Thermodynamics – Study Notes
(For GNM Admission – Chemistry)
1. Simple Explanation
Thermodynamics is the branch of chemistry that deals with the relationships among energy, heat, work, and the properties of matter.
- Energy can be stored in a system as internal energy (U).
- Heat (q) is energy transferred by temperature difference.
- Work (w) is energy transferred by mechanical or other non‑thermal means.
- The state of a system (temperature, pressure, volume, composition) determines its internal energy and other properties.
The discipline is governed by the laws of thermodynamics which set the rules for how energy can be transformed or transferred.
2. Key Points
| Concept | What it means | Formula / Symbol |
|---|---|---|
| System & Surroundings | The part of the universe we study (system) vs. everything else (surroundings). | – |
| State Variables | Temperature (T), Pressure (P), Volume (V), Composition (n), Internal Energy (U). | – |
| Extensive vs. Intensive | Extensive depends on system size (U, V, n); Intensive does not (T, P). | – |
| First Law of Thermodynamics | Energy conservation in a closed system. | ΔU = q + w |
| Second Law of Thermodynamics | Entropy of an isolated system never decreases. | ΔS_total ≥ 0 |
| Entropy (S) | Measure of disorder or number of microstates. | dS = δq_rev / T |
| Heat Capacity (C) | Rate of change of energy with temperature. | C = dQ/dT |
| Enthalpy (H) | Heat content at constant pressure. | H = U + PV |
| Gibbs Free Energy (G) | Useful for predicting spontaneity at constant T, P. | G = H – TS |
| Helmholtz Free Energy (A or F) | Useful for constant T, V. | A = U – TS |
| Spontaneity Criterion | ΔG < 0 → spontaneous; ΔG > 0 → non‑spontaneous. | – |
| Phase Rule | F = C – P + 2 (degrees of freedom). | – |
| Chemical Reaction: | ΔG° = ΔH° – TΔS° | – |
3. Worked Example
Problem:
Calculate the Gibbs free energy change (ΔG°) for the reaction
[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) ]
at 298 K using the standard enthalpy and entropy changes:
- ΔH° = –92.4 kJ mol⁻¹
- ΔS° = –198 J mol⁻¹ K⁻¹
Solution Steps
-
Convert units
ΔS° must be in kJ mol⁻¹ K⁻¹ to match ΔH°.
[ ΔS° = -198\ \text{J mol}^{-1}\text{K}^{-1} = -0.198\ \text{kJ mol}^{-1}\text{K}^{-1} ] -
Apply Gibbs free‑energy equation
[ ΔG° = ΔH° - TΔS° ] Insert the numbers: [ ΔG° = (-92.4\ \text{kJ mol}^{-1}) - (298\ \text{K})(-0.198\ \text{kJ mol}^{-1}\text{K}^{-1}) ] -
Calculate
[ TΔS° = 298 \times (-0.198) = -58.9\ \text{kJ mol}^{-1} ] [ ΔG° = -92.4 - (-58.9) = -92.4 + 58.9 = -33.5\ \text{kJ mol}^{-1} ] -
Interpretation
ΔG° = –33.5 kJ mol⁻¹ < 0 → the reaction is spontaneous at 298 K under standard conditions.
Answer:
[
\boxed{ΔG° = -33.5\ \text{kJ mol}^{-1}\ \text{(spontaneous at 298 K)}}
]
4. Quick Revision Summary
| Item | Take‑away |
|---|---|
| Energy conservation | ΔU = q + w (first law) |
| Heat vs. work | Heat flows by temperature difference; work is any other energy transfer |
| Entropy increase | ΔS_total ≥ 0 (second law) |
| Spontaneity | ΔG = ΔH – TΔS; ΔG < 0 → spontaneous |
| Enthalpy | H = U + PV; convenient at constant pressure |
| Helmholtz Free Energy | A = U – TS; convenient at constant volume |
| State functions | Dependent only on initial/final states (U, H, G, S) |
| Extensive vs. Intensive | Size‑dependent vs. size‑independent variables |
| Phase rule | F = C – P + 2 (helps predict phase equilibria) |
| Key equations | ΔG = ΔH – TΔS; ΔS = δq_rev/T; ΔU = q + w |
Final Tip
When tackling thermodynamics problems, always:
- Identify what is being asked (enthalpy, free energy, entropy).
- Choose the correct equation.
- Convert units consistently (J ↔ kJ, K, mol).
- Check the sign of ΔG to determine spontaneity.
Good luck with your GNM Chemistry admission!