Free Energy and Equilibrium

AP Chemistry· AP Chemistry CED — Applications of Thermodynamics· 14 min read

1. The Core Relationship Between $\Delta G^\circ$ and $K$★★☆☆☆⏱ 5 min

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 $Δ G = 0$ . 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, $Δ G = 0$ and $Q = K$ , so substituting gives the core relationship connecting thermodynamics and equilibrium:

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

📘 Definition

Thermodynamic Equilibrium Constant

K

The unitless ratio of product to reactant activities at equilibrium, used directly in the $Δ G^{\circ} = - R T ln K$ relationship. For dilute solutions and ideal gases, it equals the $K_{c}$ or $K_{p}$ calculated from concentrations/partial pressures.

Example:

For $a A + b B ⇌ c C + d D$ , $K = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$

If $Δ G^{\circ} < 0$ , $ln K > 0$ so $K > 1$ : products are favored at equilibrium
If $Δ G^{\circ} > 0$ , $ln K < 0$ so $K < 1$ : reactants are favored at equilibrium
If $Δ G^{\circ} = 0$ , $ln K = 0$ 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 $Δ G^{\circ} = - 46.0 kJ/mol$ at 298 K. Calculate the thermodynamic equilibrium constant $K$ for this reaction at 298 K.

1
List all known values and match units: $Δ G^{\circ} = - 46.0 kJ/mol$ , $R = 0.008314 kJ/(mol \cdotp K)$ , $T = 298 K$ .
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: $Δ G^{\circ}$ is negative, so $K > 1$ , which matches our result.

Exam tip:

Always convert $R$ to match the energy units of $Δ G^{\circ}$ : use 0.008314 kJ/(mol·K) when $Δ G^{\circ}$ is in kJ/mol, and 8.314 J/(mol·K) when $Δ G^{\circ}$ is in J/mol to avoid 1000× errors in $K$ .

2. Non-Standard $\Delta G$ and Spontaneity Prediction★★★☆☆⏱ 6 min

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 $Δ G$ directly indicates the direction of spontaneity:

$Δ G < 0$ : forward reaction is spontaneous
$Δ G > 0$ : reverse reaction is spontaneous
$Δ G = 0$ : the reaction is at equilibrium

We can derive a useful shortcut by substituting $Δ G^{\circ} = - R T ln K$ into the non-standard $Δ G$ 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 $Δ G$ matches the sign of $ln (\frac{Q}{K})$ , so we can predict spontaneity directly from comparing $Q$ and $K$ without calculating $Δ G$ .

📐 Worked Example

For the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , $K = 0.36$ at 298 K, and $Δ G^{\circ} = 2.5 kJ/mol$ 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:
$Q_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(0.50)^2}{0.10} = 2.5$
2
Compare $Q$ to $K$ : $Q = 2.5 > K = 0.36$ , so $ln (Q / K)$ is positive, meaning $Δ G$ is positive.
3
Confirm with full $Δ G$ 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
$Δ G$ 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 $Δ G^{\circ}$ and $Δ G$ : $Δ G^{\circ}$ tells you about the equilibrium position (whether $K$ is greater than or less than 1), while $Δ G$ tells you the direction of spontaneity under your specific non-standard conditions.

3. Temperature Dependence of $K$ and the van't Hoff Equation★★★★☆⏱ 7 min

Equilibrium constants change with temperature. We can derive the relationship by combining two expressions for $Δ G^{\circ}$ : $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ and $Δ G^{\circ} = - R T ln K$ . 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 ( $Δ H^{\circ} > 0$ ): increasing temperature increases $K$ , shifting equilibrium right
Exothermic reactions ( $Δ H^{\circ} < 0$ ): increasing temperature decreases $K$ , shifting equilibrium left

📐 Worked Example

The solubility product constant $K_{s p}$ for AgCl is $1.8 \times 1 0^{- 10}$ at 298 K. Dissolution of AgCl is endothermic with $Δ H^{\circ} = 65.5 kJ/mol$ . Calculate $K_{s p}$ for AgCl at 320 K.

1
Assign values and match units: $K_{1} = 1.8 \times 1 0^{- 10}$ , $T_{1} = 298 K$ , $T_{2} = 320 K$ , $Δ H^{\circ} = 65500 J/mol$ , $R = 8.314 J/(mol \cdotp K)$ .
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: $- 7878 \times (- 0.000231) \approx 1.82$ .
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.

