Entropy and Gibbs free energy — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Second law of thermodynamics, entropy change prediction, standard entropy calculation, Gibbs free energy formula, spontaneity criteria, temperature dependence of spontaneity, and the relationship between ΔG° and equilibrium constant K.

You should already know: Enthalpy change (ΔH) calculation from standard enthalpies of formation, absolute temperature measurement in Kelvin, equilibrium constant definition and interpretation.

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 Entropy and Gibbs free energy?

Entropy (symbol: $S$) is a thermodynamic state function that quantifies the degree of disorder, or the number of possible microstates available to a system. Gibbs free energy (symbol: $G$) is another state function that combines enthalpy (heat content at constant pressure) and entropy to predict if a process is spontaneous, meaning it can proceed without continuous external energy input. According to the AP Chemistry CED, Unit 9 (Applications of Thermodynamics) makes up 14-18% of the total exam score, and this subtopic accounts for roughly half of that unit, or 7-9% of your total exam score. Questions on this topic appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often integrating with equilibrium concepts, thermochemical calculations, and real-world industrial scenarios. The core question this topic answers is: how do we predict whether a given reaction will occur spontaneously, and what does that mean for its equilibrium behavior?

2. Entropy, the Second Law, and Predicting Entropy Changes

Entropy is defined by Boltzmann’s formula $S=k\ln W$, where $k$ is the Boltzmann constant and $W$ is the number of distinct microstates a system can occupy. More microstates = higher entropy = greater disorder. The second law of thermodynamics states that for any spontaneous process, the total entropy change of the universe $\Delta S_{\text{univ}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}>0$. The third law of thermodynamics tells us that a perfect crystalline substance at 0 K has zero entropy, so all pure substances above 0 K have a positive standard molar entropy $S^{\circ }$, the entropy of 1 mole at 1 atm and 298 K. To predict the sign of $\Delta S_{\text{sys}}$ for a process, use the following general rules: gases have far higher entropy than liquids, which have higher entropy than solids; an increase in moles of gas increases entropy; dissolving a crystalline solid increases entropy; increasing temperature increases entropy.

Worked Example

Predict the sign of $\Delta S_{\text{sys}}$ for each process and justify your answer: (a) Freezing of liquid water to form ice (b) Reaction: $2{\text{SO}}_2(g)+{\text{O}}_2(g)\to 2{\text{SO}}_3(g)$ (c) Dissolving solid potassium nitrate in pure water

Steps:

1. For (a): Freezing converts liquid water to solid ice. Solids have highly ordered molecular arrangements with fewer microstates than liquids, so disorder decreases. $\Delta S_{\text{sys}}<0$ (negative).
2. For (b): Count moles of gas on each side: 3 moles of gas reactants → 2 moles of gas products. A decrease in total moles of gas reduces the number of microstates, so disorder decreases. $\Delta S_{\text{sys}}<0$ (negative).
3. For (c): Crystalline ${\text{KNO}}_3$ has ordered, fixed molecules; when dissolved, molecules disperse throughout the solvent, increasing the number of available microstates. Disorder increases. $\Delta S_{\text{sys}}>0$ (positive).

Exam tip: Always check the change in moles of gas first when predicting $\Delta S$ sign. Changes in gas moles almost always dominate over changes in solids or liquids, so this will give you the correct sign 90% of the time on exam questions.

3. Calculating Standard Entropy of Reaction

The standard entropy change for a reaction $\Delta S_{\text{rxn}}^{\circ }$ is calculated from tabulated standard molar entropies of reactants and products. A key difference from standard enthalpy of formation: unlike $\Delta H_f^{\circ }$, which is zero for elements in their standard state, $S^{\circ }$ is always positive for all substances (including elements) above 0 K. Never skip including $S^{\circ }$ for elements in your calculation. The formula for $\Delta S_{\text{rxn}}^{\circ }$ is:

$$\Delta S_{\text{rxn}}^{\circ }=\sum n{S}_{\text{products}}^{\circ }-\sum m{S}_{\text{reactants}}^{\circ }$$

where $n$ and $m$ are the stoichiometric coefficients of products and reactants, respectively. $S^{\circ }$ is almost always reported in units of $\text{J/(molcdotpK)}$, so keep this in mind for later calculations involving $\Delta G$, which uses units of kJ for enthalpy.

