Absolute entropy and the second law of thermodynamics

AP Chemistry· AP Chemistry CED — Applications of Thermodynamics· 14 min read

1. Absolute Entropy and the Third Law of Thermodynamics★★☆☆☆⏱ 4 min

The third law of thermodynamics establishes the reference point needed to calculate absolute entropy: it states that the entropy of a perfect crystalline substance at absolute zero (0 K) is exactly zero. Because all substances gain thermal motion as temperature rises above 0 K, all absolute entropies at 298 K (standard temperature) are positive values. This is a critical distinction from standard enthalpy of formation, where elements in their standard state have $Δ H_{f}^{\circ} = 0$ ; elements have positive, non-zero absolute entropy.

Gases have much higher $S^{\circ}$ than liquids, which have higher $S^{\circ}$ than solids, due to greater molecular freedom and more possible microstates in higher-energy phases.
For substances in the same phase, larger, more complex molecules have higher $S^{\circ}$ than smaller, simpler molecules, because they have more atoms leading to more vibrational and rotational degrees of freedom that increase disorder.
$S^{\circ}$ increases with increasing temperature, as higher temperature increases average molecular kinetic energy and disorder.

📐 Worked Example

Without doing a calculation, rank the following substances in order of increasing standard absolute entropy at 298 K: $C_{2} H_{6} (g)$ , $C_{3} H_{8} (g)$ , $C_{3} H_{8} (l)$ . Justify your ranking.

1
First, separate substances by phase: $C_{3} H_{8} (l)$ is a liquid, while the other two substances are gases. Liquids have less molecular disorder and fewer possible microstates than gases at the same temperature, so $C_{3} H_{8} (l)$ has the lowest $S^{\circ}$ .
2
Next, compare the two gaseous alkanes: both are in the gas phase, but $C_{2} H_{6}$ has a smaller molecular size (8 total atoms per molecule) than $C_{3} H_{8}$ (11 total atoms per molecule).
3
Larger, more complex molecules have more rotational and vibrational degrees of freedom, leading to more possible microstates and higher absolute entropy than smaller molecules in the same phase.
4
Final order (increasing $S^{\circ}$ ):
5
$C_3H_8(l) < C_2H_6(g) < C_3H_8(g)$

Exam tip:

When ranking absolute entropy, always sort by phase first. Phase differences produce much larger changes in entropy than differences between molecules of the same phase, so a liquid will always have lower entropy than any gas at the same temperature, even if the liquid molecule is larger.

2. Calculating Standard Reaction Entropy Change ($\Delta S^\circ_{\text{rxn}}$)★★★☆☆⏱ 4 min

Once we have tabulated absolute entropy values for all reactants and products, we can calculate the total entropy change of the system for a reaction at standard conditions. The formula for standard reaction entropy change is derived directly from the definition of absolute entropy: the total entropy of the products minus the total entropy of the reactants, adjusted for stoichiometry.

Delta S^\circ_{\text{rxn}} = \sum n S^\circ(\text{products}) - \sum m S^\circ(\text{reactants})

where $n$ and $m$ are the stoichiometric coefficients of products and reactants from the balanced chemical equation, respectively. A common point of confusion is the treatment of elements: unlike enthalpy, where elements contribute nothing to $Δ H_{rxn}^{\circ}$ because their $Δ H_{f}^{\circ} = 0$ , elements contribute their full positive $S^{\circ}$ to the calculation, because all substances above 0 K have non-zero absolute entropy. The sign of $Δ S_{rxn}^{\circ}$ tells us whether the system becomes more disordered (positive $Δ S^{\circ}$ ) or more ordered (negative $Δ S^{\circ}$ ) when the reaction proceeds.

📐 Worked Example

Calculate $Δ S_{rxn}^{\circ}$ for the combustion of 1 mole of methane: $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ . Use the following tabulated $S^{\circ}$ values: $S^{\circ} (C H_{4} (g)) = 186.3$ J/(mol·K), $S^{\circ} (O_{2} (g)) = 205.2$ J/(mol·K), $S^{\circ} (C O_{2} (g)) = 213.8$ J/(mol·K), $S^{\circ} (H_{2} O (l)) = 69.9$ J/(mol·K).

