Thermodynamics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Thermodynamics in JEE covers three laws, enthalpy, entropy, free energy, and their applications to chemical reactions. It’s a high-weightage chapter with consistent 6–10 questions across papers. The key is understanding the sign conventions and the relationships between ΔU, ΔH, ΔS, and ΔG.

Key Formulae for Quick Recall:

• First Law: ΔU = qᵥ (constant volume) = qᵥ + w; w = −PΔV (expansion work)
• Enthalpy: H = U + PV; ΔH = ΔU + Δn_g RT (for gaseous reactions)
• Hess’s Law: ΔH is path-independent. ΔH_total = Σ ΔH_step. Always holds because H is a state function.
• Born-Haber Cycle: Lattice energy = ΔH_atomization + IE − EA + ΔH_sublimation − ΔH_f (sign convention matters!)
• Kirchhoff’s Law: ΔH₂ − ΔH₁ = ΔCₚ(T₂ − T₁); ΔS₂ − ΔS₁ = ΔCₚ ln(T₂/T₁)
• Entropy: ΔS = qᵣₑᵥ/T = ΔHᵣₑᵥ/T for reversible processes. Total entropy (system + surroundings) increases for spontaneous processes.
• Free Energy: ΔG = ΔH − TΔS; ΔG = −RT ln K (Gibbs-Helmholtz equation); Also ΔG = −nF E°_cell
• Spontaneity: ΔS_total > 0 (or ΔG < 0) for spontaneous process.

⚡ Exam tip: For ΔH from bond enthalpies: ΔH_reaction = Σ ΔH(bonds broken) − Σ ΔH(bonds formed). Bonds broken requires energy (+), bonds formed releases energy (−). Always count bonds carefully — a double bond is NOT the same as two single bonds.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Thermodynamics — JEE Chemistry Study Guide

1. System, Surroundings, and Types of Processes

A thermodynamic system is the part of the universe being studied. The surroundings are everything else.

System type	Exchange of matter	Exchange of energy
Open	Yes	Yes
Closed	No	Yes
Isolated	No	No

Process types:

• Isothermal (ΔT = 0): ΔU = 0, so q = −w = nRT ln(V₂/V₁)
• Adiabatic (q = 0): ΔU = w = nCᵥΔT; for adiabatic expansion of ideal gas, PV^γ = constant
• Isobaric (ΔP = 0): w = −PΔV, ΔH = qₚ
• Isochoric (ΔV = 0): w = 0, ΔU = qᵥ

2. Work in Reversible Isothermal Expansion of Ideal Gas

For a reversible isothermal expansion of n moles of ideal gas: w_rev = −∫ᵥ₁^ᵥ₂ P_ext dV = −∫ᵥ₁^ᵥ₂ (nRT/V) dV = −nRT ln(V₂/V₁) = −nRT ln(P₁/P₂)

For free expansion (against zero external pressure): w = 0

For irreversible expansion against constant external pressure P_ext: w = −P_ext(V₂ − V₁)

⚡ Exam tip: In JEE, free expansion is a common trap. Students sometimes calculate w = −PΔV for free expansion, but since P_ext = 0, w = 0. Internal energy doesn’t change in free expansion of ideal gas (U depends only on T), so ΔU = 0 for free expansion regardless of whether it’s isothermal or adiabatic.

3. Enthalpy — Types and Calculations

Standard enthalpy of formation (Δ_f H°): Enthalpy change when 1 mole of compound forms from its constituent elements in their standard states. By definition, Δ_f H° of any element in its standard state = 0.

Enthalpy of combustion (Δ_c H°): Enthalpy change when 1 mole burns in O₂ completely.

Enthalpy of neutralization: For strong acid + strong base, ΔH = −57.1 kJ per equivalent. For weak acid or weak base, it’s less exothermic because some energy is used to dissociate the weak electrolyte.

