Unit 4: Thermodynamics

Introduction

Thermodynamics is the branch of chemistry that deals with energy changes accompanying physical and chemical processes. In Class 12, we focus on how energy is stored, transferred, and transformed in chemical reactions, and how these changes predict the direction and extent of processes. The chapter builds on the concepts of internal energy (U), enthalpy (H), entropy (S), and Gibbs free energy (G), linking them through the first and second laws of thermodynamics.

1. Internal Energy (U)

Internal energy is the total energy contained within a system due to the kinetic and potential energies of its molecules. It is a state function, meaning its value depends only on the initial and final states of the system, not on the path taken.

• Definition: U = Σ (kinetic energy of molecules) + Σ (potential energy of intermolecular forces)
• State function property: ΔU = U_final – U_initial (independent of pathway)
• Units: Joules (J) or kilojoules (kJ)

2. First Law of Thermodynamics

The first law is a statement of energy conservation: energy cannot be created or destroyed, only transferred as heat (q) or work (w).

ΔU = q + w

• Heat (q): Energy transferred due to temperature difference (positive when absorbed by the system).
• Work (w): Energy transferred when a force moves an object; for expansion/compression work at constant external pressure:

w = -PΔV where P = external pressure, ΔV = V_final – V_initial. The negative sign indicates that work done by the system (expansion) reduces its internal energy.

Special cases:

1. Constant volume (ΔV = 0): w = 0 → ΔU = q_v (heat at constant volume).
2. Constant pressure: The heat absorbed equals the change in enthalpy (see Section 3): q_p = ΔH.

3. Enthalpy (H) and Enthalpy Changes

Enthalpy combines internal energy with the PV term to simplify heat flow at constant pressure.

H = U + PV

For a process at constant pressure:

ΔH = ΔU + PΔV = q_p

Thus, measuring heat change at constant pressure directly gives ΔH.

Types of Enthalpy Changes

• Endothermic: ΔH > 0 (system absorbs heat).
• Exothermic: ΔH < 0 (system releases heat).
• Enthalpy of reaction (ΔH_rxn): Heat change when reactants convert to products at constant pressure.
• Enthalpy of solution (ΔH_sol): Heat change when 1 mol of solute dissolves in a solvent to form an infinitely dilute solution.
• Enthalpy of formation (ΔH_f°): Heat change when 1 mol of a compound is formed from its constituent elements in their standard states.
• Enthalpy of combustion (ΔH_c°): Heat change when 1 mol of a substance undergoes complete combustion in excess oxygen.

Standard Enthalpy Values (selected)

4. Laws of Thermochemistry

Lavoisier‑Laplace Law

The enthalpy change of a reaction is equal in magnitude but opposite in sign to the enthalpy change of the reverse reaction.

ΔH_forward = –ΔH_reverse

Hess’s Law of Constant Heat Summation

The total enthalpy change for a reaction is independent of the pathway; it depends only on the initial and final states.

Application: ΔH_rxn can be calculated from known enthalpies of formation:

ΔH_rxn = Σ ΔH_f°(products) – Σ ΔH_f°(reactants)

Example: Using Hess’s Law

Calculate ΔH for the reaction:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)

Using the table above:

ΣΔH_f°(products) = (–393.5) + 2×(–285.8) = –965.1 kJ

ΣΔH_f°(reactants) = (–74.8) + 2×(0) = –74.8 kJ

ΔH_rxn = –965.1 – (–74.8) = –890.3 kJ mol⁻¹ (matches ΔH_c° of methane).

5. Entropy (S) and Spontaneity

Entropy quantifies the disorder or randomness of a system. The second law states that for any spontaneous process, the total entropy of the universe (system + surroundings) increases.

• Definition: S = k_B ln W (where W = number of microstates, k_B = Boltzmann constant).
• Change in entropy: ΔS = S_final – S_initial.
• Sign convention: ΔS > 0 → increase in disorder (favors spontaneity); ΔS < 0 → decrease in disorder.

For a reversible isothermal process:

ΔS = q_rev / T

Where q_rev is the heat exchanged reversibly and T is absolute temperature (K).

Example: Entropy of Melting Ice

At 0 °C (273.15 K), the enthalpy of fusion of ice is ΔH_fus = +6.01 kJ mol⁻¹.

ΔS_fus = ΔH_fus / T = 6010 J mol⁻¹ / 273.15 K ≈ 22.0 J mol⁻¹ K⁻¹ (positive, indicating increased disorder).

