Standard entropy S° and ΔS_rxn = Σ S°(products) − Σ S°(reactants): Worked examples

Intuition What this page is for

The parent note gave you the tool: $Δ S_{r x n}^{\circ} = \sum ν_{p} S^{\circ} (products) - \sum ν_{r} S^{\circ} (reactants)$ . Here we drill it against every kind of situation a problem can throw at you — every sign, the tricky "no gas change" case, the degenerate case where the answer is exactly zero, limiting behaviour, a real-world word problem, and an exam twist. If you finish this page, no exam question about entropy of reaction can surprise you.

Before we start, one reminder of what the symbols mean, from zero:

$S^{\circ}$ = the standard molar entropy of ONE mole of a substance, a count of messiness in units of $J K^{- 1} mol^{- 1}$ (joules per kelvin per mole).
$ν$ (Greek "nu") = the stoichiometric coefficient, the number in front of a species in the balanced equation. It tells us "how many moles," and because entropy scales with amount, we multiply by it.
$Δ$ (Greek "delta") = "change in" = (after) − (before) = products − reactants.

The scenario matrix

Every entropy-of-reaction problem falls into one of these case classes. Each row is a "cell" we must cover. The worked examples below are tagged with the cell they hit.

Cell	Case class	What makes it tricky	Example
A	Gas moles increase → $Δ S > 0$	expect positive; verify sign	Ex 1
B	Gas moles decrease → $Δ S < 0$	expect negative	Ex 2
C	No change in gas moles	sign NOT obvious — must compute	Ex 3
D	Degenerate / near-zero answer	tiny number, watch rounding & sign	Ex 4
E	Solid ↔ solid / liquid only (no gas)	small $Δ S$ , gas rule useless	Ex 5
F	Limiting behaviour — find the $T$ where $Δ G$ flips	couples $Δ S$ with $Δ H$	Ex 6
G	Real-world word problem	translate words → equation	Ex 7
H	Exam twist — reverse a reaction / find an unknown $S^{\circ}$	algebra rearrangement	Ex 8

Figure — Standard entropy S° and ΔS_rxn = Σ S°(products) − Σ S°(reactants)

Cell A — gas moles increase

Worked example Ex 1 · Water gas shift, forward split

Reaction: $2 H_{2} O (g) \to 2 H_{2} (g) + O_{2} (g)$ Data ( $J K^{- 1} mol^{- 1}$ ): $S^{\circ} (H_{2} O, g) = 188.8$ , $S^{\circ} (H_{2}) = 130.7$ , $S^{\circ} (O_{2}) = 205.2$ .

Forecast: count moles of gas. Left = $2$ . Right = $2 + 1 = 3$ . Gas goes $2 \to 3$ — we make gas → mess up → $Δ S > 0$ . Guess before reading on. (This is the cyan bar of the chart above.)

Weight products by coefficients. $S_{prod} = 2 (130.7) + 1 (205.2)$ . Why this step? Entropy is extensive: 2 mol of $H_{2}$ carries twice the entropy of 1 mol, so multiply by $ν = 2$ . $S_{prod} = 261.4 + 205.2 = 466.6$
Weight reactants by coefficients. $S_{react} = 2 (188.8) = 377.6$ . Why this step? Same rule; there is one species, coefficient $2$ .
Subtract: products − reactants. $Δ S^{\circ} = 466.6 - 377.6 = + 89.0 J K^{- 1}$ Why this step? $Δ$ means (after) − (before); "Products Party, Reactants Rest."

Verify: sign is positive, matching the forecast (we created a mole of gas). Units: $J K^{- 1}$ for the reaction as written (each term was $mol \times J K^{- 1} mol^{- 1}$ , moles cancel). ✅

Cell B — gas moles decrease

Cell C — no change in moles of gas

Worked example Ex 3 · Water gas shift (equal gas on both sides)

Reaction: $CO (g) + H_{2} O (g) \to CO_{2} (g) + H_{2} (g)$ Data: $S^{\circ} (CO) = 197.7$ , $S^{\circ} (H_{2} O, g) = 188.8$ , $S^{\circ} (CO_{2}) = 213.8$ , $S^{\circ} (H_{2}) = 130.7$ .

