Free energy and equilibrium — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: The relationship between standard Gibbs free energy change and equilibrium constant ($\Delta G^{\circ }=-RT\ln K$), non-standard free energy calculation, spontaneity prediction from Q and K, temperature dependence of K, and the van't Hoff equation connecting thermodynamics to Le Chatelier’s principle.

You should already know: Gibbs free energy definition and spontaneity rules, equilibrium constant ($K$) and reaction quotient ($Q$), basic enthalpy and entropy 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 driving force of a reaction to its final equilibrium position, uniting two core AP Chemistry concepts: thermodynamics (which predicts reaction spontaneity) and equilibrium (which describes the system when net macroscopic change stops). Aligned with the College Board AP Chemistry CED for Unit 9: Applications of Thermodynamics, this topic accounts for approximately 7-9% of the total AP exam score, and is tested in both multiple-choice (MCQ) and free-response (FRQ) sections. It regularly appears as part of multi-part FRQs that combine this content with acid-base equilibria, solubility equilibria, or thermochemical stoichiometry. At its core, this topic answers two key practical questions: What will the final equilibrium composition of reactants and products be, given a reaction’s thermodynamic properties? And how will changing reaction conditions shift that equilibrium based on free energy changes? Unlike introductory treatments of spontaneity that only consider standard conditions, this topic extends reasoning to the non-standard conditions common in real laboratory and industrial reactions.

2. The Relationship Between ΔG° and K

At equilibrium, the total free energy of the system is at a minimum, so there is no net driving force for forward or reverse reaction, meaning $\Delta G=0$. To relate the standard free energy change to the equilibrium constant, start with the general expression for free energy under any conditions: $\Delta G=\Delta G^{\circ }+RT\ln Q$. At equilibrium, $\Delta G=0$ and $Q=K$, so substituting gives: $$\Delta G^{\circ }=-RT\ln K$$ Where $\Delta G^{\circ }$ is the standard Gibbs free energy change (usually reported in kJ/mol), $R$ is the gas constant, $T$ is absolute temperature in Kelvin, and $K$ is the thermodynamic equilibrium constant. The relationship between the sign of $\Delta G^{\circ }$ and the value of $K$ is intuitive: if $\Delta G^{\circ }<0$, $\ln K>0$, so $K>1$, meaning products are favored at equilibrium. If $\Delta G^{\circ }>0$, $\ln K<0$, so $K<1$, meaning reactants are favored at equilibrium. If $\Delta G^{\circ }=0$, $K=1$, so reactants and products are equally favored under standard conditions.

Worked Example

The oxidation of sulfur dioxide to sulfur trioxide has a standard Gibbs free energy change $\Delta G^{\circ }=-46.0\text{kJ/mol}$ at 298 K. Calculate the thermodynamic equilibrium constant $K$ for this reaction at 298 K.

1. List all known values and match units: $\Delta G^{\circ }=-46.0\text{kJ/mol}$, $R=0.008314\text{kJ/(molcdotpK)}$ (to match kJ units for $\Delta G^{\circ }$), $T=298\text{K}$.
2. Rearrange the core formula to solve for $\ln K$: $\ln K=\frac{-\Delta G^{\circ }}{RT}$.
3. Substitute values: $\ln K=\frac{-(-46.0\text{kJ/mol})}{(0.008314\text{kJ/(molcdotpK)})(298\text{K})}=\frac{46.0}{2.478}\approx 18.56$.
4. Exponentiate both sides to solve for $K$: $K=e^{18.56}\approx 1.2\times 10^8$.
5. Verify against intuition: $\Delta G^{\circ }$ is negative, so $K>1$, which matches our result.

Exam tip: Always convert R to match the energy units of ΔG°: use 0.008314 kJ/(mol·K) when ΔG° is in kJ/mol, and 8.314 J/(mol·K) when ΔG° is in J/mol to avoid 1000x errors in K.

3. Non-Standard ΔG and Spontaneity Prediction

Most reactions do not occur under standard conditions (1 M concentration, 1 atm pressure, pure solids/liquids), so we need to calculate $\Delta G$ for non-standard conditions to predict the direction of spontaneous change. The general formula for non-standard free energy change is: $$\Delta G=\Delta G^{\circ }+RT\ln Q$$ Where $Q$ is the reaction quotient calculated from the current non-standard concentrations or partial pressures. The sign of $\Delta G$ directly gives the direction of spontaneity: $\Delta G<0$ means forward reaction is spontaneous, $\Delta G>0$ means reverse reaction is spontaneous, and $\Delta G=0$ means the reaction is at equilibrium. We can derive a useful shortcut by substituting $\Delta G^{\circ }=-RT\ln K$ into the non-standard $\Delta G$ formula: $\Delta G=-RT\ln K+RT\ln Q=RT\ln \left(\frac{Q}{K}\right)$. Since $R$ and $T$ are always positive, the sign of $\Delta G$ matches the sign of $\ln \left(\frac{Q}{K}\right)$. This means we can predict spontaneity directly from $Q$ and $K$ without calculating $\Delta G$ at all.

