What is Gibbs Free Energy and Spontaneity?

Master Gibbs free energy, predict reaction spontaneity, and understand the relationship between ΔG, ΔH, ΔS, and temperature.

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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

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Gibbs Free Energy and Spontaneity is part of the AP Chemistry course on Study Mondo, specifically in the Thermodynamics section. You can explore the full course for more related topics and practice resources.

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Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Gibbs Free Energy and Spontaneity

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ΔG	Process
ΔG < 0	Spontaneous (favorable)
ΔG = 0	At equilibrium
ΔG > 0	Non-spontaneous (unfavorable)

ΔH	ΔS	ΔG	Spontaneous?
-	+	Always -	Always
+	-	Always +	Never
-	-	- at low T, + at high T	Low T only
+	+	+ at low T, - at high T	High T only

Gibbs Free Energy and Spontaneity | Study Mondo

📚 Practice Problems

1Problem 1easy

❓ Question:

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = -92.2 kJ and ΔS° = -198.7 J/K. (a) Calculate ΔG° at 25°C. (b) Is the reaction spontaneous? (c) At what temperature does ΔG = 0?

💡 Show Solution

Given:

ΔH° = -92.2 kJ
ΔS° = -198.7 J/K = -0.1987 kJ/K
T = 25°C = 298 K

(a) Calculate ΔG° at 298 K

Use: ΔG° = ΔH° - TΔS°

Convert ΔS° to kJ/K: ΔS° = -198.7 J/K × (1 kJ/1000 J) = -0.1987 kJ/K

Calculate: $\Delta G° = -92.2 - (298)(-0.1987)$ $\Delta G° = -92.2 + 59.2$ $\Delta G° = -33.0 \text{ kJ}$

Answer (a): ΔG° = -33.0 kJ

(b) Is reaction spontaneous at 25°C?

ΔG° = -33.0 kJ < 0

Answer (b): Yes, spontaneous at 25°C (ΔG° < 0)

(c) Temperature where ΔG = 0

At equilibrium: ΔG = 0

$0 = \Delta H° - T_{\text{eq}}\Delta S°$

$T_{\text{eq}} = \frac{\Delta H°}{\Delta S°}$

Use consistent units (both kJ):

$T_{\text{eq}} = \frac{-92.2 \text{ kJ}}{-0.1987 \text{ kJ/K}}$

$T_{\text{eq}} = 464 \text{ K}$

Convert to °C: 464 - 273 = 191°C

Answer (c): T = 464 K or 191°C

Interpretation:

Signs: ΔH° < 0 (exothermic), ΔS° < 0 (less disorder)

Temperature effect:

Low T: ΔG < 0 (spontaneous) - ΔH term dominates
High T: ΔG > 0 (non-spontaneous) - TΔS term dominates
Below 464 K: spontaneous
Above 464 K: non-spontaneous

This explains why Haber process runs at moderate temperature!

2Problem 2medium

❓ Question:

For a reaction at 298 K, ΔH° = -92.3 kJ and ΔS° = -198 J/K. (a) Calculate ΔG° and determine if the reaction is spontaneous. (b) At what temperature does the reaction become non-spontaneous?

💡 Show Solution

Solution:

(a) Calculate ΔG°: ΔG° = ΔH° - TΔS°

Convert ΔS° to kJ/K: -198 J/K = -0.198 kJ/K

ΔG° = -92.3 kJ - (298 K)(-0.198 kJ/K) ΔG° = -92.3 kJ + 59.0 kJ ΔG° = -33.3 kJ

Since ΔG° < 0, the reaction is spontaneous at 298 K.

(b) Temperature where reaction becomes non-spontaneous: At equilibrium: ΔG° = 0 0 = ΔH° - TΔS° T = ΔH° / ΔS°

T = -92.3 kJ / (-0.198 kJ/K) = 466 K (or 193°C)

Above 466 K, ΔG° > 0 (non-spontaneous) Below 466 K, ΔG° < 0 (spontaneous)

Note: This reaction has ΔH° < 0 and ΔS° < 0, so it's spontaneous at low temperatures but non-spontaneous at high temperatures.

