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Chemistry · Ch 7 — Thermodynamics

Second Law of Thermodynamics

7.10

Second Law of Thermodynamics

The first law, on its own, only guarantees that the TOTAL energy of the universe stays conserved through any process -- it has nothing at all to say about which DIRECTION a process is allowed to run.

Two familiar examples make the gap obvious. A glass of hot water, left alone, loses heat to its surroundings over time and cools down. Mixing hydrochloric acid with sodium hydroxide produces sodium chloride and water while evolving heat. Both of these are perfectly consistent with the first law -- total energy is conserved either way. But the REVERSE of either process never happens spontaneously: cold water sitting in a room never spontaneously absorbs heat from that same room and heats itself back up, even though that reverse process would ALSO conserve total energy and so would not violate the first law at all. Supply heat energy directly to the cold water, however, and it WILL become hot -- so the change is one that does not occur spontaneously on its own, but CAN be driven by deliberately supplying energy.

Similarly, a solution of sodium chloride never spontaneously absorbs heat energy on its own in order to reform hydrochloric acid and sodium hydroxide -- and unlike the water example, this particular reverse process cannot even be driven simply by supplying energy; it needs a fundamentally different mechanism entirely. …

Spontaneity and Randomness

Look closely at two everyday spontaneous processes -- ice melting, and water evaporating -- and a common pattern emerges: in both, the RANDOMNESS or disorder of the system increases. In solid ice, the water molecules are locked into a highly organised crystal pattern, permitting almost no independent movement. As the ice melts, those molecules become disorganised and gain freedom to move; the freedom grows further still once the molecules escape into the vapour phase. Put another way: the randomness of the water molecules keeps increasing as ice melts into water, and again as water evaporates into vapour. Both of these are spontaneous processes, and both result in an INCREASE in randomness -- that is, in entropy. …

Figure 7.8Illustration showing an increase in disorder

What this figure shows. Three conical flasks side by side under a rightward arrow labelled 'Increasing entropy'. Left flask: yellow spheres packed in a tight, ordered cluster, labelled 'crystalline solid'. Middle flask: the same spheres more loosely scattered, labelled 'Liquid'. Right flask: the spheres spread thinly with motion trails, labelled 'Gas'. Double-headed arrows between each pair of flasks are marked 'ΔS>0\Delta S>0' (forward, increasing disorder) and 'ΔS<0\Delta S<0' …

Standard Entropy Change and Standard Entropy of Formation

It is possible to calculate the actual, absolute entropy of a substance at any temperature above 0 K -- unlike internal energy or enthalpy, where only CHANGES are ever measurable, entropy has a genuine absolute zero point to measure from. The absolute entropy of a substance at 298 K and one bar pressure is called its standard entropy, S0S^0. The third law of thermodynamics (Section 7.12) states, following Nernst, that the absolute entropy of elements is exactly zero ONLY at 0 K in a perfect crystal, and that standard entropies of all substances at any temperature above 0 K always come out positive.

Once the entropies of the individual substances involved are known, the standard entropy change, ΔSr0\Delta S_r^0, for a chemical reaction can be calculated directly:

ΔSr0=∑Sproducts0−∑Sreactants0(7.30)\Delta S_r^0 = \sum S_{products}^0 - \sum S_{reactants}^0 \qquad (7.30)

Standard entropy of formation, ΔSf0\Delta S_f^0, is defined as "the entropy of formation of 1 mole of a compound from the elements under standard conditions" -- calculable from the individual S0S^0 values of the elements involved, exactly the way ΔHf0\Delta H_f^0 is built from elemental enthalpies. …

Misc 7.6Problem 7.6 -- standard entropy change of carbon burning to CO$_2$

Worked out. C(s)+O2(g)→CO2(g)C(s)+O_2(g)\rightarrow CO_2(g), given S0S^0: CO2(g)=213.6_2(g)=213.6, C(s)=5.740C(s)=5.740, O2(g)=205O_2(g)=205 JK−1^{-1}. ΔSr0={SCO20}−{SC0+SO20}=213.6−[5.74+205]=213.6−210.74=2.86\Delta S_r^0=\{S_{CO_2}^0\}-\{S_C^0+S_{O_2}^0\}=213.6-[5.74+205]=213.6-210.74=2.86 JK−1^{-1}. …

Misc evaluate-yourself-6Evaluate Yourself 6 -- entropy change of urea hydrolysis

Worked out. Book's practice box (no printed solution): urea hydrolyses to ammonia and carbon dioxide; standard entropies of urea, H2_2O, CO2_2, NH3_3 are 173.8, 70, 213.5, 192.5 J mol−1^{-1}K−1^{-1}. Working it through (for NH2CONH2+H2O→2NH3+CO2NH_2CONH_2+H_2O\rightarrow 2NH_3+CO_2): ΔSr0=[2(192.5)+213.5]−[173.8+70]=[385+213.5]−[243.8]=598.5−243.8=354.7\Delta S_r^0=[2(192.5)+213.5]-[173.8+70]=[385+213.5]-[243.8]=598.5-243.8=354.7 J K−1^{-1} (own solution, not printed in the textbook). …

Entropy Change Accompanying Change of Phase

When a substance changes state -- solid to liquid (melting), liquid to vapour (evaporation), or solid to vapour (sublimation) -- there is necessarily a change in entropy. Because the two phases are in equilibrium with each other during such a change, it can be treated as occurring reversibly at constant temperature, giving the general relation:

ΔS=qrevT=ΔHrevT(7.31)\Delta S = \frac{q_{rev}}{T} = \frac{\Delta H_{rev}}{T} \qquad (7.31)

Entropy of fusion. The heat absorbed when one mole of a solid melts reversibly at its melting point is the molar heat of fusion, ΔHfusion\Delta H_{fusion}. The associated entropy change is

ΔSfusion=ΔHfusionTf(7.32)\Delta S_{fusion} = \frac{\Delta H_{fusion}}{T_f} \qquad (7.32)

where TfT_f is the melting point.

Entropy of vapourisation. The heat absorbed when one mole of liquid boils reversibly at its boiling point is the molar heat of vapourisation, ΔHv\Delta H_v. The entropy change is

ΔSv=ΔHvTb(7.33)\Delta S_v = \frac{\Delta H_v}{T_b} \qquad (7.33)

where TbT_b is the boiling point.

Entropy of transition. The heat change when one mole of a solid changes reversibly from one allotropic form to another, at its transition temperature, is the enthalpy of transition, ΔHt\Delta H_t. The entropy change is

ΔSt=ΔHtTt(7.34)\Delta S_t = \frac{\Delta H_t}{T_t} \qquad (7.34)

where TtT_t is the transition temperature. …

Misc 7.7Problem 7.7 -- entropy of fusion of ice

Worked out. 1 mole of ice melts at 0∘0^\circC (273 K) and 1 atm, ΔHfusion=6008\Delta H_{fusion}=6008 J mol−1^{-1}. ΔSfusion=ΔHfusion/Tf=6008/273=22.007\Delta S_{fusion}=\Delta H_{fusion}/T_f=6008/273=22.007 J K−1^{-1}mol−1^{-1}. …

Misc evaluate-yourself-7Evaluate Yourself 7 -- entropy change on evaporating ethanol

Worked out. Book's practice box (no printed solution): 1 mole of ethanol evaporates at 351 K, molar heat of vaporisation =39.84=39.84 kJ mol−1^{-1}. Working it through: ΔSv=39,840/351=113.5\Delta S_v=39{,}840/351=113.5 J K−1^{-1}mol−1^{-1} (own solution, not printed in the textbook). …