Thermodynamics — AP Chemistry Chem Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Enthalpy and Hess's Law, entropy and the second law of thermodynamics, Gibbs free energy and reaction spontaneity, and the relationship between thermodynamics and chemical equilibrium.

You should already know: High-school chemistry, Algebra 2.

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 papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Thermodynamics?

Thermodynamics is the study of energy transfers and transformations in chemical systems, and how these transfers determine whether a reaction will proceed spontaneously or require external energy input. This topic accounts for 7-9% of your AP Chemistry exam score, per the College Board CED, and uses standard notation distinguishing the system (the reaction or process you are studying) from the surroundings (all other matter in the universe). You may also see the subset of thermodynamics focused on heat transfer referred to as thermochemistry.

2. Enthalpy and Hess's Law

Enthalpy ($H$) is the total heat content of a system at constant pressure. We almost exclusively measure changes in enthalpy ($\Delta H$) for reactions, calculated as: $$\Delta H=H_{\text{products}}-H_{\text{reactants}}$$ A negative $\Delta H$ indicates an exothermic reaction, which releases heat to the surroundings (e.g. combustion of fossil fuels). A positive $\Delta H$ indicates an endothermic reaction, which absorbs heat from the surroundings (e.g. dissolving ammonium nitrate in water for cold packs). Standard enthalpy of formation ($\Delta H_f^{\circ }$) is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states (298 K, 1 atm); the $\Delta H_f^{\circ }$ of any element in its standard state is 0.

Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken, so you can sum $\Delta H$ values for adjusted intermediate reactions to calculate the net $\Delta H$ for the target reaction.

Worked Example

Calculate the standard enthalpy change for the combustion of methane: ${\text{CH}}_4(g)+2{\text{O}}_2(g)\to {\text{CO}}_2(g)+2{\text{H}}_2\text{O}(l)$, given the following $\Delta H_f^{\circ }$ values: ${\text{CH}}_4(g)=-74.8$ kJ/mol, ${\text{CO}}_2(g)=-393.5$ kJ/mol, ${\text{H}}_2\text{O}(l)=-285.8$ kJ/mol, ${\text{O}}_2(g)=0$ kJ/mol.

1. Use the standard enthalpy formula: $$\Delta H^{\circ }=\sum \Delta H_f^{\circ }(\text{products})-\sum \Delta H_f^{\circ }(\text{reactants})$$
2. Substitute values: $$\Delta H^{\circ }=\left[(-393.5)+2(-285.8)\right]-\left[(-74.8)+2(0)\right]$$ $$\Delta H^{\circ }=-965.1+74.8=-890.3\text{ kJ/mol}$$ This negative value confirms combustion is an exothermic process, as expected. Exam tip: When rearranging intermediate reactions for Hess's Law questions, flip the sign of $\Delta H$ if you reverse a reaction, and multiply $\Delta H$ by the same factor you use to scale reaction coefficients.

3. Entropy and 2nd Law

Entropy ($S$) is a measure of the disorder or randomness of a system, with units of J/mol·K. Higher disorder corresponds to higher entropy: gases have far higher entropy than liquids, which have higher entropy than solids. Reactions that produce more moles of gas than they consume have a positive entropy change ($\Delta S$), as do reactions that dissolve solids or increase system temperature.

The Second Law of Thermodynamics states that for any spontaneous process, the total entropy of the universe (system + surroundings) always increases: $$\Delta S_{\text{univ}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}>0\text{ (spontaneous processes)}$$ The entropy change of the surroundings is calculated as: $$\Delta S_{\text{surr}}=\frac{-\Delta H_{\text{sys}}}{T}$$ Heat released by the system increases disorder in the surroundings, and this effect is amplified at lower temperatures, where the surroundings have less pre-existing randomness.

Worked Example

Calculate $\Delta S_{\text{sys}}^{\circ }$ for the combustion of methane, given standard molar entropy values: ${\text{CH}}_4(g)=186.3$ J/mol·K, ${\text{O}}_2(g)=205.2$ J/mol·K, ${\text{CO}}_2(g)=213.8$ J/mol·K, ${\text{H}}_2\text{O}(l)=70.0$ J/mol·K.

1. Use the standard entropy formula: $$\Delta S^{\circ }=\sum S^{\circ }(\text{products})-\sum S^{\circ }(\text{reactants})$$
2. Substitute values: $$\Delta S^{\circ }=\left[213.8+2(70.0)\right]-\left[186.3+2(205.2)\right]$$ $$\Delta S^{\circ }=353.8-596.7=-242.9\text{ J/molcdotpK}$$ This negative value makes sense: the reaction reduces moles of gas from 3 to 1, increasing order in the system. However, the large exothermic $\Delta H$ makes $\Delta S_{\text{surr}}$ highly positive, so total $\Delta S_{\text{univ}}$ is positive and the reaction is spontaneous.

