Chapters

Everything NCERT Class 11 Unit 5 (Thermodynamics) says that can be asked, plus what JEE Main adds beyond the book (marked Beyond NCERT). The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Every formula in the chapter on one screen, for the last pass. Details and examples are in the sections below.

Quantity	Formula	Watch
First law	ΔU = q + w	q, w positive when into the system
Work, constant pₑₓ	w = −pₑₓ(V₂ − V₁)	Free expansion: pₑₓ = 0, w = 0
Work, reversible isothermal	w = −2.303 nRT log(V₂/V₁)	= −2.303 nRT log(p₁/p₂)
Isothermal ideal gas	ΔU = 0, ΔH = 0, q = −w	Any path, rev or not
Adiabatic	q = 0, ΔU = w = nCᵥΔT
Enthalpy	H = U + pV; ΔH = qₚ	ΔU = qᵥ
ΔH and ΔU	ΔH = ΔU + Δn(g) RT	Gas moles only
Heat	q = C ΔT = c m ΔT	Cₘ = C/n
Ideal gas	Cₚ − Cᵥ = R	Monoatomic Cᵥ = 3R/2; diatomic 5R/2
Reaction enthalpy	Σ a ΔfH°(prod) − Σ b ΔfH°(react)	ΔfH° of an element in its reference state = 0
From bond enthalpies	Σ BE(reactants) − Σ BE(products)	Gases only; reverse order to ΔfH
Solution	ΔsolH = ΔlatticeH + ΔhydH	Lattice H is positive
Entropy	ΔS = q(rev)/T	Phase change: ΔS = ΔH/T
Spontaneity	ΔS(total) = ΔS(sys) + ΔS(surr) > 0	ΔS(surr) = −ΔH(sys)/T
Gibbs	ΔG = ΔH − TΔS	ΔG < 0 spontaneous (const T, p)
Crossover T	T = ΔH/ΔS	ΔG = 0 there
ΔG° and K	ΔG° = −RT ln K = −2.303 RT log K	R in J, ΔG° in J

0 · What JEE Main asks from this chapter

178 questions in JEE Main 2019–2026, 1 or 2 per paper, mostly easy and medium. More than half are numerical-answer questions. Grouped by type (counts are approximate):

Type	Share	Typical question
Work, heat, first law: expansions, adiabatic, cycles, heat capacity	~25%	5 mol He, 10 → 20 L reversibly at 300 K: maximum work
Thermochemistry: formation, combustion, Hess, phase changes, neutralisation, solution	~25%	ΔcH of C₂H₆, C and H₂ given: find ΔfH of ethane
Spontaneity: ΔS, ΔG, crossover T, ΔG° and K	~23%	ΔH = 400 kJ, ΔS = 0.2 kJ K⁻¹: spontaneous above?
ΔH ↔ ΔU and bomb calorimetry	~12%	Bomb calorimeter gives ΔU = −742 kJ: find ΔH
Bond enthalpy, atomisation, Born–Haber	~10%	Lattice enthalpy of LiF from a Born–Haber cycle
Theory: state/path functions, intensive/extensive	~5%	Number of intensive properties in a list

NCERT's own problems and exercises come back nearly word for word: exercise 5.17 in 2024, Problem 5.12 in 2024, exercise 5.8 in 2021, Problem 5.6 in 2021 and 2022. Do every one of them.

1 · System, surroundings and state

• System: the part of the universe being studied. Surroundings: the part of the rest that can interact with it. Boundary: the real or imaginary wall between them. Universe = system + surroundings.

System	Exchanges	Example
Open	matter and energy	reactants in an open beaker
Closed	energy only	reactants in a sealed copper or steel vessel
Isolated	neither	reactants in a thermos flask

• State is fixed by measurable bulk properties: p, V, T, n and composition. Only a few need to be fixed; the rest then follow.
• State function: its value depends only on the state, not on how the state was reached. p, V, T, U, H, S, G are state functions; their change is path independent.
• Path functions: q and w. Their sum q + w = ΔU is a state-function change.
• Thermodynamics deals with macroscopic systems moving between equilibrium states. It says nothing about the rate.

