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

A reference table of standard molar entropies (all in $J K^{- 1} mol^{- 1}$ at 298 K) is used throughout. Values are given inside each problem so you never have to guess.

Level 1 — Recognition

Here you only need to spot the sign of $Δ S$ from the physical picture: count moles of gas, feel the mess.

Problem 1.1

Without any numbers, predict the sign of $Δ S^{\circ}$ for each:

(a) $H_{2} O (l) \to H_{2} O (g)$
(b) $2 NO_{2} (g) \to N_{2} O_{4} (g)$
(c) $Ag^{+} (a q) + Cl^{-} (a q) \to AgCl (s)$

Recall Solution 1.1

The rule: more spread-out matter (especially more moles of gas) → higher entropy → $Δ S > 0$ .

(a) Liquid → gas. A liquid has molecules touching; a gas has them flying free with vastly more arrangements. $Δ S > 0$ (positive).
(b) Gas moles go $2 \to 1$ . Fewer gas particles = fewer ways to arrange them. $Δ S < 0$ (negative).
(c) Two freely-roaming dissolved ions lock together into a solid crystal. Order goes UP, mess goes DOWN. $Δ S < 0$ (negative).

Problem 1.2

True or false: "The element $O_{2} (g)$ has $S^{\circ} = 0$ because elements have a zero reference, just like $Δ H_{f}^{\circ}$ ."

Recall Solution 1.2

False. $S^{\circ}$ is an absolute quantity built from the Third Law ( $S = 0$ only for a perfect crystal at 0 K). Every real substance at 298 K — element or compound — has a positive $S^{\circ}$ . In fact $S^{\circ} (O_{2}) = 205 J K^{- 1} mol^{- 1}$ , nowhere near zero. See Third Law of Thermodynamics.

Level 2 — Application

Now plug numbers into the working equation, remembering the coefficients.

Problem 2.1

Compute $Δ S^{\circ}$ for $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$ Data ( $J K^{- 1} mol^{- 1}$ ): $S^{\circ} (H_{2}) = 130.7$ , $S^{\circ} (O_{2}) = 205.0$ , $S^{\circ} (H_{2} O, l) = 69.9$ .

Recall Solution 2.1

Forecast first: gas moles go $2 + 1 = 3 \to 0$ (product is liquid). Big drop → expect a large negative $Δ S$ . Apply the equation, weighting each species by its coefficient (entropy is extensive — 2 mol carries twice the entropy of 1 mol): $Δ S^{\circ} = products [2 (69.9)] - reactants [2 (130.7) + 205.0]$ $= 139.8 - [261.4 + 205.0] = 139.8 - 466.4 = - 326.6 J K^{- 1}$ Large and negative, exactly as forecast — three moles of gas collapsed into liquid.

Problem 2.2

Compute $Δ S^{\circ}$ for the combustion $C (s) + O_{2} (g) \to CO_{2} (g)$ Data: $S^{\circ} (C, s) = 5.7$ , $S^{\circ} (O_{2}) = 205.0$ , $S^{\circ} (CO_{2}) = 213.8$ .

Recall Solution 2.2

Forecast: gas moles $1 \to 1$ (unchanged), but a solid is consumed and its entropy replaced by a gas's. Expect a small positive change. $Δ S^{\circ} = [213.8] - [5.7 + 205.0] = 213.8 - 210.7 = + 3.1 J K^{- 1}$ Tiny, because gas-in ≈ gas-out; the only nudge is the low-entropy solid carbon vanishing.

Level 3 — Analysis

Reverse the machine: reason from a known $Δ S$ back to an unknown $S^{\circ}$ , or interpret the size.

Problem 3.1

For $N_{2} O_{4} (g) \to 2 NO_{2} (g)$ the measured $Δ S^{\circ} = + 175.8 J K^{- 1}$ . Given $S^{\circ} (NO_{2}) = 240.1 J K^{- 1} mol^{- 1}$ , find $S^{\circ} (N_{2} O_{4})$ .

