Chemical Thermodynamics

Physical Chemistry Weightage: 2-3 Questions (8-12 Marks) JEE Unit 4

“Welcome back! These notes assume you have already read the NCERT chapter and know the definitions of system, state function, enthalpy and entropy, so they skip the textbook story and keep what JEE Main actually asks. Thermodynamics usually gives two to three questions per paper, and almost every one is a short calculation: work in a reversible or irreversible expansion, the gap between enthalpy and internal energy, a Hess's law or bond enthalpy sum, a Born-Haber cycle, or the temperature at which a reaction turns spontaneous. Learn the sign convention, the process table and the five worked patterns below and this chapter becomes fast, reliable marks.”

— SCORECHEM ACADEMIC TEAM

1. Signs, State Functions, Ideal-Gas Rules

First law (IUPAC): ΔU = q + w. Heat absorbed by the system: q > 0. Work done ON the system (compression): w > 0. Work done BY the system (expansion): w < 0.
State functions: U, H, S, G, P, T, V (cyclic integral is zero). Path functions: q and w. Extensive: U, H, S, G, C. Intensive: T, P, density, molarity, specific heat. The ratio of two extensive properties is intensive.
Ideal gas: U and H depend on T only. So any isothermal change has ΔU = ΔH = 0 and q = −w.
Units: 1 L atm = 101.3 J; R = 8.314 J mol⁻¹ K⁻¹ = 0.0821 L atm mol⁻¹ K⁻¹. Bring pressure in Pa and volume in m³ to get joules directly (3 kPa × 2 m³ = 6 kJ).

2. Work and the Process Table

Fig. 1: q, w, ΔU and ΔH for an ideal gas (w by IUPAC sign: work done ON the gas is positive).

Reversible isothermal: w = −nRT ln(V₂/V₁) = −2.303 nRT log(V₂/V₁) = −2.303 nRT log(P₁/P₂). When only P and V are given, nRT = P₁V₁.
Single-step irreversible: w = −P_ext(V₂ − V₁). Multi-step: add the steps, each with its own P_ext.
Worked pattern 1: 20 dm³ of ideal gas at 600 K and 0.5 MPa expands isothermally and reversibly to 0.2 MPa. w = −(0.5 × 10⁶)(20 × 10⁻³) ln(0.5/0.2) = −10 kJ × 0.916 = −9.2 kJ; ΔU = ΔH = 0; q = +9.2 kJ.
Worked pattern 2: 3 mol compressed isothermally from 60 L to 20 L at constant 5 atm: w = −5(20 − 60) = +200 L atm, so q = −200 L atm.

Fig. 2: Expansion from A to the same final volume. Shaded area under the isotherm = |w|.

Comparison rules: for the same end states, |w_rev| > |w_irrev| in expansion, and w_irrev > w_rev in compression (the reversible path needs the least work). On a P–V graph, work is the area under the curve; a cyclic process gives the enclosed area; an isochoric line gives zero.
Matching trick: reversible expansion → −nRT ln(V_f/V_i); free expansion → w = 0; irreversible expansion or compression → −p_ex(V_f − V_i) (positive for compression).

⚠️ JEE Trap: Sign of work and heat. A cup of water heated in a microwave until it boils has q > 0, w < 0 (the vapour pushes the atmosphere back), and ΔU > 0 because q outweighs |w|. Do not write w = 0 just because the volume change looks small.

3. Adiabatic Processes

Reversible adiabatic (q = 0): TV^γ−1 = constant, PV^γ = constant, T^γP^1−γ = constant, w = ΔU = nC_v(T₂ − T₁) = (P₂V₂ − P₁V₁)/(γ − 1). Expansion cools the gas, compression heats it. ΔS_sys = 0.
Irreversible adiabatic (constant P_ext): the relations above do NOT hold. Set nC_v(T₂ − T₁) = −P_ext(V₂ − V₁), write V = nRT/P and solve for T₂.
Worked pattern 3: 1 mol monatomic gas (C_v = 1.5R), 300 K and 10 atm, expands against 1 atm to 1 atm. 1.5(T₂ − 300) = −(T₂ − 30), so 2.5T₂ = 480 and T₂ = 192 K. (The reversible route would give 119 K; the irreversible gas cools less because it does less work.)
Same idea, different numbers: C_v = 2.5R, T₁ = 298 K, P₁ = 5 atm, expanding against 1 atm until the volume doubles. Then ΔV = V₁ = RT₁/P₁ per mole, so 2.5(T₂ − 298) = −(1/5)(298), giving T₂ ≈ 274 K.

