Chemical thermodynamics for JEE Main

Chemical thermodynamics answers two questions about a reaction: how much energy it releases or absorbs, and whether it can happen on its own. JEE Main tests both with short calculations. A Hess's-law sum, a ΔH from bond enthalpies, a temperature at which a reaction becomes spontaneous, or an equilibrium constant from ΔG° are all regular question types, and several make natural numerical-value questions.

The official 2026 syllabus covers system and surroundings, extensive and intensive properties, state functions, entropy and types of processes; the first law with work, heat, internal energy, enthalpy, heat capacity and molar heat capacity; Hess's law; enthalpies of bond dissociation, combustion, formation, atomisation, sublimation, phase transition, hydration, ionisation and solution; and the second law with ΔS of the universe and ΔG of the system as criteria for spontaneity, and ΔG° and the equilibrium constant.

The language of thermodynamics

The first law and the sign convention

ΔU = q + w, where q is the heat absorbed by the system and w is the work done on the system. This is the IUPAC convention used in chemistry.

Work of expansion:

• against a constant external pressure: w = −P_ext ΔV
• reversible isothermal expansion of an ideal gas: w = −2.303 nRT log(V₂/V₁)
• free expansion into a vacuum: w = 0
• for an ideal gas in an isothermal process, ΔU = 0, so q = −w

Enthalpy, ΔH and ΔU

Enthalpy H = U + PV. At constant pressure, ΔH equals the heat exchanged, and at constant volume ΔU does. A bomb calorimeter (constant volume) measures ΔU; an open calorimeter measures ΔH. They are related by

ΔH = ΔU + Δn_g RT

where Δn_g is the moles of gaseous products minus gaseous reactants. Liquids and solids do not count.

Heat and temperature: q = nCΔT, where C is the molar heat capacity. For an ideal gas, C_p − C_v = R.

Enthalpies of different kinds

The standard enthalpy of formation of an element in its standard state is zero: graphite, not diamond; O₂, not O₃.

Hess's law and bond enthalpies

Because H is a state function, enthalpy changes add along any route (Hess's law). Two working forms follow:

• ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants)
• ΔrH = Σ bond enthalpies (broken) − Σ bond enthalpies (formed), for gas-phase reactions

Bond-enthalpy answers are approximate, because tabulated values are averages over many molecules.

Entropy and the second law

Entropy measures how widely energy and matter are dispersed. For a reversible change at temperature T, ΔS = q_rev/T; for a phase change, ΔS = ΔH/T.

Entropy increases on melting, boiling and dissolving, and when the moles of gas increase.

A process is spontaneous when ΔS_total = ΔS_system + ΔS_surroundings > 0. At constant T and P, ΔS_surroundings = −ΔH_system/T.

Gibbs energy and spontaneity

At constant temperature and pressure, ΔG = ΔH − TΔS, and a process is spontaneous when ΔG < 0.

Gibbs energy links to equilibrium:

• ΔG = ΔG° + RT ln Q; at equilibrium ΔG = 0 and Q = K
• so ΔG° = −RT ln K = −2.303 RT log K
• ΔG° < 0 means K > 1: products are favoured at equilibrium

Spontaneity says nothing about speed. Diamond turning into graphite is spontaneous but immeasurably slow.

Worked problems

Problem 1: ΔH from ΔU. For C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l), ΔU = −1406 kJ/mol at 298 K. Find ΔH.

Δn_g = 2 − 4 = −2 (water is liquid).
ΔH = −1406 + (−2)(8.314 × 10⁻³)(298) = −1406 − 4.96 ≈ −1411 kJ/mol.

Problem 2: Hess's law. Given ΔfH° of CH₄(g) = −74.8, CO₂(g) = −393.5 and H₂O(l) = −285.8 kJ/mol, find ΔcH° of methane.

CH₄ + 2O₂ → CO₂ + 2H₂O(l). O₂ has ΔfH° = 0.
ΔcH° = [−393.5 + 2(−285.8)] − (−74.8) = −965.1 + 74.8 = −890.3 kJ/mol.

Problem 3: bond enthalpies. Find ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g), given C=C 614, C–C 348, C–H 414 and H–H 436 kJ/mol.

Broken: one C=C and one H–H = 614 + 436 = 1050.
Formed: one C–C and two new C–H = 348 + 2 × 414 = 1176. (The four original C–H bonds cancel.)
ΔH = 1050 − 1176 = −126 kJ/mol.

Problem 4 (numerical answer): K from ΔG°. For a reaction at 300 K, ΔG° = −11.49 kJ/mol. Find K (R = 8.314 J K⁻¹ mol⁻¹).

log K = −ΔG°/(2.303RT) = 11,490/(2.303 × 8.314 × 300) = 11,490/5744 ≈ 2.00.
K = 100.

Traps to watch

• Sign of w. In chemistry, work done on the system is positive.
• Δn_g counts gases only. Water in liquid form does not count.
• Units in ΔG = ΔH − TΔS. ΔH is usually in kJ and ΔS in J/K; convert one before combining.
• ΔG° versus ΔG. ΔG° fixes K; ΔG decides the direction under the actual conditions.
• ΔfH of elements. Zero only for the standard state form.

Practice set

1. For C(s) + O₂(g) → CO₂(g), is ΔH equal to ΔU?
2. What is the sign of ΔS for 2H₂(g) + O₂(g) → 2H₂O(l)?
3. For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ/mol and ΔS = +160 J K⁻¹ mol⁻¹. Above what temperature is it spontaneous?
4. Find w when 1 mol of an ideal gas expands reversibly and isothermally at 300 K from 10 L to 100 L.
5. A gas expands from 1 L to 5 L against a constant external pressure of 1 bar. Find w in joules.
6. The enthalpy of vaporisation of water is 40.79 kJ/mol at 373 K. Find the entropy of vaporisation.
7. Find ΔH for H₂(g) + Cl₂(g) → 2HCl(g), given H–H 436, Cl–Cl 242 and H–Cl 431 kJ/mol.
8. 100 mL of 1 M HCl is mixed with 100 mL of 1 M NaOH. Taking the enthalpy of neutralisation as −57.1 kJ/mol, how much heat is released?

Answers

1. Yes. Δn_g = 1 − 1 = 0.
2. Negative: three moles of gas become liquid.
3. T = 178,000/160 ≈ 1113 K.
4. w = −2.303 × 1 × 8.314 × 300 × log 10 ≈ −5.74 kJ.
5. w = −P_ext ΔV = −(10⁵ Pa)(4 × 10⁻³ m³) = −400 J.
6. ΔS = 40,790/373 ≈ 109 J K⁻¹ mol⁻¹.
7. (436 + 242) − 2 × 431 = −184 kJ/mol.
8. 0.1 mol of water forms: 0.1 × 57.1 = 5.71 kJ released.

What to do next

• Make a one-page sheet with the sign convention, the four ΔG cases and the ΔG°–K relation.
• Solve ten Hess's-law problems by both formation enthalpies and bond enthalpies.
• Revise chemical bonding for bond strengths, and compare with thermodynamics in physics, which uses the other sign convention.
• Move on to chemical equilibrium, where ΔG° = −RT ln K is used constantly.