The Third Law of Thermodynamics
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.
\(S=k\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}W=k\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(1)=0\)
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.
We can make careful calorimetric measurements to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies are given the label \({S}_{298}^{°}\) for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 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:
\(\text{Δ}S\text{°}=\sum \nu {S}_{298}^{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu {S}_{298}^{°}\text{(reactants)}\)
Here, ν represents stoichiometric coefficients in the balanced equation representing the process. For example, ΔS° for the following reaction at room temperature
\(m\text{A}+n\text{B}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}x\text{C}+y\text{D,}\)
is computed as the following:
\(=[x{S}_{298}^{°}(\text{C})+y{S}_{298}^{°}(\text{D})]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}[m{S}_{298}^{°}(\text{A})+n{S}_{298}^{°}(\text{B})]\)
The table below lists some standard entropies at 298.15 K. You can find additional standard entropies in this appendix.
| Standard Entropies (at 298.15 K, 1 atm) | |
|---|---|
| Substance | \({S}_{298}^{°}\) (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 for the following process:
\({\text{H}}_{2}\text{O}(g)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(l)\)
Solution
The value of the standard entropy change at room temperature, \(\text{Δ}{S}_{298}^{°},\) is the difference between the standard entropy of the product, H2O(l), and the standard entropy of the reactant, H2O(g).\(\begin{array}{}\hfill \text{Δ}{S}_{298}^{°}& ={S}_{298}^{°}({\text{H}}_{2}\text{O}(l))\phantom{\rule{0.2em}{0ex}}-{S}_{298}^{°}({\text{H}}_{2}\text{O}(g))\hfill \\ & =(\text{70.0 J}{\phantom{\rule{0.2em}{0ex}}\text{mol}}^{\text{−1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−1}})\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}(\text{188.8 J}{\text{mol}}^{\text{−1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−1}})=-118.8\phantom{\rule{0.2em}{0ex}}\text{J}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{\text{−1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−1}}\hfill \end{array}\)
The value for \(\text{Δ}{S}_{298}^{°}\) is negative, as expected for this phase transition (condensation), which the previous section discussed.
Example
Determination of ΔS°
Calculate the standard entropy change for the combustion of methanol, CH3OH:
\(2{\text{CH}}_{3}\text{OH}(l)+3{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{CO}}_{2}(g)+4{\text{H}}_{2}\text{O}(l)\)
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.\(\text{Δ}S\text{°}=\text{Δ}{S}_{298}^{°}=\sum \nu {S}_{298}^{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu {S}_{298}^{°}\text{(reactants)}\)
\(\begin{array}{c}[2{S}_{298}^{°}({\text{CO}}_{2}(g))\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}4{S}_{298}^{°}({\text{H}}_{2}\text{O}(l))]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}[2{S}_{298}^{°}({\text{CH}}_{3}\text{OH}(l))\phantom{\rule{0.2em}{0ex}}+3{S}_{298}^{°}({\text{O}}_{2}(g))]\\ \\ =\mathbf{\{}\phantom{\rule{0.1em}{0ex}}[2(213.8)+4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}70.0]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}[2(126.8)+3(205.03)]\phantom{\rule{0.2em}{0ex}}\mathbf{\}}=-161.1\phantom{\rule{0.2em}{0ex}}\text{J/mol·K}\end{array}\)
This lesson is part of:
Thermodynamics