D19.5 Third Law of Thermodynamics

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 molecule comprising the crystal (W = 1). Therefore, the entropy of this system is zero:

S = kB·ln(W) = kB·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.

Starting with zero entropy at absolute zero, it is possible to make careful calorimetric measurements ( \dfrac{q_\text{rev}}{\text{T}}) to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values at higher temperatures. (Note that, unlike enthalpy values, the third law of thermodynamics identifies a zero point for entropy. Therefore, there is no need for formation enthalpies, and every substance, including elements in their most stable states, has an absolute entropy.)

Standard entropy (S°) values are the absolute entropies per mole of substance at a pressure of 1 bar or a concentration of 1 M. The standard entropy change (ΔrS°) for any chemical process may be computed from the standard entropy of its reactant and product species:

ΔrS° = ∑S°(products) − ∑S°(reactants)

The thermodynamics table in the appendix lists standard entropies of select compounds at 298.15 K.

Exercise: Standard Entropy Change

Suppose an exothermic chemical reaction takes place at constant atmospheric pressure. There is heat transfer of energy from the reaction system to the surroundings, qsurr = –qsys. The heat transfer for the system is the enthalpy change of the reaction because, at constant pressure, ΔrH° = q. Because the energy transfer to the surroundings is reversible, the entropy change for the surroundings can also be expressed as:

 {\Delta}S_{\text{surr}} = \dfrac{q_{\text{rev}}}{T} = \dfrac{q_{\text{surr}}}{T} = \dfrac{-q_{\text{sys}}}{T} = \dfrac{-\Delta H_{\text{sys}}}{T} = \dfrac{-\Delta _{\text{r}}H^{\circ}}{T}

The same reasoning applies to an endothermic reaction: qsys and qsurr are equal but have opposite sign.

Also, for a chemical reaction system, ΔSsys = ΔrS° (the standard entropy change for the reaction). Hence, ΔSuniv can be expressed as:

ΔSuniv = ΔSsys + ΔSsurr = ΔrS° −  \dfrac{\Delta _{\text{r}}H^{\circ}}{T}

The convenience of this equation is that, for a given reaction, ΔSuniv can be calculated from thermodynamics data for the system only. That is, from data found in the Appendix.

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Chemistry 109 Fall 2021 by John Moore, Jia Zhou, and Etienne Garand is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.