Extensive and intensive (NCERT puts this in 5.2.2):

Extensive (depends on amount)	Intensive (doesn't)
mass, volume, U, H, S, G, heat capacity C	T, p, density, any molar quantity (Vₘ, Cₘ), molarity, E°cell, molar mass, specific heat

A molar property χₘ = χ/n is intensive. Halve a gas with a partition: V halves, T doesn't.

Example (2023) · Volume, molar heat capacity, molarity, E°cell, Gibbs energy change, molar mass, mole: the intensive ones are molar heat capacity, molarity, E°cell and molar mass, so 4.

2 · Internal energy and the first law

Internal energy U is the total energy of the system (chemical, electrical, mechanical and so on). Only ΔU can be measured, never U itself. U changes when heat passes in or out, work is done on or by the system, or matter enters or leaves.

• Adiabatic: no heat exchange (adiabatic wall, thermos). Then ΔU = w(ad). Joule showed that the same work done adiabatically in any way (paddles or an immersion heater) gives the same change of state, so U is a state function.
• Heat q: energy exchanged because of a temperature difference. With no work at constant volume, ΔU = q.

IUPAC sign convention (chemistry): q and w are positive when energy goes into the system. Work done on the system is +; work done by it is −. Physics books use the old opposite sign for w; JEE chemistry papers use this one.

First law: the energy of an isolated system is constant (energy is neither created nor destroyed). For an isolated system q = 0, w = 0, ΔU = 0.

Example (NCERT Problem 5.1) · Work w done on the system, no heat: ΔU = w(ad), adiabatic wall. Heat q taken out, no work: ΔU = −q, conducting walls. Work w done by the system and heat q supplied: ΔU = q − w, closed system.

Example (2021, 2019) · A system does 200 J of work and absorbs 150 J of heat: ΔU = 150 − 200 = −50 J, magnitude 50. A spring is compressed with 10 kJ of work and 2 kJ escapes as heat: ΔU = −2 + 10 = +8 kJ.

Example (2026) · Gas goes X → Y absorbing 10 J and doing 18 J of work: ΔU = −8 J. Back Y → X releasing 6 J: ΔU = +8 = −6 + w, so w = +14 J, 14 J done on the gas.

Example (2025) · A liquid in an insulated closed vessel is stirred from outside: q = 0, w > 0 (work done on it), so ΔU > 0.

3 · Work of expansion (pV work)

Only mechanical pressure–volume work is considered. For a gas in a cylinder with a frictionless piston:

The minus sign makes w positive for compression (ΔV < 0) and negative for expansion.

• Several finite steps: w = −Σ pₑₓ ΔV, step by step.
• Reversible: pₑₓ differs from the gas pressure by only dp at every stage. The change can be reversed at any moment by an infinitesimal change; it passes through a series of equilibrium states, infinitely slowly. Every other process is irreversible.
• Reversible isothermal, ideal gas:

• Free expansion (into vacuum, pₑₓ = 0): w = 0, reversible or not. For an ideal gas Joule also found q = 0, so ΔU = 0 and T doesn't change.

For an isothermal ideal gas, ΔU = 0, so q = −w:

Process	q = −w
Free expansion	0
Irreversible, constant pₑₓ	pₑₓ(V₂ − V₁)
Reversible	2.303 nRT log(V₂/V₁)

The work done by the gas is the area under the path. Reversible expansion follows the isotherm and does the most work; a single step against the final pressure does the least (the accent strip). For compression it is the other way round: a single step at the final (high) pressure needs the most work.

Example (2025, 2026) · Order of |w| between the same two states at constant T: single-stage compression > reversible compression = reversible expansion > single-stage expansion. With multi-stage steps in between: |w|(1-stage compression) > |w|(multi-stage compression) > |w|(multi-stage expansion) > |w|(1-stage expansion).

Example (NCERT Problems 5.2–5.4) · 2 L of gas at 10 atm expands isothermally to 10 L at 25 °C. Into vacuum: q = w = 0. Against 1 atm: q = −w = 1 × 8 = 8 L atm. Reversibly (1 mol): q = −w = 2.303 × 1 × 0.08206 × 298 × log 5 = 39.37 L atm.

Example (2022) · 5 mol He, reversible isothermal, 300 K, 10 → 20 L: |w| = 2.303 × 5 × 8.3 × 300 × 0.3010 = 8630 J.

