AP Chemistry: Spontaneity and Thermodynamic Favorability

AP Chemistry – Spontaneity and Thermodynamic Favorability

AP Chemistry Study Guide: Spontaneity and Thermodynamic Favorability

What This Is

Spontaneity determines whether a reaction will occur on its own without continuous external energy input. On the AP exam, you’ll predict if a reaction is spontaneous using Gibbs free energy (?G), which balances enthalpy (?H) and entropy (?S). A real-world example: Rusting iron is spontaneous (?G < 0) because it happens naturally over time, while decomposing water into H? and O? is not spontaneous (?G > 0) and requires electricity (like in electrolysis).

Key Terms & Concepts

• Spontaneous process: A reaction that occurs without continuous external energy input (e.g., ice melting at room temperature). Not necessarily fast (rusting is slow but spontaneous).
• Gibbs free energy (?G): The "energy available to do work" in a system. ?G = ?H – T?S, where:
• ?H = enthalpy change (kJ/mol)
• T = temperature (Kelvin)
• ?S = entropy change (J/mol·K)
• ?G < 0: Reaction is spontaneous (thermodynamically favorable).
• ?G > 0: Reaction is non-spontaneous (requires energy input).
• ?G = 0: Reaction is at equilibrium (forward and reverse rates are equal).
• Enthalpy (?H): Heat absorbed or released at constant pressure. ?H < 0 = exothermic (favorable), ?H > 0 = endothermic (unfavorable).
• Entropy (?S): Measure of disorder or randomness. ?S > 0 = more disorder (favorable), ?S < 0 = less disorder (unfavorable).
• Second Law of Thermodynamics: The total entropy of the universe always increases for spontaneous processes.
• Third Law of Thermodynamics: A perfect crystal at 0 K has zero entropy (S = 0).
• Standard free energy change (?G°): ?G at standard conditions (1 atm, 1 M, 25°C). ?G° = ?H° – T?S°.
• Coupled reactions: Pairing a non-spontaneous reaction (?G > 0) with a spontaneous one (?G < 0) to drive the overall process (e.g., ATP hydrolysis driving cellular reactions).

Step-by-Step: Predicting Spontaneity

1. Identify ?H and ?S signs:
2. Look at the reaction: Does it release heat (?H < 0)? Are products more disordered (?S > 0)?

3. Example: 2H?O(l)-2H?(g) + O?(g)

   • ?H > 0 (bonds broken, endothermic)
   • ?S > 0 (liquid-gases, more disorder)

4. Calculate ?G using ?G = ?H – T?S:

5. Convert units: ?S must be in kJ/mol·K (divide J by 1000).

6. Example: At 298 K, ?H = +572 kJ/mol, ?S = +0.327 kJ/mol·K

   • ?G = 572 – (298)(0.327) = +474 kJ/mol-non-spontaneous.

7. Determine spontaneity based on ?G:

8. ?G < 0: Spontaneous (favorable).
9. ?G > 0: Non-spontaneous (unfavorable).

10. ?G = 0: Equilibrium (e.g., phase changes at melting/boiling points).

11. Adjust for temperature (if needed):

12. If ?H and ?S have the same sign, spontaneity depends on T.

    • ?H < 0, ?S > 0: Always spontaneous (e.g., combustion).
    • ?H > 0, ?S < 0: Never spontaneous (e.g., water freezing above 0°C).
    • ?H > 0, ?S > 0: Spontaneous only at high T (e.g., ice melting).
    • ?H < 0, ?S < 0: Spontaneous only at low T (e.g., water freezing).

13. Check standard conditions vs. non-standard:

14. ?G° is for 1 M/1 atm. For non-standard conditions, use: ?G = ?G° + RT ln(Q), where Q = reaction quotient.

Common Mistakes

• Mistake: Assuming exothermic reactions (?H < 0) are always spontaneous. Correction: Spontaneity depends on both ?H and ?S. A reaction with ?H < 0 but ?S < 0 (e.g., water freezing) is only spontaneous below a certain temperature.

