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Study Guide: AP Chemistry: Entropy (ΔS) and Gibbs Free Energy (ΔG = ΔH – TΔS)
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AP Chemistry: Entropy (ΔS) and Gibbs Free Energy (ΔG = ΔH – TΔS)

By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.

⏱️ ~7 min read

AP Chemistry – Entropy (ΔS) and Gibbs Free Energy (ΔG = ΔH – TΔS)



What This Is

Entropy (ΔS) and Gibbs Free Energy (ΔG) are the "decision-makers" of chemical reactions—they tell you whether a reaction will happen spontaneously (on its own) or not. Think of entropy as a measure of disorder (like a messy room getting messier over time), while Gibbs Free Energy combines entropy and enthalpy (heat energy) to predict spontaneity. On the AP exam, you’ll use these concepts to explain why some reactions occur naturally (e.g., ice melting at room temperature) while others need a push (e.g., photosynthesis requiring sunlight). Real-world example: A campfire burns spontaneously because the increase in entropy (gases spreading out) and release of heat (exothermic) make ΔG negative—nature "prefers" this outcome.


Key Terms & Concepts

  • Entropy (ΔS): A measure of disorder or randomness in a system. Units: J/(mol·K). Higher entropy = more disorder (e.g., gas > liquid > solid).
  • Second Law of Thermodynamics: The total entropy of the universe always increases for spontaneous processes. "You can’t break even—disorder always wins."
  • Gibbs Free Energy (ΔG = ΔH – TΔS): Predicts spontaneity of a reaction.
  • ΔG < 0: Spontaneous (reaction happens on its own).
  • ΔG > 0: Non-spontaneous (needs energy input).
  • ΔG = 0: At equilibrium (no net change).
  • Variables: ΔH = enthalpy change (kJ/mol), T = temperature (K), ΔS = entropy change (J/mol·K).
  • Standard Entropy (S°): Entropy of a substance at 25°C and 1 atm. Found in tables (e.g., S° for H₂O(l) = 69.9 J/mol·K).
  • Standard Free Energy Change (ΔG°): ΔG at standard conditions (1 atm, 1 M, 25°C). Calculated using ΔG° = ΔH° – TΔS° or ΔG° = ΣΔG°(products) – ΣΔG°(reactants).
  • Spontaneity vs. Speed: Spontaneous ≠ fast! A reaction can be spontaneous but slow (e.g., diamond turning to graphite—ΔG < 0, but it takes millions of years).
  • Temperature Dependence: ΔG depends on T. A reaction can be non-spontaneous at low T but spontaneous at high T (e.g., ice melting above 0°C).
  • Coupled Reactions: Non-spontaneous reactions (ΔG > 0) can occur if paired with a spontaneous reaction (e.g., ATP hydrolysis drives cellular processes).
  • Third Law of Thermodynamics: The entropy of a perfect crystal at 0 K is zero. "You can’t reach absolute zero—disorder can’t be eliminated completely."
  • ΔS_universe = ΔS_system + ΔS_surroundings: For a spontaneous process, ΔS_universe > 0. If ΔS_system decreases (e.g., water freezing), ΔS_surroundings must increase more.
  • Phase Changes: Entropy increases with phase changes (solid → liquid → gas). ΔS_vap > ΔS_fus because gases are much more disordered than liquids.


Step-by-Step: Predicting Spontaneity Using ΔG

  1. Identify ΔH and ΔS for the reaction:
  2. Use given values or calculate from tables (ΔH°_f or S°).
  3. Tip: If ΔS is positive, the reaction becomes more spontaneous at higher T; if negative, less spontaneous at higher T.

  4. Convert units to match:

  5. ΔH is usually in kJ/mol; ΔS is in J/mol·K. Convert ΔS to kJ/mol·K (divide by 1000) or ΔH to J/mol.

  6. Plug into ΔG = ΔH – TΔS:

  7. Use T in Kelvin (K = °C + 273).
  8. Example: For H₂O(l) → H₂O(g) at 100°C (373 K):
    ΔH = +40.7 kJ/mol, ΔS = +118.9 J/mol·K = +0.1189 kJ/mol·K.
    ΔG = 40.7 – (373)(0.1189) = 40.7 – 44.3 = –3.6 kJ/mol → spontaneous (boiling occurs).

  9. Interpret the sign of ΔG:

  10. ΔG < 0: Spontaneous (reaction favors products).
  11. ΔG > 0: Non-spontaneous (reaction favors reactants).
  12. ΔG = 0: Equilibrium (e.g., phase changes at melting/boiling points).

  13. Check temperature dependence:

  14. If ΔH and ΔS have the same sign, spontaneity depends on T.
    • Both positive: Spontaneous at high T (e.g., melting, vaporization).
    • Both negative: Spontaneous at low T (e.g., freezing, condensation).
  15. If ΔH and ΔS have opposite signs, spontaneity is independent of T.

  16. For multi-step reactions:

  17. ΔG is additive (like Hess’s Law). If a non-spontaneous step is coupled with a spontaneous step, the overall ΔG can be negative.

Common Mistakes

  • Mistake: Confusing ΔS_system with ΔS_universe.
    Correction: A reaction can have ΔS_system < 0 (e.g., water freezing) but still be spontaneous if ΔS_surroundings > 0 (e.g., heat released warms the surroundings, increasing their entropy).