Example (2019) · 1 L → 10 L against 1 bar: w = −1 bar × 9 L = −9 L bar = −0.9 kJ (1 L bar = 100 J).

Example (2021) · 50 g Fe + HCl → FeCl₂ + H₂ at 25 °C, 1 bar. H₂ = 50/55.85 = 0.895 mol. Work by the gas = pΔV = nRT = 0.895 × 8.314 × 298 = 2218 J.

Adiabatic changes of an ideal gas · Beyond NCERT (NCERT gives only ΔU = w(ad))

Example (2023) · 1 mol expands reversibly and adiabatically from 300 K doing 3 kJ of work; Cᵥ = 20 J mol⁻¹ K⁻¹. −3000 = 20(T₂ − 300), so T₂ = 150 K.

Example (2024) · Cᵥ = 5/2 R, 298 K, 5 atm, expands adiabatically against 1 atm until its volume doubles. nCᵥ(T₂ − 298) = −1 × V₁ and V₁ = nR(298)/5, so (5/2)(T₂ − 298) = −59.6 and T₂ = 274 K.

Cyclic process: the system returns to its start, so ΔU = 0 and q = −w. The net work is the area enclosed by the cycle on a p–V plot (2024, 2025).

4 · Enthalpy

Most reactions run at constant (atmospheric) pressure, not constant volume. At constant p, ΔU = qₚ − pΔV, which rearranges to qₚ = (U₂ + pV₂) − (U₁ + pV₁). So define:

• H is a state function (U, p, V all are), so qₚ is path independent too.
• ΔH < 0: exothermic (heat evolved). ΔH > 0: endothermic (heat absorbed).
• For solids and liquids ΔH ≈ ΔU; the volume change is tiny. With gases, use the ideal gas law:

Δn(g) = moles of gaseous products − moles of gaseous reactants. Liquids and solids don't count.

Example (NCERT Problem 5.5) · H₂O(l) → H₂O(g) at 100 °C, ΔH = 41.00 kJ mol⁻¹, Δn(g) = 1. ΔU = 41.00 − 1 × 8.3 × 373 × 10⁻³ = 37.90 kJ mol⁻¹. Asked as "38" in 2021, 2022 and 2024.

Example (NCERT Problem 5.7) · A swimmer's 18 g water film (1 mol) at 298 K, ΔvapH = 44.01 kJ mol⁻¹. Heat needed = 44.01 kJ; ΔvapU = 44.01 − 2.48 = 41.53 kJ.

Example (2019) · Combustion of heptane(l): C₇H₁₆(l) + 11O₂ → 7CO₂ + 8H₂O(l). Δn(g) = 7 − 11 = −4, so ΔH − ΔU = −4RT.

Heat capacity:

• C depends on the size and nature of the system; a large C means a small temperature rise. Water has a large heat capacity.
• Molar heat capacity Cₘ: heat to raise 1 mol by 1 K. Specific heat c: heat to raise unit mass by 1 K.
• Cᵥ = (∂U/∂T)ᵥ gives qᵥ = CᵥΔT = ΔU; Cₚ = (∂H/∂T)ₚ gives qₚ = CₚΔT = ΔH.
• Ideal gas: for 1 mol ΔH = ΔU + RΔT, so

Cₚ is always greater than Cᵥ: at constant p some heat goes into expansion work (2026 asked this as a statement). For an ideal gas U and H depend only on T, and dU = Cᵥ dT for any process (2020).

Ideal gas	Cᵥ, Cₚ	γ = Cₚ/Cᵥ
Monoatomic (He, Ar)	3R/2, 5R/2	1.67
Diatomic (N₂, O₂), room T	5R/2, 7R/2	1.40

Example (NCERT exercise 5.9) · 60.0 g Al (2.22 mol), Cₘ = 24 J mol⁻¹ K⁻¹, 35 → 55 °C: q = 2.22 × 24 × 20 = 1.07 kJ.

Example (2022) · Cₚ = 20.785 J K⁻¹ mol⁻¹, so Cᵥ = 12.471. ΔU = 5000 J for 300 → 500 K: n = 5000 / (12.471 × 200) = 2 mol.

