Free energy and equilibrium — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: The relationship between Gibbs free energy change and chemical equilibrium, calculation of ΔG° from K and K from ΔG°, prediction of reaction direction from ΔG relative to Q, and temperature dependence of K for AP Chemistry MCQ and FRQ questions.

You should already know: Gibbs free energy definition and ΔG calculation from enthalpy and entropy, equilibrium constant K and reaction quotient Q definitions, standard state conventions for thermodynamic calculations.

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 Free energy and equilibrium?

Free energy and equilibrium connects the thermodynamic concept of Gibbs free energy (from Unit 6: Thermodynamics) to the dynamic equilibrium state of chemical reactions (Unit 7: Equilibrium). According to the AP Chemistry CED, this topic makes up ~18% of Unit 7, which contributes 7-9% of the total AP exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. From a free energy perspective, equilibrium is defined as the point where the total Gibbs free energy of the system is at its minimum value: any reaction not at equilibrium will spontaneously progress toward equilibrium to reduce total free energy. This topic provides the quantitative relationship between standard free energy change ΔG° and the equilibrium constant K, and lets us predict the direction of a reaction under non-standard conditions using the reaction quotient Q. It also explains why and how equilibrium constants change with temperature, linking Le Chatelier’s principle to fundamental thermodynamics. It is one of the most heavily tested integration topics on the AP exam, as it requires combining skills from two different core units.

2. ΔG, Q, and Spontaneous Reaction Direction

Non-standard Gibbs free energy change ΔG is the value that determines whether a reaction will proceed spontaneously forward, reverse, or not at all (at equilibrium) under actual reaction conditions, which rarely match standard state (1 M concentration, 1 atm pressure, specified temperature). The core formula relating ΔG to standard free energy change ΔG° and reaction quotient Q is: $$\Delta G=\Delta G^{\circ }+RT\ln Q$$ Where $R=8.314\text{J/(molcdotpK)}$, $T$ is absolute temperature in Kelvin, and $Q$ is calculated the same way as K, using the actual concentrations/partial pressures of reactants and products instead of equilibrium values. The intuition for this formula is that ΔG° describes the free energy change when all components are at standard conditions; the $RT\ln Q$ term corrects for non-standard concentrations/pressures. The spontaneity rules are straightforward: if $\Delta G<0$, the forward reaction is spontaneous; if $\Delta G>0$, the reverse reaction is spontaneous; if $\Delta G=0$, the system is at equilibrium.

Worked Example

For the reaction ${\text{N}}_2{\text{O}}_4(g)\rightleftharpoons 2{\text{NO}}_2(g)$, $\Delta G^{\circ }=+4.8\text{kJ/mol}$ at 298 K. A reaction vessel held at 298 K has partial pressures $P_{{\text{N}}_2{\text{O}}_4}=2.0\text{atm}$ and $P_{{\text{NO}}_2}=0.10\text{atm}$. Is the reaction spontaneous forward, reverse, or at equilibrium?

1. Convert units to match R: $\Delta G^{\circ }=4.8\text{kJ/mol}=4800\text{J/mol}$.
2. Calculate Q for the given pressures: $Q=\frac{(P_{{\text{NO}}_2}{)}^2}{P_{{\text{N}}_2{\text{O}}_4}}=\frac{(0.10{)}^2}{2.0}=0.0050$.
3. Calculate $\ln Q=\ln (0.0050)\approx -5.30$.
4. Substitute into the formula: $\Delta G=4800+(8.314)(298)(-5.30)\approx 4800-13100=-8300\text{J/mol}=-8.3\text{kJ/mol}$.
5. Interpret: $\Delta G<0$, so the reaction is spontaneous in the forward direction.

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.

3. Relationship Between ΔG° and Equilibrium Constant K

By definition, when a system reaches equilibrium, $\Delta G=0$ and $Q=K$ (the equilibrium constant). We can substitute these into the ΔG formula from the previous section to derive the key relationship between standard free energy change and K: $$0=\Delta G^{\circ }+RT\ln K⟹\Delta G^{\circ }=-RT\ln K$$ This is the most important formula in this topic, and it is tested heavily on both MCQ and FRQ. The relationship between the sign of ΔG° and the magnitude of K is intuitive: if ΔG° is negative, $\ln K$ is positive, so $K>1$, meaning products are favored at equilibrium. If ΔG° is positive, $\ln K$ is negative, so $K<1$, meaning reactants are favored at equilibrium. If ΔG° = 0, $K=1$, so roughly equal amounts of reactants and products are present at equilibrium. This formula lets us calculate K from tabulated standard free energy of formation values, or calculate ΔG° from an experimentally measured K.

Worked Example

Given $\Delta G_f^{\circ }({\text{NH}}_3(g))=-16.5\text{kJ/mol}$, and $\Delta G_f^{\circ }({\text{N}}_2(g))=\Delta G_f^{\circ }({\text{H}}_2(g))=0\text{kJ/mol}$, calculate K for the reaction ${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)$ at 298 K.