Bond enthalpy: Energy required to break 1 mole of bonds in gaseous molecules. Average bond enthalpy is used because bond energy varies slightly with environment.

Example: Calculate ΔH for combustion of CH₄: CH₄ + 2O₂ → CO₂ + 2H₂O ΔH = [ΔH(C=O) in CO₂ × 2 + ΔH(O–H) × 4] − [ΔH(C–H) × 4 + ΔH(O=O) × 2] = [(2 × 805) + (4 × 463)] − [(4 × 414) + (2 × 496)] = (1610 + 1852) − (1656 + 992) = 3462 − 2648 = +814 kJ/mol

Wait: Combustion should be exothermic (−ΔH). Let me recalculate: Actually: Bonds broken = 4(C–H) + 2(O=O) = 4(414) + 2(496) = 1656 + 992 = 2648 kJ (input) Bonds formed = 2(C=O) + 4(O–H) = 2(805) + 4(463) = 1610 + 1852 = 3462 kJ (released) ΔH = bonds broken − bonds formed = 2648 − 3462 = −814 kJ/mol ✓

4. Kirchhoff’s Law — Temperature Dependence of ΔH

ΔH(T₂) = ΔH(T₁) + ∫_{T₁}^{T₂} ΔCₚ dT

If ΔCₚ is constant: ΔH(T₂) = ΔH(T₁) + ΔCₚ(T₂ − T₁)

Similarly: ΔS(T₂) = ΔS(T₁) + ∫_{T₁}^{T₂} (ΔCₚ/T) dT = ΔS(T₁) + ΔCₚ ln(T₂/T₁)

Example: Find ΔH at 500 K for a reaction where ΔH₂₉₈° = −92 kJ and ΔCₚ = −10 J/K. ΔH₅₀₀ = −92 + (−10/1000) × (500 − 298) = −92 − 2.02 = −94.02 kJ

5. Spontaneity — The Second and Third Laws

Second Law of Thermodynamics: The total entropy of an isolated system always increases for a spontaneous process: ΔS_total = ΔS_system + ΔS_surroundings > 0

ΔS_surroundings = −ΔH_system/T (at constant P)

Third Law: At absolute zero (0 K), perfect crystalline solids have S = 0 (perfect order).

Gibbs Free Energy and Spontaneity:

At constant T and P: ΔG = ΔH − TΔS

ΔH	ΔS	ΔG	Process
−	+	Always −	Always spontaneous
+	−	Always +	Never spontaneous
−	−	− at low T, + at high T	Spontaneous at low T
+	+	+ at low T, − at high T	Spontaneous at high T

Example: For water vaporization at 373 K, ΔH = +40.7 kJ/mol, ΔS = +109 J/K·mol ΔG = 40.7 − 373 × 0.109 = 40.7 − 40.7 = 0 (at equilibrium, boiling point)

6. Relationship Between ΔG and Equilibrium Constant

ΔG° = −RT ln K (at T = 298 K: ΔG° = −5.708 log₁₀ K kJ/mol) ΔG° = ΔG + RT ln Q (non-standard conditions) At equilibrium: ΔG = 0, Q = K

Example: For a reaction, Kp = 10 at 298 K. Find ΔG°. ΔG° = −RT ln K = −(8.314 × 10⁻³)(298) ln(10) = −2.478 × 2.303 = −5.71 kJ/mol Or: ΔG° = −5.708 log₁₀(10) = −5.708 kJ/mol

⚡ Exam tip: When calculating ΔS for a reaction from standard molar entropies: ΔS°_reaction = Σ nS°(products) − Σ nS°(reactants). Pay attention to stoichiometric coefficients. Unlike enthalpy, entropy values are always positive for elements in their standard states.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Thermodynamics — Comprehensive JEE Advanced Notes

1. The Three Laws of Thermodynamics — Complete Understanding

Zeroth Law: If two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other. Establishes temperature as a fundamental property.

