Chemical Thermodynamics & Thermochemistry

Thermodynamics governs whether a reaction is feasible. Advanced tests the first law with different processes, enthalpy via Hess's law and bond energies, and the Gibbs criterion linking ΔG, ΔH, ΔS and equilibrium. Sign conventions and state-function reasoning are central.

Key results

• First law: ΔU = q + w, w = −P_extΔV (work done on system positive). ΔH = ΔU + Δn_g RT.
• Hess's law: enthalpy is a state function; add reaction steps to get ΔH. ΔH_rxn = ΣΔHf(products) − ΣΔHf(reactants) = Σ BE(bonds broken) − Σ BE(bonds formed).
• Entropy: ΔS = q_rev/T; spontaneity of universe ΔS_total > 0.
• Gibbs: ΔG = ΔH − TΔS; spontaneous if ΔG < 0. ΔG° = −RT ln K = −nFE°.
• Temperature dependence: sign of ΔH and ΔS decides how spontaneity varies with T.

Quantity	Relation
Enthalpy link	ΔH = ΔU + Δn_g RT
Gibbs energy	ΔH − TΔS
ΔG° and K	−RT ln K
ΔG° and E°	−nFE°

Exam Tricks & Tips

• 🎯 Use ΔG = ΔH − TΔS to find the crossover temperature (T = ΔH/ΔS) where a reaction switches between spontaneous and non-spontaneous.
• 🎯 Enthalpy of reaction = bonds broken − bonds formed (using average bond energies) — a fast estimate when formation data are absent.
• 🎯 ΔH = ΔU + Δn_g RT: only the change in moles of gas matters; reactions with Δn_g = 0 have ΔH = ΔU.
• 🎯 ΔG° = −RT ln K links thermodynamics to equilibrium: a large negative ΔG° means K » 1 (products favoured).
• 🎯 Both ΔH < 0 and ΔS > 0 guarantee spontaneity at all T; both positive/negative means temperature decides.
• ❌ Common mistake: confusing q and ΔU — heat is path-dependent, internal energy is a state function; ΔU depends only on the endpoints.

Expected exam pattern

1–2 questions: a Hess's-law/bond-energy enthalpy numerical and a Gibbs/entropy spontaneity problem (often linking ΔG° to K or E°). Sign-convention conceptuals appear as multiple-correct.

Quick recap

ΔU = q + w; ΔH = ΔU + Δn_g RT. Hess's law and bond energies give ΔH. Spontaneity: ΔG = ΔH − TΔS < 0. ΔG° = −RT ln K = −nFE°. Crossover temperature at ΔH/ΔS.

Cover the answer, recall, then check. 12 cards on thermodynamics for JEE Advanced.

Q1. First law of thermodynamics (chemistry sign convention).
A1. ΔU = q + w, with work done on the system taken positive.

Q2. Relation between ΔH and ΔU.
A2. ΔH = ΔU + Δn_g RT (Δn_g = change in moles of gas).

Q3. Hess's law.
A3. The total enthalpy change is independent of the path (enthalpy is a state function).

Q4. Enthalpy from bond energies.
A4. ΔH = Σ(bonds broken) − Σ(bonds formed).

Q5. Gibbs free-energy criterion for spontaneity.
A5. ΔG = ΔH − TΔS < 0 for a spontaneous process.

Q6. Relation between ΔG° and equilibrium constant K.
A6. ΔG° = −RT ln K.

Q7. Relation between ΔG° and standard cell potential.
A7. ΔG° = −nFE°.

Q8. Crossover temperature where spontaneity changes.
A8. T = ΔH/ΔS (where ΔG = 0).

Q9. When is a reaction spontaneous at all temperatures?
A9. When ΔH < 0 and ΔS > 0.

Q10. Work done in an irreversible expansion against constant external pressure.
A10. w = −P_ext ΔV.

Q11. Entropy change for a reversible process.
A11. ΔS = q_rev/T.

Q12. For a reaction with Δn_g = 0, how do ΔH and ΔU compare?
A12. ΔH = ΔU (no PV work from gas-mole change).

Thermodynamics tells you whether a reaction can happen (spontaneity) and how much energy it exchanges, without saying anything about speed. JEE Advanced tests the first law with sign conventions, Hess's law manipulations, and the Gibbs-energy criterion — with a strong emphasis on state vs path functions.

Core concept: enthalpy, entropy and Gibbs energy are state functions; ΔG = ΔH − TΔS decides spontaneity, and Hess's law lets you add reactions like algebra.