Forecast: gas moles $= 2$ on both sides — this is the tiny white "rule silent" bar of the chart. You genuinely cannot guess the sign from moles; you must compute. (Still, a hint: $H_{2}$ is a tiny 2-atom molecule with few vibrations, so it holds little entropy — replacing bigger molecules with $H_{2}$ often lowers $S$ .)

Products. $S_{prod} = 213.8 + 130.7 = 344.5$ . Why this step? All coefficients are $1$ , so we just add the two product entropies (each weighted by $ν = 1$ ).
Reactants. $S_{react} = 197.7 + 188.8 = 386.5$ . Why this step? Same extensive rule: both reactants have coefficient $1$ , so add their $S^{\circ}$ values to get the total entropy we started with.
Subtract. $Δ S^{\circ} = 344.5 - 386.5 = - 42.0 J K^{- 1}$ Why this step? $Δ S$ = products − reactants; here the "before" pile is larger, so the change comes out negative.

Verify: the answer is a modest negative number — much smaller in magnitude than the $\pm 188$ swings of Cells A/B, exactly because no gas was created or destroyed. The small size is the tell-tale of a "no-gas-change" reaction; only molecular details (small $H_{2}$ ) tip it slightly negative. ✅

Cell D — degenerate / near-zero answer

Worked example Ex 4 · An almost-perfectly-balanced swap

Reaction: $H_{2} (g) + Cl_{2} (g) \to 2 HCl (g)$ Data: $S^{\circ} (H_{2}) = 130.7$ , $S^{\circ} (Cl_{2}) = 223.1$ , $S^{\circ} (HCl) = 186.9$ .

Forecast: gas moles $2 \to 2$ (no change). Two diatomic gases become two diatomic gases — very similar, so expect $Δ S \approx 0$ , sign unknown.

Products. $S_{prod} = 2 (186.9) = 373.8$ . Why this step? $HCl$ has coefficient $2$ , and entropy is extensive, so we double its molar $S^{\circ}$ to get the entropy of what we made.
Reactants. $S_{react} = 130.7 + 223.1 = 353.8$ . Why this step? $H_{2}$ and $Cl_{2}$ each have coefficient $1$ , so add their molar entropies for the total we started with.
Subtract. $Δ S^{\circ} = 373.8 - 353.8 = + 20.0 J K^{- 1}$ Why this step? $Δ S$ = products − reactants; the small leftover after cancellation is the true change, and its sign must be reported, not rounded away.

Verify: the result is tiny and positive ( $+ 20.0$ ), a near-degenerate case. Sanity: forming two identical HCl molecules from two different reactant molecules increases the "mixing" of arrangements slightly → small positive. If you'd carelessly written $Δ S \approx 0$ you'd lose the sign; the honest calculation keeps it. Watch rounding here — small numbers are sensitive, so keep one decimal. ✅

Cell E — condensed phases only (no gas at all)

Worked example Ex 5 · Graphite → diamond

Reaction: $C(graphite) \to C(diamond)$ Data: $S^{\circ} (graphite) = 5.7$ , $S^{\circ} (diamond) = 2.4$ .

Forecast: no gas anywhere, so the gas rule is useless. Physics hint: diamond is a rigid, tightly bonded 3-D lattice — atoms can barely vibrate, fewer arrangements — while graphite's loose sheets slide and vibrate more. So diamond should have lower entropy → $Δ S < 0$ .