⚠️ JEE Trap: Using TV^γ−1 for an irreversible path. That relation is for the reversible adiabatic only. Against a fixed external pressure, equate w = −P_extΔV with nC_vΔT. Also remember that the adiabatic curve is steeper than the isotherm by the factor γ.

4. Enthalpy and Heat Capacity

H = U + PV; q_p = ΔH, q_v = ΔU. For reactions ΔH = ΔU + Δn_gRT, where Δn_g counts only gaseous products minus gaseous reactants.
- Benzoic acid burning (C₆H₅COOH(s) + 7.5O₂ → 7CO₂ + 3H₂O(l)): Δn_g = −0.5, so ΔH = ΔU − 0.5R(300) = ΔU − 150R (R in kJ).
- Vaporisation of water at 373 K: Δn_g = +1, ΔU = 40.79 − (8.3 × 373)/1000 = 37.7, about 38 kJ/mol.
Mayer: C_p − C_v = R. γ = C_p/C_v.

Gas type	Degrees of freedom	C_v	C_p	γ
Monoatomic (He, Ar)	3	3R/2	5R/2	5/3
Diatomic, linear (O₂, CO₂)	5	5R/2	7R/2	7/5
Non-linear polyatomic (H₂O, CH₄)	6	3R	4R	4/3

Calorimetry: a bomb calorimeter measures ΔU (constant volume): ΔU = −C_totalΔT × (M/m), then convert to ΔH with Δn_gRT.
Kirchhoff: ΔH_T2 = ΔH_T1 + ΔC_p(T₂ − T₁).

5. Entropy and the Second Law

Second law: ΔS_univ = ΔS_sys + ΔS_surr > 0 for a spontaneous process. dS = dq_rev/T.
Ideal gas: ΔS = nC_v ln(T₂/T₁) + nR ln(V₂/V₁) = nC_p ln(T₂/T₁) − nR ln(P₂/P₁). Heating a solid or liquid: ΔS = ms ln(T₂/T₁).

Fig. 6: Entropy bookkeeping. ΔS of the system depends only on the end states, not on the path.

Phase changes at the transition temperature: ΔS = ΔH/T. Trouton's rule: ΔS_vap is about 85 to 88 J K⁻¹ mol⁻¹ for non-associated liquids and higher for H-bonded ones (water about 109). Order of entropy: gas >> liquid > solid.
Third law: the entropy of a perfect crystal is zero at 0 K. Residual entropy remains for CO, N₂O, NO, ice and HD (orientational or isotopic disorder).

6. Gibbs Energy and Equilibrium

Fig. 3: Spontaneity at constant T and P. At the switch temperature ΔG = 0, so T = ΔH/ΔS.

G = H − TS; ΔG_sys = −TΔS_univ. Spontaneous: ΔG < 0; equilibrium: ΔG = 0; at equilibrium ΔG° = −RT ln K = −2.303 RT log K. Non-standard: ΔG = ΔG° + RT ln Q.
Switch temperature: T = ΔH/ΔS. Worked: ΔH = +400 kJ and ΔS = 0.2 kJ K⁻¹ gives spontaneity above 2000 K; boiling with ΔH_vap = 30 kJ and ΔS_vap = 75 J K⁻¹ gives T_b = 400 K.
van't Hoff plot: log K = −ΔH°/(2.303RT) + ΔS°/(2.303R). So a graph of log K against 1/T is a straight line with slope = −ΔH°/2.303R and intercept = ΔS°/2.303R. Two temperatures: ln(K₂/K₁) = (ΔH°/R)(1/T₁ − 1/T₂).
Worked pattern 4 (mixed data): for A₂ + B₂ → 2AB at 500 K with log K = 2.2: ΔG° = −2.303 × 8.3 × 500 × 2.2 = −21.0 kJ. ΔS° = 2S(AB) − S(A₂) − S(B₂), ΔH° = ΔG° + TΔS°. Keep J and kJ consistent.

7. Thermochemistry and Hess's Law

Fig. 5: Types of enthalpy change, with the point JEE usually tests.

