Absolute entropy and the second law of thermodynamics — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: This chapter covers absolute entropy (S°) from the third law of thermodynamics, calculation of entropy changes for systems and surroundings, the second law of thermodynamics, and application of second law criteria to identify spontaneous physical and chemical changes.

You should already know: Entropy as a measure of molecular disorder; calculation of reaction enthalpy from standard formation values; definition of a spontaneous process.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Absolute entropy and the second law of thermodynamics?

Absolute entropy (symbol: $S^{\circ }$, units: J/(mol·K)) is the total entropy of a pure substance at a given temperature and pressure, measured relative to the reference point of a perfect crystal at absolute zero. Unlike standard reaction entropy change ($\Delta S_{\text{rxn}}^{\circ }$), which is only the difference between product and reactant entropies, absolute entropy gives a direct, non-arbitrary value for the total entropy of a substance. This topic makes up approximately 20% of the content of AP Chemistry Unit 9 (Applications of Thermodynamics), and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam, often as a foundation for questions on Gibbs free energy and reaction spontaneity. 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. This topic is the gateway to predicting reaction favorability, a key high-weight skill for the AP exam.

2. Absolute Entropy and the Third Law of Thermodynamics

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 have increasing thermal motion as temperature rises above 0 K, all absolute entropies at standard temperature (298 K) are positive values. This is a critical distinction from standard enthalpy of formation, where elements in their standard state have $\Delta H_f^{\circ }=0$; elements have positive, non-zero absolute entropy. Standard absolute entropy values ($S^{\circ }$) are tabulated for most pure substances at 1 atm and 298 K, and follow predictable trends that are frequently tested on the AP exam:

1. 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.
2. 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.
3. $S^{\circ }$ 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 $S^{\circ }$): $\boxed{C_3{H}_8(l)<C_2{H}_6(g)<C_3{H}_8(g)}$

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.

3. Calculating Standard Reaction Entropy Change ($\Delta S_{\text{rxn}}^{\circ }$)

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: $$\Delta S_{\text{rxn}}^{\circ }=\sum n{S}^{\circ }(\text{products})-\sum m{S}^{\circ }(\text{reactants})$$ 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 $\Delta H_{\text{rxn}}^{\circ }$ because their $\Delta H_f^{\circ }=0$, elements contribute their full positive $S^{\circ }$ to the calculation, because all substances above 0 K have non-zero absolute entropy. The sign of $\Delta S_{\text{rxn}}^{\circ }$ tells us whether the system becomes more disordered (positive $\Delta S^{\circ }$) or more ordered (negative $\Delta S^{\circ }$) when the reaction proceeds from reactants to products.

Worked Example

Calculate $\Delta S_{\text{rxn}}^{\circ }$ for the combustion of 1 mole of methane: $C{H}_4(g)+2O_2(g)\to C{O}_2(g)+2H_{2}O(l)$. 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: $\Delta S_{\text{rxn}}^{\circ }=\left[1\times S^{\circ }(C{O}_2(g))+2\times S^{\circ }(H_{2}O(l))\right]-\left[1\times S^{\circ }(C{H}_4(g))+2\times S^{\circ }(O_2(g))\right]$
2. Substitute the given values into the formula: $\Delta S_{\text{rxn}}^{\circ }=\left[(1\times 213.8)+(2\times 69.9)\right]-\left[(1\times 186.3)+(2\times 205.2)\right]$
3. 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
4. Subtract to get the final result: $\Delta S_{\text{rxn}}^{\circ }=353.6-596.7=\boxed{-243.1}$ J/(mol·K)

Exam tip: Always check units after calculation. Absolute entropy has units of J/(mol·K), so $\Delta S_{\text{rxn}}^{\circ }$ 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.

4. The Second Law of Thermodynamics and Spontaneity

The second law of thermodynamics states that for any spontaneous process, the total entropy change of the universe is positive. The universe is divided into two parts: the system (the reaction or process we are studying) and the surroundings (everything outside the system). 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 the fact that 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. Combining these relationships gives the second law criterion for spontaneity: a process is spontaneous at constant T and P if $\Delta S_{\text{univ}}>0$, non-spontaneous if $\Delta S_{\text{univ}}<0$, and at equilibrium if $\Delta S_{\text{univ}}=0$.