Recall Solution 3.1

Set up the equation with the unknown as a symbol: $Δ S^{\circ} = 2 S^{\circ} (NO_{2}) - S^{\circ} (N_{2} O_{4})$ $175.8 = 2 (240.1) - S^{\circ} (N_{2} O_{4}) = 480.2 - S^{\circ} (N_{2} O_{4})$ Rearrange: $S^{\circ} (N_{2} O_{4}) = 480.2 - 175.8 = 304.4 J K^{- 1} mol^{- 1}$ Sanity check: $N_{2} O_{4}$ is one big molecule with more atoms and vibrational modes than a single $NO_{2}$ , so its $S^{\circ}$ (304) exceeds that of one $NO_{2}$ (240) — physically reasonable.

Problem 3.2

Two reactions each produce 1 mol of gas from solids:

(i) $CaCO_{3} (s) \to CaO (s) + CO_{2} (g)$ , $Δ S^{\circ} = + 160.6 J K^{- 1}$
(ii) $MgCO_{3} (s) \to MgO (s) + CO_{2} (g)$ , $Δ S^{\circ} = + 175.0 J K^{- 1}$

Both liberate the same gas ( $CO_{2}$ ). Explain qualitatively why the two $Δ S^{\circ}$ differ, even though the "gas released" term is identical.

Recall Solution 3.2

The $+ 213.8$ from $CO_{2}$ is common to both. The difference lives entirely in the solids: $Δ S^{\circ} = [S^{\circ} (oxide) + 213.8] - S^{\circ} (carbonate)$ A change of solid $S^{\circ}$ values (which depend on molar mass, crystal packing, and lattice vibrations) shifts the result. Magnesium's smaller, differently-packed lattice gives a different balance between the carbonate's entropy and the oxide's entropy, so $Δ S_{(ii)}^{\circ} > Δ S_{(i)}^{\circ}$ . Lesson: the gas dominates the sign, but the solids fine-tune the magnitude.

Level 4 — Synthesis

Combine $Δ S$ with $Δ H$ to reach spontaneity via Gibbs free energy ΔG = ΔH − TΔS.

Problem 4.1

For $N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g)$ : $Δ H^{\circ} = - 92.4 kJ$ and $Δ S^{\circ} = - 198.7 J K^{- 1}$ .

(a) Compute $Δ G^{\circ}$ at 298 K.
(b) Above what temperature does the reaction become non-spontaneous?

Recall Solution 4.1

(a) Unit-match first. $Δ H$ is in kJ, $Δ S$ in J — convert $Δ S$ to kJ: $- 198.7 J K^{- 1} = - 0.1987 kJ K^{- 1}$ . $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = - 92.4 - (298) (- 0.1987)$ $= - 92.4 + 59.2 = - 33.2 kJ$ Negative → spontaneous at 298 K. ✅

(b) As $T$ rises, the $- T Δ S$ term (with $Δ S < 0$ , i.e. $+ ∣ Δ S ∣ T$ ) grows more positive and eventually overwhelms the negative $Δ H$ . The crossover is where $Δ G = 0$ : $T = \frac{Δ H ^{\circ}}{Δ S ^{\circ}} = \frac{- 92.4}{- 0.1987} = 465 K$ Above ~465 K, $Δ G > 0$ and ammonia synthesis becomes non-spontaneous — which is why the Haber process is run at moderate temperature and squeezed with high pressure rather than pushed hot.

Problem 4.2

A reaction has $Δ H^{\circ} = + 55.0 kJ$ and $Δ S^{\circ} = + 140.0 J K^{- 1}$ . Classify its spontaneity type, and find the temperature at which it becomes spontaneous.