Worked Example

Calculate $\Delta S_{\text{rxn}}^{\circ }$ for the reaction $2\text{NO}(g)+{\text{O}}_2(g)\to 2{\text{NO}}_2(g)$, given the following standard molar entropies: $S^{\circ }(\text{NO}(g))=210.8\text{ J/(molcdotpK)}$, $S^{\circ }({\text{O}}_2(g))=205.2\text{ J/(molcdotpK)}$, $S^{\circ }({\text{NO}}_2(g))=240.1\text{ J/(molcdotpK)}$.

Steps:

1. Write the formula with stoichiometric coefficients substituted: $\Delta S_{\text{rxn}}^{\circ }=\left[2\times S^{\circ }({\text{NO}}_2(g))\right]-\left[2\times S^{\circ }(\text{NO}(g))+1\times S^{\circ }({\text{O}}_2(g))\right]$
2. Plug in the given values: $\Delta S_{\text{rxn}}^{\circ }=(2\times 240.1)-(2\times 210.8+205.2)=480.2-(421.6+205.2)=480.2-626.8=-146.6\text{ J/(molcdotpK)}$
3. Check sign against prediction: 3 moles of gas → 2 moles of gas, so $\Delta S$ should be negative. The result matches, so we can round to 3 significant figures: $\Delta S_{\text{rxn}}^{\circ }=-147\text{ J/(molcdotpK)}$.

Exam tip: Always write down units for every value when doing entropy calculations. This will help you catch unit mismatch errors when you later calculate $\Delta G$.

4. Gibbs Free Energy and Spontaneity

At constant temperature and pressure (the conditions for most chemical reactions), Gibbs free energy change combines enthalpy and entropy into a single value that directly predicts spontaneity. The core formula is:

$$\Delta G=\Delta H-T\Delta S$$

where $T$ is absolute temperature in Kelvin (always positive). The spontaneity rules are:

• $\Delta G<0$: Spontaneous in the forward direction
• $\Delta G=0$: Reaction is at equilibrium
• $\Delta G>0$: Non-spontaneous in the forward direction (spontaneous in reverse)

For standard state conditions, the formula becomes $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$. We can use this to find the threshold temperature where a reaction switches spontaneity by setting $\Delta G^{\circ }=0$, giving $T=\frac{\Delta H^{\circ }}{\Delta S^{\circ }}$. This is only useful when $\Delta H$ and $\Delta S$ have the same sign; if they have opposite signs, the reaction is always or never spontaneous regardless of temperature.

Worked Example

For the decomposition of magnesium carbonate: ${\text{MgCO}}_3(s)\to \text{MgO}(s)+{\text{CO}}_2(g)$, $\Delta H^{\circ }=+117.3\text{ kJ/mol}$ and $\Delta S^{\circ }=+174.8\text{ J/(molcdotpK)}$. What temperature range is the reaction spontaneous?

Steps:

1. Convert $\Delta S^{\circ }$ to kJ to match $\Delta H^{\circ }$ units: $174.8\text{ J/(molcdotpK)}=0.1748\text{ kJ/(molcdotpK)}$.
2. The reaction has $\Delta H^{\circ }$ positive and $\Delta S^{\circ }$ positive, so it is spontaneous at high temperatures. Find the threshold temperature by setting $\Delta G^{\circ }=0$: $T=\frac{\Delta H^{\circ }}{\Delta S^{\circ }}=\frac{117.3}{0.1748}\approx 671\text{ K}$.
3. Confirm: Above 671 K (≈ 398°C), $\Delta G^{\circ }<0$, so the decomposition is spontaneous. Below 671 K, $\Delta G^{\circ }>0$ and the reaction is non-spontaneous.

