AP Chemistry · AP Chemistry CED Unit 7: Equilibrium · 14 min read
1. ΔG, Q, and Spontaneous Reaction Direction★★☆☆☆⏱ 4 min
Non-standard Gibbs free energy change ΔG determines whether a reaction will proceed spontaneously forward, reverse, or be at equilibrium under actual reaction conditions, which rarely match standard state (1 M concentration, 1 atm pressure, specified temperature).
Delta G = \Delta G^\circ + RT \ln Q
Where $R = 8.314\ \text{J/(mol·K)}$, $T$ is absolute temperature in Kelvin, and $Q$ is calculated the same way as K, using actual concentrations/partial pressures of reactants and products instead of equilibrium values. ΔG° describes the free energy change when all components are at standard conditions; the $RT \ln Q$ term corrects for non-standard conditions. The spontaneity rules are:
If $\Delta G < 0$: forward reaction is spontaneous
If $\Delta G > 0$: reverse reaction is spontaneous
If $\Delta G = 0$: the system is at equilibrium
Exam tip: Always convert ΔG° from kJ/mol to J/mol before plugging into this formula. R is almost always given as 8.314 J/(mol·K) on the AP exam, so a unit mismatch will throw your result off by a factor of 1000 every time.
2. Relationship Between ΔG° and Equilibrium Constant K★★★☆☆⏱ 5 min
By definition, when a system reaches equilibrium, $\Delta G = 0$ and $Q = K$ (the equilibrium constant). Substituting these into the non-standard ΔG formula gives the key quantitative relationship between standard free energy change and the equilibrium constant:
0 = \Delta G^\circ + RT \ln K \implies \Delta G^\circ = -RT \ln K
This relationship is tested heavily on both MCQ and FRQ, and follows intuitive rules:
If ΔG° < 0: $\ln K > 0$, so $K > 1$, products are favored at equilibrium
If ΔG° > 0: $\ln K < 0$, so $K < 1$, reactants are favored at equilibrium
If ΔG° = 0: $K = 1$, roughly equal amounts of reactants and products at equilibrium
Exam tip: Double-check your negative sign after calculating K. If you have a negative ΔG°, you must get a K > 1. If your result is the opposite, you flipped the negative sign during algebra, an extremely common easy-to-fix mistake.
3. Temperature Dependence of K and the van't Hoff Equation★★★★☆⏱ 5 min
We can combine the two expressions for ΔG° ($\Delta G^\circ = -RT \ln K$ and $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$) to derive how K changes with temperature, assuming ΔH° and ΔS° are approximately constant over small to moderate temperature ranges (the standard AP Chemistry assumption). Rearranging gives the two-point van't Hoff equation:
This equation lets you calculate K at a new temperature if you know K at one temperature and ΔH° for the reaction. The result always matches Le Chatelier’s principle:
Endothermic reactions (ΔH° > 0): increasing temperature increases K
Exothermic reactions (ΔH° < 0): increasing temperature decreases K
Exam tip: Always confirm your result matches Le Chatelier’s principle after calculation. If you have an endothermic reaction, K must increase when T increases. If it does not, you flipped the sign of ΔH° in the equation.
4. AP-Style Worked Practice Problems★★★☆☆⏱ 5 min
Common Pitfalls
Why: Students confuse constant standard-state ΔG° with non-standard ΔG that changes with Q.
Why: Tabulated values are almost always given in kJ, so students forget to convert units.
Why: Students rush algebraic rearrangement and miss the negative sign.
Why: Questions often give temperature in °C, and students forget to convert to absolute temperature.
Why: Students forget the standard AP Chemistry approximation for this topic.