1
Write the formula matching the balanced reaction stoichiometry:
2
$Delta S^\circ_{\text{rxn}} = \left[1 \times S^\circ(CO_2(g)) + 2 \times S^\circ(H_2O(l))\right] - \left[1 \times S^\circ(CH_4(g)) + 2 \times S^\circ(O_2(g))\right]$
3
Substitute the given values into the formula:
4
$Delta S^\circ_{\text{rxn}} = \left[(1 \times 213.8) + (2 \times 69.9)\right] - \left[(1 \times 186.3) + (2 \times 205.2)\right]$
5
Calculate the sum of product and reactant entropies: Sum of products = $213.8 + 139.8 = 353.6$ J/K; Sum of reactants = $186.3 + 410.4 = 596.7$ J/K
6
Subtract to get the final result:
7
$Delta S^\circ_{\text{rxn}} = 353.6 - 596.7 = -243.1 \text{ J/(mol·K)}$

Exam tip:

Always check units after calculation. Absolute entropy has units of J/(mol·K), so $Δ S_{rxn}^{\circ}$ will have units of J/K for the reaction as written, or J/(mol·K) when reported per mole of limiting reactant. If you end up with units of kJ, that is a red flag that you confused entropy units with enthalpy units.

3. The Second Law of Thermodynamics and Spontaneity★★★☆☆⏱ 4 min

The second law of thermodynamics is the core physical law that governs whether any process occurs spontaneously (without continuous external energy input). It links entropy changes of the system (the process being studied) and its surroundings (everything outside the system) to process spontaneity.

The second law states that for any spontaneous process, the total entropy change of the universe is positive. This gives the relationship:

Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}

For any process occurring at constant pressure and temperature, the entropy change of the surroundings is related to the enthalpy change of the system by the formula:

Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T}

This relationship comes from heat transfer: any heat released by the system is absorbed by the surroundings, increasing the surroundings' entropy, and any heat absorbed by the system is removed from the surroundings, decreasing the surroundings' entropy. A process is spontaneous at constant T and P if $Δ S_{univ} > 0$ , non-spontaneous if $Δ S_{univ} < 0$ , and at equilibrium if $Δ S_{univ} = 0$ .

📐 Worked Example

For a certain reaction at 298 K, $Δ S_{sys} = - 150$ J/K and $Δ H_{sys} = - 40$ kJ. Is the reaction spontaneous at this temperature?

1
Convert all values to consistent units: $Δ H_{sys} = - 40$ kJ = $- 40000$ J, T = 298 K.
2
Calculate $Δ S_{surr}$ using the second law relationship:
3
$Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T} = -\frac{(-40000 \text{ J})}{298 \text{ K}} = +134.2 \text{ J/K}$
4
Calculate the total entropy change of the universe:
5
$Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = (-150 \text{ J/K}) + 134.2 \text{ J/K} = -15.8 \text{ J/K}$
6
Apply the second law criterion: since $Δ S_{univ} < 0$ , the reaction is not spontaneous at 298 K.

✓ Quick check

Test your understanding with these AP-style questions:

Which of the following correctly ranks the substances in order of increasing standard absolute entropy at 298 K?
- A) $H_{2} O (s) < H_{2} O (g) < H_{2} O (l)$
- B) $N a F (s) < H C l (g) < C_{6} H_{12} O_{6} (s)$
- C) $C H_{3} O H (l) < C_{2} H_{5} O H (l) < C_{3} H_{7} O H (l)$
- D) $O_{2} (g) < N_{2} (g) < A r (g)$
Reveal answer
C —
All three alcohols are in the liquid phase, and molecular complexity increases from methanol to ethanol to propanol, so absolute entropy increases in this order. Other options are incorrect: A misorders liquid and gas, B misorders solid and gas, D misorders gas entropy by molecular mass.
Iron(III) oxide is reduced by carbon monoxide: $F e_{2} O_{3} (s) + 3 C O (g) \to 2 F e (s) + 3 C O_{2} (g)$ . Given the tabulated $S^{\circ}$ values and $Δ H_{r x n}^{\circ} = - 23.5$ kJ at 298 K, is the reaction spontaneous?
- A) Not spontaneous, $Δ S_{sy s}^{\circ}$ is positive
- B) Spontaneous, $Δ S_{u ni v}^{\circ}$ is positive
- C) Not spontaneous, $Δ H_{r x n}^{\circ}$ is negative
- D) Spontaneous, $Δ S_{sy s}^{\circ}$ is negative
Reveal answer
B —
Calculating $Δ S_{sy s}^{\circ} = 15.5$ J/K, $Δ S_{s u r r}^{\circ} = + 78.9$ J/K, so $Δ S_{u ni v}^{\circ} = + 94.4$ J/K $> 0$ , meaning the reaction is spontaneous per the second law.