Worked Example

For the reaction $N_2{O}_4(g)\rightleftharpoons 2N{O}_2(g)$, $K=0.36$ at 298 K, and $\Delta G^{\circ }=2.5\text{kJ/mol}$ at 298 K. A reaction mixture contains 0.10 atm $N_2{O}_4$ and 0.50 atm $N{O}_2$. Is the reaction spontaneous forward, reverse, or at equilibrium under these conditions?

1. Calculate $Q_p$ from the given partial pressures: $Q_p=\frac{(P_{N{O}_2}{)}^2}{P_{N_2{O}_4}}=\frac{(0.50{)}^2}{0.10}=2.5$.
2. Compare $Q$ to $K$: $Q=2.5>K=0.36$, so $\ln (Q/K)$ is positive, meaning $\Delta G$ is positive.
3. Confirm with full $\Delta G$ calculation: $\Delta G=2.5\text{kJ/mol}+(0.008314\text{kJ/(molcdotpK)})(298\text{K})\ln (2.5)\approx 2.5+2.27=4.77\text{kJ/mol}$.
4. $\Delta G$ is positive, so the reverse reaction is spontaneous: the reaction will shift to form more reactants until $Q$ equals $K$.

Exam tip: Do not confuse ΔG° and ΔG: ΔG° tells you about the equilibrium position (whether K is greater than or less than 1), while ΔG tells you the direction of spontaneity under your specific non-standard conditions.

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

Equilibrium constants change with temperature, and we can derive this relationship by combining two expressions for $\Delta G^{\circ }$: $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$ and $\Delta G^{\circ }=-RT\ln K$. Setting these equal and rearranging gives the linear form of the van't Hoff equation: $$\ln K=-\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T}\right)+\frac{\Delta S^{\circ }}{R}$$ For calculating $K$ at a new temperature when we know $K$ at an initial temperature, we use the two-point form of the 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 formula confirms our earlier Le Chatelier reasoning for temperature changes: for endothermic reactions ($\Delta H^{\circ }>0$), increasing temperature increases $K$, shifting equilibrium right. For exothermic reactions ($\Delta H^{\circ }<0$), increasing temperature decreases $K$, shifting equilibrium left.

Worked Example

The solubility product constant $K_{sp}$ for AgCl is $1.8\times 10^{-10}$ at 298 K. Dissolution of AgCl is endothermic with $\Delta H^{\circ }=65.5\text{kJ/mol}$. Calculate $K_{sp}$ for AgCl at 320 K.

1. Assign values and match units: $K_1=1.8\times 10^{-10}$, $T_1=298\text{K}$, $T_2=320\text{K}$, $\Delta H^{\circ }=65500\text{J/mol}$, $R=8.314\text{J/(molcdotpK)}$.
2. Substitute into the two-point van't Hoff equation: $\ln \left(\frac{K_2}{1.8\times 10^{-10}}\right)=-\frac{65500}{8.314}\left(\frac{1}{320}-\frac{1}{298}\right)$.
3. Simplify the right-hand side: $-7878\times (-0.000231)\approx 1.82$.
4. Solve for $K_2$: $\frac{K_2}{1.8\times 10^{-10}}=e^{1.82}\approx 6.17$, so $K_2\approx 1.1\times 10^{-9}$.
5. Verify intuition: The reaction is endothermic, so increasing temperature increases $K$, which matches our result.

Exam tip: After calculating K at a new temperature, always cross-check against Le Chatelier's principle. If your result contradicts Le Chatelier, you have a sign error in the van't Hoff equation.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $R=8.314\text{kJ/(molcdotpK)}$ instead of $0.008314\text{kJ/(molcdotpK)}$ when calculating $K$ from $\Delta G^{\circ }$. Why: Most students memorize $R$ as 8.314 but forget $\Delta G$ is usually reported in kJ, leading to a 1000x error in $\ln K$ and an incorrect $K$ by many orders of magnitude. Correct move: Always write units for all values before plugging in, and convert $R$ to match the energy units of $\Delta G^{\circ }$.
• Wrong move: Concluding that a reaction with $\Delta G^{\circ }>0$ can never proceed forward spontaneously. Why: Students confuse $\Delta G^{\circ }$ (standard conditions) with $\Delta G$ (non-standard conditions). Correct move: Always check $Q$ relative to $K$: even if $\Delta G^{\circ }>0$, if $Q<<K$ (e.g., only reactants present initially), $\Delta G$ will be negative and the reaction proceeds forward spontaneously.
• Wrong move: Using $\Delta G^{\circ }=0$ to conclude $K=0$. Why: Students mix up logarithm rules: $\ln 1=0$, not $\ln 0=0$. Correct move: Remember that $0=-RT\ln K$ simplifies to $\ln K=0$, so $K=e^0=1$, meaning $K=1$ when $\Delta G^{\circ }=0$.
• Wrong move: Accepting a calculation that gives an increased $K$ for an exothermic reaction at higher temperature. Why: Sign errors in the van't Hoff equation are common, and students do not cross-check their result. Correct move: Always check your result against Le Chatelier: endothermic → $T$ up → $K$ up; exothermic → $T$ up → $K$ down. If it doesn't match, you have a sign error.
• Wrong move: Concluding that $K>1$ because you calculated a negative $\Delta G$ for non-standard conditions. Why: Students mix up what $\Delta G$ vs $\Delta G^{\circ }$ tells you about $K$. Correct move: Only $\Delta G^{\circ }$ determines the value of $K$. A negative $\Delta G$ only means the reaction is spontaneous forward under those specific non-standard conditions, which can happen even if $K<1$.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