4. Gibbs Free Energy — Spontaneity

Gibbs free energy ($G$) combines enthalpy and entropy into a single system property that eliminates the need to calculate surroundings entropy to determine spontaneity. It is defined as: $$\Delta G=\Delta H-T\Delta S$$ where $T$ is temperature in Kelvin. The sign of $\Delta G$ directly tells you reaction spontaneity at a given temperature:

• $\Delta G<0$: spontaneous in the forward direction
• $\Delta G=0$: system is at equilibrium
• $\Delta G>0$: non-spontaneous in the forward direction, spontaneous in the reverse direction

Standard Gibbs free energy change ($\Delta G^{\circ }$) applies to reactions at 298 K, 1 atm, and 1 M concentration, and can also be calculated from standard free energies of formation using the same formula as $\Delta H^{\circ }$.

Worked Example

Calculate $\Delta G^{\circ }$ for methane combustion at 298 K, using $\Delta H^{\circ }=-890.3$ kJ/mol and $\Delta S^{\circ }=-242.9$ J/mol·K.

1. Convert $\Delta S$ to kJ to match units: $-0.2429$ kJ/mol·K
2. Substitute into the Gibbs free energy formula: $$\Delta G^{\circ }=-890.3-(298\times -0.2429)$$ $$\Delta G^{\circ }=-890.3+72.4=-817.9\text{ kJ/mol}$$ The negative $\Delta G^{\circ }$ confirms the reaction is spontaneous under standard conditions. Exam tip: Examiners frequently ask about temperature dependence of spontaneity: if $\Delta H$ and $\Delta S$ have the same sign, you can solve for the threshold temperature where $\Delta G=0$ to find when the reaction becomes spontaneous.

5. Thermodynamics + Equilibrium

Gibbs free energy directly links to chemical equilibrium via the reaction quotient ($Q$, the ratio of product to reactant concentrations at non-equilibrium conditions) and the equilibrium constant ($K$, the ratio at equilibrium). The formula for non-standard Gibbs free energy is: $$\Delta G=\Delta G^{\circ }+RT\ln Q$$ where $R=8.314$ J/mol·K. At equilibrium, $\Delta G=0$ and $Q=K$, so rearranging gives the key relationship between standard free energy and equilibrium: $$\Delta G^{\circ }=-RT\ln K$$ This relationship explains the equilibrium position of a reaction:

• If $\Delta G^{\circ }<0$, $\ln K>0$ so $K>1$, and products are favored at equilibrium
• If $\Delta G^{\circ }=0$, $K=1$, and equal amounts of products and reactants are present
• If $\Delta G^{\circ }>0$, $\ln K<0$ so $K<1$, and reactants are favored at equilibrium

Worked Example

Calculate the equilibrium constant $K$ for methane combustion at 298 K, given $\Delta G^{\circ }=-817900$ J/mol.

1. Rearrange the equilibrium formula to solve for $\ln K$: $$\ln K=\frac{-\Delta G^{\circ }}{RT}$$
2. Substitute values: $$\ln K=\frac{-(-817900)}{8.314\times 298}\approx 330$$ $$K=e^{330}$$ This extremely large $K$ value confirms the reaction goes almost to completion, as expected for combustion.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to convert $\Delta S$ from J/mol·K to kJ/mol·K when calculating $\Delta G$, leading to order-of-magnitude errors. Why students do it: Entropy values are almost always given in J, while enthalpy values are given in kJ, creating a unit mismatch. Correct move: Write unit labels next to every value during calculations, and convert all values to consistent units before plugging into formulas.
• Wrong move: Assuming exothermic reactions ($\Delta H<0$) are always spontaneous, ignoring entropy effects. Why students do it: Most common spontaneous reactions are exothermic, so students generalize incorrectly. Correct move: Always use $\Delta G$ to judge spontaneity, not just $\Delta H$ or $\Delta S$ alone; endothermic reactions can be spontaneous at high temperatures if entropy increases enough.
• Wrong move: Using $\Delta G^{\circ }$ to judge spontaneity under non-standard conditions. Why students do it: Standard free energy values are easy to look up, but only apply to 1 M concentration, 1 atm pressure, and 298 K. Correct move: Use $\Delta G=\Delta G^{\circ }+RT\ln Q$ for non-standard conditions; $\Delta G^{\circ }$ only tells you the equilibrium position of a reaction, not real-time spontaneity under arbitrary concentrations.
• Wrong move: Flipping or scaling reaction coefficients for Hess's Law but forgetting to apply the same change to $\Delta H$. Why students do it: Rushing through intermediate reaction rearrangements. Correct move: Write each adjusted intermediate reaction clearly next to its modified $\Delta H$ value before summing.
• Wrong move: Counting moles of solids and liquids when predicting the sign of $\Delta S$. Why students do it: Students assume all state changes affect entropy equally, but solids and liquids have negligible entropy compared to gases. Correct move: First compare moles of gaseous reactants vs products; this will always give you the correct $\Delta S$ sign for AP exam questions.

