6.6.1 Calculation of The Entropy Change

This topic explains how to calculate the entropy change for a chemical reaction using standard entropy values of reactants and products.

Definition

Entropy Change (ΔS°): The difference between the standard entropies of the products and the standard entropies of the reactants.

Main Idea

The entropy change for a chemical reaction can be calculated by subtracting the sum of the standard entropies of the reactants from the sum of the standard entropies of the products. The formula is:

Δ S^° = Σ S^°_products - Σ S^°_reactants

Core Concept

• ΔS° > 0: Entropy increases (more disorder).
• ΔS° < 0: Entropy decreases (more order).
• **ΔS° = 0:** No change in entropy.
• More gaseous molecules in products: Entropy increases.
• Fewer gaseous molecules in products: Entropy decreases.

Detailed Explanation

Formula for Entropy Change

The entropy change for a chemical reaction can be calculated by the following equation:

Δ S^° = Σ S^°_products - Σ S^°_reactants

• Δ S^° = standard entropy change
• Σ S^°_products = sum of standard entropies of products
• Σ S^°_reactants = sum of standard entropies of reactants

Example Calculation

Suppose we want to calculate ΔS° for the following reaction:

A + 2B → C

• S^°(A) = 40 kJ/mol
• S^°(B) = 50 kJ/mol
• S^°(C) = 100 kJ/mol

Required: - Δ S^°

Formula: Δ S^° = Σ S^°_products - Σ S^°_reactants

Substitution: Δ S^° = [1(100)] - [1(40) + 2(50)]

Calculation: Δ S^° = 100 - [40 + 100] Δ S^° = 100 - 140 Δ S^° = -40 kJ/mol

Final Answer: Δ S^° = -40 kJ/mol

Since ΔS° is negative, the entropy of the system decreases.

Key Point

• As the number of gaseous molecules in a chemical reaction increases, the entropy also increases.
• Conversely, as the number of gaseous molecules decreases, the entropy decreases.

Concept Assessment Exercise 6.9

For which of the following reactions is ΔS° positive or negative?

(Exercise based on the textbook.)

Formula

Δ S^° = Σ S^°_products - Σ S^°_reactants

Example

• S^°(A) = 40 kJ/mol
• S^°(B) = 50 kJ/mol
• S^°(C) = 100 kJ/mol

Reaction: A + 2B → C

Solution: Δ S^° = [1(100)] - [1(40) + 2(50)] Δ S^° = 100 - 140 = -40 kJ/mol

Result: ΔS° = −40 kJ/mol (entropy decreases).

Key Point

• More gaseous molecules in products → ΔS° positive.
• Fewer gaseous molecules in products → ΔS° negative.

Concept Assessment Exercise 6.9

For which of the following reactions is ΔS° positive or negative?

(Exercise based on the textbook.)

Important Points

• 1. Formula: ΔS° = ΣS°(products) − ΣS°(reactants).
• 2. ΔS° > 0: Entropy increases.
• 3. ΔS° < 0: Entropy decreases.
• 4. Example: A + 2B → C, ΔS° = −40 kJ/mol.
• 5. More gaseous molecules → higher entropy.

Remember

• - Formula: ΔS° = ΣS°(products) − ΣS°(reactants)
• - ΔS° > 0: Entropy increases
• - ΔS° < 0: Entropy decreases
• - **Example:** A + 2B → C, ΔS° = −40 kJ/mol
• - More gaseous molecules: Higher entropy

Common Mistakes

• 1. Forgetting the formula: ΔS° = ΣS°(products) − ΣS°(reactants).
• 2. Mixing up reactants and products: Products − Reactants.
• 3. Ignoring the effect of gaseous molecules: More gases → higher entropy.
• 4. Forgetting the sign of ΔS°: Positive = increase; Negative = decrease.

Quick Revision

• Formula: ΔS° = ΣS°(products) − ΣS°(reactants).
• ΔS° > 0: Entropy increases.
• ΔS° < 0: Entropy decreases.
• Example: A + 2B → C, ΔS° = −40 kJ/mol.
• More gaseous molecules → higher entropy.

60-Second Revision

• Formula: ΔS° = ΣS°(products) − ΣS°(reactants)
• ΔS° > 0: Entropy increases
• ΔS° < 0: Entropy decreases
• **Example:** A + 2B → C, ΔS° = −40 kJ/mol
• More gaseous molecules: Higher entropy

QUALITY VALIDATION

• Topic: 6.6.1 Calculation of The Entropy Change
• Chapter: 6 (Energetics)

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• Examples checked: PASS
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• Unnecessary repetition: NO
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