Spontaneity — second law; entropy ΔS: Worked examples

Before anything: three symbols you will see over and over, each defined once in plain words.

The scenario matrix

Every entropy problem in this chapter falls into one of these cells. The final column names the worked example below that lands on it.

Cell	What makes it that case	Sign / behaviour to watch	Example
A. Exothermic, gas made	$Δ H < 0$ , $Δ n_{g a s} > 0$	both terms $+$ → always spontaneous	Ex 1
B. Endothermic, disorder up	$Δ H > 0$ , $Δ S_{sy s} > 0$	fight; $T$ decides	Ex 2
C. Exothermic, order up	$Δ H < 0$ , $Δ S_{sy s} < 0$	fight the other way; low $T$ wins	Ex 3
D. Phase change AT equilibrium	reversible, $Δ S_{u ni v} = 0$	the "zero" case	Ex 4
E. Ideal-gas volume change	isothermal, $Δ U = 0$	sign follows $V_{2} / V_{1}$ ; compression $\to$ negative	Ex 5
F. Degenerate input ( $W = 1$ )	perfect crystal at $0 K$	$S = 0$ exactly	Ex 6
G. Limiting behaviour	$T \to \infty$ and $T \to 0$	which term dominates	Ex 7
H. Real-world word problem	mixing / dissolving	predict + compute	Ex 8
I. Exam twist	"find the crossover $T$ "	solve $Δ S_{u ni v} = 0$	Ex 9

Example 1 — Cell A: exothermic AND gas produced

Worked example Combustion-type: both terms push forward

A reaction has $Δ H_{sy s} = - 286 kJ mol^{- 1}$ at $T = 298 K$ and forms extra gas, giving $Δ S_{sy s} = + 120 J K^{- 1} mol^{- 1}$ . Find $Δ S_{s u r r}$ , $Δ S_{u ni v}$ , and decide spontaneity.

Forecast: Exothermic → surroundings gain heat → $Δ S_{s u r r} > 0$ . Gas made → $Δ S_{sy s} > 0$ . Both positive → guess strongly spontaneous.

$Δ S_{s u r r} = - \frac{Δ H _{sy s}}{T} = - \frac{- 286000}{298} = + 960 J K^{- 1} mol^{- 1}$ . Why this step? Heat leaving the system ( $Δ H < 0$ ) enters the surroundings, spreading energy there — a positive contribution. We convert kJ→J (×1000) so units match $Δ S_{sy s}$ .
$Δ S_{u ni v} = Δ S_{sy s} + Δ S_{s u r r} = 120 + 960 = + 1080 J K^{- 1} mol^{- 1}$ . Why this step? The master test needs the total, so we add.
$Δ S_{u ni v} > 0 \Rightarrow$ spontaneous.

Verify: Both drivers point the same way, so a large positive total is exactly expected. Units: $J K^{- 1} mol^{- 1}$ throughout — consistent.

Example 2 — Cell B: endothermic but disorder rises (the surprise case)

Worked example Endothermic yet spontaneous

Ammonium nitrate dissolving: $Δ H_{sy s} = + 25.7 kJ mol^{- 1}$ , $Δ S_{sy s} = + 108 J K^{- 1} mol^{- 1}$ , $T = 298 K$ . Spontaneous?

Forecast: It absorbs heat (the beaker feels cold), so $Δ S_{s u r r} < 0$ — that fights us. But solid salt breaking into freely-moving ions is a big rise in disorder. Guess: entropy might win → spontaneous, but let's check the numbers.

$Δ S_{s u r r} = - \frac{+ 25700}{298} = - 86.2 J K^{- 1} mol^{- 1}$ . Why this step? Heat is pulled from the surroundings, so their energy spread drops → negative.
$Δ S_{u ni v} = 108 + (- 86.2) = + 21.8 J K^{- 1} mol^{- 1}$ . Why this step? We let the two opposing terms compete by adding them.
$Δ S_{u ni v} > 0 \Rightarrow$ spontaneous — the system's disorder gain outweighs the surroundings' loss.