Worked Example

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

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

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

5. Common Pitfalls (and how to avoid them)

• Wrong move: Omitting the absolute entropy of elemental reactants/products when calculating $\Delta S_{\text{rxn}}^{\circ }$, on the basis that elements have $\Delta H_f^{\circ }=0$. Why: Students confuse the convention for standard enthalpy of formation with the definition of absolute entropy, where all substances have positive non-zero entropy above 0 K. Correct move: Always include every reactant and product (including elements) multiplied by their stoichiometric coefficient when calculating $\Delta S_{\text{rxn}}^{\circ }$.
• 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 $C{H}_4(g)$ in entropy). Why: Students focus on molecular complexity first, not phase, which leads to wrong ranking. 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 $\Delta S_{\text{sys}}$ means the process cannot be spontaneous. Why: Students confuse the entropy change of the system with the entropy change of the universe. The second law only requires $\Delta S_{\text{univ}}$ to be positive. Correct move: Always calculate $\Delta S_{\text{surr}}$ from $\Delta H$ and add it to $\Delta S_{\text{sys}}$ to get $\Delta S_{\text{univ}}$ before concluding if a process is spontaneous. A negative $\Delta S_{\text{sys}}$ can still give a positive $\Delta S_{\text{univ}}$ if $\Delta H$ is sufficiently negative.
• Wrong move: Forgetting to convert $\Delta H$ from kJ to J when calculating $\Delta S_{\text{surr}}$, leading to a $\Delta S_{\text{univ}}$ with the wrong sign. Why: $\Delta H$ is commonly reported in kJ/mol, while $\Delta S$ is reported in J/(mol·K), so unit mismatch is common. Correct move: Before plugging into $\Delta S_{\text{surr}}=-\Delta H/T$, always check units: convert $\Delta H$ to joules if $\Delta S$ is in joules.
• Wrong move: Claiming that absolute entropy can be negative for a stable substance at 298 K. Why: Students confuse absolute entropy 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 $\Delta S_{\text{surr}}$ as $\Delta H/T$ without the negative sign. Why: Students forget the sign convention for heat transfer between system and surroundings. Correct move: Memorize the relationship: if the system releases heat ($\Delta H$ negative), surroundings gain entropy, so $\Delta S_{\text{surr}}$ is positive, which requires the negative sign: $\Delta S_{\text{surr}}=-\Delta H/T$.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

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) $NaF(s)<HCl(g)<C_6{H}_{12}{O}_6(s)$ C) $C{H}_{3}OH(l)<C_2{H}_{5}OH(l)<C_3{H}_{7}OH(l)$ D) $O_2(g)<N_2(g)<Ar(g)$

Worked Solution: To solve this problem, we apply the rules for ranking absolute entropy: sort by phase first (solids < liquids < gases), then sort by molecular complexity within the same phase. Checking each option: Option A is incorrect because liquid water has higher entropy than solid water, so the order is wrong. Option B is incorrect because HCl is a gas, which has higher entropy than solid glucose, so the order is wrong. Option C: all three are alcohols in the liquid phase, and molecular complexity increases from methanol to ethanol to propanol, so absolute entropy increases in that order, which matches the given ranking. Option D is incorrect because all are gases, and molecular mass increases from N₂ < O₂ < Ar, so entropy should increase in that order, which does not match D. The correct answer is C.

Question 2 (Free Response)

Iron(III) oxide can be reduced to solid iron by reaction with carbon monoxide gas, producing carbon dioxide gas. The balanced reaction is: $F{e}_2{O}_3(s)+3CO(g)\to 2Fe(s)+3C{O}_2(g)$ Use the following standard absolute entropy values to answer the questions below:

(a) Calculate the standard entropy change of the system ($\Delta S_{\text{sys}}^{\circ }$) for this reaction. Show your working. (b) The standard enthalpy change for this reaction is $\Delta H_{\text{rxn}}^{\circ }=-23.5$ kJ at 298 K. Calculate the entropy change of the surroundings ($\Delta S_{\text{surr}}^{\circ }$) at 298 K. (c) Is this reaction spontaneous at 298 K and 1 atm? Justify your answer using second law principles.

