Gibbs Free Energy and Spontaneity

Master Gibbs free energy, predict reaction spontaneity, and understand the relationship between ΔG, ΔH, ΔS, and temperature.

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Gibbs Free Energy and Spontaneity

Gibbs Free Energy (G)

Free energy: Energy available to do work

Gibbs equation:

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

Units:

ΔG: kJ/mol
ΔH: kJ/mol
T: K (always Kelvin!)
ΔS: J/(mol·K) → convert to kJ/(mol·K)

Key conversion: 1 kJ = 1000 J

Spontaneity Criterion

ΔG predicts spontaneity:

| ΔG | Process | |----|---------| | ΔG < 0 | Spontaneous (favorable) | | ΔG = 0 | At equilibrium | | ΔG > 0 | Non-spontaneous (unfavorable) |

Important: Spontaneous ≠ fast

Thermodynamics (ΔG) vs kinetics (rate)
Diamonds → graphite: spontaneous but infinitely slow

Temperature Dependence

Four cases from ΔG = ΔH - TΔS:

Case 1: ΔH < 0, ΔS > 0

ΔG < 0 at all temperatures
Spontaneous always
Example: Combustion reactions

Case 2: ΔH > 0, ΔS < 0

ΔG > 0 at all temperatures
Never spontaneous
Example: Reverse of combustion

Case 3: ΔH < 0, ΔS < 0

ΔG < 0 at low T
ΔG > 0 at high T
Spontaneous only when cold

Case 4: ΔH > 0, ΔS > 0

ΔG > 0 at low T
ΔG < 0 at high T
Spontaneous only when hot
Example: Melting, vaporization

Summary Table

| ΔH | ΔS | ΔG | Spontaneous? | |----|----|----|--------------| | - | + | Always - | Always | | + | - | Always + | Never | | - | - | - at low T, + at high T | Low T only | | + | + | + at low T, - at high T | High T only |

Calculating ΔG°_{rxn}

Method 1: From ΔH° and ΔS°

$\Delta G° = \Delta H° - T\Delta S°$

Watch units! Convert ΔS° from J/K to kJ/K

Method 2: From ΔG°_f values

$\Delta G°_{\text{rxn}} = \sum n\Delta G°_f(\text{products}) - \sum n\Delta G°_f(\text{reactants})$

Like ΔH°_f:

Elements in standard state: ΔG°_f = 0
Tabulated for compounds

Standard Free Energy of Formation

ΔG°_f: Free energy change to form 1 mole from elements (25°C, 1 atm)

Examples:

H₂(g): ΔG°_f = 0
H₂O(l): ΔG°_f = -237.1 kJ/mol
CO₂(g): ΔG°_f = -394.4 kJ/mol

Temperature and Equilibrium

At equilibrium: ΔG = 0

$0 = \Delta H - T_{\text{eq}}\Delta S$

$T_{\text{eq}} = \frac{\Delta H}{\Delta S}$

This gives transition temperature:

Below: one direction spontaneous
Above: reverse direction spontaneous
At T_eq: equilibrium (both directions equal)

Free Energy and Equilibrium Constant

Relationship:

$\Delta G° = -RT\ln K$

Or:

$\Delta G = \Delta G° + RT\ln Q$

Where:

R = 8.314 J/(mol·K)
K = equilibrium constant
Q = reaction quotient

Interpretation:

ΔG° < 0: K > 1 (products favored)
ΔG° = 0: K = 1 (equal amounts)
ΔG° > 0: K < 1 (reactants favored)

Coupling Reactions

Non-spontaneous reaction can be driven by spontaneous one:

If ΔG₁ > 0 (unfavorable):

Couple with reaction where ΔG₂ < 0
If |ΔG₂| > ΔG₁, overall ΔG < 0

Example: ATP in biology

ATP → ADP + Pi: ΔG° = -30.5 kJ/mol
Drives many unfavorable biological reactions

📚 Practice Problems

1Problem 1easy

❓ Question:

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = -92.2 kJ and ΔS° = -198.7 J/K. (a) Calculate ΔG° at 25°C. (b) Is the reaction spontaneous? (c) At what temperature does ΔG = 0?

💡 Show Solution

Given:

ΔH° = -92.2 kJ
ΔS° = -198.7 J/K = -0.1987 kJ/K
T = 25°C = 298 K

(a) Calculate ΔG° at 298 K

Use: ΔG° = ΔH° - TΔS°

Convert ΔS° to kJ/K: ΔS° = -198.7 J/K × (1 kJ/1000 J) = -0.1987 kJ/K

Calculate: $\Delta G° = -92.2 - (298)(-0.1987)$ $\Delta G° = -92.2 + 59.2$ $\Delta G° = -33.0 \text{ kJ}$

Answer (a): ΔG° = -33.0 kJ

(b) Is reaction spontaneous at 25°C?