ΔH when Cₚ depends on T, and Kirchhoff's equation · Beyond NCERT

Example (2019) · ΔsubH of I₂ = 24 cal g⁻¹ at 200 °C; c(solid) 0.055, c(vapour) 0.031. Δc = 0.031 − 0.055 = −0.024, so at 250 °C: 24 − 0.024 × 50 = 22.8 cal g⁻¹.

5 · Measuring ΔU and ΔH: calorimetry

Bomb calorimeter (ΔU): a sealed steel bomb in a water bath. The substance burns in pure O₂. The volume can't change, so no work is done and the heat measured is qᵥ = ΔU. Heat from the reaction = −(heat capacity of calorimeter × ΔT).

Coffee-cup type calorimeter (ΔH): open to the atmosphere, so it measures qₚ = ΔrH, the heat (enthalpy) of reaction.

Example (NCERT Problem 5.6) · 1 g graphite in a bomb calorimeter, 298 → 299 K, C = 20.7 kJ K⁻¹. q = −20.7 kJ per gram, × 12 = −2.48 × 10² kJ mol⁻¹. Δn(g) = 0 (C + O₂ → CO₂), so ΔH = ΔU. 2021 asked the reverse: heat from 1 g = 248/12 ≈ 21 kJ.

Example (2023) · 0.3 g ethane (0.01 mol) in a bomb, ΔT = 0.5 K, C = 20 kJ K⁻¹. ΔU = −10 kJ / 0.01 mol = −1000 kJ mol⁻¹. C₂H₆(g) + 7/2 O₂ → 2CO₂ + 3H₂O(l), Δn(g) = −2.5. ΔH = −1000 − 2.5 × 8.3 × 300 × 10⁻³ = −1006 kJ mol⁻¹.

Example (NCERT exercise 5.8, 2021) · Cyanamide: NH₂CN(s) + 3/2 O₂ → N₂ + CO₂ + H₂O(l), ΔU = −742.7 kJ mol⁻¹. Δn(g) = 2 − 1.5 = +0.5. ΔH = −742.7 + 0.5 × 8.314 × 298 × 10⁻³ = −741.5 kJ mol⁻¹.

6 · Reaction enthalpy and Hess's law

Standard state of a substance at a given temperature: its pure form at 1 bar. Standard values carry ° (ΔH°); data are usually at 298 K. Liquid ethanol at 298 K: pure liquid at 1 bar. Iron at 500 K: pure iron at 1 bar.

Phase changes happen at constant T and p, and all three below are positive:

Change	Definition (1 mol, standard pressure)	Water / examples
Fusion ΔfusH	solid → liquid at the m.p.	6.00 kJ mol⁻¹ at 273 K
Vaporisation ΔvapH	liquid → vapour at constant T	40.79 kJ mol⁻¹ at 373 K; 44.01 at 298 K
Sublimation ΔsubH	solid → vapour directly	dry ice 25.2 (195 K); naphthalene 73.0

• ΔsubH = ΔfusH + ΔvapH (at the same T). (2021: 2.8 + 98.2 = 101.)
• The size of the enthalpy follows the strength of intermolecular forces: water (H-bonds) needs more heat to vaporise than acetone (dipole–dipole).
• Freezing releases the same heat as melting absorbs.

Example (NCERT Problem 5.8) · 1 mol water at 100 °C → ice at 0 °C. Cool the liquid: −18 × 4.2 × 100 = −7.56 kJ. Freeze: −6.00 kJ. ΔH = −13.56 kJ mol⁻¹ = ΔU (no gas involved).

Example (NCERT exercise 5.10, 2025) · Water at 10 °C → ice at −10 °C, ΔfusH 6.03, Cₚ(l) 75.3, Cₚ(s) 36.8 J mol⁻¹ K⁻¹: −0.753 − 6.03 − 0.368 = −7.15 kJ mol⁻¹. In symbols (2025): −10(100x + y + z) J, with x in kJ.

Standard enthalpy of formation ΔfH°: the enthalpy change when 1 mol of a compound forms from its elements in their reference states (most stable form at 25 °C and 1 bar: H₂(g), O₂(g), C(graphite), S(rhombic), Br₂(l)). ΔfH° of an element in its reference state is zero.

