# Free Energy and Equilibrium

> AP Chemistry · Unit 9: Applications of Thermodynamics
> Source: https://www.owlsprep.com/study/ap-chemistry-u9-free-energy-and-equilibrium/

This module unites thermodynamics and equilibrium for AP Chemistry, covering the core relationship between standard Gibbs free energy change ($\Delta G^\circ$) and $K$, non-standard $\Delta G$ calculations, spontaneity prediction, and temperature dependence of $K$ via the van't Hoff equation.

**Prerequisites:** [Gibbs free energy definition and spontaneity rules](https://www.owlsprep.com/study/ap-chemistry-u9-gibbs-free-energy/); [Equilibrium constant and reaction quotient basics](https://www.owlsprep.com/study/ap-chemistry-u7-equilibrium-introduction/); [Enthalpy and entropy calculations](https://www.owlsprep.com/study/ap-chemistry-u9-enthalpy-entropy/)

## Learning objectives

- Relate standard Gibbs free energy change to the equilibrium constant $K$
- Calculate $\text{Δ}G$ for non-standard conditions and predict spontaneity direction
- Use the van't Hoff equation to find $K$ at new temperatures
- Avoid common unit and sign errors in equilibrium-thermodynamics calculations

## The Core Relationship Between $\Delta G^\circ$ and $K$

At equilibrium, the total free energy of the system is at its minimum, so there is no net driving force for forward or reverse reaction, meaning $\Delta G = 0$. We start with the general expression for free energy under any conditions to relate standard free energy change to the equilibrium constant.

$$\Delta G = \Delta G^\circ + RT \ln Q$$

At equilibrium, $\Delta G = 0$ and $Q = K$, so substituting gives the core relationship connecting thermodynamics and equilibrium:

$$\Delta G^\circ = -RT \ln K$$

**Thermodynamic Equilibrium Constant** — The unitless ratio of product to reactant activities at equilibrium, used directly in the $\Delta G^\circ = -RT \ln K$ relationship. For dilute solutions and ideal gases, it equals the $K_c$ or $K_p$ calculated from concentrations/partial pressures.

*Notation:* K

*Example:* For $aA + bB \rightleftharpoons cC + dD$, $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

- If $\Delta G^\circ < 0$, $\ln K > 0$ so $K > 1$: products are favored at equilibrium
- If $\Delta G^\circ > 0$, $\ln K < 0$ so $K < 1$: reactants are favored at equilibrium
- If $\Delta G^\circ = 0$, $\ln K = 0$ so $K = 1$: reactants and products are equally favored

**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: $\Delta G^\circ = -46.0\ \text{kJ/mol}$, $R = 0.008314\ \text{kJ/(mol·K)}$, $T = 298\ \text{K}$.
2. Rearrange the core formula to solve for $\ln K$:

   $$\ln K = \frac{-\Delta G^\circ}{RT}$$
3. Substitute values:

   $$\ln K = \frac{-(-46.0\ \text{kJ/mol})}{(0.008314\ \text{kJ/(mol·K)}) (298\ \text{K})} = \frac{46.0}{2.478} ≈ 18.56$$
4. Exponentiate both sides to solve for $K$:

   $$K = e^{18.56} ≈ 1.2 \times 10^8$$
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 $\Delta G^\circ$: use 0.008314 kJ/(mol·K) when $\Delta G^\circ$ is in kJ/mol, and 8.314 J/(mol·K) when $\Delta G^\circ$ is in J/mol to avoid 1000× errors in $K$.

## Non-Standard $\Delta G$ and Spontaneity Prediction

Most reactions do not occur under standard conditions (1 M concentration, 1 atm pressure, pure solids/liquids). To predict the direction of spontaneous change under non-standard conditions, we use the general free energy formula:

$$\Delta G = \Delta G^\circ + RT \ln Q$$

Where $Q$ is the reaction quotient calculated from current non-standard concentrations or partial pressures. The sign of $\Delta G$ directly indicates the direction of spontaneity:

- $\Delta G < 0$: forward reaction is spontaneous
- $\Delta G > 0$: reverse reaction is spontaneous
- $\Delta G = 0$: 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 $\ln\left(\frac{Q}{K}\right)$, so we can predict spontaneity directly from comparing $Q$ and $K$ without calculating $\Delta G$.

**Worked example:** For the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$, $K = 0.36$ at 298 K, and $\Delta G^\circ = 2.5\ \text{kJ/mol}$ at 298 K. A reaction mixture contains 0.10 atm $N_2O_4$ and 0.50 atm $NO_2$. Is the reaction spontaneous forward, reverse, or at equilibrium under these conditions?

1. Calculate $Q_p$ from the given partial pressures:

   $$Q_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(0.50)^2}{0.10} = 2.5$$
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:

   $$\Delta G = 2.5\ \text{kJ/mol} + (0.008314\ \text{kJ/(mol·K)})(298\ \text{K}) \ln(2.5) ≈ 2.5 + 2.27 = 4.77\ \text{kJ/mol}$$
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 $\Delta G^\circ$ and $\Delta G$: $\Delta G^\circ$ tells you about the equilibrium position (whether $K$ is greater than or less than 1), while $\Delta G$ tells you the direction of spontaneity under your specific non-standard conditions.

