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!

Solution:

(a) Calculate ΔG°: ΔG° = ΔH° - TΔS°

Convert ΔS° to kJ/K: -198 J/K = -0.198 kJ/K

ΔG° = -92.3 kJ - (298 K)(-0.198 kJ/K) ΔG° = -92.3 kJ + 59.0 kJ ΔG° = -33.3 kJ

Since ΔG° < 0, the reaction is spontaneous at 298 K.

(b) Temperature where reaction becomes non-spontaneous: At equilibrium: ΔG° = 0 0 = ΔH° - TΔS° T = ΔH° / ΔS°

T = -92.3 kJ / (-0.198 kJ/K) = 466 K (or 193°C)

Above 466 K, ΔG° > 0 (non-spontaneous) Below 466 K, ΔG° < 0 (spontaneous)

Note: This reaction has ΔH° < 0 and ΔS° < 0, so it's spontaneous at low temperatures but non-spontaneous at high temperatures.

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

Solution:

Given: K = 5.0 × 10⁴, T = 25°C = 298 K, R = 8.314 J/(mol·K)

(a) Calculate ΔG°: ΔG° = -RT ln K ΔG° = -(8.314 J/(mol·K))(298 K) ln(5.0 × 10⁴) ΔG° = -(2477.6) ln(50,000) ΔG° = -(2477.6)(10.82) ΔG° = -26,800 J/mol = -26.8 kJ/mol

(b) Interpretation:

ΔG° < 0 and K >> 1 means:

• The reaction is product-favored (spontaneous in forward direction)
• At equilibrium, products predominate over reactants
• K = 50,000 means [products]/[reactants] ≈ 50,000

Relationship summary:

• K > 1 → ΔG° < 0 (product-favored)
• K = 1 → ΔG° = 0 (equal products and reactants)
• K < 1 → ΔG° > 0 (reactant-favored)

The large value of K (5.0 × 10⁴) indicates the reaction goes nearly to completion under standard conditions.

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