  • Mistake: Forgetting to convert units (kJ vs. J).
    Correction: ΔH is in kJ/mol; ΔS is in J/mol·K. Convert ΔS to kJ/mol·K (divide by 1000) before plugging into ΔG = ΔH – TΔS.

  • Mistake: Assuming ΔG = 0 means no reaction occurs.
    Correction: ΔG = 0 means the system is at equilibrium (forward and reverse reactions occur at equal rates). Example: Ice and water coexist at 0°C.

  • Mistake: Ignoring temperature’s role in spontaneity.
    Correction: A reaction can be non-spontaneous at low T but spontaneous at high T (or vice versa). Always check the sign of ΔH and ΔS to predict T dependence.

  • Mistake: Thinking "spontaneous" means "fast." Correction: Spontaneity is about thermodynamics (energy), not kinetics (speed). A reaction can be spontaneous but slow (e.g., rusting of iron).


AP Exam Insights

  1. Tricky Distinction: ΔG vs. ΔG°
  2. ΔG°: Standard free energy change (1 atm, 1 M, 25°C). Used for comparing reactions.
  3. ΔG: Free energy change at non-standard conditions. Calculated using ΔG = ΔG° + RT ln(Q), where Q is the reaction quotient.
  4. AP Trap: A reaction with ΔG° > 0 can still have ΔG < 0 if Q is very small (e.g., products are removed, shifting equilibrium right).

  5. Frequent FRQ Type: "Explain Spontaneity"

  6. You’ll be given ΔH and ΔS (or asked to calculate them) and must:


    1. Predict spontaneity at different temperatures.
    2. Explain using ΔG = ΔH – TΔS.
    3. Relate to real-world examples (e.g., why ice melts above 0°C).
  7. Multiple-Choice Traps:

  8. Trap: "A reaction with ΔH < 0 and ΔS < 0 is always spontaneous." False! It’s only spontaneous at low T.
  9. Trap: "Entropy always increases in a spontaneous process." False! ΔS_system can decrease if ΔS_surroundings increases more (e.g., water freezing).

  10. Coupled Reactions:

  11. AP loves testing this! Example: ATP hydrolysis (ΔG < 0) drives non-spontaneous reactions in cells (e.g., protein synthesis). Look for questions where a non-spontaneous reaction is "powered" by a spontaneous one.

Quick Check Questions

  1. Multiple Choice:
    For the reaction 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = –572 kJ/mol and ΔS = –327 J/mol·K. At what temperature (in K) will the reaction be at equilibrium?
    (A) 175 K
    (B) 1,750 K
    (C) 17,500 K
    (D) The reaction is never at equilibrium.

Answer: (B) 1,750 K.
Explanation: At equilibrium, ΔG = 0 = ΔH – TΔS. Solve for T: T = ΔH/ΔS = (–572,000 J/mol) / (–327 J/mol·K) ≈ 1,750 K.


  1. Short FRQ:
    The dissolution of ammonium nitrate in water is endothermic (ΔH > 0) and increases entropy (ΔS > 0).
    a) Predict the sign of ΔG for this process at 25°C. Justify your answer.
    b) How would increasing the temperature affect the spontaneity of this process? Explain.

Answer:
a) ΔG is negative (spontaneous) at 25°C. Although ΔH > 0, the large positive ΔS makes –TΔS dominate, so ΔG = ΔH – TΔS < 0.
b) Increasing temperature makes the process more spontaneous because ΔS > 0, so –TΔS becomes more negative, decreasing ΔG further.


  1. Multiple Choice:
    Which of the following processes has a negative ΔS?
    (A) H₂O(l) → H₂O(g)
    (B) 2H₂(g) + O₂(g) → 2H₂O(g)
    (C) NaCl(s) → Na⁺(aq) + Cl⁻(aq)
    (D) 2NH₃(g) → N₂(g) + 3H₂(g)

Answer: (B) 2H₂(g) + O₂(g) → 2H₂O(g).
Explanation: 3 moles of gas → 2 moles of gas, so disorder decreases (ΔS < 0). The other options increase disorder (ΔS > 0).


Last-Minute Cram Sheet

  1. ΔG = ΔH – TΔS → Spontaneity equation. ΔG < 0 = spontaneous.
  2. ΔS > 0 = more disorder (gas > liquid > solid).
  3. ΔH < 0, ΔS > 0 → Always spontaneous (e.g., combustion).
  4. ΔH > 0, ΔS < 0 → Never spontaneous (e.g., water freezing above 0°C).
  5. ΔH and ΔS same sign → Spontaneity depends on T.
  6. ΔG° = ΣΔG°(products) – ΣΔG°(reactants) → Standard free energy change.
  7. ΔG = ΔG° + RT ln(Q) → Non-standard conditions. ⚠️ R = 8.314 J/mol·K.
  8. Coupled reactions: Non-spontaneous reactions can occur if paired with a spontaneous one (e.g., ATP hydrolysis).
  9. Phase changes: ΔS_vap > ΔS_fus (gases are much more disordered).
  10. ⚠️ Units matter! Convert ΔS to kJ/mol·K or ΔH to J/mol before plugging into ΔG = ΔH – TΔS.


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