• Mistake: Forgetting to convert ?S from J to kJ. Correction: ?G = ?H – T?S requires consistent units. Convert ?S to kJ/mol·K (divide by 1000) before plugging in.

• Mistake: Confusing "spontaneous" with "fast." Correction: Spontaneity is about thermodynamics (will it happen?), not kinetics (how fast?). Diamond-graphite is spontaneous (?G < 0) but extremely slow.

• Mistake: Ignoring temperature’s role when ?H and ?S have opposite signs. Correction: If ?H and ?S have the same sign, spontaneity depends on T. Use T = ?H/?S to find the crossover temperature.

• Mistake: Misapplying ?G = ?G° + RT ln(Q). Correction: ?G° is for standard conditions (1 M/1 atm). For non-standard conditions, use Q (e.g., if [products] > [reactants], Q > 1 and ?G > ?G°).

AP Exam Insights

• Tricky distinction: ?G vs. ?G°
• ?G° = standard free energy (1 M, 1 atm, 25°C).

• ?G = free energy at any conditions (use ?G = ?G° + RT ln(Q)).

• FRQ favorite: Predicting spontaneity at different temperatures.

• Example: "At what temperature does the reaction become spontaneous?"-Set ?G = 0 and solve for T.

• Multiple-choice trap: Mixing up ?S signs.

• ?S > 0: More gas moles, phase changes (s-l-g), dissolving solids.

• ?S < 0: Fewer gas moles, precipitation, condensation.

• Lab connection: Calorimetry + ?G calculations.

• You might measure ?H in a lab, then use ?G = ?H – T?S to predict spontaneity.

Quick Check Questions

1. For the reaction 2H?(g) + O?(g)-2H?O(l), ?H° = –572 kJ/mol and ?S° = –0.327 kJ/mol·K. At what temperature (in K) does the reaction become non-spontaneous?
2. (A) 175 K
3. (B) 1,750 K
4. (C) 17,500 K

5. (D) The reaction is always spontaneous. Answer: (B) 1,750 K. Set ?G = 0-T = ?H/?S = (–572)/(–0.327)-1,750 K. Below this T, ?G < 0 (spontaneous); above, ?G > 0 (non-spontaneous).

6. Which of the following processes is not spontaneous at 25°C?

7. (A) NaCl(s)-Na?(aq) + Cl?(aq)
8. (B) H?O(l)-H?O(g)
9. (C) 2H?(g) + O?(g)-2H?O(l)

10. (D) CO?(s)-CO?(g) Answer: (B) H?O(l)-H?O(g). At 25°C, liquid water is more stable than vapor (?G > 0). The others are spontaneous (?G < 0).

11. Short FRQ: The reaction N?O?(g)-2NO?(g) has ?H° = +57.2 kJ/mol and ?S° = +0.176 kJ/mol·K.

12. (a) Calculate ?G° at 298 K.
13. (b) Is the reaction spontaneous at 298 K? Explain.
14. (c) At what temperature does the reaction become spontaneous? Answers:
15. (a) ?G° = ?H° – T?S° = 57.2 – (298)(0.176) = +5.0 kJ/mol.
16. (b) No, ?G° > 0-non-spontaneous at 298 K.
17. (c) Set ?G = 0-T = ?H/?S = 57.2/0.176-325 K.

Last-Minute Cram Sheet

1. ?G < 0 = spontaneous; ?G > 0 = non-spontaneous; ?G = 0 = equilibrium.
2. ?G = ?H – T?S (convert ?S to kJ/mol·K!).
3. ?H < 0, ?S > 0-always spontaneous.
4. ?H > 0, ?S < 0-never spontaneous.
5. ?H and ?S same sign-spontaneity depends on T.
6. T = ?H/?S-crossover temperature where ?G = 0.
7. ?G° = standard free energy (1 M, 1 atm, 25°C).
8. ?G = ?G° + RT ln(Q) for non-standard conditions.
9. Entropy increases with more gas moles, phase changes (s-l-g), and dissolving.
10. Spontaneous-fast (e.g., diamond-graphite is slow but ?G < 0).