The Third Law of Thermodynamics

D. Latimer; D. Vanderwel; J. Hollett

73 The Third Law of Thermodynamics

Learning Objectives

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

State and explain the third law of thermodynamics
Calculate entropy changes for chemical reactions under standard conditions, using a table of Standard Entropy values.
Understand that many factors that affect the entropy of a substance (e.g. phase, temperature, size and complexity)
Be able to predict the relative entropies of the products versus reactants in a chemical reaction.

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.

$S=k\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}W=k\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(1\right)=0$

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°) are for one mole of substance under standard conditions (a pressure of 1 bar and a temperature of 298.15 K; see details regarding standard conditions in the thermochemistry chapter of this text). The standard entropy change (ΔS°) for a reaction may be computed using standard entropies as shown below:

$\Delta}S\text{°}=\sum \nu S\text{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu S\text{°}\text{(reactants)}$

where ν represents stoichiometric coefficients in the balanced equation representing the process. For example, ΔS° for the following reaction at room temperature

$m\text{A}+n\text{B}\phantom{\rule{0.2em}{0ex}}\rightarrow\phantom{\rule{0.2em}{0ex}}x\text{C}+y\text{D,}$

is computed as:

=[x S^o(C) + y S^o(D)] – [m S^o(A) + n S^o(B)]

A partial listing of standard entropies is provided in Table 73-1, and additional values are provided in Appendix G. The example exercises that follow demonstrate the use of S° values in calculating standard entropy changes for physical and chemical processes.


Table 73-1: Standard entropies for selected substances measured at 1 atm and 298.15 K. (Values are approximately equal to those measured at 1 bar, the currently accepted standard state pressure.)
Substance	$S\text{°}$ (J mol⁻¹ K⁻¹)
CARBON
C(s, graphite)	5.740
C(s, diamond)	2.38
CO(g)	197.7
CO₂(g)	213.8
CH₄(g)	186.3
C₂H₄(g)	219.5
C₂H₆(g)	229.5
CH₃OH(l)	126.8
C₂H₅OH(l)	160.7
HYDROGEN
H₂(g)	130.57
H(g)	114.6
H₂O(g)	188.71
H₂O(l)	69.91
HCI(g)	186.8
H₂S(g)	205.7
OXYGEN
O₂(g)	205.03

EXAMPLE:

Determination of ΔS°

Calculate the standard entropy change of the system for the following process:

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

Solution

Calculate the entropy change using standard entropies as shown above:

ΔS = [1 mol x S^o(H₂O(l)] – [1 mol x S^o(H₂O(g)]

ΔS = (1 mol) (70.0 J mol^-1 K^-1) – (1 mol)(188.8 J mol^-1 K^-1) = -118.8 J K^-1

The value for ΔS° is negative, as expected for this phase transition (condensation), which the previous section discussed.

Check Your Learning

Calculate the standard entropy change of the system for the following process:

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

Answer: −120.6 J K^-1

Determination of ΔS°

Calculate the standard entropy change of the system for the combustion of methanol, CH₃OH:

$2{\text{CH}}_{3}\text{OH}\left(l\right)+3{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}\rightarrow\phantom{\rule{0.2em}{0ex}}2{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(l\right)$

Solution

Calculate the entropy change using standard entropies as shown above:

$\Delta}S\text{°}=\sum \nu S\text{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu S\text{°}\text{(reactants)}$

ΔS^o = [2 mol x S^o(CO₂) + 4 mol x S^o(H₂O)] – [2 mol x S^o(CH₃OH) + 3 mol x S^o(O₂)]

= [(2 x 213.8) + (4 x 70.0)] – [(2 x 126.8) + (3 x 205.03)] = – 161.1 J K^-1

Check Your Learning

Calculate the standard entropy change for the following reaction:

$\text{Ca}{\left(\text{OH}\right)}_{2}\left(\text{s}\right)\phantom{\rule{0.2em}{0ex}}\rightarrow\phantom{\rule{0.2em}{0ex}}\text{CaO}\left(s\right)+{\text{H}}_{2}\text{O}\left(l\right)$

Answer: 24.7 J K^-1

Key Concepts and Summary

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

$\Delta}S\text{°}=\sum \nu S\text{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu S\text{°}\text{(reactants)}$