Verify: This matches reality: cold-pack chemistry dissolves eagerly. The sign flip from Ex 1 (this time $Δ S_{s u r r} < 0$ ) is the whole point of cell B.

Example 3 — Cell C: exothermic but order increases (low- $T$ favoured)

Worked example Freezing water: exothermic, but system orders up

Water freezing: $Δ H_{sy s} = - 6.01 kJ mol^{- 1}$ , $Δ S_{sy s} = - 22.0 J K^{- 1} mol^{- 1}$ . Test at $T = 263 K$ ( $- 10^{\circ} C$ ).

Forecast: Liquid → solid is more ordered, so $Δ S_{sy s} < 0$ (fights us). But it releases heat, so $Δ S_{s u r r} > 0$ (helps). At a cold temperature the surroundings term is boosted (dividing by a small $T$ ). Guess: spontaneous below freezing.

$Δ S_{s u r r} = - \frac{- 6010}{263} = + 22.85 J K^{- 1} mol^{- 1}$ . Why this step? Small denominator $T$ makes each joule of released heat "count for more" disorder in cold surroundings — the deep reason ice forms only when it's cold.
$Δ S_{u ni v} = - 22.0 + 22.85 = + 0.85 J K^{- 1} mol^{- 1}$ . Why this step? Add the fighting terms; the surroundings just barely win.
$Δ S_{u ni v} > 0 \Rightarrow$ spontaneous: water freezes below $0^{\circ} C$ . ✔

Verify: Sanity check the crossover — at exactly $273 K$ we should get $0$ (that's Ex 4). Here $263 < 273$ , so we expect a small positive value, and $+ 0.85$ is indeed small and positive. Consistent.

Example 4 — Cell D: the ZERO case (phase change at equilibrium)

Worked example Melting ice exactly at

$0^{\circ} C$ $Δ H_{f u s} = 6.01 kJ mol^{- 1}$ , at the melting point $T = 273 K$ . Find $Δ S_{sy s}$ , $Δ S_{s u r r}$ , $Δ S_{u ni v}$ .

Forecast: At the exact melting point, ice and water coexist happily — neither direction "wins". Guess: $Δ S_{u ni v} = 0$ (equilibrium, the degenerate boundary).

$Δ S_{sy s} = \frac{Δ H _{f u s}}{T} = \frac{6010}{273} = + 22.01 J K^{- 1} mol^{- 1}$ . Why this step? At the melting point the melt is reversible, so $q_{r e v} = Δ H_{f u s}$ and $Δ S = q_{r e v} / T$ applies exactly.
$Δ S_{s u r r} = - \frac{6010}{273} = - 22.01 J K^{- 1} mol^{- 1}$ . Why this step? The same heat that melts the ice is taken from the surroundings, at the same $T$ — equal magnitude, opposite sign.
$Δ S_{u ni v} = + 22.01 - 22.01 = 0$ .

Verify: Exactly zero → equilibrium, matching the second law's " $= 0$ reversible" clause. This is the boundary that Ex 3 sat just below.

Example 5 — Cell E: ideal gas, both directions (expansion AND compression)

The figure shows why volume controls entropy: more room = more places each molecule can sit.

Figure — Spontaneity — second law; entropy ΔS

Worked example 5a. Expansion and 5b. Compression

$n = 2.0 mol$ ideal gas, isothermal at $T = 300 K$ , gas constant $R = 8.314 J K^{- 1} mol^{- 1}$ . (a) Expand $V_{1} = 1.0 L \to V_{2} = 4.0 L$ . (b) Compress $V_{1} = 4.0 L \to V_{2} = 1.0 L$ .

Forecast: Expansion spreads molecules over more space → more microstates → $Δ S > 0$ . Compression is the exact reverse → $Δ S < 0$ , same size. Guess: two answers equal and opposite.

