AP Chemistry · College Board AP Chemistry CED Unit 6: Thermodynamics · 18 min read
1. Enthalpy and Hess's Law★★☆☆☆⏱ 5 min
Hess's Law states the total enthalpy change for a reaction is independent of the reaction pathway. We can sum adjusted intermediate $Δ H$ values to get the net $Δ H$ for a target reaction. Standard enthalpy of formation ($Δ H^\circ_f$) is the enthalpy change when 1 mole of a compound forms from its elements in standard states (298 K, 1 atm); $Δ H^\circ_f$ of any element in its standard state is 0.
2. Entropy and the Second Law of Thermodynamics★★★☆☆⏱ 4 min
The Second Law of Thermodynamics states that for any spontaneous process, the total entropy of the universe (system + surroundings) always increases. We calculate total entropy change as:
This negative value makes sense: the reaction reduces moles of gas from 3 to 1, increasing order in the system. However, the large exothermic $Δ H$ makes $Δ S_{\text{surr}}$ highly positive, so total $Δ S_{\text{univ}}$ is positive and the reaction is spontaneous.
3. Gibbs Free Energy and Reaction Spontaneity★★★☆☆⏱ 4 min
Gibbs free energy combines enthalpy and entropy into a single system property that lets us determine spontaneity without calculating the entropy change of the surroundings.
Δ G = \u0394 H - T\u0394 S
Standard Gibbs free energy change ($\u0394 G^\circ$) applies to reactions at standard conditions (298 K, 1 atm, 1 M concentration) and can be calculated from standard free energies of formation using the same sum formula as $Δ H^\circ$ and $Δ S^\circ$.
4. Thermodynamics and Chemical Equilibrium★★★★☆⏱ 4 min
Gibbs free energy directly connects to chemical equilibrium, relating the standard free energy change to the equilibrium constant $K$. For non-standard conditions, we calculate $Δ G$ using the reaction quotient $Q$, the ratio of product to reactant concentrations at non-equilibrium:
Δ G = \u0394 G^\circ + RT \ln Q
At equilibrium, $Δ G = 0$ and $Q = K$, so rearranging gives the core relationship between $Δ G^\circ$ and $K$:
Δ G^\circ = -RT \ln K
If $Δ G^\circ < 0$: $\ln K > 0 \rightarrow K > 1$, products are favored at equilibrium
If $Δ G^\circ = 0$: $\ln K = 0 \rightarrow K = 1$, equal amounts of products and reactants
If $Δ G^\circ > 0$: $\ln K < 0 \rightarrow K < 1$, reactants are favored at equilibrium
This extremely large $K$ confirms methane combustion goes almost to completion, as expected.
5. AP Style Additional Worked Practice★★★★☆⏱ 5 min
Common Pitfalls
Why: Entropy values are almost always given in joules, while enthalpy values are given in kilojoules, creating a unit mismatch that leads to large order-of-magnitude errors.
Why: Most common spontaneous reactions are exothermic, so students incorrectly generalize that $Δ H$ alone determines spontaneity, ignoring entropy contributions.
Why: $Δ G^\circ$ only applies to standard conditions (1 M concentration, 1 atm, 298 K) but students often use it for all conditions.
Why: Students rush through rearranging intermediate reactions and only adjust the chemical equation, not the enthalpy value.
Why: Students assume all state changes affect entropy equally, but solids and liquids have negligible entropy compared to gases.