Thermodynamics
'Thermodynamics is the only physical theory of universal content which I am convinced will never be overthrown.' — Albert Einstein
1. Chapter Overview
Chemical THERMODYNAMICS deals with ENERGY CHANGES accompanying chemical reactions. Unlike the Physics thermodynamics chapter (which focuses on heat engines), this chapter focuses on REACTION ENTHALPY, BOND ENERGIES, HESS'S LAW, ENTROPY, GIBBS FREE ENERGY, and the CRITERION OF SPONTANEITY. These concepts tell you WHETHER a reaction will occur and how MUCH heat is involved.
2. Thermodynamic Terms
- System: The PART of the universe under study
- Open: exchanges matter AND energy
- Closed: exchanges energy ONLY
- Isolated: exchanges NEITHER
- Surroundings: Everything OUTSIDE the system
- State Functions: Depend ONLY on the state, NOT the path (P, V, T, U, H, S, G)
- Path Functions: Depend on the PATH taken (Heat Q, Work W)
3. Internal Energy (U) and Enthalpy (H)
First Law of Thermodynamics
- ΔU = Q + W (Chemistry convention: W = work done ON the system)
- ΔU = Q — PΔV (for pressure-volume work only)
Enthalpy Definition
- H = U + PV (Enthalpy = Internal energy + Pressure-Volume work)
- ΔH = ΔU + Δn_gRT (at constant P)
- ΔH = Q_p (heat at constant pressure)
- ΔU = Q_v (heat at constant volume)
Exothermic vs Endothermic
| Type | ΔH sign | Heat | Examples |
|---|---|---|---|
| Exothermic | Negative | Heat released | Combustion, respiration |
| Endothermic | Positive | Heat absorbed | Melting, decomposition |
4. Enthalpy Changes in Reactions
Standard Enthalpy of Reaction (Δ_rH°)
- Enthalpy change when reaction occurs with ALL substances in their STANDARD states
- Standard state: 1 bar pressure, pure substance at specified T
Standard Enthalpy of Formation (Δ_fH°)
- Enthalpy change when ONE MOLE of compound is formed from its ELEMENTS in their standard states
- Δ_fH° of elements in standard state = 0 (reference)
- Example: C(s) + O₂(g) → CO₂(g), Δ_fH° = -393.5 kJ/mol
Standard Enthalpy of Combustion (Δ_cH°)
- Enthalpy change when ONE MOLE of substance is completely burned in oxygen
Bond Enthalpy
- AVERAGE energy required to break a specific type of bond
- Δ_rH° = Σ(Bond enthalpies of reactants) — Σ(Bond enthalpies of products)
- Average values: C—H = 413, C≡C = 839, O=O = 498 kJ/mol
5. Hess's Law of Constant Heat Summation
- Statement: The total enthalpy change for a reaction is INDEPENDENT of the PATH
- ΔH for the overall reaction = ΣΔH of individual steps
- Application: Used to calculate ΔH for reactions that are DIFFICULT to measure directly
Worked Problem
Q: Calculate Δ_fH° of CO using: (1) C + O₂ → CO₂, ΔH₁ = -393.5 kJ; (2) CO + ½O₂ → CO₂, ΔH₂ = -283 kJ. A: Target: C + ½O₂ → CO. ΔH = ΔH₁ — ΔH₂ = -393.5 — (-283) = -110.5 kJ/mol.
6. Entropy (S) and the Second Law
- Entropy: Measure of RANDOMNESS or DISORDER
- Second Law: Total entropy of the universe ALWAYS INCREASES for a spontaneous process
- ΔS_system + ΔS_surroundings > 0 for spontaneous process
Entropy Changes
- Solids < Liquids < Gases (increasing entropy)
- S° increases with temperature
- S° of a PERFECT crystal at 0 K = 0 (Third Law of Thermodynamics)
7. Gibbs Free Energy (G) and Spontaneity
Definition
- G = H — TS
- ΔG = ΔH — TΔS
Criterion of Spontaneity
| ΔH | ΔS | ΔG = ΔH — TΔS | Result |
|---|---|---|---|
| - | + | ALWAYS negative | Spontaneous at ALL T |
| + | - | ALWAYS positive | Non-spontaneous at ALL T |
| - | - | Negative at LOW T | Spontaneous at low T |
| + | + | Negative at HIGH T | Spontaneous at high T |
Key Points
- If ΔG < 0: Process is SPONTANEOUS (feasible)
- If ΔG = 0: System is at EQUILIBRIUM
- If ΔG > 0: Process is NON-SPONTANEOUS (reverse is spontaneous)
Standard Gibbs Free Energy
- Δ_rG° = ΣΔ_fG°(products) — ΣΔ_fG°(reactants)
- Δ_rG° = -RT ln K (K = equilibrium constant)
- At equilibrium: Δ_rG = 0
Worked Problem
Q: For a reaction, ΔH = 100 kJ and ΔS = 200 J/K. Is it spontaneous at 300 K? A: ΔG = 100 — 0.2 × 300 = 100 — 60 = 40 kJ. ΔG > 0 → NON-SPONTANEOUS at 300 K. At T > 500 K: ΔG becomes negative → spontaneous.