7. Practice Questions (AP Chemistry Style)

Question 1

Use the following intermediate reactions to calculate $\Delta H^{\circ }$ for the target reaction: $2\text{C}(s)+{\text{H}}_2(g)\to {\text{C}}_2{\text{H}}_2(g)$

1. ${\text{C}}_2{\text{H}}_2(g)+\frac{5}{2}{\text{O}}_2(g)\to 2{\text{CO}}_2(g)+{\text{H}}_2\text{O}(l)$ $\Delta H^{\circ }=-1299.6$ kJ/mol
2. $\text{C}(s)+{\text{O}}_2(g)\to {\text{CO}}_2(g)$ $\Delta H^{\circ }=-393.5$ kJ/mol
3. ${\text{H}}_2(g)+\frac{1}{2}{\text{O}}_2(g)\to {\text{H}}_2\text{O}(l)$ $\Delta H^{\circ }=-285.8$ kJ/mol

Solution

1. Adjust intermediate reactions to match the target:

• Multiply reaction 2 by 2 to get 2 moles of C(s) on the left: $\Delta H=2(-393.5)=-787$ kJ/mol
• Keep reaction 3 unchanged to get 1 mole of H₂(g) on the left: $\Delta H=-285.8$ kJ/mol
• Flip reaction 1 to get C₂H₂(g) on the right: $\Delta H=+1299.6$ kJ/mol

2. Sum the adjusted reactions and cancel common terms on both sides: $2\text{C}(s)+{\text{H}}_2(g)\to {\text{C}}_2{\text{H}}_2(g)$
3. Sum the adjusted $\Delta H$ values: $\Delta H^{\circ }=-787-285.8+1299.6=226.8$ kJ/mol

Question 2

For the decomposition of limestone: ${\text{CaCO}}_3(s)\to \text{CaO}(s)+{\text{CO}}_2(g)$, $\Delta H^{\circ }=+178.3$ kJ/mol and $\Delta S^{\circ }=+160.6$ J/mol·K. a) Calculate $\Delta G^{\circ }$ at 298 K b) Find the minimum temperature (in °C) where the reaction becomes spontaneous.

Solution

a) Convert $\Delta S$ to kJ: $0.1606$ kJ/mol·K $$\Delta G^{\circ }=178.3-(298\times 0.1606)=178.3-47.86=130.4\text{ kJ/mol}$$ The positive value means the reaction is non-spontaneous at room temperature, as expected.

b) Set $\Delta G=0$ to find the threshold temperature: $$T=\frac{\Delta H}{\Delta S}=\frac{178.3}{0.1606}\approx 1110\text{ K}$$ Convert to °C: $1110-273=837$ °C

Question 3

For the reaction ${\text{N}}_2{\text{O}}_4(g)\rightleftharpoons 2{\text{NO}}_2(g)$, $\Delta G^{\circ }=+4.8$ kJ/mol at 298 K. a) Calculate the equilibrium constant $K$ at 298 K b) If the partial pressures of ${\text{N}}_2{\text{O}}_4=2.0$ atm and ${\text{NO}}_2=0.3$ atm, is the reaction spontaneous in the forward direction under these conditions?

Solution

a) Convert $\Delta G^{\circ }$ to J: 4800 J/mol $$\ln K=\frac{-\Delta G^{\circ }}{RT}=\frac{-4800}{8.314\times 298}\approx -1.937$$ $$K=e^{-1.937}\approx 0.144$$ The $K<1$ confirms reactants are favored at standard conditions.

b) First calculate the reaction quotient $Q$: $$Q=\frac{(P_{{\text{NO}}_2}{)}^2}{P_{{\text{N}}_2{\text{O}}_4}}=\frac{(0.3{)}^2}{2.0}=0.045$$ Calculate non-standard $\Delta G$: $$\Delta G=4800+(8.314\times 298\times \ln (0.045))$$ $$\Delta G=4800-7684=-2884\text{ J/mol}=-2.88\text{ kJ/mol}$$ The negative $\Delta G$ means the reaction is spontaneous forward under these non-standard conditions, even though $\Delta G^{\circ }$ is positive.

8. Quick Reference Cheatsheet

9. What's Next

Thermodynamics is a foundational topic that connects to nearly 30% of the remaining AP Chemistry syllabus. You will use enthalpy calculations to analyze activation energy in kinetics, enthalpy of neutralization in acid-base reactions, and the link between Gibbs free energy and cell potential ($\Delta G=-nF{E}^{\circ }$) in electrochemistry. Entropy and spontaneity rules also explain why some redox reactions generate electricity spontaneously, while others require external power for electrolysis. Mastering this unit will make all subsequent reaction chemistry content significantly easier to understand.

If you struggle with any of the calculations, sign conventions, or application questions covered in this guide, you can ask Ollie, our AI tutor, for custom practice problems, step-by-step walkthroughs, or targeted feedback on your work 24/7. You can also head to the homepage to access more AP Chemistry study guides, full-length practice exams, and flashcards aligned to the College Board CED to prepare for your test.