For the reaction $2HI(g)\rightleftharpoons H_2(g)+I_2(g)$, $\Delta G^{\circ }=-13.0\text{kJ/mol}$ at 500 K. What is the value of $K$ at 500 K? A) $1.0\times 10^{-2}$ B) $2.4$ C) $22$ D) $1.0\times 10^2$

Worked Solution: We use the relationship $\ln K=-\Delta G^{\circ }/(RT)$. Convert units correctly: $\Delta G^{\circ }=-13.0$ kJ/mol, $R=0.008314$ kJ/(mol·K), $T=500$ K. Plugging in gives $\ln K=-(-13.0)/(0.008314\ast 500)=13.0/4.157\approx 3.13$. Exponentiating gives $K=e^{3.13}\approx 22$. Common wrong answers come from unit errors or sign errors. The correct answer is C.

Question 2 (Free Response)

Consider the weak acid dissociation of hydrocyanic acid: $HCN(aq)+H_{2}O(l)\rightleftharpoons H_3{O}^{+}(aq)+C{N}^{-}(aq)$. At 298 K, $K_a=4.9\times 10^{-10}$, and $\Delta H^{\circ }$ for this reaction is $+4.5$ kJ/mol. (a) Calculate $\Delta G^{\circ }$ for the dissociation reaction at 298 K. (b) Is the reaction spontaneous under standard conditions at 298 K? Justify your answer. (c) Calculate $K_a$ for HCN at 310 K. Predict whether the acid becomes stronger or weaker as temperature increases, and justify your prediction with Le Chatelier’s principle.

Worked Solution: (a) Use $\Delta G^{\circ }=-RT\ln K_a$. Substitute values: $R=0.008314$ kJ/(mol·K), $T=298$ K, $\ln (4.9\times 10^{-10})\approx -21.44$. $\Delta G^{\circ }=-(0.008314\ast 298)(-21.44)\approx +53.1$ kJ/mol. (b) The reaction is not spontaneous under standard conditions. $\Delta G^{\circ }$ is positive, which means $K_a<1$, so reactants are favored at equilibrium, and no net forward reaction occurs when all solutes are at 1 M concentration. (c) Use the two-point van't Hoff equation: $\ln \left(\frac{K_2}{K_1}\right)=-\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$. Substituting values gives $\ln (K_2/K_1)\approx 0.0704$, so $K_2\approx 1.073\ast 4.9\times 10^{-10}\approx 5.3\times 10^{-10}$. The acid becomes stronger as temperature increases: dissociation is endothermic ($\Delta H^{\circ }>0$), so increasing temperature shifts equilibrium right to absorb added heat, increasing $K_a$.

Question 3 (Application / Real-World Style)

The Haber process for ammonia synthesis follows $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$, with $K=1.4\times 10^{-5}$ at 500 K. A reactor is operated at 500 K with initial partial pressures: $P(N_2)=1.0$ atm, $P(H_2)=3.0$ atm, $P(N{H}_3)=0.5$ atm. Calculate $\Delta G$ for this mixture at 500 K, and predict if the reaction proceeds forward or reverse to reach equilibrium.

Worked Solution: First calculate the reaction quotient $Q$: $Q=\frac{(P_{N{H}_3}{)}^2}{P_{N_2}(P_{H_2}{)}^3}=\frac{(0.5{)}^2}{(1.0)(3.0{)}^3}\approx 9.26\times 10^{-3}$. Use the spontaneity shortcut $\Delta G=RT\ln (Q/K)$. Substitute values: $\Delta G=(0.008314\ast 500)\ast \ln \left(\frac{9.26\times 10^{-3}}{1.4\times 10^{-5}}\right)\approx 4.157\ast 6.49\approx +27.0$ kJ/mol. $\Delta G$ is positive, so the reaction proceeds spontaneously in the reverse direction. In context, ammonia will decompose into nitrogen and hydrogen under these operating conditions until $Q$ equals $K$.

7. Quick Reference Cheatsheet

8. What's Next

This topic unites thermodynamics and equilibrium, and is a prerequisite for all applied equilibrium topics in AP Chemistry, as well as for understanding coupled reactions and biological thermodynamics. Next, you will apply the relationship between free energy and equilibrium to solubility equilibria, where you will calculate ΔG° from Ksp and predict the solubility of ionic compounds at different temperatures. You will also use this relationship to calculate pKa and pH of weak acid solutions at non-standard temperatures, connecting this topic directly to core acid-base chemistry. Without mastering sign rules and unit consistency in this chapter, you will not be able to correctly predict equilibrium shifts or calculate equilibrium constants in these applied topics.

• Solubility equilibria
• Acid-base equilibria
• Thermodynamics of coupled reactions
• Entropy and Gibbs free energy