Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Quick Summary
Thermodynamics is one of the highest-weightage chapters for NEET (3–4 questions per year). Key areas: laws of thermodynamics, enthalpy calculations, Hess's law applications, entropy concept, and Gibbs energy criteria for spontaneity. Board exams test 5-mark derivations of ΔG and Hess's law.
Key Concepts and Definitions
- System: Part of universe under study
- Surroundings: Everything outside the system
- Open system: Exchanges both matter and energy with surroundings
- Closed system: Exchanges only energy
- Isolated system: Exchanges neither matter nor energy
- State functions: Properties depending only on current state (U, H, S, G) — not path
- Path functions: Depend on the path (q and w)
- Exothermic: ΔH < 0 (heat released)
- Endothermic: ΔH > 0 (heat absorbed)
- Entropy (S): Measure of disorder; always increases for isolated systems (2nd law)
Formulae
| Concept | Formula |
|---|---|
| First law | ΔU = q + w |
| Work at constant P | w = −PΔV |
| Enthalpy | H = U + PV |
| ΔH at constant P | ΔH = qₚ |
| ΔH vs ΔU | ΔH = ΔU + ΔngRT |
| Gibbs free energy | ΔG = ΔH − TΔS |
| Spontaneity | ΔG < 0 (spontaneous) |
| Entropy change | ΔS = qᵣₑᵥ / T |
| Bond enthalpy | ΔH = Σ(bonds broken) − Σ(bonds formed) |
Types of Enthalpy
| Type | Definition |
|---|---|
| ΔH_f° (formation) | Enthalpy when 1 mol formed from elements in standard state |
| ΔH_c° (combustion) | Enthalpy when 1 mol burns completely in O₂ |
| ΔH_sol (solution) | Enthalpy when 1 mol dissolves in large amount of solvent |
| ΔH_neutralisation | Enthalpy when 1 mol H⁺ reacts with 1 mol OH⁻ (−57.3 kJ/mol) |
| ΔH_atomisation | Enthalpy to convert 1 mol substance to gaseous atoms |
Spontaneity and Gibbs Energy
| ΔH | ΔS | ΔG | Spontaneity |
|---|---|---|---|
| − | + | Always − | Always spontaneous |
| + | − | Always + | Never spontaneous |
| − | − | −/+ | Spontaneous at low T |
| + | + | −/+ | Spontaneous at high T |
At equilibrium: ΔG = 0 Relationship with equilibrium constant: ΔG° = −RT ln K
Hess's Law Application
Calculate ΔH for C(s) + ½O₂(g) → CO(g) using: 1. C + O₂ → CO₂; ΔH₁ = −393.5 kJ 2. CO + ½O₂ → CO₂; ΔH₂ = −283 kJ
Important Questions
One Mark (5)
- State the first law of thermodynamics.
- Define an isolated system.
- What is the sign of ΔH for an exothermic reaction?
- Define standard enthalpy of formation.
- State the second law of thermodynamics.
Short Answer 3-mark (3)
- Distinguish between ΔH and ΔU. Derive the relationship ΔH = ΔU + ΔngRT.
- State Hess's law. What is its significance?
- For the reaction N₂ + 3H₂ → 2NH₃, ΔH = −92 kJ. Is this reaction exothermic or endothermic? What happens to entropy?
Long Answer 5-mark (2)
- Explain Gibbs free energy. Derive ΔG = ΔH − TΔS. State the conditions for spontaneity.
- State Hess's law and use it to calculate the enthalpy of formation of CO from given thermochemical equations.
Common Mistakes
- Using ΔG < 0 for feasibility but forgetting that kinetics (rate) may still be slow.
- Forgetting to convert T to Kelvin in ΔG = ΔH − TΔS.
- Confusing ΔH < 0 with spontaneous — entropy must also be considered.
- Sign error in ΔngRT: Δng = moles of gaseous products − moles of gaseous reactants.
Board Exam Tips
- ΔG = ΔH − TΔS spontaneity table is asked as a 3 or 5-mark question every year.
- Hess's law numerical always appears in the question paper.
- NEET: Bond enthalpy calculations and ΔG = −RT ln K are regular questions.
- Know standard enthalpy of neutralisation: −57.3 kJ/mol (strong acid + strong base).
Revision Checklist
- [ ] First and second laws stated correctly
- [ ] All enthalpy types defined with one example each
- [ ] ΔG = ΔH − TΔS spontaneity table completed
- [ ] Hess's law numerical practised (3 examples)
- [ ] ΔH = ΔU + ΔngRT formula applied in a numerical
Frequently Asked Questions
Energy is neither created nor destroyed; it is only converted from one form to another. Mathematically: ΔU = q + w, where q is heat absorbed and w is work done on the system.