## Temperature Dependence of $K$ and the van't Hoff Equation

Equilibrium constants change with temperature. We can derive the 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:

$$\ln K = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T}\right) + \frac{\Delta S^\circ}{R}$$

To calculate $K$ at a new temperature when $K$ is known 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 relationship confirms Le Chatelier's principle for temperature changes:

- Endothermic reactions ($\Delta H^\circ > 0$): increasing temperature increases $K$, shifting equilibrium right
- 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: $K_1 = 1.8 \times 10^{-10}$, $T_1 = 298\ \text{K}$, $T_2 = 320\ \text{K}$, $\Delta H^\circ = 65500\ \text{J/mol}$, $R = 8.314\ \text{J/(mol·K)}$.
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: $-7878 \times (-0.000231) ≈ 1.82$.
4. Solve for $K_2$:

   $$\frac{K_2}{1.8 \times 10^{-10}} = e^{1.82} ≈ 6.17 \implies K_2 ≈ 1.1 \times 10^{-9}$$
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.

## Exam-Style Concept Check

**Check your understanding**

Test your understanding with these AP-style questions:

1. 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?

   - $1.0 \times 10^{-2}$
   - $2.4$
   - $22$
   - $1.0 \times 10^2$

   *Answer:* $22$

   *Why:* Correct! Unit matching gives $\ln K ≈ 3.13$, so $K = e^{3.13} ≈ 22$. Common errors come from incorrect $R$ units or sign mistakes.

2. Consider the weak acid dissociation of hydrocyanic acid: $HCN(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CN^-(aq)$. 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.

   *Why:* Model solution: (a) $\Delta G^\circ = -RT \ln K_a ≈ +53.1$ kJ/mol. (b) Not spontaneous: $\Delta G^\circ > 0$ so $K_a < 1$, and reactants are favored at 1 M standard conditions. (c) $K_a ≈ 5.3 \times 10^{-10}$; acid becomes stronger because endothermic dissociation means increasing T shifts equilibrium right, increasing $K_a$.

## Common pitfalls

- **Wrong:** Using $R = 8.314\ \text{kJ/(mol·K)}$ instead of $0.008314\ \text{kJ/(mol·K)}$ when calculating $K$ from $\Delta G^\circ$.
  - Why it fails: Most students memorize $R$ as 8.314 but forget $\Delta G$ is usually reported in kJ, leading to a 1000× error in $\ln K$ and an incorrect $K$ by many orders of magnitude.
  - Correct: Always write units for all values before plugging in, and convert $R$ to match the energy units of $\Delta G^\circ$.
- **Wrong:** Concluding that a reaction with $\Delta G^\circ > 0$ can never proceed forward spontaneously.
  - Why it fails: Students confuse $\Delta G^\circ$ (standard conditions) with $\Delta G$ (non-standard conditions).
  - Correct: 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:** Using $\Delta G^\circ = 0$ to conclude $K = 0$.
  - Why it fails: Students mix up logarithm rules: $\ln 1 = 0$, not $\ln 0 = 0$.
  - Correct: 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:** Accepting a calculation that gives an increased $K$ for an exothermic reaction at higher temperature.
  - Why it fails: Sign errors in the van't Hoff equation are common, and students do not cross-check their result.
  - Correct: 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:** Concluding that $K > 1$ because you calculated a negative $\Delta G$ for non-standard conditions.
  - Why it fails: Students mix up what $\Delta G$ vs $\Delta G^\circ$ tells you about $K$.
  - Correct: 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 $K < 1$.

## Cheatsheet

| Category | Formula | Key Notes |
| --- | --- | --- |
| ΔG° vs K | $\Delta G^\circ = -RT \ln K$ | Use R = 0.008314 kJ/(mol·K) for ΔG° in kJ/mol; ΔG° < 0 → K > 1 (products favored) |
| Non-standard ΔG | $\Delta G = \Delta G^\circ + RT \ln Q$ | Q = reaction quotient for non-standard conditions; ΔG < 0 → forward spontaneous |
| Q/K Spontaneity Shortcut | $\Delta G = RT \ln\left(\frac{Q}{K}\right)$ | Sign of ΔG matches ln(Q/K); Q < K → ΔG < 0 → forward spontaneous |
| van't Hoff Linear Form | $\ln K = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T}\right) + \frac{\Delta S^\circ}{R}$ | Slope of ln K vs 1/T = -ΔH°/R; use R = 8.314 J/(mol·K) |
| van't Hoff Two-Point Form | $\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ | ΔH° assumed constant; use ΔH° in J/mol to match R units |
| Combined ΔG° Relation | $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -RT \ln K$ | Links reaction thermodynamics to equilibrium constant |
| Temperature Dependence Rule |  | Endothermic (ΔH° > 0): T↑ → K↑; Exothermic (ΔH° < 0): T↑ → K↓; matches Le Chatelier |

## What's next

This topic unites two foundational concepts of AP Chemistry: thermodynamics (which predicts reaction spontaneity) and equilibrium (which describes a system's final composition). Mastery of sign rules, unit consistency, and core relationships here is required for all applied equilibrium topics, and this content regularly appears in multi-part AP Chemistry FRQs, often combined with acid-base or solubility topics. Next, you will apply the $\Delta G^\circ-K$ relationship to solubility equilibria, where you will calculate $\Delta G^\circ$ from $K_{sp}$ and predict how solubility changes with temperature. You will also extend this relationship to acid-base equilibria to calculate $pK_a$ and pH at non-standard temperatures.

- [Galvanic (voltaic) and electrolytic cells](https://www.owlsprep.com/study/ap-chemistry-u9-galvanic-and-electrolytic-cells/)
- [Cell Potential and Free Energy](https://www.owlsprep.com/study/ap-chemistry-u9-cell-potential-and-free-energy/)
- [Cell potential under nonstandard conditions](https://www.owlsprep.com/study/ap-chemistry-u9-cell-potential-under-nonstandard-conditions/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ap-chemistry-u9-free-energy-and-equilibrium/