Entropy and Gibbs Free Energy

This lesson covers entropy, entropy changes, Gibbs free energy and the feasibility of reactions. These concepts extend the study of energetics beyond enthalpy alone and are essential for understanding why some endothermic reactions occur spontaneously (AQA specification 3.1.4).

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What Is Entropy?

Key Definition: Entropy (S) is a measure of the disorder or randomness in a system. The greater the disorder, the higher the entropy.

Entropy is measured in J K⁻¹ mol⁻¹ (note: joules, not kilojoules).

Examples of Entropy Values

Substance	State	S° (J K⁻¹ mol⁻¹)
Diamond	s	2.4
Graphite	s	5.7
NaCl	s	72.1
Ice, H₂O	s	48.0
Water, H₂O	l	69.9
Steam, H₂O	g	188.7
CO₂	g	213.6
O₂	g	205.0
N₂	g	191.6
NH₃	g	192.3
H₂	g	130.6
Ar	g	154.7
C₂H₅OH	l	160.7
CH₄	g	186.2

Key Trends in Entropy

1. State of matter: S(solid) < S(liquid) << S(gas)

Solids have low entropy (particles in fixed, ordered positions). Liquids have higher entropy (particles can move). Gases have much higher entropy (particles widely spaced, moving randomly).

2. Complexity of molecule: More atoms → higher entropy (more ways to arrange atoms and more vibrational modes).

Compare: Ar (154.7) < CH₄ (186.2) < C₂H₅OH(g) (280.6)

3. Temperature: Entropy increases with temperature (particles have more kinetic energy).

4. Dissolving: Dissolving a solid in a solvent generally increases entropy (ions/molecules become dispersed).

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Entropy Changes (ΔS)

The standard entropy change for a reaction:

ΔS° = Σ S°(products) − Σ S°(reactants)

When Is ΔS Positive (Entropy Increases)?

• Solid → liquid → gas (change of state to a more disordered state)
• Fewer moles of gas → more moles of gas
• Dissolving a solid in water
• Decomposition reactions (one substance → several)

When Is ΔS Negative (Entropy Decreases)?

• Gas → liquid → solid (change of state to a more ordered state)
• More moles of gas → fewer moles of gas
• Precipitation (ions in solution → solid)

Worked Example 1: Calculating ΔS°

Question: Calculate the standard entropy change for:

CaCO₃(s) → CaO(s) + CO₂(g)

Given: S°(CaCO₃) = 92.9, S°(CaO) = 39.7, S°(CO₂) = 213.6 (all in J K⁻¹ mol⁻¹)

Answer:

ΔS° = Σ S°(products) − Σ S°(reactants)

ΔS° = [39.7 + 213.6] − [92.9]

ΔS° = 253.3 − 92.9 = +160.4 J K⁻¹ mol⁻¹

The positive ΔS makes sense: a solid decomposes to produce a gas, increasing disorder.

Worked Example 2

Question: Calculate the standard entropy change for:

N₂(g) + 3H₂(g) → 2NH₃(g)

Given: S°(N₂) = 191.6, S°(H₂) = 130.6, S°(NH₃) = 192.3 (all in J K⁻¹ mol⁻¹)

Answer:

ΔS° = [2 × 192.3] − [191.6 + 3 × 130.6]

ΔS° = 384.6 − [191.6 + 391.8]

ΔS° = 384.6 − 583.4 = −198.8 J K⁻¹ mol⁻¹

The negative ΔS makes sense: 4 moles of gas become 2 moles of gas — a decrease in disorder.

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Gibbs Free Energy (ΔG)

Enthalpy alone does not determine whether a reaction is feasible. Some endothermic reactions occur spontaneously (e.g., dissolving ammonium nitrate in water). We need to consider both enthalpy and entropy.

Key Equation:

ΔG = ΔH − TΔS

Where:

• ΔG = Gibbs free energy change (kJ mol⁻¹)
• ΔH = enthalpy change (kJ mol⁻¹)
• T = temperature (K) — must be in kelvin
• ΔS = entropy change (kJ K⁻¹ mol⁻¹) — convert from J by dividing by 1000

CRITICAL: ΔS is usually given in J K⁻¹ mol⁻¹ but ΔH is in kJ mol⁻¹. You MUST convert ΔS to kJ K⁻¹ mol⁻¹ (divide by 1000) before substituting into ΔG = ΔH − TΔS, or convert ΔH to J mol⁻¹.

Feasibility

• If ΔG < 0 (negative): the reaction is thermodynamically feasible (spontaneous)
• If ΔG > 0 (positive): the reaction is not thermodynamically feasible at that temperature
• If ΔG = 0: the system is at equilibrium

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Four Scenarios

ΔH	ΔS	ΔG = ΔH − TΔS	Feasibility
− (exothermic)	+ (entropy increases)	Always negative	Always feasible at all temperatures
− (exothermic)	− (entropy decreases)	Negative at low T, positive at high T	Feasible at low temperatures
+ (endothermic)	+ (entropy increases)	Positive at low T, negative at high T	Feasible at high temperatures
+ (endothermic)	− (entropy decreases)	Always positive	Never feasible (at any temperature)