Hess's law: add, reverse (change sign) and multiply equations; the enthalpy follows. ΔH°_r = ΣΔH_f°(products) − ΣΔH_f°(reactants) = ΣΔH_c°(reactants) − ΣΔH_c°(products).
Worked pattern 5 (combustion from formation data): benzene(l) with ΔH_f = +48.5, CO₂ = −393.5, H₂O(l) = −286 kJ/mol: ΔH_c = 6(−393.5) + 3(−286) − 48.5 = −3267.5 kJ per mole, so 2 mol gives −6535 kJ.
Formation zero convention: C(graphite) = 0 but diamond = +1.9 kJ; S(rhombic) = 0; white P = 0; H⁺(aq) = 0. Elements in their standard states only.
Bond enthalpies: ΔH_r = Σ(bonds broken) − Σ(bonds formed). C₂H₄ + H₂ → C₂H₆ with C=C 615, H–H 435, C–C 347, C–H 414: ΔH = 615 + 435 − 347 − 2(414) = −125 kJ (the two extra C–H bonds are formed). For an atomisation question, 4ε(C–H) = −ΔH_f(CH₄) + ΔH_sub(C) + 2ΔH_diss(H₂).
Resonance energy: hydrogenation of benzene should give 3 × (−119) = −357 kJ, experiment gives −205 kJ, so resonance energy is 152 kJ/mol.
Neutralisation: strong acid + strong base = −57.1 kJ per mole of water (mono-acid, mono-base). H₂SO₄ with excess NaOH liberates 2x that of HCl for equal volumes of 1 M solution. Weak acid or base: ΔH_neut = −57.1 + ΔH_ionisation, so the magnitude is smaller.
Solution and hydration: CuSO₄(s) → solution: −70 kJ; CuSO₄·5H₂O → solution: +12 kJ. Hydration enthalpy = −70 − 12 = −82 kJ.

Fig. 4: Born-Haber cycle for LiF (kJ/mol). Lattice enthalpy magnitude = 1031.

Born–Haber: ΔH_f = ΔH_sub + ½ΔH_diss + IE + EGE + U(lattice, energy released on forming the solid). For LiF: −594 = 155 + 75 + 520 − 313 + U, so |U| = 1031 kJ/mol.

⚠️ JEE Trap: Half-mole of F₂ in Born–Haber. One mole of F atoms needs only ½ of the F₂ dissociation enthalpy (150 → 75), and electron gain enthalpy is negative. Missing the ½ is the commonest way to lose this question.

8. How JEE Frames Questions

Numerical (integer answer): work in L atm or kJ for isothermal and adiabatic changes; ΔH − ΔU; Hess/bond-enthalpy sums; switch temperature; hydration and neutralisation data.
Match the column: work expressions (reversible, free, irreversible expansion and compression); process versus q, w, ΔU; quantities such as ΔU = nC_vΔT and ΔH = nC_pΔT evaluated to kJ.
Statements / assertion-reason: signs of q, w, ΔU; van't Hoff slope and intercept; neutralisation being constant only for strong–strong.
Graph: P–V area for maximum work; reaction-coordinate profiles (an endothermic overall reaction ends higher than it starts).

⚠️ JEE Trap: Mixing J and kJ, atm and Pa. ΔS is usually given in J K⁻¹ mol⁻¹ and ΔH in kJ mol⁻¹. Convert before you divide or subtract, and use R = 8.314 J (or 8.3 if the question gives it).

9. Quick Sheet and Checklist

Idea	Rule
First law	ΔU = q + w, w = −P_extΔV
Reversible isothermal	w = −nRT ln(V₂/V₁), ΔU = 0
Adiabatic (irreversible)	nC_vΔT = −P_extΔV
Adiabatic (reversible)	TV^γ−1 = constant
Enthalpy	ΔH = ΔU + Δn_gRT
Gibbs	ΔG = ΔH − TΔS, ΔG° = −RT ln K
van't Hoff plot	slope −ΔH°/2.303R, intercept ΔS°/2.303R

Before the exam, check you can:

Compute w, q, ΔU and ΔH for an isothermal reversible, isothermal irreversible and adiabatic process from given data.
Find T₂ for an irreversible adiabatic expansion using V = nRT/P.
Convert between ΔH and ΔU and use the sign of Δn_g correctly.
Solve a Born–Haber, Hess or bond-enthalpy problem in under two minutes, and state the switch temperature of a reaction.