ΔG° = -33.0 kJ < 0

Answer (b): Yes, spontaneous at 25°C (ΔG° < 0)

(c) Temperature where ΔG = 0

At equilibrium: ΔG = 0

$0 = \Delta H° - T_{\text{eq}}\Delta S°$

$T_{\text{eq}} = \frac{\Delta H°}{\Delta S°}$

Use consistent units (both kJ):

$T_{\text{eq}} = \frac{-92.2 \text{ kJ}}{-0.1987 \text{ kJ/K}}$

$T_{\text{eq}} = 464 \text{ K}$

Convert to °C: 464 - 273 = 191°C

Answer (c): T = 464 K or 191°C

Interpretation:

Signs: ΔH° < 0 (exothermic), ΔS° < 0 (less disorder)

Temperature effect:

Low T: ΔG < 0 (spontaneous) - ΔH term dominates
High T: ΔG > 0 (non-spontaneous) - TΔS term dominates
Below 464 K: spontaneous
Above 464 K: non-spontaneous

This explains why Haber process runs at moderate temperature!

2Problem 2medium

❓ Question:

Calculate ΔG°_{rxn} for: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) using ΔG°_f values (kJ/mol): C₃H₈(g) = -23.5, CO₂(g) = -394.4, H₂O(l) = -237.1, O₂(g) = 0

💡 Show Solution

Reaction: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

ΔG°_f values (kJ/mol):

C₃H₈(g): -23.5
O₂(g): 0 (element)
CO₂(g): -394.4
H₂O(l): -237.1

Formula:

$\Delta G°_{\text{rxn}} = \sum n\Delta G°_f(\text{products}) - \sum n\Delta G°_f(\text{reactants})$

Products:

3 mol CO₂: 3(-394.4) = -1183.2 kJ 4 mol H₂O: 4(-237.1) = -948.4 kJ

Sum: -1183.2 + (-948.4) = -2131.6 kJ

Reactants:

1 mol C₃H₈: 1(-23.5) = -23.5 kJ 5 mol O₂: 5(0) = 0 kJ

Sum: -23.5 kJ

Calculate ΔG°_{rxn}:

$\Delta G°_{\text{rxn}} = -2131.6 - (-23.5)$ $\Delta G°_{\text{rxn}} = -2131.6 + 23.5$ $\Delta G°_{\text{rxn}} = -2108.1 \text{ kJ}$

Answer: ΔG°_{rxn} = -2108 kJ

Interpretation:

Highly negative ΔG°:

Very spontaneous
This is combustion of propane
Large energy release
Why propane is excellent fuel

Compare to other fuels:

All combustions have large negative ΔG°
Spontaneous and exothermic
Release useful energy

3Problem 3hard

❓ Question:

For the vaporization of water at 100°C: H₂O(l) → H₂O(g), ΔH°{vap} = 40.7 kJ/mol. (a) Calculate ΔS°{vap}. (b) Calculate ΔG° at 90°C, 100°C, and 110°C. (c) Explain results.

💡 Show Solution

Given:

Process: H₂O(l) → H₂O(g)
ΔH°_{vap} = 40.7 kJ/mol
Normal boiling point: 100°C = 373 K

(a) Calculate ΔS°_{vap}

At boiling point: ΔG = 0 (equilibrium)

$\Delta G = \Delta H - T\Delta S = 0$

$\Delta S = \frac{\Delta H}{T}$

At T = 373 K:

$\Delta S°_{\text{vap}} = \frac{40.7 \text{ kJ/mol}}{373 \text{ K}}$

$\Delta S°_{\text{vap}} = 0.109 \text{ kJ/(mol·K)} = 109 \text{ J/(mol·K)}$

Answer (a): ΔS°_{vap} = 109 J/(mol·K)

(b) Calculate ΔG° at different temperatures

Use: ΔG° = ΔH° - TΔS°

ΔH° = 40.7 kJ/mol
ΔS° = 0.109 kJ/(mol·K)

At 90°C (363 K): $\Delta G° = 40.7 - (363)(0.109)$ $\Delta G° = 40.7 - 39.6 = +1.1 \text{ kJ/mol}$

At 100°C (373 K): $\Delta G° = 40.7 - (373)(0.109)$ $\Delta G° = 40.7 - 40.7 = 0 \text{ kJ/mol}$

At 110°C (383 K): $\Delta G° = 40.7 - (383)(0.109)$ $\Delta G° = 40.7 - 41.7 = -1.0 \text{ kJ/mol}$

Answer (b):

90°C: ΔG° = +1.1 kJ/mol
100°C: ΔG° = 0 kJ/mol
110°C: ΔG° = -1.0 kJ/mol

(c) Explain results

At 90°C (below boiling point):

ΔG° = +1.1 kJ (positive)
Non-spontaneous
Liquid is stable phase
Water won't boil at 1 atm

At 100°C (boiling point):

ΔG° = 0
Equilibrium
Liquid ⇌ gas
Both phases coexist at 1 atm
Normal boiling point!

At 110°C (above boiling point):

ΔG° = -1.0 kJ (negative)
Spontaneous
Gas is stable phase
Water vaporizes spontaneously
Can't maintain liquid at 1 atm

General principle:

Phase transitions:

ΔH > 0, ΔS > 0 (liquid → gas)
Low T: ΔG > 0 (liquid favored)
T = T_{boiling}: ΔG = 0 (equilibrium)
High T: ΔG < 0 (gas favored)

Boiling point is where ΔG = 0 for vaporization at 1 atm

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