First Law: Energy can neither be created nor destroyed, only converted from one form to another. Mathematical: ΔU = q + w. This defines internal energy U as a state function — path independent. ΔU depends only on initial and final states, not how you got there.

Second Law:

• Kelvin-Planck: No process is possible whose sole result is complete conversion of heat into work.
• Clausius: No process is possible whose sole result is transfer of heat from a colder body to a hotter body. Both statements are equivalent and define the direction of spontaneous processes.

Third Law: Absolute entropy can be defined because S → 0 as T → 0 for a perfect crystal. Enables calculation of absolute entropy values.

2. Heat Capacity — Detailed Treatment

Molar heat capacity at constant volume: Cᵥ = (∂U/∂T)ᵥ For monatomic ideal gas: Cᵥ = 3/2 R (only translational modes) For diatomic: Cᵥ = 5/2 R (translational + rotational at moderate T) For polyatomic: Cᵥ depends on number of atoms and molecular complexity

Molar heat capacity at constant pressure: Cₚ = Cᵥ + R For ideal gas: H = U + PV = U + RT → dH = dU + RdT → Cₚ − Cᵥ = R ✓

Ratio γ = Cₚ/Cᵥ:

• Monatomic: γ = (5/2R)/(3/2R) = 5/3 ≈ 1.667
• Diatomic (no vib): γ = (7/2R)/(5/2R) = 7/5 = 1.4
• Used in adiabatic equations: PV^γ = const, TV^{γ−1} = const, TV^{γ−1} = const

Adiabatic reversible expansion of ideal gas: w = (P₁V₁ − P₂V₂)/(γ − 1) = nR(T₁ − T₂)/(γ − 1)

3. Calorimetry — Measurement of ΔU and ΔH

Coffee cup calorimeter (constant pressure): Measures ΔH. Used for solution reactions. q_reaction = −(m × c × ΔT) where m = total mass, c = specific heat capacity ΔH = qₚ/mol of limiting reagent

Bomb calorimeter (constant volume): Measures ΔU. Used for combustion reactions. q_reaction = −(C_cal × ΔT) where C_cal = heat capacity of calorimeter ΔH = ΔU + Δn_g RT

Example: Combustion of naphthalene (C₁₀H₈) in bomb calorimeter: ΔU_comb = −5153 kJ/mol C₁₀H₈(l) + 12O₂(g) → 10CO₂(g) + 4H₂O(l) Δn_g = 10 − 12 = −2 ΔH_comb = ΔU + Δn_g RT = −5153 + (−2)(8.314 × 10⁻³)(298) = −5153 + (−4.96) ≈ −5158 kJ/mol

4. Thermochemistry of Phase Changes

Fusion (melting): ΔH_fus = T_fus × ΔS_fus Vaporization: ΔH_vap = T_b × ΔS_vap Sublimation: ΔH_sub = ΔH_fus + ΔH_vap

Clausius-Clapeyron Equation (relates vapor pressure to temperature): ln(P₂/P₁) = −(ΔH_vap/R)(1/T₂ − 1/T₁) This is crucial for calculating vapor pressure at any temperature, boiling point at different pressures.

** Trouton’s Rule**: ΔH_vap/T_b ≈ 85–88 J/mol·K for many liquids Exceptions: H₂O (109 J/mol·K — due to strong H-bonding), HF, NH₃ (higher — molecular association)

5. Advanced Enthalpy Calculations

Enthalpy of solution: ΔH_sol = ΔH_lattice + ΔH_hydration For NaCl: Lattice energy = +787 kJ/mol, ΔH_hydration(Na⁺) = −406, ΔH_hydration(Cl⁻) = −364 ΔH_sol = +787 − 406 − 364 = +17 kJ/mol (endothermic, explains why NaCl dissolves better at higher T)

Bond Enthalpy from Heat of Formation (per bond): For methane: CH₄(g) → C(g) + 4H(g): ΔH = 1665 kJ/mol Each C–H bond: 1665/4 = 416 kJ/mol