Core ideas

Beginner — first law and enthalpy

First law: ΔU = q + w (chemistry convention: w = work done on the system = −PΔV for expansion). At constant volume q_v = ΔU; at constant pressure q_p = ΔH. They relate by ΔH = ΔU + Δn_g RT, where Δn_g is the change in moles of gas — a frequent numerical link. Enthalpy of reaction is negative for exothermic, positive for endothermic.

Intermediate — Hess's law and standard enthalpies

Because H is a state function, Hess's law says ΔH for a reaction is the same by any route: add/subtract known thermochemical equations. Useful standard enthalpies: formation (ΔH_f°), combustion, neutralisation (constant ~−57 kJ/mol for strong acid + strong base), bond enthalpy. ΔH_reaction = Σ(bonds broken) − Σ(bonds formed) = ΣΔH_f°(products) − ΣΔH_f°(reactants). Born–Haber cycles apply Hess's law to lattice energies.

Advanced — entropy, Gibbs energy and spontaneity

Second law: a spontaneous process increases the entropy of the universe (ΔS_universe > 0). For the system, ΔG = ΔH − TΔS combines both: ΔG < 0 spontaneous, = 0 equilibrium, > 0 non-spontaneous. The sign table by temperature: ΔH<0,ΔS>0 always spontaneous; ΔH>0,ΔS<0 never; the mixed cases flip at T = ΔH/ΔS. Also ΔG° = −RT ln K links thermodynamics to equilibrium, and ΔG° = −nFE°cell to electrochemistry. Third law sets S = 0 for a perfect crystal at 0 K, giving absolute entropies.

Worked example

For a reaction ΔH = +30 kJ/mol and ΔS = +100 J/(mol·K). Above what temperature is it spontaneous? Spontaneity needs ΔG = ΔH − TΔS < 0, i.e. T > ΔH/ΔS = 30000 J / 100 J·K⁻¹ = 300 K. So above 300 K the entropy term wins and the reaction proceeds; below it, ΔG is positive. This "crossover temperature" reasoning — endothermic but entropy-driven — is exactly the conceptual test JEE sets.

How JEE Advanced tests this

ΔH ↔ ΔU conversions via Δn_gRT; Hess's-law and Born–Haber-cycle calculations; bond-enthalpy estimates of ΔH; spontaneity sign-analysis and crossover temperature; ΔG° = −RT ln K linking to equilibrium; and identifying state vs path functions.

Exam tricks & shortcuts

• ΔH = ΔU + Δn_gRT (Δn_g = gaseous products − gaseous reactants).
• ΔG = ΔH − TΔS; find crossover T = ΔH/ΔS for the mixed-sign cases.
• Bond enthalpy: ΔH = Σ(broken) − Σ(formed).
• Mnemonic: "Negative ΔG goes."

Key formulas

• First law: ΔU = q + w; w = −P_extΔV; reversible isothermal w = −nRT ln(V₂/V₁).
• Enthalpy: H = U + PV; ΔH = ΔU + Δn_g RT; ΔH = q_p.
• Hess's law: ΔH_rxn = ΣΔH_f(products) − ΣΔH_f(reactants) = bonds broken − bonds formed.
• Entropy: ΔS = q_rev/T; ΔS_total > 0 (spontaneous).
• Gibbs: ΔG = ΔH − TΔS; ΔG° = −RT ln K = −nFE°.

Gibbs energy combines enthalpy and entropy to predict spontaneity; ΔG° links to the equilibrium constant.

Worked Example

Problem: A reaction has ΔH = +30 kJ/mol and ΔS = +100 J/mol·K. (a) Find ΔG at 298 K and at 400 K. (b) Determine the temperature above which the reaction becomes spontaneous.

Solution:
Use the Gibbs relation ΔG = ΔH − TΔS (keep units consistent; ΔS = 0.100 kJ/mol·K).

(a) At 298 K:
ΔG = 30 − (298)(0.100) = 30 − 29.8 = +0.2 kJ/mol (> 0 → non-spontaneous).
At 400 K:
ΔG = 30 − (400)(0.100) = 30 − 40 = −10 kJ/mol (< 0 → spontaneous).

(b) The changeover is at ΔG = 0:
0 = ΔH − TΔS → T = ΔH/ΔS = 30 000 J / 100 J·K⁻¹ = 300 K.
Above 300 K the −TΔS term outweighs the positive ΔH, so the reaction is spontaneous.

Answer: ΔG = +0.2 kJ/mol at 298 K, −10 kJ/mol at 400 K; spontaneous above T = 300 K.