Unit 9 · Topic 9.5 Beta

Free Energy and Equilibrium

The standard free energy change and the equilibrium constant are linked by ΔG° = −RT ln K.

Practice 4: Model AnalysisPractice 5: Mathematical Routines

Part 1 · Hook

Why this matters

Making ammonia for fertilizer feeds about half the people on Earth. Engineers need to know how much ammonia an equilibrium mixture can hold before they build a plant. They do not have to run the reaction to find out: a single free energy value, ΔG°, tells them the equilibrium constant.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What does K > 1 tell you about an equilibrium?

Products are favored at equilibrium
Reactants are favored at equilibrium
The reaction is fast
The reaction has stopped

Show the answer

A large K means the equilibrium mixture is mostly products.

Correct: Products are favored at equilibrium:
Reactants are favored at equilibrium:
The reaction is fast:
The reaction has stopped:

2. What is ln(1)?

0
1
2.303
e

Show the answer

e⁰ = 1, so ln 1 = 0.

Correct: 0:
1:
2.303:
e:

3. Which value of R is used with energy in joules?

8.314 J/(mol·K)
0.08206 L·atm/(mol·K)
96,485 C/mol
6.022 × 10²³ /mol

Show the answer

The energy form of the gas constant is 8.314 J/(mol·K).

Correct: 8.314 J/(mol·K):
0.08206 L·atm/(mol·K):
96,485 C/mol:
6.022 × 10²³ /mol:

Part 4 · See it

See it first

Two matched scales. Negative ΔG° on the left lines up with K greater than 1, products favored; positive ΔG° on the right lines up with K less than 1, reactants favored; ΔG° of zero lines up with K equal to 1. At 298 K, −20 kJ/mol gives K about 3 × 10³ and +20 kJ/mol gives K about 3 × 10⁻⁴. — Negative ΔG° goes with K > 1 (products favored); positive ΔG° goes with K < 1. ΔG° = 0 means K = 1. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

ΔG° compares products and reactants under standard conditionsit measures how far the reaction goes before reaching equilibrium
The two are linked by ΔG° = −RT ln Ka negative ΔG° gives ln K > 0 and K > 1
The relation is exponentiala change of a few kJ/mol in ΔG° changes K by a large factor
At equilibrium Q = K and there is no net driveΔG = 0, while ΔG° keeps its fixed value

Part 6 · Key ideas

Key ideas

ΔG° = −RT ln K, with R = 8.314 J/(mol·K), T in kelvin, and ΔG° in J/mol.
ΔG° < 0 ⇔ K > 1 (products favored). ΔG° > 0 ⇔ K < 1 (reactants favored). ΔG° = 0 ⇔ K = 1.
To find K: K = e^(−ΔG°/RT). Use ln and e, not log and 10.
ΔG° near zero means K near 1: both reactants and products are present in large amounts.
Reverse a reaction: ΔG° changes sign and K inverts. Double it: ΔG° doubles and K is squared.

Part 7 · Misconception

A common mistake

The wrong idea: A reaction with a positive ΔG° makes no product at all.

What actually happens: A positive ΔG° means K < 1, not K = 0. Starting from pure reactants, Q < K, so some product forms until Q = K; reactants are simply favored at equilibrium.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Five reactions at 298 K

Standard free energy changes for five reactions, A to E, all at 298 K. R = 8.314 J/(mol·K).

ΔG° of five reactions at 298 K
Reaction	ΔG° (kJ/mol)
A	−20.0
B	−5.0
C	0.0
D	+5.0
E	+20.0

1. For which reaction is K closest to 1?

Show the answer

ΔG° = −RT ln K. When ΔG° = 0, ln K = 0 and K = 1.

A: ΔG° = −20.0 kJ/mol gives K of about 3 × 10³, far above 1.
Correct: C: Right: ΔG° = 0 gives ln K = 0, so K = 1.
E: ΔG° = +20.0 kJ/mol gives K of about 3 × 10⁻⁴, far below 1.
B: ΔG° = −5.0 kJ/mol gives K of about 7.5: close-ish, but not as close as reaction C.

2. Calculate K for reaction A at 298 K.

Type a number.

Show the answer

ln K = −ΔG°/RT = −(−20,000 J/mol)/[(8.314 J/(mol·K))(298 K)] = 8.072. K = e^8.072 = 3.2 × 10³ (two significant figures). K > 1, as expected for ΔG° < 0.