Chemistry End of Chapter Exercises

1. What is the difference between ΔS and ΔS° for a chemical change?

2. Calculate ΔS^o for the following changes.

(a) SnCl₄(l) → SnCl₄(g)

(b) CS₂ (g) → CS₂ (l)

(c) Cu (s) → Cu (g)

(d) H₂O (l) → H₂O (g)

(e) 2H₂(g) + O₂ (g) → 2 H₂O (l)

(f) 2HCl (g) + Pb (s) → PbCl₂ (s) + H₂ (g)

(g) Zn (s) + CuSO₄ (s) → Cu (s) + ZnSO₄ (s)

Answer(s): (a) 107 J/K; (b) −86.4 J/K; (c) 133.2 J/K; (d) 118.8 J/K; (e) −326.6 J/K; (f) −171.9 J/K; (g) −7.2 J/K

3. The balanced chemical equation for the combustion of liquid ethanol is as follows:

C₂H₅OH (l) + 3 O₂(g) → 2 CO₂ (g) + 2 H₂O (l)

Determine the entropy change for the combustion of liquid ethanol under the standard conditions to give gaseous carbon dioxide and liquid water.

4. Determine the entropy change for the combustion of gaseous propane, C₃H₈, under the standard conditions to give gaseous carbon dioxide and water. The balanced chemical equation for the combustion of propane is as follows:

C₃H₈(g) + 5 O₂(g) → 3 CO₂ (g) + 4 H₂O (l)

Answer(s): 100.6 J K^-1

5. “Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is Fe₂O₃ (s) + 2Al (s) → Al₂O₃(s) + 2Fe(s). Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ mol^-1 of heat.

6. Using the relevant S^ovalues listed in Appendix G, calculate ΔS^o for the following changes:

(a) N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)

(b) N₂ (g) + 5/2 O₂ (g) → N₂O₅ (g)

Answer(s): (a) −198.1 J K^-1; (b) −348.9 J K^-1.

7. Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.

(a) MnO₂ (s) → Mn (s) + O₂ (g)

(b) H₂ (g) + Br₂ (l) → 2HBr (g)

(c) Cu (s) + S (g) → CuS (s)

(d) 2 LiOH (s) + CO₂ (g) → Li₂CO₃ (s) + H₂O (g)

(e) CH₄ (g) + O₂ (g) → C (s, graphite) + 2 H₂O (g)

(f) CS₂ (g) + 3 Cl₂ (g) → CCl₄ (g) + S₂Cl₂ (g)

8. Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.

(a) C (s, graphite) + O₂ → CO₂ (g)
(b) O₂ (g) + N₂ (g) → 2NO (g)
(c) 2 Cu (s) + S (g) → Cu₂S (s)
(d) CaO (s) + H₂O (l) → Ca(OH)₂ (s)
(e) Fe₂O₃ (s) + 3 CO (g) → 2 Fe (s) + 3 CO₂ (g)
(f) CaSO₄•2H₂O (s) → CaSO₄ (s) + 2 H₂O (g)

Answer(s): (a) 2.86 J/K; (b) 24.8 J/K; (c) −113.2 J/K; (d) −24.7 J/K; (e) 15.5 J/K; (f) 290.0 J/K

9. By calculating ΔS_univ at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500 °C and at 700 °C.

S_NaCl(s) = 72.11 J mol^-1K^-1
S_NaCl(l) = 95.06 J mol^-1K^-1
ΔH_fusion = 27.95 kJ mol^-1

What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?

Answer(s): As ΔS_univ < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.

Glossary

absolute zero (0 K)

the temperature at which the entropy of a perfect crystal is zero. Theoretically, all molecules stop moving and there is no thermal motion.

standard entropy (S°)

entropy for one mole of a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K

standard entropy change (ΔS°): change in entropy for a reaction calculated using the standard entropies

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

73 The Third Law of Thermodynamics

Learning Objectives

EXAMPLE:

Determination of ΔS°

Check Your Learning

Determination of ΔS°

Check Your Learning

Key Concepts and Summary

Key Equations

Chemistry End of Chapter Exercises

Glossary

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