Worked Solution: (a) Use the standard reaction entropy formula: $$\Delta S_{\text{sys}}^{\circ }=\sum n{S}^{\circ }(\text{products})-\sum m{S}^{\circ }(\text{reactants})$$ Substitute values: $\Delta S_{\text{sys}}^{\circ }=\left[(2\times 27.3)+(3\times 213.8)\right]-\left[87.4+(3\times 197.7)\right]=(54.6+641.4)-(87.4+593.1)=696-680.5=\boxed{15.5}$ J/K

(b) Convert $\Delta H^{\circ }$ to joules: $-23.5$ kJ = $-23500$ J. Use the formula for $\Delta S_{\text{surr}}^{\circ }$: $$\Delta S_{\text{surr}}^{\circ }=-\frac{\Delta H_{\text{rxn}}^{\circ }}{T}=-\frac{(-23500\text{J})}{298\text{K}}=\boxed{+78.9}\text{J/K}$$

(c) Calculate $\Delta S_{\text{univ}}^{\circ }=\Delta S_{\text{sys}}^{\circ }+\Delta S_{\text{surr}}^{\circ }=15.5+78.9=94.4$ J/K. The second law states that a process is spontaneous at constant T and P if $\Delta S_{\text{univ}}>0$. Since 94.4 J/K is positive, this reaction is spontaneous at 298 K and 1 atm.

Question 3 (Application / Real-World Style)

At normal human body temperature (310 K), the folding of a denatured protein into its native functional structure is a spontaneous process. The denatured protein has a more disordered structure than the folded protein, so the entropy change of the system (the protein) is $\Delta S_{\text{sys}}=-300$ J/K. The folding process releases 100 kJ of heat to the surroundings at constant pressure. Use the second law of thermodynamics to confirm that folding is spontaneous at 310 K, and explain why a spontaneous process can have a negative entropy change for the system.

Worked Solution: First, since the process releases 100 kJ of heat at constant pressure, $\Delta H_{\text{sys}}=-100$ kJ = $-100000$ J. Calculate $\Delta S_{\text{surr}}$: $$\Delta S_{\text{surr}}=-\frac{\Delta H_{\text{sys}}}{T}=-\frac{(-100000\text{J})}{310\text{K}}=+322.6\text{J/K}$$ Next calculate $\Delta S_{\text{univ}}$: $\Delta S_{\text{univ}}=-300\text{J/K}+322.6\text{J/K}=+22.6$ J/K. Since $\Delta S_{\text{univ}}>0$, folding is spontaneous at 310 K, matching experimental observation. The entropy of the universe still increases overall: the negative entropy change of the protein (system) is more than offset by a larger positive entropy change of the surroundings caused by the heat released during folding.

7. Quick Reference Cheatsheet

8. What's Next

This chapter is the foundational prerequisite for Gibbs free energy, the next core topic in Unit 9 Applications of Thermodynamics. You will use the relationships between $\Delta S_{\text{sys}}$, $\Delta H_{\text{sys}}$, and absolute temperature derived here to develop the Gibbs free energy equation, which simplifies the spontaneity criterion to a calculation based only on system properties, eliminating the need to calculate $\Delta S_{\text{surr}}$ separately. Without mastering the calculation of absolute entropy and $\Delta S_{\text{rxn}}^{\circ }$, and understanding the second law criteria for spontaneity, you will not be able to correctly interpret Gibbs free energy values or predict reaction spontaneity, a skill that makes up a large portion of the AP Chemistry exam. Follow-up topics that build on this chapter are: Gibbs free energy and spontaneity Free energy and thermodynamic equilibrium Entropy and temperature dependence of spontaneity