4. Exam-Style Concept Check★★★☆☆⏱ 4 min

✓ Quick check

Test your understanding with these AP-style questions:

For the reaction $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$ , $Δ G^{\circ} = - 13.0 kJ/mol$ at 500 K. What is the value of $K$ at 500 K?
- $1.0 \times 1 0^{- 2}$
- $2.4$
- $22$
- $1.0 \times 1 0^{2}$
Reveal answer
2 —
Correct! Unit matching gives $ln K \approx 3.13$ , so $K = e^{3.13} \approx 22$ . Common errors come from incorrect $R$ units or sign mistakes.
Consider the weak acid dissociation of hydrocyanic acid: $H C N (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + C N^{-} (a q)$ . At 298 K, $K_{a} = 4.9 \times 1 0^{- 10}$ , and $Δ H^{\circ}$ for this reaction is $+ 4.5$ kJ/mol. (a) Calculate $Δ G^{\circ}$ 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.

5. Common Pitfalls

Wrong move:

Using $R = 8.314 kJ/(mol \cdotp K)$ instead of $0.008314 kJ/(mol \cdotp K)$ when calculating $K$ from $Δ G^{\circ}$ .

Why:

Most students memorize $R$ as 8.314 but forget $Δ G$ is usually reported in kJ, leading to a 1000× 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 $Δ G^{\circ}$ .

Wrong move:

Concluding that a reaction with $Δ G^{\circ} > 0$ can never proceed forward spontaneously.

Why:

Students confuse $Δ G^{\circ}$ (standard conditions) with $Δ G$ (non-standard conditions).

Correct move:

Always check $Q$ relative to $K$ : even if $Δ G^{\circ} > 0$ , if $Q << K$ (e.g., only reactants present initially), $Δ G$ will be negative and the reaction proceeds forward spontaneously.

Wrong move:

Using $Δ G^{\circ} = 0$ to conclude $K = 0$ .

Why:

Students mix up logarithm rules: $ln 1 = 0$ , not $ln 0 = 0$ .

Correct move:

Remember that $0 = - R T ln K$ simplifies to $ln K = 0$ , so $K = e^{0} = 1$ , meaning $K = 1$ when $Δ G^{\circ} = 0$ .

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 $Δ G$ for non-standard conditions.

Why:

Students mix up what $Δ G$ vs $Δ G^{\circ}$ tells you about $K$ .

Correct move:

Only $Δ G^{\circ}$ determines the value of $K$ . A negative $Δ G$ only means the reaction is spontaneous forward under those specific non-standard conditions, which can happen even if $K < 1$ .

6. Quick Reference Cheatsheet

Category	Formula	Key Notes
ΔG° vs K	$Δ G^{\circ} = - R T ln K$	Use R = 0.008314 kJ/(mol·K) for ΔG° in kJ/mol; ΔG° < 0 → K > 1 (products favored)
Non-standard ΔG	$Δ G = Δ G^{\circ} + R T ln Q$	Q = reaction quotient for non-standard conditions; ΔG < 0 → forward spontaneous
Q/K Spontaneity Shortcut	$Δ G = R T ln (\frac{Q}{K})$	Sign of ΔG matches ln(Q/K); Q < K → ΔG < 0 → forward spontaneous
van't Hoff Linear Form	$ln K = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T}) + \frac{Δ S ^{\circ}}{R}$	Slope of ln K vs 1/T = -ΔH°/R; use R = 8.314 J/(mol·K)
van't Hoff Two-Point Form	$ln (\frac{K _{2}}{K _{1}}) = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$	ΔH° assumed constant; use ΔH° in J/mol to match R units
Combined ΔG° Relation	$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = - R T ln K$	Links reaction thermodynamics to equilibrium constant
Temperature Dependence Rule		Endothermic (ΔH° > 0): T↑ → K↑; Exothermic (ΔH° < 0): T↑ → K↓; matches Le Chatelier

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · FRQ
Thermodynamics-equilibrium connection
2022 · MCQ
van't Hoff equation calculation
2021 · FRQ
Non-standard spontaneity prediction

Going deeper

official syllabusAP Chemistry CED Unit 9Official learning objectives for this unit

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 $Δ G^{\circ} - K$ relationship to solubility equilibria, where you will calculate $Δ G^{\circ}$ from $K_{s p}$ and predict how solubility changes with temperature. You will also extend this relationship to acid-base equilibria to calculate $p K_{a}$ and pH at non-standard temperatures.