Delta {S}_{\text{surr}}=\Delta {S}_{\text{sys}}+\dfrac{{q}_{\text{surr}}}{T}$

We may use this equation to predict the spontaneity of a process as illustrated in Example 1.

### Example 13.3.1: Will Ice Spontaneously Melt?

The entropy change for the process is $\ce{H2O}\left(s\right)\longrightarrow \ce{H2O}\left(l\right)$

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?

Show Solution

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

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

$\begin{array}{cc}\hfill \Delta {S}_{\text{univ}}& =\Delta {S}_{\text{sys}}+\Delta {S}_{\text{surr}}=\Delta {S}_{\text{sys}}+\dfrac{{q}_{\text{surr}}}{T}\hfill \\ & =\text{22.1 J/K}+\dfrac{-6.00\times {10}^{3}\text{J}}{\text{263.15 K}}=-0.7\text{ J/K}\hfill \end{array}$

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

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

$\begin{array}{l}\\ \Delta {S}_{\text{univ}}&=&\Delta {S}_{\text{sys}}+\dfrac{{q}_{\text{surr}}}{T}\\{}&=&22.1\text{J/K}+\dfrac{-6.00\times {10}^{3}\text{J}}{\text{283.15 K}}=\text{+0.9 J/K}\end{array}$

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

## The Third Law of Thermodynamics

The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal (W = 1). According to the Boltzmann equation, the entropy of this system is zero.

$S=k\text{ln}W=k\text{ln}\left(1\right)=0$

This limiting condition for a system’s entropy represents the third law of thermodynamics: the entropy of a pure, perfect crystalline substance at 0 K is zero.

Careful calorimetric measurements can be made to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies (${S}^{\circ}$) are for one mole of substance under standard conditions (a pressure of 1 bar and a temperature of 298.15 K. The standard entropy change (ΔS°) for any process may be computed from the standard entropies of its reactant and product species like the following:

$\Delta S ^{\circ} =\sum \nu {S}^{\circ }\text{(products)}-\sum \nu {S}^{\circ }\text{(reactants)}$

Here, $\nu$ represents stoichiometric coefficients in the balanced equation representing the process. For example, ΔS° for the following reaction at room temperature $m\text{A}+n\text{B}\longrightarrow x\text{C}+y\text{D}$ is computed as the following:

$=\left[x{S}^{\circ }\left(\text{C}\right)+y{S}^{\circ }\left(\text{D}\right)\right]-\left[m{S}^{\circ }\left(\text{A}\right)+n{S}^{\circ }\left(\text{B}\right)\right]$

Table 13.3.2 lists some standard entropies at 298.15 K. You can find additional standard entropies in Standard Thermodynamic Properties for Selected Substances.

Table 13.3.2. Standard Entropies (at 298.15 K, 1 atm)
Substance ${S}^{\circ }$ (J mol−1 K−1)
carbon
$\ce{C}$(s, graphite) 5.740
$\ce{C}$(s, diamond) 2.38
$\ce{CO}$(g) 197.7
$\ce{CO2}$(g) 213.8
$\ce{CH4}$(g) 186.3
$\ce{C2H4}$(g) 219.5
$\ce{C2H6}$(g) 229.5
$\ce{CH3OH}$(l) 126.8
$\ce{C2H5OH}$(l) 160.7
hydrogen
$\ce{H2}$(g) 130.57
$\ce{H}$(g) 114.6
$\ce{H2}$O(g) 188.71
$\ce{H2O}$(l) 69.91
$\ce{HCl}$(g) 186.8
$\ce{H2S}$(g) 205.7
oxygen
$\ce{O2}$(g) 205.03

### Example 13.3.2: Determination of ΔS°

Calculate the standard entropy change for the following process:

$\ce{H2O}\left(g\right)\longrightarrow \ce{H2O}\left(l\right)$

Show Solution

The value of the standard entropy change at room temperature, $\Delta {S}^{\circ }$, is the difference between the standard entropy of the product, $\ce{H2O}$(l), and the standard entropy of the reactant, $\ce{H2O}$(g).

$\begin{array}{l}\\ \hfill \Delta {S}^{\circ }& ={S}^{\circ }\left(\ce{H2O}\left(l\right)\right)-{S}^{\circ }\left(\ce{H2O}\left(g\right)\right)\hfill \\ & =\left(\text{70.0 J}{\text{mol}}^{-1}{\text{K}}^{-1}\right)-\left(\text{188.8 J}{\text{mol}}^{-1}{\text{K}}^{-1}\right)=-118.8\text{ J}{\text{mol}}^{-1}{\text{K}}^{-1}\hfill \end{array}$

The value for $\Delta {S}^{\circ }$ is negative, as expected for this phase transition (condensation), which the previous section discussed.

### Example 13.3.3: Determination of ΔS°

Calculate the standard entropy change for the combustion of methanol, $\ce{CH3OH}$:

$\ce{2CH3OH}\left(l\right)+\ce{3O2}\left(g\right)\longrightarrow \ce{2CO2}\left(g\right)+\ce{4H2O}\left(l\right)$

Show Solution

The value of the standard entropy change is equal to the difference between the standard entropies of the products and the entropies of the reactants scaled by their stoichiometric coefficients.