• H₂(g) + ½O₂(g) → H₂O(l): ΔfH° = −285.8 kJ mol⁻¹. C(graphite) + 2H₂ → CH₄(g): −74.81. 2C + 3H₂ + ½O₂ → C₂H₅OH(l): −277.7.
• Not formation enthalpies: CaO + CO₂ → CaCO₃ (made from compounds, not elements); H₂ + Br₂(l) → 2HBr (two moles formed; ΔfH°(HBr) = ½ × −72.8 = −36.4).

Example (NCERT) · CaCO₃ → CaO + CO₂: (−635.1) + (−393.5) − (−1206.9) = +178.3 kJ mol⁻¹, endothermic, so limestone must be heated. Fe₂O₃ + 3H₂ → 2Fe + 3H₂O(l): 3(−285.83) − (−824.2) = −33.3 kJ mol⁻¹.

Thermochemical equations: a balanced equation with its ΔrH and the physical (and allotropic) state of each substance.

1. Coefficients are moles, never molecules.
2. ΔrH (kJ mol⁻¹) is per mole of reaction as written. Halve the equation and ΔrH halves (−33.3 → −16.6 for the Fe₂O₃ equation above). Enthalpy is extensive.
3. Reverse the equation and ΔrH changes sign: N₂ + 3H₂ → 2NH₃, −91.8; 2NH₃ → N₂ + 3H₂, +91.8.

Hess's law of constant heat summation: if a reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the steps, at the same temperature. It follows from H being a state function.

Example (NCERT) · C + O₂ → CO₂, −393.5 (i); CO + ½O₂ → CO₂, −283.0 (ii). Reverse (ii) and add to (i): C + ½O₂ → CO, ΔH = −393.5 + 283.0 = −110.5 kJ mol⁻¹. This can't be measured directly because some CO₂ always forms. 2019 asked it as x = y + z.

The method: write the target equation; for each species in it, pick the given equation that contains it; multiply or reverse that equation to match; add. Species that appear in no target cancel.

7 · Enthalpies of different types of reactions

Combustion ΔcH°: per mole of substance burnt completely, everything in standard states. Always negative.

• C₄H₁₀(g) + 13/2 O₂ → 4CO₂ + 5H₂O(l): ΔcH° = −2658.0 kJ mol⁻¹ (cooking gas).
• C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O: −2802.0 kJ mol⁻¹. The body gets energy from food by the same overall reaction, through many enzyme steps.

From combustion to formation (most common JEE type): the formation reaction = (combustion of the elements) − (combustion of the compound).

Example (NCERT Problem 5.9) · Benzene: ΔcH = −3267.0; CO₂ −393.5, H₂O(l) −285.83. ΔfH = 6(−393.5) + 3(−285.83) + 3267.0 = +48.51 kJ mol⁻¹ (asked in 2025).

Example (NCERT exercise 5.5) · CH₄: −393.5 + 2(−285.8) + 890.3 = −74.8 kJ mol⁻¹. Ethane (2022): 2(−394) + 3(−286) + 1560 = −86 kJ mol⁻¹. Propane (2022): 3(−393.5) + 4(−285.8) + 2220 = −104 kJ mol⁻¹.

Example (2022) · 3C₂H₂(g) → C₆H₆(l), ΔcH of benzene −3268 and acetylene −1300: ΔH = 3(−1300) − (−3268) = −632 kJ mol⁻¹. Reactants' combustion minus products' combustion.

Atomisation ΔaH°: the enthalpy change to break 1 mol of a substance completely into gaseous atoms.

• H₂(g) → 2H(g): 435.0 kJ mol⁻¹. CH₄(g) → C(g) + 4H(g): 1665 kJ mol⁻¹. Na(s) → Na(g): 108.4 kJ mol⁻¹, the same as its sublimation enthalpy.
• For a diatomic molecule, ΔaH = bond dissociation enthalpy. For Br₂(l), atomisation also includes vaporisation, so it is larger than the Br–Br bond enthalpy (2020).

Bond enthalpy ΔbondH:

• Bond dissociation enthalpy: the change when 1 mol of a covalent bond in a gaseous compound breaks into gaseous products. H–H 435.0, Cl–Cl 242, O=O 428 kJ mol⁻¹ (NCERT's text; its table gives 498 for O=O).
• Mean bond enthalpy: in a polyatomic molecule, successive bonds differ. CH₄ → CH₃ + H needs 427; then 439, 452 and 347 (sum 1665). The mean C–H = 1665/4 = 416 kJ mol⁻¹. Mean values change only a little between compounds.

