Thermochemistry and Thermodynamics

Core Review

Thermochemistry: Enthalpy (ΔH)

Exothermic: Heat is released to surroundings; ΔH < 0; temperature of surroundings increases. Combustion, neutralization, condensation. Endothermic: Heat is absorbed from surroundings; ΔH > 0; temperature of surroundings decreases. Melting, vaporization, photosynthesis.

Standard enthalpy of formation (ΔH°_f): Enthalpy change when 1 mole of a compound forms from its elements in their standard states. ΔH°_f of any element in its standard state = 0.

Standard enthalpy of reaction:

ΔH°_rxn = Σ n·ΔH°_f(products) − Σ n·ΔH°_f(reactants)

Bond enthalpy approach: ΔH ≈ Σ (bond energies broken) − Σ (bond energies formed). Bonds broken require energy (+); bonds formed release energy (−). Less accurate but conceptually useful.

Calorimetry

q = m × c × ΔT

q = heat (J or cal), m = mass (g), c = specific heat capacity (J/g·°C), ΔT = T_final − T_initial.

For water: c = 4.184 J/g·°C (often used as 4.18 or 1 cal/g·°C).

Worked Example: A 50.0 g piece of metal at 100.0°C is placed in 100.0 g of water at 20.0°C. The final temperature is 24.5°C. What is the specific heat of the metal?

• q_metal = −q_water (heat lost = heat gained)
• m_metal × c_metal × ΔT_metal = −(m_water × c_water × ΔT_water)
• 50.0 × c_metal × (24.5 − 100.0) = −(100.0 × 4.184 × (24.5 − 20.0))
• 50.0 × c_metal × (−75.5) = −(100.0 × 4.184 × 4.5)
• −3775 × c_metal = −1883 → c_metal = 0.499 J/g·°C

Hess's Law

The enthalpy change for a reaction is independent of the pathway — it depends only on initial and final states. You can:

1. Reverse a reaction → flip the sign of ΔH.
2. Multiply a reaction by a factor → multiply ΔH by the same factor.
3. Add reactions together → add ΔH values.

Worked Example: Find ΔH for C(s) + ½O₂(g) → CO(g) given: (1) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ (2) CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ

Reverse (2): CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ. Add to (1): C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g). Cancel CO₂ and ½O₂: C(s) + ½O₂(g) → CO(g), ΔH = −393.5 + 283.0 = −110.5 kJ.

Thermodynamics: Entropy (ΔS)

Entropy (S) measures molecular disorder or the number of accessible microstates. Units: J/mol·K.

ΔS°_rxn = Σ n·S°(products) − Σ n·S°(reactants)

• ΔS > 0: increasing disorder (solid → liquid → gas; fewer moles → more moles of gas).
• ΔS < 0: decreasing disorder (gas → liquid → solid; more moles → fewer moles of gas).
• S° is always positive for any substance at T > 0 K (Third Law).

Gibbs Free Energy (ΔG)

ΔG = ΔH − TΔS   (T in Kelvin)

• ΔG < 0: spontaneous in the forward direction.
• ΔG = 0: at equilibrium.
• ΔG > 0: non-spontaneous in the forward direction (spontaneous in reverse).

Temperature dependence of spontaneity:

The crossover temperature where ΔG = 0: T = ΔH / ΔS.

Standard Free Energy and Equilibrium

ΔG° = −RT ln K

R = 8.314 J/mol·K (energy units here!). T in Kelvin.

• K > 1 → ln K > 0 → ΔG° < 0 (products favored, spontaneous under standard conditions).
• K < 1 → ln K < 0 → ΔG° > 0 (reactants favored).
• K = 1 → ΔG° = 0.

This connects thermodynamics (ΔG°) to equilibrium position (K).

Non-standard conditions: ΔG = ΔG° + RT ln Q. At equilibrium, Q = K, ΔG = 0, so ΔG° = −RT ln K.

Thermodynamics vs. Kinetics

ΔG determines whether a reaction CAN occur (spontaneity); activation energy (E_a) determines how FAST it occurs (rate). Diamond → graphite has ΔG < 0 but is kinetically stable — it doesn't happen at room temperature because E_a is enormous. The DAT will test this distinction explicitly.

Key Equations

Common Traps

• Using ΔH sign alone to predict spontaneity — must also consider TΔS.
• Forgetting to convert ΔS (J/mol·K) and ΔH (kJ/mol) to the same units in ΔG = ΔH − TΔS.
• Using R = 0.08206 in ΔG° = −RT ln K (use R = 8.314 J/mol·K).
• Thinking catalysts affect ΔG or K (they affect rate only, lowering E_a for forward and reverse equally).
• Confusing ΔG° (standard conditions, all 1 M / 1 atm) with ΔG (actual conditions).