Spontaneity — second law; entropy ΔS: Exercises — Chemistry deep dive

Quick symbol reminder, all earned in the parent:

$S$ = entropy, "how many microscopic arrangements the state has", units $J K^{- 1}$ .
$Δ S$ = change in entropy = (entropy of end state) − (entropy of start state).
$q_{r e v}$ = heat exchanged along the gentle reversible path; $T$ = absolute temperature in kelvin.
$Δ S_{u ni v} = Δ S_{sy s} + Δ S_{s u r r}$ , and $Δ S_{s u r r} = - \frac{Δ H _{sy s}}{T}$ at constant pressure.

Level 1 — Recognition

Recall Solution L1.1

The single idea: more spread-out matter/energy = more microstates = higher $S$ . Gases have far more microstates than liquids, liquids more than solids.

(a) Vapour → liquid: gas becomes liquid, molecules lose freedom → $Δ S_{sy s} < 0$ . (b) Solid → gas: huge jump in freedom → $Δ S_{sy s} > 0$ . (c) Count gas moles: left $= 1 + 3 = 4$ mol gas; right $= 2$ mol gas. Fewer gas moles → $Δ S_{sy s} < 0$ . (d) A cold object gaining heat: its molecules jiggle over more energy levels → $Δ S_{sy s} > 0$ .

Rule of thumb used: $Δ n_{g a s} > 0 \Rightarrow Δ S > 0$ ; heating something $\Rightarrow Δ S > 0$ .

Level 2 — Application

Recall Solution L2.1

WHAT: the surroundings receive the heat the system dumps. WHY this formula: at constant $P$ , $q_{P} = Δ H$ , and heat into surroundings is $- Δ H_{sy s}$ , so $Δ S_{s u r r} = \frac{- Δ H _{sy s}}{T} = \frac{- ( - 145000 J )}{310 K} = + 468 J K^{- 1} .$ Positive: exothermic reaction warms the surroundings, spreading energy → more microstates for them.

Recall Solution L2.2

At the boiling point liquid and vapour coexist reversibly, so the phase change is isothermal and reversible: $q_{r e v} = Δ H_{v a p}$ . $Δ S_{sy s} = \frac{q _{r e v}}{T} = \frac{40700}{373} = + 109 J K^{- 1} mol^{- 1} .$ Positive because liquid → gas is a big jump in microstates. (This value near $+ 109$ is Trouton's rule — most liquids sit near $85$ – $110 J K^{- 1} mol^{- 1}$ .)

Recall Solution L2.3

WHY this formula: for an ideal gas at constant $T$ , $Δ U = 0$ , so all the reversible heat becomes expansion work and $Δ S = n R ln \frac{V _{2}}{V _{1}}$ . Here $V_{2} / V_{1} = 3$ . $Δ S = n R ln \frac{V _{2}}{V _{1}} = 2.0 \times 8.314 \times ln 3 = 16.628 \times 1.0986 = + 18.3 J K^{- 1} .$ Positive: bigger box = more positions per molecule = more microstates.

Level 3 — Analysis

Recall Solution L3.1

WHAT: we need the total entropy change, because the second law tests the universe, not the system. First the surroundings: $Δ S_{s u r r} = \frac{- Δ H _{sy s}}{T} = \frac{55000}{298} = + 184.6 J K^{- 1} .$ Then the total: $Δ S_{u ni v} = Δ S_{sy s} + Δ S_{s u r r} = - 110 + 184.6 = + 74.6 J K^{- 1} > 0.$ Spontaneous. The system got more ordered ( $Δ S_{sy s} < 0$ ), but the large heat released disordered the surroundings even more — the sum still climbs. This is exactly the energy-bookkeeping point: the system can order itself if it pays the surroundings enough heat.

Recall Solution L3.2

Spontaneity fails when $Δ S_{u ni v}$ falls to zero: $Δ S_{u ni v} = Δ S_{sy s} - \frac{Δ H _{sy s}}{T} = 0 \Rightarrow T = \frac{Δ H _{sy s}}{Δ S _{sy s}} .$ $T = \frac{- 55000}{- 110} = 500 K .$ Below $500 K$ : the surroundings term (which grows as $T$ shrinks) wins → spontaneous. Above $500 K$ : the $\frac{Δ H}{T}$ term shrinks, the system's own order ( $Δ S_{sy s} < 0$ ) dominates → non-spontaneous. See the figure — the two contributions cross exactly at $500 K$ .