Use $Δ S = n R ln \frac{V _{2}}{V _{1}}$ . Why this formula and not $q_{r e v} / T$ directly? For an ideal gas at constant $T$ , $Δ U = 0$ , so $q_{r e v} = - w_{r e v} = n R T ln (V_{2} / V_{1})$ ; dividing by $T$ cancels it to $n R ln (V_{2} / V_{1})$ . The $ln$ appears because volume "spreads" multiply, and entropy adds — logs turn multiplying into adding (same reason as $S = k_{B} ln W$ ).
(a) $Δ S = 2.0 \times 8.314 \times ln 4 = + 23.05 J K^{- 1}$ .
(b) $Δ S = 2.0 \times 8.314 \times ln \frac{1}{4} = - 23.05 J K^{- 1}$ .

Verify: $ln 4 = - ln \frac{1}{4}$ , so the two answers are equal and opposite — exactly the forecast. Compression giving a negative $Δ S$ is fine: the master test only forbids negative $Δ S_{u ni v}$ , and squeezing gas requires work that dumps heat outward, raising $Δ S_{s u r r}$ .

Example 6 — Cell F: degenerate input, $W = 1$

Example 7 — Cell G: limiting behaviour ( $T \to \infty$ and $T \to 0$ )

Worked example Who wins as temperature runs to the extremes?

Take a reaction with fixed $Δ H_{sy s} = - 100 kJ mol^{- 1}$ and fixed $Δ S_{sy s} = - 50 J K^{- 1} mol^{- 1}$ (cell C style). Examine $Δ S_{u ni v} = Δ S_{sy s} - \frac{Δ H _{sy s}}{T}$ as $T \to 0$ and $T \to \infty$ .

Forecast: The surroundings term has $T$ in the denominator. Small $T$ blows it up (surroundings dominate); huge $T$ kills it (system dominates). Guess: spontaneous when cold, not when hot.

Write it out: $Δ S_{u ni v} = - 50 + \frac{100000}{T}$ . Why this step? $- Δ H_{sy s} = + 100000 J$ ; keep everything in joules.
$T \to \infty$ : the $\frac{100000}{T}$ term $\to 0$ , leaving $Δ S_{u ni v} \to - 50 < 0$ → non-spontaneous at very high $T$ . Why? When surroundings are hot, a fixed dump of heat barely disturbs them, so the system's own ordering ( $Δ S_{sy s} < 0$ ) dominates.
$T \to 0$ : $\frac{100000}{T} \to + \infty$ , so $Δ S_{u ni v} \to + \infty > 0$ → spontaneous at very low $T$ . Why? Cold surroundings treasure every joule; the released heat spreads huge disorder there.

Verify: Crossover at $Δ S_{u ni v} = 0 \Rightarrow T = 100000/50 = 2000 K$ . Below $2000 K$ spontaneous, above it not — a clean, checkable limit that sets up Ex 9's method.

Example 8 — Cell H: real-world word problem (heat spreading between bodies)

Worked example Hot metal dropped into cool water (tiny transfer limit)

A large water bath at $T_{co l d} = 290 K$ receives $q = 500 J$ from a hot block held at $T_{h o t} = 350 K$ . The bodies are big enough that their temperatures barely change. Find $Δ S_{u ni v}$ and confirm heat flows hot→cold.

Forecast: Heat leaves the hot body (loses entropy) and enters the cold one (gains entropy). Dividing the same $500 J$ by a smaller $T$ gives more entropy, so the cold body should gain more than the hot one loses → total $> 0$ . That's why heat flows this way at all.

Hot body loses heat: $Δ S_{h o t} = \frac{- 500}{350} = - 1.4286 J K^{- 1}$ . Why this step? Energy leaving a reservoir at $T_{h o t}$ reduces its spread.
Cold body gains heat: $Δ S_{co l d} = \frac{+ 500}{290} = + 1.7241 J K^{- 1}$ . Why this step? Same joules, smaller $T$ → bigger entropy gain (each joule "spreads further" when it's cold).
$Δ S_{u ni v} = - 1.4286 + 1.7241 = + 0.2955 J K^{- 1} > 0$ → the flow is spontaneous.