Answer: 3.2e+3

3. At equilibrium, which reaction has mostly reactants and very little product?

Show the answer

The more positive ΔG°, the smaller K, and the more the equilibrium mixture is reactants.

A: ΔG° < 0 gives K > 1: products are favored.
C: K = 1: neither side is strongly favored, so both are present in large amounts.
D: K ≈ 0.13: reactants are favored, but much less strongly than in reaction E.
Correct: E: Right: ΔG° = +20.0 kJ/mol gives K ≈ 3 × 10⁻⁴, so reactants are favored.

4. A student says reaction D, with ΔG° = +5.0 kJ/mol, "cannot happen at all." Which reply is best?

K ≈ 0.13, so from pure reactants some product forms until Q = K.
The student is right, because a reaction with ΔG° > 0 makes no product.
The reaction goes to completion, because ΔG° is so close to zero.
The reaction happens once a catalyst is added, which makes ΔG° negative.

Show the answer

ΔG° refers to standard conditions. A positive value means K < 1: reactants are favored at equilibrium, but some product is still present.

Correct: K ≈ 0.13, so from pure reactants some product forms until Q = K.: Right: a positive ΔG° means K < 1, not K = 0. An equilibrium mixture still contains product.
The student is right, because a reaction with ΔG° > 0 makes no product.: ΔG° > 0 means K < 1. Starting from reactants, Q = 0 < K, so the net reaction goes forward a little.
The reaction goes to completion, because ΔG° is so close to zero.: Close to zero means K is close to 1, not that the reaction goes to completion.
The reaction happens once a catalyst is added, which makes ΔG° negative.: A catalyst does not change ΔG° or K.

5. For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), ΔG° = −33.2 kJ/mol at 298 K. Calculate K. (R = 8.314 J/(mol·K))

Type a number.

Show the answer

ln K = −ΔG°/RT = 33,200 J/mol ÷ [(8.314 J/(mol·K))(298 K)] = 13.40. K = e^13.40 = 6.6 × 10⁵ (two significant figures). The large K matches the strongly negative ΔG°.

Answer: 6.6 × 10⁵

6. If ΔG° for a reaction is large and negative, what is true of K?

K is much greater than 1, so products are favored at equilibrium.
K is much less than 1, so reactants are favored at equilibrium.
K is equal to 1, so equal amounts of reactants and products are present.
K is negative, because ΔG° is negative.

Show the answer

K = e^(−ΔG°/RT). A large negative ΔG° gives a large positive exponent and K ≫ 1.

Correct: K is much greater than 1, so products are favored at equilibrium.: Right: ln K = −ΔG°/RT is large and positive.
K is much less than 1, so reactants are favored at equilibrium.: That is the case for a large positive ΔG°.
K is equal to 1, so equal amounts of reactants and products are present.: K = 1 goes with ΔG° = 0.
K is negative, because ΔG° is negative.: K is an exponential, e^(−ΔG°/RT), so it is positive whatever the sign of ΔG°.

7. A reaction mixture is at equilibrium. Which statement is correct?

ΔG° is zero, because nothing in the mixture is changing.
ΔG for the system is zero, while ΔG° keeps its fixed value set by K.
ΔG is negative, because the forward reaction is still running.
K equals zero, because the reaction has stopped.

Show the answer

ΔG measures the drive under the actual conditions; it is zero at equilibrium. ΔG° refers to standard conditions and is fixed for a given T.

ΔG° is zero, because nothing in the mixture is changing.: ΔG° is the standard value and equals −RT ln K; it is zero only when K = 1.
Correct: ΔG for the system is zero, while ΔG° keeps its fixed value set by K.: Right: at equilibrium Q = K and ΔG = 0; ΔG° = −RT ln K does not change.
ΔG is negative, because the forward reaction is still running.: Forward and reverse rates are equal, so there is no net drive in either direction: ΔG = 0.
K equals zero, because the reaction has stopped.: The reaction has not stopped (it is dynamic), and K is a fixed positive number.

Part 9 · Summary

Summary

The standard free energy change and the equilibrium constant are linked by ΔG° = −RT ln K. Negative ΔG° means K > 1 and products are favored; positive means K < 1. Because the link is exponential, small changes in ΔG° make big changes in K. Use R = 8.314 J/(mol·K), kelvin and ln.

Part 10 · Up next

What comes next

Part 11 · Connections