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.

$$S=k\cdot lnW=k\cdot ln(1)=0$$

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:

$$ΔS^°=∑νS^°\text{(products)}-∑νS^°\text{(reactants)}$$

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

$$m\text{A}+n\text{B}⟶x\text{C}+y\text{D}$$

is computed as:

$$[xΔS^°\text{(C)}+yΔS^°\text{(D)}]-[mΔS^°\text{(A)}+nΔS^°\text{(B)}]$$

A partial listing of standard entropies is provided in the table below, and additional values are provided in the Appendices. The example exercises that follow demonstrate the use of S° values in calculating standard entropy changes for physical and chemical processes.

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 $S^°$ (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

Determination of ΔS°
Calculate the standard entropy change for the following process:

$$H_2O(g)⟶H_2O(l)$$

Solution
Calculate the entropy change using standard entropies as shown above:

$$ΔS^°=\text{(1 mol)}(70.0\; \text{J mol}^{-1}\; \text{K}^{-1})-\text{(1 mol)}(188.8\; \text{J mol}^{-1}\; \text{K}^{-1})=-118.8\;\text{J/K}$$

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 for the following process:

$$H_2(g)+C_2H_4(g)⟶C_2H_6(g)$$
Answer:

−120.6 J K–1 mol–1

Determination of ΔS°
Calculate the standard entropy change for the combustion of methanol, CH3OH:

$$2CH_3OH(l)+3O_2(g)⟶2CO_2(g)+4H_2O(l)$$

Solution
Calculate the entropy change using standard entropies as shown above:

$$ΔS^°=∑νS^°\text{(products)}-∑νS^°\text{(reactants)}$$ $$[2\;mol \times S^°(CO_2(g))+4\;mol \times S^°(H_2O(l))]-[2\;mol \times S^°(CH_3OH(l))+3\;mol \times S^°(O_2(g))]$$ $$[2(213.8)+4 \times [2(126.8)+3(205.03)]=-161.1\;J/K$$

Check Your Learning
Calculate the standard entropy change for the following reaction:

$$2Ca(OH)_2(s)⟶CaO(s)+H_2O(l)$$
Answer:

24.7 J K–1 mol–1