Verify: Reverse the flow (cold→hot) and every sign flips → $Δ S_{u ni v} = - 0.2955 < 0$ → forbidden. So heat cannot climb from cold to hot on its own. This is the second law in one line.

Example 9 — Cell I: exam twist ("find the crossover temperature")

Worked example At what temperature does a reaction switch from spontaneous to not?

For $CaCO_{3} (s) \to CaO (s) + CO_{2} (g)$ : $Δ H_{sy s} = + 178 kJ mol^{- 1}$ , $Δ S_{sy s} = + 161 J K^{- 1} mol^{- 1}$ (both roughly constant). Find the temperature above which decomposition becomes spontaneous.

Forecast: Endothermic ( $Δ H > 0$ , surroundings hurt) but big positive $Δ S_{sy s}$ (a gas is born). At low $T$ the surroundings' penalty $- Δ H / T$ is huge and negative → not spontaneous. At high $T$ it shrinks and $Δ S_{sy s}$ wins. So there's a crossover; guess it's a high temperature (kilns are hot!).

Set $Δ S_{u ni v} = 0$ : $Δ S_{sy s} - \frac{Δ H _{sy s}}{T} = 0$ . Why this step? The switch happens exactly at equilibrium, where the total entropy change is zero.
Solve for $T$ : $T = \frac{Δ H _{sy s}}{Δ S _{sy s}} = \frac{178000}{161} = 1105.6 K$ . Why this step? Rearranging isolates the crossover temperature; keep $Δ H$ in joules to match $Δ S$ 's units.
Above $\approx 1106 K$ ( $\approx 833^{\circ} C$ ) the reaction is spontaneous; below it, not.

Verify: Real limestone kilns run near $900^{\circ} C$ – $1000^{\circ} C$ — right in this ballpark. Units cancel: $\frac{J mol ^{- 1}}{J K ^{- 1} mol ^{- 1}} = K$ . ✔ (This is exactly $T = Δ H /Δ S$ , the $Δ G = 0$ condition you'll meet in Gibbs Free Energy ΔG.)

Recall Which cell was which?

Which examples had $Δ S_{s u r r} < 0$ ? ::: Ex 2 (endothermic dissolving) and Ex 4's surroundings term — because heat was pulled from the surroundings. Which single example gave $Δ S_{u ni v} = 0$ and why? ::: Ex 4, melting at exactly 273 K — reversible phase change at equilibrium. In Ex 5, why are expansion and compression equal and opposite? ::: $ln 4 = - ln \frac{1}{4}$ ; volume ratios are reciprocals so the logs negate. Why is $S = 0$ for the perfect crystal in Ex 6? ::: $W = 1$ and $ln 1 = 0$ — one arrangement means no dispersal (Third Law). In Ex 7, why does high $T$ favour the ordered (cell-C) product? ::: The $- Δ H / T$ surroundings term shrinks toward zero, so the system's negative $Δ S_{sy s}$ dominates. The crossover temperature formula from Ex 9? ::: $T = \frac{Δ H _{sy s}}{Δ S _{sy s}}$ , from setting $Δ S_{u ni v} = 0$ .

Connections

Parent: 2.5.12 Spontaneity — second law; entropy ΔS (Hinglish)
Gibbs Free Energy ΔG — the crossover $T = Δ H /Δ S$ is exactly $Δ G = 0$ .
Enthalpy ΔH — supplies $Δ H_{sy s}$ for every surroundings term.
First Law of Thermodynamics — gives $q_{r e v} = - w_{r e v}$ used in Ex 5.
Reversible vs Irreversible Processes — why we use $q_{r e v}$ at all.
Boltzmann distribution & microstates — the $S = k_{B} ln W$ behind Ex 6.
Third Law of Thermodynamics — the $W = 1 \Rightarrow S = 0$ anchor.