Thermodynamics – Study Notes (GUJCET Chemistry)
Goal – Understand the basic principles of thermodynamics that are frequently asked in GUJCET, and be able to calculate enthalpy, Gibbs free energy, and equilibrium constants.
1. Simple Explanation
Thermodynamics studies how energy changes form and moves in chemical processes.
Key ideas:
| Concept | What it means | Why it matters |
|---|---|---|
| System | The part of the universe you’re studying (e.g., a reaction vessel). | Keeps focus on the process. |
| Surroundings | Everything else outside the system. | Determines how energy is exchanged. |
| Energy | Kinetic + potential (heat, work, chemical). | Drives reactions. |
| Entropy (S) | Measure of disorder/randomness. | Tells if a process is spontaneous. |
| Enthalpy (H) | Heat content at constant pressure. | Useful for predicting heat changes. |
| Gibbs Free Energy (G) | G = H – T S. |
Determines spontaneity (ΔG < 0 ⇒ spontaneous). |
| Equilibrium Constant (K) | Ratio of product to reactant activities at equilibrium. | Connects to ΔG via ΔG° = –RT ln K. |
2. Key Points
2.1 Laws of Thermodynamics
| Law | Statement | Implication for Chemistry |
|---|---|---|
| 0th | If A↔B and B↔C are in equilibrium, then A↔C are in equilibrium. | Temperature is a state function. |
| 1st | ΔU = q + w (energy conservation). At constant pressure, ΔH = q_p. |
Heat evolved/absorbed is measured as ΔH. |
| 2nd | ΔS_universe ≥ 0 (entropy of isolated system never decreases). |
A reaction is spontaneous if ΔS_universe > 0. |
| 3rd | Entropy of a perfect crystal at 0 K = 0. | Provides a reference for absolute entropy values. |
2.2 Enthalpy (ΔH)
- Exothermic: ΔH < 0 (heat released).
- Endothermic: ΔH > 0 (heat absorbed).
- Hess’s Law: ΔH is path‑independent.
ΔH_total = Σ ΔH(steps)
2.3 Entropy (ΔS)
- Increase in randomness → ΔS > 0.
- Decrease in randomness → ΔS < 0.
- Typical order: solid < liquid < gas.
2.4 Gibbs Free Energy (ΔG)
ΔG = ΔH – TΔS(at constant T,P).- Spontaneity:
- ΔG < 0 → spontaneous.
- ΔG = 0 → equilibrium.
- ΔG > 0 → non‑spontaneous.
- Relation to equilibrium:
ΔG° = –RT ln K→K = exp(–ΔG°/RT)
2.5 Calorimetry
- Bomb calorimeter – measures ΔHₑ (enthalpy of reaction).
- Coffee‑cup calorimeter – measures ΔHₑ at constant pressure.
- Heat capacity:
q = C ΔT.
2.6 Other Useful Relations
| Symbol | Meaning | Formula |
|---|---|---|
q_p |
Heat at constant pressure | q_p = ΔH |
q_v |
Heat at constant volume | q_v = ΔU |
ΔG |
Gibbs free energy change | ΔG = ΔH – TΔS |
ΔG° |
Standard Gibbs free energy | ΔG° = –RT ln K° |
ΔS° |
Standard entropy change | ΔS° = (ΔH° – ΔG°)/T |
3. Worked Example
Problem
Calculate the Gibbs free energy change (ΔG°) for the reaction at 298 K:
C₂H₄(g) + H₂O(g) → C₂H₅OH(g)
Given standard enthalpies (ΔH°) and entropies (S°) (kJ mol⁻¹ and J mol⁻¹ K⁻¹):
| Species | ΔH° (kJ mol⁻¹) | S° (J mol⁻¹ K⁻¹) |
|---|---|---|
| C₂H₄ (g) | 52.3 | 186.3 |
| H₂O (g) | –241.8 | 188.7 |
| C₂H₅OH (g) | –277.5 | 257.5 |
Solution
- ΔH° reaction
[ \Delta H^\circ = \sum \Delta H^\circ_{\text{products}} - \sum \Delta H^\circ_{\text{reactants}} ]
[ \Delta H^\circ = (-277.5) - [52.3 + (-241.8)] = -277.5 - (52.3 - 241.8) ]
[ \Delta H^\circ = -277.5 - (-189.5) = -88.0 \text{ kJ mol}^{-1} ]
- ΔS° reaction
[ \Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}} ]
[ \Delta S^\circ = 257.5 - [186.3 + 188.7] = 257.5 - 375.0 = -117.5 \text{ J mol}^{-1} \text{K}^{-1} ]
Convert ΔS° to kJ:
ΔS° = –0.1175 kJ mol⁻¹ K⁻¹.
- ΔG° at 298 K
[ \Delta G