Product − reactant (both coefficients $1$ ): $Δ S^{\circ} = S^{\circ} (diamond) - S^{\circ} (graphite) = 2.4 - 5.7$ Why this step? Same subtraction rule; only two solids, so it's a one-line calculation. $Δ S^{\circ} = - 3.3 J K^{- 1}$

Verify: negative and very small — exactly what condensed-phase-only reactions give (no gas to flood the system with microstates). Matches the "rigid diamond = neat stack" picture. Notice all these numbers are single-digit, dwarfed by the gas cases; that's the signature of solid↔solid chemistry. ✅

Cell F — limiting behaviour: the temperature where spontaneity flips

Worked example Ex 6 · When does dolomite decompose? (couples

$Δ S$ with $Δ H$ ) Reaction: $CaCO_{3} (s) \to CaO (s) + CO_{2} (g)$ , with $Δ S^{\circ} = + 160.6 J K^{- 1}$ and $Δ H^{\circ} = + 178 kJ$ (from the parent note). Question: above what temperature does this become spontaneous?

Forecast: $Δ H > 0$ (needs heat) but $Δ S > 0$ (makes a gas). At low $T$ , enthalpy wins → non-spontaneous. As $T$ grows, the $T Δ S$ term grows and eventually beats $Δ H$ . So there is a crossover temperature where $Δ G = 0$ ; above it, spontaneous.

Write the spontaneity condition. Spontaneous means $Δ G < 0$ , where $Δ G = Δ H - T Δ S$ . See Gibbs free energy ΔG = ΔH − TΔS. Why this step? $Δ S$ alone can't decide spontaneity of the system; we need $Δ G$ , which weighs entropy by temperature.
Match units. $Δ H = 178 kJ$ , $Δ S = 0.1606 kJ K^{- 1}$ . Why this step? $Δ S$ is in J, $Δ H$ in kJ; mixing them is the classic error. Convert $160.6 J = 0.1606 kJ$ .
Set $Δ G = 0$ to find the crossover. $0 = Δ H - T Δ S \Rightarrow T = \frac{Δ H}{Δ S} = \frac{178}{0.1606} = 1108 K$ Why this step? At the flip point the two terms exactly cancel; solving for $T$ gives the threshold.

Verify (limiting checks):

As $T \to 0$ : $Δ G \to Δ H = + 178 kJ > 0$ → non-spontaneous. ✅ (matches "cold kiln = no reaction")
As $T \to \infty$ : the $- T Δ S$ term dominates and $Δ G \to - \infty$ → spontaneous. ✅
At exactly $1108 K$ : $Δ G = 178 - 1108 (0.1606) = 178 - 178.0 = 0$ . ✅ crossover confirmed. That's why lime kilns run near $1100 K$ . ✅

The plot below is exactly this argument drawn out: the cyan line is $Δ G$ falling as $T$ rises, crossing zero at the amber dot ( $1108 K$ ). Left of the dot $Δ G > 0$ (non-spontaneous); right of it $Δ G < 0$ (spontaneous). The two limiting checks are the far-left and far-right ends of that line.

Cell G — real-world word problem

Worked example Ex 7 · "Why does frost form but never un-form?" (translate words → numbers)

A cook watches water vapour turn to ice on a freezer wall: $H_{2} O (g) \to H_{2} O (s)$ . Data: $S^{\circ} (H_{2} O, g) = 188.8$ , $S^{\circ} (H_{2} O, s) = 48.0$ .

Forecast: a gas becomes a solid — the messiest phase collapses to the neatest → big negative $Δ S$ for the system.

System entropy change. $Δ S_{sy s} = S^{\circ} (ice) - S^{\circ} (vapour) = 48.0 - 188.8 = - 140.8 J K^{- 1}$ Why this step? Products − reactants, one mole each.
Doesn't the 2nd law forbid this? The 2nd law is about the universe, not the system alone. See Second Law of Thermodynamics. Freezing releases heat to the cold surroundings, and cold surroundings gain a lot of entropy from that heat (see Entropy of surroundings ΔS_surr = −ΔH/T). Why this step? To resolve the paradox we must include the surroundings; the system may lose entropy if the surroundings gain more.
Interpret. The system's $- 140.8 J K^{- 1}$ is overwhelmed by a larger positive $Δ S_{s u r r}$ , so $Δ S_{u ni v} > 0$ and frost forms — never the reverse.