1. Calculate ΔG° for the reaction using standard free energies of formation: $\Delta G^{\circ }=\sum \Delta G_f^{\circ }(\text{products})-\sum \Delta G_f^{\circ }(\text{reactants})=(2\times -16.5)-(0+3\times 0)=-33.0\text{kJ/mol}=-33000\text{J/mol}$.
2. Rearrange $\Delta G^{\circ }=-RT\ln K$ to solve for $\ln K$: $\ln K=\frac{-\Delta G^{\circ }}{RT}$.
3. Substitute values: $\ln K=\frac{-(-33000)}{(8.314)(298)}\approx \frac{33000}{2478}\approx 13.3$.
4. Exponentiate to solve for K: $K=e^{13.3}\approx 5.9\times 10^5$.
5. Confirm: $K>1$ matches the negative ΔG°, so products are favored at 298 K, as expected.

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, which is an extremely common easy-to-fix mistake.

4. Temperature Dependence of K and the van't Hoff Equation

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 Chem assumption). Rearranging gives the two-point van't Hoff equation: $$\ln \left(\frac{K_2}{K_1}\right)=-\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$ This equation lets us calculate K at a new temperature if we know K at one temperature and ΔH° for the reaction. The result always matches Le Chatelier’s principle: for endothermic reactions (ΔH° > 0), increasing T increases K; for exothermic reactions (ΔH° < 0), increasing T decreases K. This is a key connection between qualitative equilibrium rules and quantitative thermodynamics, often tested in FRQ reasoning questions.

Worked Example

The acid dissociation of acetic acid has $\Delta H^{\circ }=-0.84\text{kJ/mol}$, and $K_a=1.8\times 10^{-5}$ at 298 K. What is $K_a$ for acetic acid at 310 K (human body temperature)?

1. Convert ΔH° to J/mol to match R units: $\Delta H^{\circ }=-840\text{J/mol}$.
2. Assign values: $K_1=1.8\times 10^{-5}$, $T_1=298\text{K}$, $T_2=310\text{K}$, solve for $K_2$.
3. Substitute into the van't Hoff equation: $$\ln \left(\frac{K_2}{1.8\times 10^{-5}}\right)=-\frac{(-840)}{8.314}\left(\frac{1}{310}-\frac{1}{298}\right)\approx 101(-0.00013)\approx -0.013$$
4. Exponentiate both sides: $\frac{K_2}{1.8\times 10^{-5}}=e^{-0.013}\approx 0.987$, so $K_2\approx 1.8\times 10^{-5}$.
5. Interpret: The small magnitude of ΔH° means $K_a$ changes very little over this small temperature range.

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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using ΔG° instead of ΔG to determine spontaneity for non-standard conditions, or concluding that a reaction with positive ΔG° will never proceed forward. Why: Students confuse constant standard-state ΔG° with non-standard ΔG that changes with Q. Correct move: Always use ΔG = ΔG° + RT ln Q for non-standard conditions, and remember even positive ΔG° reactions can be spontaneous forward if Q is small enough to make ΔG negative.
• Wrong move: Leaving ΔG or ΔH in kJ/mol when plugging into equations with R = 8.314 J/mol·K, leading to a result off by a factor of 1000. Why: Tabulated values are almost always given in kJ, so students forget to convert. Correct move: Convert all energy values to joules per mole before substituting into any equation with R.
• Wrong move: Flipping the negative sign in ΔG° = -RT ln K, getting K < 1 for a negative ΔG° (or vice versa). Why: Students rush rearrangement and miss the negative sign. Correct move: After calculating K, cross-check that the sign of ΔG° matches the magnitude of K, and correct the sign if it does not.
• Wrong move: Using Celsius temperature instead of Kelvin in all free energy/equilibrium equations. Why: Questions often give temperature in °C, and students forget to convert. Correct move: Immediately add 273 to any Celsius temperature before plugging into equations, and round to match the given significant figures.
• Wrong move: Assuming ΔH° changes significantly with small temperature changes, leading to overcomplicating van't Hoff calculations. Why: Students forget the AP Chem approximation. Correct move: Use the given ΔH° at 298 K for van't Hoff calculations unless the question explicitly provides a different value.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

For the reaction $\text{A}(g)\rightleftharpoons \text{B}(g)$, $\Delta G^{\circ }=-11.7\text{kJ/mol}$ at 27°C. Which of the following is correct at 27°C when partial pressures are $P_A=0.1\text{atm}$ and $P_B=15\text{atm}$? A) ΔG = -11.7 kJ/mol, reaction is at equilibrium B) ΔG = 0 kJ/mol, reaction is spontaneous forward C) ΔG ≈ +0.8 kJ/mol, reaction is spontaneous reverse D) ΔG ≈ -24 kJ/mol, reaction is spontaneous forward

Worked Solution: First, convert temperature to Kelvin: $T=27+273=300\text{K}$, and convert ΔG° to joules: $\Delta G^{\circ }=-11700\text{J/mol}$. Calculate the reaction quotient: $Q=\frac{P_B}{P_A}=\frac{15}{0.1}=150$, so $\ln Q\approx 5.01$. Substitute into $\Delta G=\Delta G^{\circ }+RT\ln Q$: $\Delta G=-11700+(8.314)(300)(5.01)\approx -11700+12500=+800\text{J/mol}=+0.8\text{kJ/mol}$. A positive ΔG means the reaction is spontaneous in the reverse direction. The correct answer is C.