Recall Solution 4.2

Both positive → this is the "entropy-driven, needs heat" class: unfavourable enthalpy, favourable entropy. It flips to spontaneous when $- T Δ S$ beats $+ Δ H$ , i.e. at high $T$ . Crossover ( $Δ G = 0$ ), with $Δ S = 0.140 kJ K^{- 1}$ : $T = \frac{Δ H ^{\circ}}{Δ S ^{\circ}} = \frac{55.0}{0.140} = 392.9 K$ For $T > 392.9 K$ , $Δ G < 0$ → spontaneous. (This is the general "endothermic + entropy-increasing" pattern, like melting ice or decomposing limestone.)

Level 5 — Mastery

Full picture: system + surroundings + universe, tying together Second Law of Thermodynamics and Entropy of surroundings ΔS_surr = −ΔH/T.

Problem 5.1

For freezing water at $- 10^{\circ} C$ (263 K), $H_{2} O (l) \to H_{2} O (s)$ , take $Δ H = - 6.01 kJ mol^{- 1}$ and $Δ S_{sy s} = - 22.0 J K^{- 1} mol^{- 1}$ . Show, using the universe, that freezing is spontaneous at 263 K even though the system's entropy drops.

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

Recall Solution 5.1

The Second Law demands $Δ S_{u ni v} = Δ S_{sy s} + Δ S_{s u r r} \geq 0$ for a spontaneous change — the universe, not the system alone.

Step 1 — entropy of surroundings. Freezing releases heat ( $Δ H < 0$ ); that heat flows into the surroundings, spreading energy there. Using $Δ S_{s u r r} = - Δ H / T$ with $T = 263 K$ and $Δ H = - 6010 J mol^{- 1}$ : $Δ S_{s u r r} = - \frac{- 6010}{263} = + 22.85 J K^{- 1} mol^{- 1}$ Step 2 — add them (look at the figure: the surroundings' orange gain outweighs the system's violet loss): $Δ S_{u ni v} = system - 22.0 + surroundings 22.85 = + 0.85 J K^{- 1} mol^{- 1} > 0$ Positive → spontaneous. The system got more ordered (ice), but it dumped enough heat to disorder the surroundings by even more. This is the rigorous answer to " $Δ S_{sy s} < 0$ means it can't happen" — false, because the universe still wins.

Problem 5.2

At what temperature does water freezing reach equilibrium ( $Δ S_{u ni v} = 0$ )? Use the same $Δ H = - 6.01 kJ mol^{- 1}$ and $Δ S_{sy s} = - 22.0 J K^{- 1} mol^{- 1}$ , and check the result against everyday experience.

Recall Solution 5.2

Equilibrium is the balance point where $Δ S_{u ni v} = Δ S_{sy s} + Δ S_{s u r r} = 0$ : $Δ S_{sy s} - \frac{Δ H}{T} = 0 \Rightarrow T = \frac{Δ H}{Δ S _{sy s}}$ $T = \frac{- 6010}{- 22.0} = 273.2 K = 0^{\circ} C$ Reality check: water and ice coexist at exactly $0^{\circ} C$ — the thermodynamics reproduces the freezing point we measure with a thermometer. Below this $T$ the surroundings term wins (freezing spontaneous, as in 5.1); above it, melting is spontaneous instead.

Wrap-up recall

Recall One-line takeaways per level

L1 :::- Sign of $Δ S$ comes from moles of gas + phase changes; $S^{\circ}$ of any real substance is positive.
L2 :::- Always weight each $S^{\circ}$ by its coefficient $ν$ before subtracting.
L3 :::- Solve backwards for an unknown $S^{\circ}$ ; it can never come out negative.
L4 :::- To reach $Δ G$ , convert $Δ S$ to kJ, then $Δ G = Δ H - T Δ S$ ; crossover at $T = Δ H /Δ S$ .
L5 :::- Spontaneity is judged by the universe: $Δ S_{u ni v} = Δ S_{sy s} - Δ H / T$ , with $T$ = the actual temperature.

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