#Thermodynamic Favorability: Is It Going to Happen? 🤔

Hey, future AP Chem master! Let's dive into thermodynamic favorability, which is all about predicting whether a reaction will occur spontaneously. Think of it like this: is the reaction naturally inclined to happen, or does it need a push? Let's find out!

This topic is crucial for understanding reaction spontaneity and equilibrium, which are major themes in AP Chemistry. Expect to see these concepts in both multiple-choice and free-response questions.

#Explaining Thermodynamic Favorability

Thermodynamic favorability helps us determine if a reaction is spontaneous (happens on its own) or nonspontaneous (needs external help). Spontaneous reactions are thermodynamically favorable, while non-spontaneous ones are thermodynamically unfavorable.

#Enthalpy (ΔH°) and Entropy (ΔS°): The Dynamic Duo

Two main factors determine spontaneity:

1. Enthalpy Change (ΔH°): This is the heat change in a reaction.

   • Exothermic (ΔH° < 0): System loses heat to surroundings. Think of it as 'heat exiting'. 🔥

   • Endothermic (ΔH° > 0): System gains heat from surroundings. Think of it as 'heat entering'. 🧊

*Caption: Visualizing exothermic and endothermic reactions. Exothermic reactions release heat, while endothermic reactions absorb it.*

2. Entropy Change (ΔS°): This is the change in disorder or randomness of a system. * Positive ΔS°: Disorder increases. More chaos! 🎉 * Negative ΔS°: Disorder decreases. More order. 😥

#Gibbs Free Energy (ΔG°): The Spontaneity Decider

Gibbs Free Energy (ΔG°) combines enthalpy and entropy to determine spontaneity. The formula is:

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

• ΔG° < 0: Reaction is spontaneous (exergonic). Energy is released. ✅
• ΔG° > 0: Reaction is nonspontaneous (endergonic). Energy is absorbed. ❌

#Practice Problem: Let's Calculate ΔG°

Reaction: 2H₂ + N₂ ⇌ N₂H₄

Given:

• ΔH° = 50.6 kJ/mol
• ΔS° = -0.332 kJ/(mol·K)
• T = 25°C (298 K)

Let's plug these values into the Gibbs Free Energy equation:

$$\Delta G^\circ = 50.6 , kJ/mol - (298 , K)(-0.332 , kJ/(mol \cdot K))$$

$$\Delta G^\circ = 50.6 + 98.936 = 149.5 , kJ/mol$$

Since ΔG° is positive, this reaction is nonspontaneous.

#Calculating ΔG° Using Standard Free Energies of Formation

Just like with ΔH° and ΔS°, we can calculate ΔG° using standard free energies of formation:

$$\Delta G^\circ_{rxn} = \sum n\Delta G^\circ_f (products) - \sum n\Delta G^\circ_f (reactants)$$

Caption: Standard free energies of formation are used to calculate the overall Gibbs free energy change for a reaction.

#Conditions for ΔG°: When is it Positive or Negative?

To be spontaneous, a reaction must either:

1. Be exothermic (ΔH° < 0)
2. Increase entropy (ΔS° > 0)

Let’s analyze the different scenarios:

• ΔH° < 0 and ΔS° > 0: ΔG° is always negative, reaction is always spontaneous. 🎉
• ΔH° > 0 and ΔS° < 0: ΔG° is always positive, reaction is never spontaneous. 😥
• ΔH° < 0 and ΔS° < 0: Spontaneous at low temperatures. 🔥
• ΔH° > 0 and ΔS° > 0: Spontaneous at high temperatures. 🧊

Caption: A handy chart summarizing the conditions for spontaneity based on enthalpy and entropy changes.

#Entropy-Driven vs. Enthalpy-Driven Reactions

• Entropy-Driven: Spontaneity is mainly due to an increase in disorder (ΔS°). Example: Dissolving NaCl (endothermic but spontaneous due to increased entropy). 🧂

• Enthalpy-Driven: Spontaneity is mainly due to a release of heat (ΔH°). Example: Combustion reactions (exothermic and usually spontaneous). 🔥

#Final Exam Focus

• High-Priority Topics: Gibbs Free Energy calculations, understanding the relationship between ΔG°, ΔH°, and ΔS°, and predicting spontaneity based on these values.
• Common Question Types: Calculating ΔG° using the equation or standard free energies of formation, determining spontaneity based on given ΔH° and ΔS° values, and identifying whether a reaction is entropy- or enthalpy-driven.
• Time Management: Practice calculations to build speed. Focus on understanding the concepts rather than just memorizing formulas. Be careful with unit conversions!
• Common Pitfalls: Forgetting to convert temperature to Kelvin, mixing up signs in the Gibbs Free Energy equation, and not understanding the conditions for spontaneity.

#Practice Questions

Practice Question

#Multiple Choice Questions

1. For a reaction to be spontaneous at all temperatures, which of the following conditions must be met? (A) ΔH > 0, ΔS > 0 (B) ΔH < 0, ΔS < 0 (C) ΔH < 0, ΔS > 0 (D) ΔH > 0, ΔS < 0

2. A reaction has a positive enthalpy change and a positive entropy change. Under what conditions will this reaction be spontaneous? (A) Always spontaneous (B) Never spontaneous (C) Spontaneous at high temperatures (D) Spontaneous at low temperatures

3. Which of the following processes is most likely to be entropy-driven? (A) Freezing of water (B) Combustion of methane (C) Dissolving of sugar in water (D) Condensation of steam

#Free Response Question

Consider the following reaction:

$$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$

Given the following thermodynamic data at 298 K:

Substance	ΔH°f (kJ/mol)	S° (J/mol·K)
N₂(g)	0	192
H₂(g)	0	131
NH₃(g)	-46	193

(a) Calculate the standard enthalpy change, ΔH° for the reaction. (b) Calculate the standard entropy change, ΔS° for the reaction. (c) Calculate the standard Gibbs free energy change, ΔG° for the reaction. (d) Is this reaction spontaneous under standard conditions? Explain. (e) Is this reaction enthalpy-driven or entropy-driven? Explain.

#Scoring Breakdown for FRQ

(a) Calculating ΔH° (2 points)

• 1 point for correct setup: ΔH° = [2(-46)] - [0 + 3(0)]
• 1 point for correct answer: ΔH° = -92 kJ/mol

(b) Calculating ΔS° (2 points)

• 1 point for correct setup: ΔS° = [2(193)] - [192 + 3(131)]
• 1 point for correct answer: ΔS° = -199 J/mol·K or -0.199 kJ/mol·K

(c) Calculating ΔG° (2 points)

• 1 point for correct setup: ΔG° = -92 - (298)(-0.199)
• 1 point for correct answer: ΔG° = -32.6 kJ/mol

(d) Spontaneity (1 point)

• 1 point for stating that the reaction is spontaneous because ΔG° is negative.

(e) Enthalpy-Driven or Entropy-Driven (1 point)

• 1 point for stating that the reaction is enthalpy-driven because both ΔH and ΔG are negative and ΔS is negative.

You've got this! Keep practicing, and you'll ace the AP Chemistry exam! 🎉