8. Comparison: Physics vs Chemistry Thermodynamics
| Aspect | Physics (Class 11) | Chemistry (Class 11) |
|---|---|---|
| Focus | Heat engines, efficiency | Reaction spontaneity, ΔH |
| W sign | ΔU = Q — W | ΔU = Q + W (or Q — PΔV) |
| Key concept | Carnot efficiency | ΔG = ΔH — TΔS |
| Applications | Engine cycles | Chemical reactions |
9. Common Mistakes
- Sign convention confusion: Physics: ΔU = Q — W. Chemistry: ΔU = Q + W (W = work ON system). Be CAREFUL about the convention used
- ΔH°_f of elements is ZERO only in standard state: O(g) ≠ O₂(g)! Δ_fH° of O(g) is +249 kJ/mol
- Entropy of vaporisation >> entropy of fusion: Gas has much higher disorder than liquid
- ΔG < 0 means spontaneous, NOT fast: Thermodynamics tells us FEASIBILITY, not RATE (kinetics)
- Hess's law applies because H is a state function: The path doesn't matter, only initial and final states
10. CBSE Exam Focus
- Enthalpy, internal energy — relation ΔH = ΔU + ΔnRT (3-mark)
- Hess's law numerical problems (5-mark)
- Bond enthalpy calculations (3/5-mark)
- ΔG, ΔH, ΔS — numerical problems on spontaneity (5-mark)
- Standard enthalpy of formation, combustion, neutralisation
- Relationship between ΔG° and equilibrium constant
11. Key Formulas
- ΔU = Q + W = Q — PΔV (at constant P)
- ΔH = ΔU + Δn_gRT
- Δ_rH° = ΣΔ_fH°(products) — ΣΔ_fH°(reactants)
- Δ_rH = Σ(Bond energy)_reactants — Σ(Bond energy)_products
- ΔG = ΔH — TΔS
- ΔG° = -RT ln K
- ΔG = ΔG° + RT ln Q (Q = reaction quotient)
12. Self-Test (5+ Q&A)
Q1: For the reaction N₂ + 3H₂ → 2NH₃, ΔH = -92.4 kJ at 298 K. Find ΔU if Δn_g = -2. A: ΔH = ΔU + Δn_gRT → -92.4 = ΔU + (-2)(8.314×298)/1000 → -92.4 = ΔU — 4.96 → ΔU = -87.44 kJ.
Q2: Calculate Δ_fH° of C₂H₆ given: C(s) + O₂ → CO₂ ΔH = -393.5; H₂ + ½O₂ → H₂O ΔH = -285.8; C₂H₆ + 7/2O₂ → 2CO₂ + 3H₂O ΔH = -1560 kJ. A: Δ_fH°[C₂H₆] = 2(-393.5) + 3(-285.8) — (-1560) = -787 — 857.4 + 1560 = -84.4 kJ/mol.
Q3: For a reaction, ΔH = 30 kJ, ΔS = 100 J/K. Find temperature at which it becomes spontaneous. A: At equilibrium ΔG = 0. T = ΔH/ΔS = 30000/100 = 300 K. Spontaneous ABOVE 300 K (ΔH +, ΔS +).
Q4: Is bond breaking endothermic or exothermic? A: ENDOTHERMIC — energy is absorbed to break bonds. Bond FORMATION is exothermic (energy released).
Q5: What is the significance of ΔG = 0? A: ΔG = 0 means the system is at EQUILIBRIUM. No net change occurs. The forward and reverse rates are equal.
13. Conclusion
Chemical thermodynamics provides TOOLS for predicting reaction feasibility and energy changes. Enthalpy tells us about heat flow; Hess's law allows CALCULATION of otherwise unmeasurable reaction enthalpies. Entropy introduces the DIRECTION of spontaneous change. Gibbs free energy COMBINES enthalpy and entropy into a single SPONTANEITY criterion. These concepts are ESSENTIAL for understanding chemical equilibrium, electrochemistry, and biochemistry.
