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Concept 06 / 06Thermodynamics

Entropy and the Second Law of Thermodynamics

Spontaneous processes and why enthalpy alone cannot predict them, entropy as a measure of disorder and energy dispersal, ΔS = qrev/T, entropy changes in phase transitions and ideal gas processes, the second law in terms of total entropy, and the third law.

Figure 1
Spontaneity. A spontaneous process occurs on its own without continuous outside help (it may still be slow). Many spontaneous processes are exothermic, but not all: ice melts above 0∘C0^{\circ}\mathrm{C}, NH4Cl\mathrm{NH_{4}Cl} dissolves in water with cooling, and gases mix without any heat change. So ΔH\Delta H alone cannot decide spontaneity. A second factor is needed: entropy. Entropy SS. A state function that measures the dispersal of energy and matter, loosely the disorder or randomness of a system. Statistically, S=kln⁡WS = k\ln W, where WW is the number of microscopic arrangements consistent with the macroscopic state. Entropy increases when: a solid melts or a liquid vaporises (Ssolid<Sliquid≪SgasS_{\text{solid}} < S_{\text{liquid}} \ll S_{\text{gas}}); temperature rises; a gas expands; substances mix or dissolve; the number of moles of gas increases in a reaction; molecules become larger or more complex. Quantitative definition. For a reversible transfer of heat at temperature TT: ΔS=qrevT\Delta S = \frac{q_{\text{rev}}}{T}. The same heat causes a larger entropy change at low temperature, where the system is more ordered to begin with. Phase transitions (reversible at the transition temperature): ΔSfus=ΔHfusTm\Delta S_{\text{fus}} = \frac{\Delta H_{\text{fus}}}{T_{m}} and ΔSvap=ΔHvapTb\Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T_{b}}. Trouton's rule: for many liquids, ΔSvap≈85\Delta S_{\text{vap}} \approx 85 to 88 J K−1 mol−188\,\mathrm{J\,K^{-1}\,mol^{-1}}; hydrogen-bonded liquids such as water (109109) are higher, because the liquid is more ordered. Ideal gas: ΔS=nCVln⁡T2T1+nRln⁡V2V1\Delta S = nC_{V}\ln\frac{T_{2}}{T_{1}} + nR\ln\frac{V_{2}}{V_{1}}. For isothermal change, ΔS=nRln⁡V2V1=nRln⁡P1P2\Delta S = nR\ln\frac{V_{2}}{V_{1}} = nR\ln\frac{P_{1}}{P_{2}}. For a reversible adiabatic process, ΔS=0\Delta S = 0 (isentropic). For a reaction: ΔrS∘=∑S∘(products)−∑S∘(reactants)\Delta_{r}S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants}), using absolute molar entropies. Second law. The total entropy of an isolated system (or of system plus surroundings) increases in every spontaneous process and stays constant at equilibrium: ΔStotal=ΔSsystem+ΔSsurroundings>0\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0 (spontaneous). At constant TT and PP, the surroundings receive heat −ΔHsystem-\Delta H_{\text{system}}, so ΔSsurr=−ΔHsysT\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T}. An exothermic reaction raises the entropy of the surroundings, more so at low temperature. Third law. The entropy of a perfectly crystalline pure substance is zero at 0 K0\,\mathrm{K}. This fixes absolute entropies, which (unlike absolute enthalpies) can be tabulated. Elements do not have S∘=0S^{\circ} = 0 at 298 K298\,\mathrm{K}. Key Takeaways: - ΔS=qrevT\Delta S = \frac{q_{\text{rev}}}{T}; ΔStrans=ΔHtransTtrans\Delta S_{\text{trans}} = \frac{\Delta H_{\text{trans}}}{T_{\text{trans}}}. - Isothermal ideal gas: ΔS=nRln⁡V2V1\Delta S = nR\ln\frac{V_{2}}{V_{1}}. - Spontaneous: ΔStotal>0\Delta S_{\text{total}} > 0; equilibrium: ΔStotal=0\Delta S_{\text{total}} = 0. - ΔSsurr=−ΔHsysT\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T}. - Third law: S=0S = 0 for a perfect crystal at 0 K0\,\mathrm{K}. - Pitfall: entropy of the system can decrease in a spontaneous process (water freezing below 0∘C0^{\circ}\mathrm{C}); only the total must increase. Brain-Teaser 1: ΔvapH\Delta_{vap}H of water is 40.7 kJ mol−140.7\,\mathrm{kJ\,mol^{-1}} at 373 K373\,\mathrm{K}. Find ΔvapS\Delta_{vap}S and comment. Solution: ΔS=40700373=109 J K−1 mol−1\Delta S = \frac{40700}{373} = 109\,\mathrm{J\,K^{-1}\,mol^{-1}}. This exceeds the Trouton value (about 8585) because hydrogen bonding makes liquid water unusually ordered, so vaporisation increases disorder more. Brain-Teaser 2: One mole of an ideal gas expands isothermally and reversibly to twice its volume at 300 K300\,\mathrm{K}. Find ΔSsys\Delta S_{\text{sys}}, ΔSsurr\Delta S_{\text{surr}} and ΔStotal\Delta S_{\text{total}}. What if the same expansion occurs freely into a vacuum? Solution: ΔSsys=Rln⁡2=5.76 J K−1\Delta S_{\text{sys}} = R\ln 2 = 5.76\,\mathrm{J\,K^{-1}} in both cases (state function). Reversible: the surroundings supply q=RTln⁡2q = RT\ln 2, so ΔSsurr=−5.76\Delta S_{\text{surr}} = -5.76 and ΔStotal=0\Delta S_{\text{total}} = 0. Free expansion: no heat is exchanged, ΔSsurr=0\Delta S_{\text{surr}} = 0, so ΔStotal=+5.76 J K−1\Delta S_{\text{total}} = +5.76\,\mathrm{J\,K^{-1}}: an irreversible, spontaneous process. Brain-Teaser 3: Predict the sign of ΔS\Delta S: (a) 2SO2(g)+O2(g)→2SO3(g)\mathrm{2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)}; (b) CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_{3}(s) \rightarrow CaO(s) + CO_{2}(g)}; (c) I2(s)→I2(g)\mathrm{I_{2}(s) \rightarrow I_{2}(g)}. Solution: (a) Negative: gas moles fall from 33 to 22. (b) Positive: a gas is produced from a solid. (c) Positive: sublimation. Changes in the number of gas moles usually dominate the sign of ΔS\Delta S.

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