Verify: sign is a large negative for the system, exactly as forecast for gas→solid; and the everyday observation (frost forms, never spontaneously sublimes back in a freezer) confirms the surroundings' gain dominates. ✅

Cell H — exam twist: reversing a reaction & finding an unknown $S^{\circ}$

Worked example Ex 8 · Two twists in one

Part (a): The parent note found $Δ S^{\circ} = - 198.7 J K^{- 1}$ for $N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g)$ . What is $Δ S^{\circ}$ for the reverse reaction $2 NH_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$ ?

Forecast (a): reversing swaps "products" and "reactants," so the whole subtraction flips sign → $+ 198.7$ . (Also matches physics: reverse makes gas $2 \to 4$ , mess up.)

Since $Δ S_{rev} = S_{react} - S_{prod} = - (S_{prod} - S_{react}) = - Δ S_{fwd}$ : $Δ S_{rev}^{\circ} = - (- 198.7) = + 198.7 J K^{- 1}$ Why this step? $Δ S$ is a state function; reversing start↔end negates it — no re-calculation needed.

Verify (a): positive, gas moles rise $2 \to 4$ . ✅

Part (b): For $2 NH_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$ you're told $Δ S^{\circ} = + 198.7 J K^{- 1}$ , $S^{\circ} (N_{2}) = 191.6$ , $S^{\circ} (H_{2}) = 130.7$ . Find the unknown $S^{\circ} (NH_{3})$ .

Forecast (b): ammonia is a small molecule; a value near $190 J K^{- 1} mol^{- 1}$ is plausible.

Write the master equation with the unknown. $Δ S^{\circ} = [1 \cdot S^{\circ} (N_{2}) + 3 \cdot S^{\circ} (H_{2})] - [2 \cdot S^{\circ} (NH_{3})]$ Why this step? Plug in everything you know; leave $S^{\circ} (NH_{3})$ as the one algebra symbol.
Substitute numbers. $198.7 = [191.6 + 3 (130.7)] - 2 S^{\circ} (NH_{3}) = 583.7 - 2 S^{\circ} (NH_{3})$ Why this step? Turning every known symbol into its number collapses the equation to a single unknown, so ordinary algebra can now isolate it.
Solve for the unknown. $2 S^{\circ} (NH_{3}) = 583.7 - 198.7 = 385.0 \Rightarrow S^{\circ} (NH_{3}) = 192.5 J K^{- 1} mol^{- 1}$ Why this step? Isolate the unknown by algebra — the exam twist is just rearranging the same formula.

Verify (b): $192.5 J K^{- 1} mol^{- 1}$ matches the tabulated value in the parent note. ✅ And it landed near our $\sim 190$ guess. ✅

Active Recall

Recall Which cell is hardest to guess, and why?

Cell C (and D): equal moles of gas on both sides. The fast "count gas" rule goes silent, so you must actually compute — molecular size/mass decides a small sign.

Recall Reverse-reaction rule for

$Δ S^{\circ}$ ? Reversing a reaction negates $Δ S^{\circ}$ (state function: start↔end swap flips the sign).

Recall How do you find a crossover temperature?

Set $Δ G = Δ H - T Δ S = 0$ and solve $T = Δ H /Δ S$ — matching units first (kJ with kJ).

Connections

Yeh note Hinglish mein padho →
Gibbs free energy ΔG = ΔH − TΔS — Cell F couples $Δ S$ into spontaneity
Second Law of Thermodynamics — resolves Ex 7's "how can frost form?"
Third Law of Thermodynamics — why $S^{\circ}$ values are absolute to begin with
Boltzmann distribution and microstates — why gas > liquid > solid in $S^{\circ}$
Standard enthalpy of reaction ΔH_rxn — the $Δ H$ we paired with $Δ S$ in Ex 6
Entropy of surroundings ΔS_surr = −ΔH/T — the missing half in Ex 7

#flashcards/chemistry