Resonance energy (from thermochemical data): Actual ΔH_f° of benzene = +49.8 kJ/mol (endothermic) Calculated (hypothetical, Kekulé structure with 3 double bonds): Expected ~ 3 × 28.5 = 85.5 kJ/mol more negative Resonance energy = actual − calculated = 49.8 − (−36) ≈ +86 kJ/mol (stabilization from delocalization)

6. Gibbs-Helmholtz Equation — Temperature Dependence of ΔG

(∂ΔG/T)_P = −ΔH/T²

Integrated form: ΔG(T₂)/T₂ − ΔG(T₁)/T₁ = ΔH(1/T₂ − 1/T₁)/T₁

Alternatively: ΔG(T) = ΔH° − TΔS° (with ΔH° and ΔS° at 298 K, assumes ΔH and ΔS are temperature-independent — valid when phase change not involved in the T range).

Example problem: For a reaction, ΔH° = +75 kJ/mol, ΔS° = +150 J/mol·K. Find the temperature above which the reaction is spontaneous. ΔG < 0 → T > ΔH/ΔS = 75000/150 = 500 K

7. Thermodynamics of Electrochemical Cells

For a cell: ΔG = −nFE_cell E°_cell = (RT/nF) ln K = (2.303 RT/nF) log₁₀ K = (0.0591/n) log₁₀ K at 298 K

Nernst Equation (concentration dependence): E_cell = E°_cell − (0.0591/n) log₁₀ Q at 298 K

Temperature dependence of E°: E°(T₂)/E°(T₁) ≈ ΔG°(T₂)/ΔG°(T₁) = [ΔH° − T₂ΔS°]/[ΔH° − T₁ΔS°]

8. Non-Ideal Behavior — Van der Waals Gas

For real gases, the equation is: (P + an²/V²)(V − nb) = nRT

Joule-Thomson Effect — Real gas cooling on expansion: μ_JT = (∂T/∂P)_H = [T(∂V/∂T)_P − V]/Cₚ

At inversion temperature (μ_JT = 0), a gas neither heats nor cools on expansion. For N₂, T_inv ≈ 621 K (above room temperature, so N₂ cools on expansion at room T). For H₂, T_inv ≈ 202 K (below room T, so H₂ heats on expansion at room T — requires pre-cooling for liquefaction).

9. Entropy from Statistical Thermodynamics

S = k_B ln W where W = number of microstates, k_B = Boltzmann constant

For a perfect crystal at 0 K: W = 1 (unique ground state) → S = 0 ✓ (Third Law)

For mixing of two ideal gases: ΔS_mix = −nR(x₁ ln x₁ + x₂ ln x₂) > 0 (spontaneous mixing)

For expansion of ideal gas: ΔS = nR ln(V₂/V₁) (free or reversible isothermal)

⚡ Exam tip: Remember that ΔS_surroundings = −ΔH_system/T only when heat transfer to surroundings is reversible (which it is at constant T and P for a system at equilibrium). For irreversible processes, you must calculate ΔS_total separately. Common JEE trap: students use the same formula for ΔS_surroundings in both reversible and irreversible cases.

10. Gibbs Paradox and Entropy of Mixing

When two identical gases mix, there’s no entropy change (no disorder increase since gases were already indistinguishable). But for different gases, ΔS_mix > 0. The apparent discontinuity at the point of becoming identical is the “Gibbs paradox.”

This has implications in calculations: make sure you check whether gases are the same or different when computing entropy change of mixing.

⚡ Exam tip: In enthalpy calculations from combustion data, use the formula: ΔH°_f = ΔH°_comb(products) − ΔH°_comb(reactants). This is Hess’s law applied to combustion enthalpies. For C(s, graphite), ΔH°_f is defined as zero by convention — so all carbon combustion energy goes into CO₂ formation.

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