Figure — Spontaneity — second law; entropy ΔS

Level 4 — Synthesis

Recall Solution L4.1

Step 1 — surroundings at 298 K. Melting absorbs heat, so heat leaves the surroundings: $Δ H_{sy s} = + 6010 J$ (endothermic), thus $Δ S_{s u r r} = \frac{- Δ H _{sy s}}{T} = \frac{- 6010}{298} = - 20.17 J K^{- 1} .$ Step 2 — total at 298 K. $Δ S_{u ni v} = + 22.0 + (- 20.17) = + 1.83 J K^{- 1} > 0 \Rightarrow spontaneous .$ Step 3 — why 273 K is different. Redo the surroundings at the melting point: $Δ S_{s u r r} (273) = \frac{- 6010}{273} = - 22.0 J K^{- 1},$ so $Δ S_{u ni v} = 22.0 - 22.0 = 0$ → equilibrium, ice and water coexist. The deep point: $Δ S_{sy s}$ (a property of the substances) barely changes with $T$ , but $Δ S_{s u r r} = - Δ H_{sy s} / T$ shrinks in size as $T$ grows. At the melting point they cancel; heat the system a little and the surroundings term weakens, tipping $Δ S_{u ni v}$ positive — melting wins. This is a reversible vs irreversible boundary crossing.

Recall Solution L4.2

WHY an integral, not $q / T$ ? Here $T$ is changing, so we cannot pull it out of $\int d q_{r e v} / T$ . At constant volume $d q_{r e v} = C_{V} d T$ , so $Δ S = \int_{T_{1}}^{T_{2}} \frac{C _{V} d T}{T} = C_{V} ln \frac{T _{2}}{T _{1}} .$ WHAT the $ln$ does: summing up $\frac{1}{T}$ across every tiny heating step gives a logarithm — each degree added near a cold start raises entropy more than the same degree near a hot end. $Δ S = 12.47 \times ln \frac{600}{300} = 12.47 \times 0.6931 = + 8.64 J K^{- 1} .$ Positive: hotter gas accesses more energy microstates (a Boltzmann spreading).

Level 5 — Mastery

Recall Solution L5.1

(a) Free expansion: gas pushes against nothing, so $w = 0$ ; insulated, so $q = 0$ . Indeed $Δ U = 0$ too (ideal gas, $T$ unchanged). (b) Entropy is a state function — it depends only on the start and end states, not the path taken. So to compute $Δ S$ we imagine a reversible path between the same endpoints and use it: $Δ S_{sy s} = n R ln \frac{V _{2}}{V _{1}} = 1 \times 8.314 \times ln 2 = + 5.76 J K^{- 1} .$ (c) No contradiction. The definition uses $q_{rev}$ , not the actual $q$ . The real path had $q = 0$ , but the reversible replacement path would exchange $q_{r e v} = n R T ln 2 > 0$ . The gas ends up more spread out either way, so $Δ S_{sy s} > 0$ . Second-law check: surroundings unchanged ( $q_{s u r r} = 0 \Rightarrow Δ S_{s u r r} = 0$ ), so $Δ S_{u ni v} = + 5.76 > 0$ — irreversible and spontaneous, exactly as expected.

Recall Solution L5.2

(a) $S = k_{B} ln 1 = k_{B} \times 0 = 0$ . This is the Third Law: a perfect crystal at absolute zero has zero entropy — the one degenerate, perfectly-ordered case that anchors the entropy scale. (b) Take the log of a product: $ln (4^{N_{A}}) = N_{A} ln 4$ . So $S = k_{B} N_{A} ln 4 = R ln 4 = 8.314 \times 1.3863 = + 11.5 J K^{- 1} mol^{- 1} .$ (We used $k_{B} N_{A} = R$ .) Real disordered solids like CO ice do have a small residual $S$ — an experimental fingerprint that $S = k_{B} ln W$ is literally counting arrangements.

Recall Self-test summary (reveal after attempting all)

L1.1 signs ::: (a) − (b) + (c) − (d) + L2.1 $Δ S_{s u r r}$ ::: $+ 468 J K^{- 1}$ L2.2 $Δ S_{v a p}$ ::: $+ 109 J K^{- 1} mol^{- 1}$ L2.3 $Δ S$ (expansion) ::: $+ 18.3 J K^{- 1}$ L3.1 $Δ S_{u ni v}$ ::: $+ 74.6 J K^{- 1}$ → spontaneous L3.2 crossover $T$ ::: $500 K$ L4.1 $Δ S_{u ni v}$ at 298 K ::: $+ 1.83 J K^{- 1}$ → spontaneous L4.2 $Δ S$ (heating) ::: $+ 8.64 J K^{- 1}$ L5.1 $Δ S_{sy s}$ (free expansion) ::: $+ 5.76 J K^{- 1}$ L5.2b residual $S$ ::: $R ln 4 = + 11.5 J K^{- 1} mol^{- 1}$

Connections

Parent: Spontaneity — second law; entropy ΔS
Gibbs Free Energy ΔG — packages $Δ S_{u ni v}$ into a system-only test.
Enthalpy ΔH — supplies $Δ H_{sy s}$ for the surroundings term.
First Law of Thermodynamics — where $q_{r e v} = - w_{r e v}$ in L2.3 comes from.
Reversible vs Irreversible Processes — the heart of L5.1.
Boltzmann distribution & microstates — why heating raises $S$ (L4.2).
Third Law of Thermodynamics — the $S = 0$ anchor in L5.2.