Question 2 (Free Response)

The dissolution of calcium hydroxide in water is given by: ${\text{Ca(OH)}}_2(s)\rightleftharpoons {\text{Ca}}^{2+}(aq)+2{\text{OH}}^{-}(aq)$. The following standard free energy of formation values are available at 298 K: ΔG°f (Ca(OH)₂(s)) = -898.5 kJ/mol ΔG°f (Ca²⁺(aq)) = -553.6 kJ/mol ΔG°f (OH⁻(aq)) = -157.2 kJ/mol (a) Calculate ΔG° for the dissolution reaction at 298 K. (b) Calculate $K_{sp}$ for Ca(OH)₂ at 298 K, and state whether reactants or products are favored at equilibrium. (c) The dissolution reaction is endothermic, ΔH° = +16.2 kJ/mol. Predict whether Ksp will increase, decrease, or stay the same when the solution is heated to 350 K. Justify your answer.

Worked Solution: (a) Use the formula for ΔG° of reaction: $$\Delta G^{\circ }=\sum \Delta G_f^{\circ }(\text{products})-\sum \Delta G_f^{\circ }(\text{reactants})$$ $$\Delta G^{\circ }=\left[\Delta G_f^{\circ }({\text{Ca}}^{2+})+2\Delta G_f^{\circ }({\text{OH}}^{-})\right]-\Delta G_f^{\circ }({\text{Ca(OH)}}_2)$$ $$\Delta G^{\circ }=(-553.6+2(-157.2))-(-898.5)=-868.0+898.5=+30.5\text{kJ/mol}$$

(b) Rearrange ΔG° = -RT ln K to solve for Ksp: $$\ln K_{sp}=\frac{-\Delta G^{\circ }}{RT}=\frac{-30500\text{J/mol}}{(8.314\text{J/molcdotpK})(298\text{K})}\approx -12.31$$ $$K_{sp}=e^{-12.31}\approx 4.5\times 10^{-6}$$ Since Ksp < 1, reactants (solid Ca(OH)₂) are favored at equilibrium.

(c) Ksp will increase. For an endothermic reaction, ΔH° is positive. From the van't Hoff relationship $\ln K=-\frac{\Delta H^{\circ }}{RT}+\frac{\Delta S^{\circ }}{R}$, as T increases, the term $-\frac{\Delta H^{\circ }}{RT}$ becomes less negative, so ln K increases, so Ksp increases. This matches Le Chatelier's principle: adding heat to an endothermic reaction shifts equilibrium right, increasing solubility.

Question 3 (Application / Real-World Style)

Hemoglobin (Hb) binds oxygen in the lungs to form oxyhemoglobin (HbO₂) according to the reaction: $\text{Hb}(aq)+{\text{O}}_2(g)\rightleftharpoons {\text{HbO}}_2(aq)$. At human body temperature (310 K), ΔG° for this reaction is -70 kJ/mol. A climber at high altitude has an oxygen partial pressure of 0.08 atm, and equal concentrations of Hb and HbO₂ ($1.0\times 10^{-6}\text{M}$ each). Calculate ΔG for the reaction under these conditions and state whether oxygen binding is spontaneous. Assume standard state for solutes is 1 M.

Worked Solution: Convert ΔG° to J/mol: $\Delta G^{\circ }=-70000\text{J/mol}$, T = 310 K. Calculate Q: $$Q=\frac{[{\text{HbO}}_2]}{[\text{Hb}]P_{{\text{O}}_2}}=\frac{1.0\times 10^{-6}}{(1.0\times 10^{-6})(0.08)}=12.5$$ $\ln Q\approx 2.526$. Substitute into ΔG = ΔG° + RT ln Q: $$\Delta G=-70000+(8.314)(310)(2.526)\approx -70000+6510=-63500\text{J/mol}=-63.5\text{kJ/mol}$$ ΔG is negative, so oxygen binding to hemoglobin remains spontaneous at high altitude, even though the driving force is smaller than at sea level (where oxygen partial pressure is ~0.21 atm).

7. Quick Reference Cheatsheet

8. What's Next

This topic bridges thermodynamics (Unit 6) and core equilibrium (Unit 7), and is a prerequisite for all applied equilibrium topics that follow, including acid-base equilibrium, solubility equilibrium, and the relationship between cell potential and equilibrium in electrochemistry. Without mastering the relationship between ΔG°, ΔG, K, and Q, you will not be able to solve multi-step integration problems that are common in AP Chemistry FRQs. This topic also reinforces the key distinction between standard and non-standard conditions, a reasoning skill tested across all units. Next you will apply the $\Delta G^{\circ }=-RT\ln K$ relationship directly to solubility product and acid dissociation constant calculations.

• Acid-base equilibrium
• Solubility equilibrium
• Gibbs free energy and thermodynamics
• Electrochemical cell potential