Exam tip:

Always convert $Δ H$ to joules when calculating $Δ S_{surr}$ , because $Δ S$ is almost always reported in J/K. Failing to convert kJ to J gives a $Δ S_{surr}$ that is 1000 times too small, leading to the wrong conclusion about spontaneity.

4. Common Pitfalls

Wrong move:

Omitting the absolute entropy of elemental reactants/products when calculating $Δ S_{rxn}^{\circ}$ , because elements have $Δ H_{f}^{\circ} = 0$ .

Why:

Students confuse the standard enthalpy of formation convention with the definition of absolute entropy, where all substances above 0 K have non-zero positive entropy.

Correct move:

Always include every reactant and product (including elements) multiplied by their stoichiometric coefficient when calculating $Δ S_{rxn}^{\circ}$ .

Wrong move:

Ranking a larger molecule in a lower-entropy phase above a smaller molecule in a higher-entropy phase (e.g. ranking $C_{10} H_{22} (s)$ higher than $C H_{4} (g)$ ).

Why:

Students prioritize molecular complexity over phase when ranking, but phase has a much larger effect on entropy.

Correct move:

Always sort by phase first (solids < liquids < gases) when ranking absolute entropy, then compare molecular size/complexity within the same phase.

Wrong move:

Claiming that a negative $Δ S_{sys}$ means the process cannot be spontaneous.

Why:

Students confuse the entropy change of the system with the total entropy change of the universe. The second law only requires $Δ S_{univ}$ to be positive.

Correct move:

Always calculate $Δ S_{surr}$ from $Δ H$ and add it to $Δ S_{sys}$ to get $Δ S_{univ}$ before concluding spontaneity. A negative $Δ S_{sys}$ can still give a positive $Δ S_{univ}$ if $Δ H$ is sufficiently negative.

Wrong move:

Forgetting to convert $Δ H$ from kJ to J when calculating $Δ S_{surr}$ , leading to a $Δ S_{univ}$ with the wrong sign.

Why:

$Δ H$ is commonly reported in kJ/mol, while $Δ S$ is reported in J/(mol·K), so unit mismatch is extremely common.

Correct move:

Before plugging into $Δ S_{surr} = - Δ H / T$ , always check units and convert $Δ H$ to joules to match $Δ S$ units.

Wrong move:

Claiming that absolute entropy can be negative for a stable substance at 298 K.

Why:

Students confuse absolute entropy (a total value) with entropy change (which can be positive or negative).

Correct move:

Remember the third law: entropy is zero at 0 K for a perfect crystal, and all substances gain entropy as temperature increases, so all absolute entropies at 298 K are positive.

Wrong move:

Writing the formula for $Δ S_{surr}$ as $Δ H / T$ without the negative sign.

Why:

Students forget the sign convention for heat transfer between the system and surroundings.

Correct move:

Memorize that if the system releases heat ( $Δ H$ negative), surroundings gain entropy ( $Δ S_{surr}$ positive), which requires the negative sign: $Δ S_{surr} = - Δ H / T$ .

5. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Note
Absolute entropy ( $S^{\circ}$ )	S = 0 for perfect crystal at 0 K	All $S^{\circ}$ at 298 K are positive, even for elements
Ranking $S^{\circ}$	Sort by phase first: solid < liquid < gas, then molecular size	Phase differences are larger than molecular size differences
$Δ S_{r x n}^{\circ}$	$\sum n S^{\circ} (p r o d u c t s) - \sum m S^{\circ} (r e a c t an t s)$	Include all species, even elements in standard state
Second Law Criterion	Spontaneous if $Δ S_{u ni v} = Δ S_{sy s} + Δ S_{s u r r} > 0$	Only $Δ S_{u ni v}$ needs to be positive, not $Δ S_{sy s}$
$Δ S_{s u r r}$ (constant T,P)	$Δ S_{s u r r} = - Δ H_{sy s} / T$	Convert $Δ H$ to joules to match $Δ S$ units

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Rank absolute entropies by phase
2022 · FRQ
Calculate ΔS° and spontaneity

What's Next

This sub-topic is the foundation for predicting reaction favorability, the core of AP Chemistry Unit 9. The second law and entropy change skills you learned here directly lead to the definition of Gibbs free energy, which simplifies spontaneity predictions to a single system property, eliminating the need to calculate separate entropy changes for the system and surroundings. You will use the skills of calculating $Δ S_{r x n}^{\circ}$ in every subsequent thermodynamics topic on the AP exam, from Gibbs free energy to entropy of dissolution to temperature dependence of spontaneity.