Unit 9 · Topic 9.5

9.5 Free Energy and Equilibrium

ΔG° and the equilibrium constant are linked by ΔG° = −RT ln K. A negative ΔG° means K > 1 (products favored at equilibrium), a positive ΔG° means K < 1, and a ΔG° close to zero means K is close to 1.

Key terms

ΔG° = −RT ln K
equilibrium constant (K)
gas constant R = 8.314 J/(mol·K)
product-favored

The equation

ΔG° = −RT ln K, or equivalently K = e^(−ΔG°/RT). R is the gas constant, 8.314 J/(mol·K), and T is in kelvins. Because R is in joules, ΔG° must be in joules per mole here, not kilojoules.

For gas reactions, this K is written with partial pressures (like Kp); for reactions in solution, with molar concentrations (like Kc).

What the signs mean

ΔG°	ln K	K	At equilibrium
Negative	Positive	Greater than 1	Products favored
Zero	Zero	Equal to 1	Neither strongly favored
Positive	Negative	Less than 1	Reactants favored

Estimating without a calculator

At 298 K, RT is about 2.48 kJ/mol. When ΔG° is small compared with RT (within a few kJ/mol of zero), K is close to 1. When ΔG° is many times RT, K is far from 1.

Because of the exponential, K changes very quickly with ΔG°. At 298 K, ΔG° = −10 kJ/mol gives K ≈ 57, while ΔG° = +10 kJ/mol gives K ≈ 0.018. A ΔG° of −100 kJ/mol gives a K of about 10¹⁷, meaning the reaction essentially goes to completion.

A handy rule at 298 K: every 5.7 kJ/mol change in ΔG° changes K by a factor of 10, because RT ln 10 ≈ 5.7 kJ/mol. So ΔG° = −11.4 kJ/mol gives K ≈ 100, and ΔG° = +17.1 kJ/mol gives K ≈ 0.001.

Thermodynamically favored means K > 1

When ΔG° is negative, K is greater than 1, so an equilibrium mixture contains more products than reactants (in the K-expression sense). That's all 'thermodynamically favored' promises. It doesn't mean every reactant molecule is converted, and it says nothing about speed.

Combined with ΔG° = ΔH° − TΔS°, this equation shows why K changes with temperature: changing T changes ΔG°, and so changes K. For an endothermic reaction, raising T makes ΔG° less positive (or more negative), so K increases, which matches topic 7.10.

What ΔG° = 0 means

ΔG° = 0 doesn't mean nothing happens. It means K = 1, so at equilibrium neither side is strongly favored. If you start with only reactants, the reaction still runs forward until Q = K.

Also keep ΔG° and K tied to the same equation. If you reverse an equation, ΔG° changes sign and K becomes 1/K, and both changes are consistent with ΔG° = −RT ln K, because ln(1/K) = −ln K.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
K from ΔG°
For 2NO₂(g) ⇌ N₂O₄(g), ΔG° = −4.7 kJ/mol at 298 K. Calculate K.
Show the solutionHide the solution
1. Step 1: Convert to joules: ΔG° = −4700 J/mol.
2. Step 2: ln K = −ΔG° / RT = 4700 ÷ (8.314 × 298) = 4700 ÷ 2478 = 1.90.
3. Step 3: K = e^1.90 = 6.7.
4. Step 4: ΔG° is small and negative, so K is a bit greater than 1. Both NO₂ and N₂O₄ are present in significant amounts at equilibrium.
Answer: K ≈ 6.7
Example 2Calculator allowed
ΔG° from K
Acetic acid has Ka = 1.8 × 10⁻⁵ at 298 K. Find ΔG° for its ionization in water.
Show the solutionHide the solution
1. Step 1: ΔG° = −RT ln K = −(8.314 J/(mol·K))(298 K) ln(1.8 × 10⁻⁵).
2. Step 2: ln(1.8 × 10⁻⁵) = −10.93.
3. Step 3: ΔG° = −(2478)(−10.93) = +27,100 J/mol = +27.1 kJ/mol.
4. Step 4: Positive, consistent with Ka being much less than 1: acetic acid is a weak acid, and the un-ionized form is favored.
Answer: ΔG° ≈ +27.1 kJ/mol
Example 3Calculator allowed
Trap: joules versus kilojoules
A student calculates K for a reaction with ΔG° = −10.0 kJ/mol at 298 K by computing e^(10.0 / (8.314 × 298)) and gets K ≈ 1.004. What went wrong, and what is the correct K?
Show the solutionHide the solution
1. Step 1: R is in J/(mol·K), but ΔG° was left in kJ/mol, so the exponent came out 1000 times too small.
2. Step 2: Correct exponent: 10,000 ÷ (8.314 × 298) = 4.04.
3. Step 3: K = e^4.04 = 57.
4. Step 4: Sanity check: a ΔG° of −10 kJ/mol is about four times RT, so K should be clearly greater than 1, not almost exactly 1.
Answer: K ≈ 57 (the student forgot to convert kJ to J)

Common mistakes

Leaving ΔG° in kJ when R is in J.
Dropping the minus sign: ΔG° = −RT ln K.
Using log (base 10) instead of ln (natural log).
Saying K > 1 means the reaction is fast.

On the exam

Many questions ask only for the relationship: given the sign of ΔG°, is K greater than, less than or equal to 1? Answer with the sign of ln K.
When you do calculate, show the unit conversion and the rearranged equation.

Connected topics

Videos

Free energy and equilibrium | Applications of thermodynamics | AP Chemistry | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Kinetic Control - ΔG and the Equilibrium Constant - AP Chem Unit 9, Topics 4-5
Jeremy Krug (krugslist)Watch on YouTube (opens in a new tab)
Free Energy and the Equilibrium Constant
Bozeman ScienceWatch on YouTube (opens in a new tab)
18.5 Gibbs Free Energy and the Equilibrium Constant
Chad's PrepWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 9.5 Free Energy and Equilibrium. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A reaction has ΔG° = +17.1 kJ/mol at 298 K. What is the approximate value of K at 298 K? (R = 8.314 J/(mol·K))

Question 2 of 4Calculator allowed

At 298 K, reaction 1 has ΔG° = −10.0 kJ/mol and reaction 2 has ΔG° = −20.0 kJ/mol. Which of the following is closest to the ratio K₂/K₁? (R = 8.314 J/(mol·K))

Question 3 of 4Calculator allowed

Ka for acetic acid is 1.8 × 10⁻⁵ at 298 K. What is ΔG° for the acid's ionization at 298 K? (R = 8.314 J/(mol·K))

Question 4 of 4

A reaction A(aq) ⇌ B(aq) has ΔG° = +5.0 kJ/mol at 298 K. A solution of pure A is prepared at 298 K. Which of the following describes what happens?

0 of 4 answered