Exam tip: Memorize the four sign combinations: (1) ΔH -, ΔS + = always spontaneous; (2) ΔH +, ΔS - = never spontaneous; (3) ΔH -, ΔS - = spontaneous at low T; (4) ΔH +, ΔS + = spontaneous at high T. This is a very common MCQ question, so memorizing it saves time.

5. Relationship Between ΔG° and Equilibrium Constant K

Gibbs free energy connects thermodynamics to chemical equilibrium: the standard Gibbs free energy change is directly related to the equilibrium constant $K$ of a reaction by:

$$\Delta G^{\circ }=-RT\ln K$$

where $R=8.314\text{ J/(molcdotpK)}$ (or $0.008314\text{ kJ/(molcdotpK)}$) and $T$ is absolute temperature. The key interpretations are:

• If $\Delta G^{\circ }<0$, $\ln K>0$, so $K>1$: Products are favored at equilibrium
• If $\Delta G^{\circ }=0$, $\ln K=0$, so $K=1$: Products and reactants are equally favored at equilibrium
• If $\Delta G^{\circ }>0$, $\ln K<0$, so $K<1$: Reactants are favored at equilibrium

Worked Example

At 298 K, $\Delta G^{\circ }$ for the dissociation of acetic acid in water is +27.1 kJ/mol. Calculate $K_a$ for acetic acid at 298 K.

Steps:

1. Rearrange the formula to solve for $\ln K_a$: $\ln K_a=-\frac{\Delta G^{\circ }}{RT}$.
2. Convert $\Delta G^{\circ }$ to J to match $R$ units: $+27.1\text{ kJ/mol}=+27100\text{ J/mol}$.
3. Plug in values: $\ln K_a=-\frac{27100}{(8.314)(298)}\approx -\frac{27100}{2478}\approx -10.94$.
4. Exponentiate to get $K_a$: $K_a=e^{-10.94}\approx 1.8\times 10^{-5}$. This matches the known $K_a$ of acetic acid, confirming the calculation.

Exam tip: Always confirm that $R$ units match $\Delta G^{\circ }$ units: if $\Delta G^{\circ }$ is in kJ, use $R=0.008314\text{ kJ/(molcdotpK)}$ to avoid 1000x magnitude errors in $K$.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to convert $\Delta S$ from $\text{J/(molcdotpK)}$ to $\text{kJ/(molcdotpK)}$ before plugging into $\Delta G=\Delta H-T\Delta S$. Why: $\Delta S$ is always reported in J per mole-K, while $\Delta H$ is almost always reported in kJ per mole, leading to a 1000x error in $\Delta G$. Correct move: Write all units down for every value, and convert units so $\Delta H$ and $\Delta S$ match before doing any calculation.
• Wrong move: Treating $S^{\circ }$ of elements in standard state as zero, following the $\Delta H_f^{\circ }$ rule. Why: Students confuse the convention for enthalpy of formation with entropy, which is non-zero for all substances above 0 K. Correct move: Always include the full $S^{\circ }$ value of elements in your $\Delta S_{\text{rxn}}^{\circ }$ calculation, never assume it is zero.
• Wrong move: Predicting $\Delta S$ sign based on total moles of all species instead of moles of gas. Why: Students count all moles, but solids and liquids have negligible entropy compared to gases, so total moles can give the wrong sign. Correct move: Always calculate the change in moles of gas first to get the sign of $\Delta S$.
• Wrong move: Concluding that a reaction with $\Delta G<0$ will happen quickly. Why: Students confuse thermodynamic spontaneity with reaction kinetics. Correct move: Remember that $\Delta G$ only tells you if a reaction is thermodynamically favorable, not how fast it will proceed.
• Wrong move: Assuming a reaction with $\Delta G^{\circ }>0$ can never produce products. Why: Students confuse standard state $\Delta G^{\circ }$ with non-standard $\Delta G$, which depends on reactant/product concentrations. Correct move: $\Delta G^{\circ }>0$ only means $K<1$, so reactants are favored at equilibrium, but products can still form if you start with pure reactants.

7. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

For which of the following reactions is $\Delta S^{\circ }$ negative? A) ${\text{BaCl}}_2(s)\to {\text{Ba}}^{2+}(aq)+2{\text{Cl}}^{-}(aq)$ B) $2{\text{C}}_2{\text{H}}_6(g)+7{\text{O}}_2(g)\to 4{\text{CO}}_2(g)+6{\text{H}}_2\text{O}(g)$ C) ${\text{CaCO}}_3(s)\to \text{CaO}(s)+{\text{CO}}_2(g)$ D) $2{\text{SO}}_2(g)+{\text{O}}_2(g)\to 2{\text{SO}}_3(g)$

Worked Solution: To find $\Delta S^{\circ }$ sign, we check the change in moles of gas for each option. Option A: 1 mole of solid becomes 3 moles of aqueous ions, disorder increases, $\Delta S^{\circ }$ positive. Option B: 9 moles of gas → 10 moles of gas, increase, $\Delta S^{\circ }$ positive. Option C: 0 moles of gas → 1 mole of gas, increase, $\Delta S^{\circ }$ positive. Option D: 3 moles of gas → 2 moles of gas, decrease, $\Delta S^{\circ }$ negative. Correct answer: D.

Question 2 (Free Response)

Given the reaction ${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)$ with the following data:

(a) Calculate $\Delta H_{\text{rxn}}^{\circ }$ for the reaction. (b) Calculate $\Delta S_{\text{rxn}}^{\circ }$ for the reaction. (c) Calculate $\Delta G_{\text{rxn}}^{\circ }$ at 298 K and state if the reaction is spontaneous at this temperature.

Worked Solution: (a) $\Delta H_{\text{rxn}}^{\circ }=2\Delta H_f^{\circ }({\text{NH}}_3)-\left[\Delta H_f^{\circ }({\text{N}}_2)+3\Delta H_f^{\circ }({\text{H}}_2)\right]=2(-46.1)-(0+0)=-92.2\text{ kJ/mol}$. (b) $\Delta S_{\text{rxn}}^{\circ }=2S^{\circ }({\text{NH}}_3)-\left[S^{\circ }({\text{N}}_2)+3S^{\circ }({\text{H}}_2)\right]=2(192.8)-(191.6+3(130.7))=385.6-583.7=-198.1\text{ J/(molcdotpK)}=-0.1981\text{ kJ/(molcdotpK)}$. (c) $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }=-92.2-(298)(-0.1981)=-92.2+59.0=-33.2\text{ kJ/mol}$. Since $\Delta G^{\circ }<0$, the reaction is spontaneous at 298 K.

Question 3 (Application / Real-World Style)

The Haber process for ammonia production is run industrially at ~700 K. Given that $\Delta G^{\circ }$ for the reaction in Question 2 is +51.1 kJ/mol at 700 K, calculate $K$ for the Haber process at 700 K, and explain what this means for the industrial process.

Worked Solution: Use $\Delta G^{\circ }=-RT\ln K$ → $\ln K=-\Delta G^{\circ }/(RT)=-(51100\text{ J/mol})/(8.314\text{ J/(molcdotpK)}\times 700\text{ K})\approx -8.77$. $K=e^{-8.77}\approx 1.6\times 10^{-4}$. Since $K<<1$, equilibrium favors reactants at 700 K, so industrial Haber process uses continuous removal of ammonia to drive the reaction toward product, despite the unfavorable equilibrium constant.

8. Quick Reference Cheatsheet

9. What's Next

This topic is the foundation for all further study of thermodynamic favorability and equilibrium in AP Chemistry. Next you will apply the concepts of ΔG and spontaneity to solubility equilibria, electrochemistry (cell potential), and free energy changes under non-standard conditions. Without mastering the sign conventions, unit conversions, and core relationships between ΔG, ΔH, ΔS, and K, you will struggle to connect thermodynamics to these downstream topics, which make up a large portion of the AP Chemistry exam. Entropy and Gibbs free energy also tie together the two big pillars of AP Chemistry: thermodynamics and equilibrium, showing that equilibrium is a natural consequence of the second law of thermodynamics.

Thermodynamic favorability and temperature Gibbs free energy and equilibrium Free energy and cell potential Introduction to entropy