6. Gibbs Free Energy (G)

Gibbs free energy combines enthalpy and entropy to predict spontaneity at constant temperature and pressure.

G = H – TS

For a process:

ΔG = ΔH – TΔS

• ΔG < 0: Process is spontaneous (exergonic).
• ΔG > 0: Process is non‑spontaneous (endergonic).
• ΔG = 0: System is at equilibrium.

Relation to Equilibrium Constant

At equilibrium, ΔG = 0 leads to:

ΔG° = –RT ln K

Where:

• ΔG° = standard Gibbs free energy change (J mol⁻¹)
• R = 8.314 J mol⁻¹ K⁻¹ (universal gas constant)
• T = temperature in kelvin
• K = equilibrium constant (dimensionless)

Re‑arranged:

K = e^{–ΔG°/(RT)}

Example: Calculating ΔG and Predicting Spontaneity

For the reaction N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) at 298 K, ΔH° = –92.4 kJ mol⁻¹ and ΔS° = –198.3 J mol⁻¹ K⁻¹.

Convert ΔS to kJ: ΔS° = –0.1983 kJ mol⁻¹ K⁻¹.

ΔG° = ΔH° – TΔS° = (–92.4) – 298×(–0.1983) = –92.4 + 59.1 = –33.3 kJ mol⁻¹.

Since ΔG° < 0, the formation of ammonia is spontaneous under standard conditions at 298 K.

Equilibrium constant:

K = e^{–ΔG°/(RT)} = e^{–(–33300 J)/(8.314×298)} = e^{13.44} ≈ 6.9×10⁵.

Numerical Problems (Integrated)

Problem 1: Internal Energy Change at Constant Volume

A gas absorbs 450 J of heat at constant volume. Calculate ΔU.

Solution: At constant volume, w = 0 → ΔU = q_v = +450 J.

Problem 2: Work Done During Expansion

One mole of an ideal gas expands isothermally and reversibly from 2.0 L to 5.0 L at 300 K against external pressure of 1.0 atm. Calculate the work done.

Solution: For reversible isothermal expansion, w = –nRT ln(V₂/V₁).
w = –(1 mol)(8.314 J mol⁻¹ K⁻¹)(300 K) ln(5.0/2.0)
w = –2494.2 J × ln(2.5) ≈ –2494.2 J × 0.9163 ≈ –2286 J (≈ –2.29 kJ). Negative sign indicates work done by the system.

Problem 3: Enthalpy of Reaction Using Hess’s Law

Calculate ΔH° for the reaction: C₂H₄(g) + H₂(g) → C₂H₆(g) given:

• ΔH_f°[C₂H₄(g)] = +52.3 kJ mol⁻¹
• ΔH_f°[H₂(g)] = 0
• ΔH_f°[C₂H₆(g)] = –84.0 kJ mol⁻¹

Solution: ΣΔH_f°(products) = –84.0 kJ
ΣΔH_f°(reactants) = (+52.3) + 0 = +52.3 kJ
ΔH° = –84.0 – 52.3 = –136.3 kJ mol⁻¹ (exothermic).

Problem 4: Entropy Change for Vaporization

The enthalpy of vaporization of water at 100 °C is ΔH_vap = +40.7 kJ mol⁻¹. Calculate ΔS_vap.

Solution: T = 373.15 K.
ΔS_vap = ΔH_vap / T = 40700 J mol⁻¹ / 373.15 K ≈ 109.1 J mol⁻¹ K⁻¹.

Problem 5: Gibbs Free Energy and Equilibrium Constant

For the dissociation of N₂O₄(g) ⇌ 2 NO₂(g), ΔG° = +5.4 kJ mol⁻¹ at 298 K. Calculate K.

Solution: ΔG° = 5400 J mol⁻¹
K = e^{–ΔG°/(RT)} = e^{–5400/(8.314×298)} = e^{–2.18} ≈ 0.113.
Since K < 1, the equilibrium lies toward reactants (N₂O₄) at this temperature.

Summary

This chapter has presented the core thermodynamic quantities—internal energy, enthalpy, entropy, and Gibbs free energy—along with the laws that govern their changes. Mastery of the formulas, sign conventions, and problem‑solving strategies (especially Hess’s law and the ΔG = –RT ln K relationship) is essential for predicting the direction and extent of chemical processes. Practice with the numerical examples provided will reinforce understanding and prepare students for examinations and real‑world applications.