$\begin{array}{rll}{}\Delta S^{\circ }=\Delta {S}^{\circ }&=&\sum \nu {S}^{\circ }\text{(products)}-\sum \nu {S}^{\circ }\text{(reactants)}\\{}&=&\left[2\times{S}^{\circ}\left({\ce{CO}}_{2}\left(g\right)\right)+4\times{S}^{\circ}\left(\ce{H2O}\left(l\right)\right)\right]-\left[2\times{S}^{\circ}\left(\ce{CH3OH}\left(l\right)\right)+3\times{S}^{\circ}\left(\ce{O2}\left(g\right)\right)\right]\\{}&=&\left\{\left[2(213.8)+4\times 70.0\right]-\left[2\left(126.8\right)+3\left(205.03\right)\right]\right\}=-161.1\text{J/mol}\cdot\text{ K}\end{array}$

### Key Concepts and Summary

The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, Suniv > 0. If $\Delta$Suniv < 0, the process is nonspontaneous, and if $\Delta$Suniv = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.

#### Key Equations

• $\Delta S^{\circ }=\Delta {S}^{\circ }=\sum \nu {S}^{\circ }\text{(products)}-\sum \nu {S}^{\circ }\text{(reactants)}$
• $\Delta S=\dfrac{{q}_{\text{rev}}}{T}$
• $\Delta {S}_{\text{univ}}=\Delta {S}_{\text{sys}}+\Delta {S}_{\text{surr}}$
• $\Delta {S}_{\text{univ}}=\Delta {S}_{\text{sys}}+\Delta {S}_{\text{surr}}=\Delta {S}_{\text{sys}}+\dfrac{{q}_{\text{surr}}}{T}$

### Try It

1. Calculate $\Delta {S}_{298}^{\circ }$ for the following changes.
1. $\ce{2H2}\left(g\right)+\ce{O2}\left(g\right)\longrightarrow \ce{2H2O}\left(l\right)$
2. By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of $\ce{NaCl}$(s) is spontaneous at 500 °C and at 700 °C.
${S}_{\ce{NaCl}\left(s\right)}^{\circ }=72.11\frac{\text{J}}{\text{mol}\cdot\text{ K}}{S}_{\ce{NaCl}\left(l\right)}^{\circ }=95.06\frac{\text{J}}{\text{mol}\cdot\text{ K}}\Delta {H}_{\text{fusion}}^{\circ }=\text{27.95 kJ/mol}$
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
Show Selected Solutions

1. $\Delta {S}_{298}^{\circ }$ for each change is as follows:

1. $\begin{array}{rll}\\ \\ \Delta {S}_{298}^{\circ }&=&\sum \nu \Delta {S}_{298}^{\circ }\text{(products)}-\sum \nu \Delta {S}_{298}^{\circ }\text{(reactants)}\\{}&=&2\Delta {S}_{298}^{\circ }\ce{H2O}\left(l\right)-\left[1\Delta {S}_{298}^{\circ }\ce{O2}\left(g\right)+2\Delta {S}_{298}^{\circ }\ce{H2}\left(g\right)\right]\\{}&=&\left[\text{2 mol}\left(70.0\frac{\text{J}}{\text{mol K}}\right)\right]-\left[\text{1 mol}\left(205.2\frac{\text{J}}{\text{mol K}}\right)+\text{2 mol}\left(130.7\frac{\text{J}}{\text{mol K}}\right)\right]=-\text{326.6 J/K}\end{array}$

2. The process is $\ce{NaCl}\left(s\right)\longrightarrow \ce{NaCl}\left(l\right)$. At 500 °C, the following is true:

$\Delta {S}_{\text{univ}}=\Delta {S}_{\text{sys}}+\frac{{q}_{\text{surr}}}{T}=\left(95.06 - 72.11\right)\frac{\text{J}}{\text{mol}\cdot\text{ K}}+\frac{-27.95\times {10}^{3}\frac{\text{J}}{\text{mol}}}{500+273.15}=-13.2\frac{\text{J}}{\text{mol}\cdot\text{ K}}$

At 700 °C, the following is true:

$\Delta {S}_{\text{univ}}=\Delta {S}_{\text{sys}}+\frac{{q}_{\text{surr}}}{T}=\left(95.06 - 72.11\right)\frac{\text{J}}{\text{mol}\cdot\text{ K}}+\frac{-27.95\times {10}^{3}\frac{\text{J}}{\text{mol}}}{700+273.15}=-5.8\frac{\text{J}}{\text{mol}\cdot\text{ K}}$

As ΔSuniv < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for $\ce{NaCl}$ at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.

## Glossary

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 S298°

standard entropy change ($\Delta$S°): change in entropy for a reaction calculated using the standard entropies, usually at room temperature and denoted $\Delta {S}_{298}^{\circ }$

third law of thermodynamics: entropy of a perfect crystal at absolute zero (0 K) is zero