Thermodynamics — Free AP Chemistry Review Games.
This unit covers enthalpy, entropy, Gibbs free energy and Hess's law — essential concepts for AP Chemistry. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers enthalpy, entropy, Gibbs free energy and Hess's law — essential concepts for AP Chemistry. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Enthalpy
Enthalpy (H) measures heat flow at constant pressure; ΔH = q_p. Exothermic reactions release heat (ΔH < 0) and endothermic reactions absorb heat (ΔH > 0). Students must be able to calculate ΔH using bond energies, standard enthalpies of formation, or calorimetry data.
Key Points
- ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants); ΔH°f of any pure element in standard state = 0
- Bond breaking is endothermic; bond forming is exothermic. ΔH ≈ Σ(bonds broken) − Σ(bonds formed)
- ΔH is extensive: doubling coefficients doubles ΔH; reversing a reaction flips the sign of ΔH
- Calorimetry: q = mcΔT; for a coffee-cup calorimeter, q_rxn = −q_solution
Calculate ΔH°rxn for: N₂(g) + 3H₂(g) → 2NH₃(g), given ΔH°f[NH₃(g)] = −46.0 kJ/mol.
Apply ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants). Products: 2 mol NH₃ × (−46.0 kJ/mol) = −92.0 kJ. Reactants: N₂ and H₂ are pure elements, so their ΔH°f = 0. Therefore ΔH°rxn = −92.0 − 0 = −92.0 kJ. The negative value confirms the reaction is exothermic.
2 Entropy
Entropy (S) is a measure of the dispersal of energy and matter; systems spontaneously move toward higher entropy (Second Law). Students must predict the sign of ΔS for a reaction from physical clues and know that absolute entropy increases with temperature and complexity. The Third Law states that perfect crystalline solids at 0 K have S = 0.
Key Points
- ΔS > 0 when: gases are produced, moles of gas increase, solids dissolve, or temperature increases
- ΔS°rxn = Σ S°(products) − Σ S°(reactants); unlike ΔH°f, S° of elements ≠ 0
- Gas phase > liquid phase > solid phase in terms of molar entropy
- Larger, more complex molecules have higher standard molar entropy than smaller, simpler ones
Predict the sign of ΔS for: CaCO₃(s) → CaO(s) + CO₂(g).
One mole of solid reactant produces one mole of solid and one mole of gas. The number of moles of gas increases from 0 to 1, which greatly increases the dispersal of matter and energy. Therefore ΔS > 0 (positive), meaning entropy increases for this reaction.
3 Gibbs Free Energy
Gibbs free energy (G) determines spontaneity: ΔG = ΔH − TΔS. A process is spontaneous when ΔG < 0, nonspontaneous when ΔG > 0, and at equilibrium when ΔG = 0. Students must analyze how temperature affects spontaneity when ΔH and ΔS have the same sign, and connect ΔG° to the equilibrium constant K.
Key Points
- ΔG < 0: spontaneous (product-favored); ΔG > 0: nonspontaneous; ΔG = 0: equilibrium
- Temperature dependence: if ΔH and ΔS are both positive, reaction becomes spontaneous at high T (T > ΔH/ΔS)
- ΔG° = −RT ln K; if ΔG° < 0 then K > 1 (products favored); if ΔG° > 0 then K < 1 (reactants favored)
- ΔG°rxn = Σ ΔG°f(products) − Σ ΔG°f(reactants); also calculated via ΔG = ΔH − TΔS at standard conditions
A reaction has ΔH° = +120 kJ and ΔS° = +300 J/K. At what temperature does the reaction become spontaneous?
Set ΔG = 0 at the crossover temperature: 0 = ΔH − TΔS → T = ΔH/ΔS. Convert units: ΔH = 120,000 J, ΔS = 300 J/K. T = 120,000 J ÷ 300 J/K = 400 K. Above 400 K, the TΔS term dominates and ΔG becomes negative, so the reaction is spontaneous only at temperatures greater than 400 K.
4 Hess's Law
Hess's Law states that ΔH for an overall reaction equals the sum of ΔH values for any series of steps that add up to that reaction, because enthalpy is a state function. Students must manipulate given equations (reversing, scaling) to construct a target reaction and combine their ΔH values correctly. This principle also underlies the formation enthalpy method.
Key Points
- Reversing a reaction: multiply ΔH by −1
- Multiplying coefficients by a factor n: multiply ΔH by n
- Cancel species that appear on both sides after algebraic combination; what remains is the target equation
- All algebraic manipulations applied to the equation must also be applied to ΔH
Given: (1) C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ; (2) CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ. Find ΔH for: C(s) + ½O₂(g) → CO(g).
The target reaction needs CO as a product, but CO appears as a reactant in equation (2), so reverse equation (2): CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ. Now add reversed equation (2) to equation (1): C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g). Cancel CO₂ and ½O₂ from both sides to get C(s) + ½O₂(g) → CO(g), and ΔH = −393.5 + 283.0 = −110.5 kJ.
Questions, answered.
What is Thermodynamics?
Thermodynamics is Unit 6 of AP Chemistry, covering enthalpy, entropy, Gibbs free energy and Hess's law.
How to study for AP Chemistry Unit 6?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 30+ review questions across 5 different game modes.