3Problem 3medium

❓ Question:

Calculate ΔG°_{rxn} for: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) using ΔG°_f values (kJ/mol): C₃H₈(g) = -23.5, CO₂(g) = -394.4, H₂O(l) = -237.1, O₂(g) = 0

💡 Show Solution

Reaction: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

ΔG°_f values (kJ/mol):

C₃H₈(g): -23.5
O₂(g): 0 (element)
CO₂(g): -394.4
H₂O(l): -237.1

Formula:

$Δ G °_{rxn} = \sum n Δ G °_{f} (products$

4Problem 4hard

❓ Question:

Consider the relationship between ΔG° and K (equilibrium constant): ΔG° = -RT ln K. For a reaction at 25°C with K = 5.0 × 10⁴, (a) calculate ΔG°, and (b) explain what this tells you about the position of equilibrium.

💡 Show Solution

Solution:

Given: K = 5.0 × 10⁴, T = 25°C = 298 K, R = 8.314 J/(mol·K)

(a) Calculate ΔG°: ΔG° = -RT ln K ΔG° = -(8.314 J/(mol·K))(298 K) ln(5.0 × 10⁴) ΔG° = -(2477.6) ln(50,000) ΔG° = -(2477.6)(10.82) ΔG° = -26,800 J/mol = -26.8 kJ/mol

(b) Interpretation:

ΔG° < 0 and K >> 1 means:

The reaction is product-favored (spontaneous in forward direction)
At equilibrium, products predominate over reactants
K = 50,000 means [products]/[reactants] ≈ 50,000

Relationship summary:

5Problem 5hard

❓ Question:

For the vaporization of water at 100°C: H₂O(l) → H₂O(g), ΔH°{vap} = 40.7 kJ/mol. (a) Calculate ΔS°{vap}. (b) Calculate ΔG° at 90°C, 100°C, and 110°C. (c) Explain results.

💡 Show Solution

Given:

Process: H₂O(l) → H₂O(g)
ΔH°_{vap} = 40.7 kJ/mol
Normal boiling point: 100°C = 373 K

(a) Calculate ΔS°_{vap}

At boiling point: ΔG = 0 (equilibrium)

$\Delta G = \Delta H - T\Delta S = 0$

Explain using:

📋 AP Chemistry — Exam Format Guide

⏱ 3 hours 15 minutes📝 67 questions📊 3 sections

Section	Format	Questions	Time	Weight	Calculator
Multiple Choice	MCQ	60	90 min	50%	✅
Free Response (Long)	FRQ	3	69 min	30%	✅
Free Response (Short)	FRQ	4	36 min	20%	✅

📊 Scoring: 1-5

Extremely Qualified

~12%

Well Qualified

~16%

Qualified

~24%

Possibly Qualified

~24%

No Recommendation

~24%

💡 Key Test-Day Tips

✓Memorize common polyatomic ions
✓Practice dimensional analysis
✓Know your gas laws

⚠️ Common Mistakes: Gibbs Free Energy and Spontaneity

Avoid these 3 frequent errors

🌍 Real-World Applications: Gibbs Free Energy and Spontaneity

See how this math is used in the real world

📝 Worked Example: Stoichiometry — Limiting Reagent

Problem:

$2$ mol of $H_2$ reacts with $1$ mol of $O_2$ . How many grams of water are produced? Which is the limiting reagent? ( $2H_2 + O_2 \to 2H_2O$ )

2Determine the limiting reagent

3Calculate moles of product

4Convert moles to grams

)

-

\sum

n

Δ

G

°_{f}

(

reactants

)

\Delta G°_{\text{rxn}} = \sum n\Delta G°_f(\text{products}) - \sum n\Delta G°_f(\text{reactants})

Gibbs Free Energy and Spontaneity

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Gibbs Free Energy and Spontaneity

Gibbs Free Energy (G)

Spontaneity Criterion

Temperature Dependence

Case 1: ΔH < 0, ΔS > 0

Case 2: ΔH > 0, ΔS < 0

Case 3: ΔH < 0, ΔS < 0

Case 4: ΔH > 0, ΔS > 0

Summary Table

Calculating ΔG°_{rxn}

Method 1: From ΔH° and ΔS°

Method 2: From ΔG°_f values

Standard Free Energy of Formation

Temperature and Equilibrium

Free Energy and Equilibrium Constant

Coupling Reactions

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2medium

❓ Question:

3Problem 3medium

❓ Question:

4Problem 4hard

❓ Question:

5Problem 5hard

❓ Question:

📋 AP Chemistry — Exam Format Guide

📊 Scoring: 1-5

💡 Key Test-Day Tips

⚠️ Common Mistakes: Gibbs Free Energy and Spontaneity

🌍 Real-World Applications: Gibbs Free Energy and Spontaneity

📝 Worked Example: Stoichiometry — Limiting Reagent

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