Approximate, and valid only when every species is gaseous. If a liquid or solid is involved, first add its vaporisation or sublimation.

Example (2021) · C₂H₆ → C₂H₄ + H₂ with C–C 347, C=C 611, C–H 414, H–H 436. Broken: C–C + 6 C–H = 2831. Formed: C=C + 4 C–H + H–H = 2703. ΔrH = +128 kJ mol⁻¹.

Example (NCERT exercise 5.15) · CCl₄(g) → C(g) + 4Cl(g). ΔfH(CCl₄, g) = −135.5 + 30.5 = −105. ΔH = 715 + 2 × 242 − (−105) = 1304 kJ mol⁻¹; C–Cl = 1304/4 = 326 kJ mol⁻¹.

Example (2025) · ΔfH of H(g) 220, O(g) 250, H₂O(g) −242. H₂O(g) → 2H + O: 2(220) + 250 + 242 = 932, so O–H = 466 kJ mol⁻¹.

Example (2021) · SF₆: ΔfH SF₆(g) −1100, S(g) 275, F(g) 80. SF₆ → S + 6F: 275 + 480 + 1100 = 1855, so S–F = 309 kJ mol⁻¹.

Lattice enthalpy ΔlatticeH: the enthalpy change when 1 mol of an ionic solid dissociates into its gaseous ions. NaCl(s) → Na⁺(g) + Cl⁻(g): +788 kJ mol⁻¹. It can't be measured directly, so it is found from a Born–Haber cycle: by Hess's law, the enthalpy changes round a cycle add to zero.

NaCl: 411.2 + 108.4 + 496 + 121 − 348.6 = +788 kJ mol⁻¹. The internal energy of the same change is smaller by 2RT (Δn(g) = 2): +783 kJ mol⁻¹.

Example (2021) · KCl: 436.7 + 89.2 + 419.0 + ½(243.0) − 348.6 = 717.8, so 718 kJ mol⁻¹. LiF (2026): 594 + 155 + 520 + ½(150) − 313 = 1031 kJ mol⁻¹. Halve the bond enthalpy: one Cl atom per formula unit.

Ionisation energy vs ionisation enthalpy (NCERT's box): IE and electron affinity are defined at 0 K; the enthalpies at T differ by 5/2 RT. ΔiH = IE + 5/2 RT; ΔegH = −A − 5/2 RT.

Enthalpy of solution ΔsolH: the change when 1 mol dissolves in a specified amount of solvent. At infinite dilution, ion–ion interactions are negligible. For an ionic solid, the lattice breaks up (endothermic) and the ions are hydrated (exothermic):

• NaCl: +788 − 784 = +4 kJ mol⁻¹; very little heat change (2020 asked for ΔhydH = −784).
• For most salts ΔsolH > 0, so solubility rises with temperature. A very high lattice enthalpy can stop dissolution; many fluorides are less soluble than the chlorides.

Enthalpy of dilution: depends on the starting concentration and the solvent added. HCl(g) in 10, 25, 40 and ∞ mol water: −69.01, −72.03, −72.79, −74.85 kJ mol⁻¹. The value approaches a limit at infinite dilution. HCl·25aq + 15aq → HCl·40aq: −72.79 − (−72.03) = −0.76 kJ mol⁻¹. So ΔsolH depends on the amount of solvent (2025).

Example (2024) · ΔsolH of CuSO₄ −70 and CuSO₄·5H₂O +12. CuSO₄ + 5H₂O → CuSO₄·5H₂O: −70 − 12 = −82 kJ. (Anhydrous dissolves; reverse the hydrate's dissolution.)

Enthalpy of neutralisation · Beyond NCERT

• Strong acid + strong base: H⁺(aq) + OH⁻(aq) → H₂O, ΔH = −57.1 kJ per mol of water (−57.3 in some papers), whatever the acid and base. Per mol of water formed, not per mol of acid.
• A weak acid or base gives less heat: part goes into ionising it. ΔionH = 57.3 − 55.3 = 2 kJ mol⁻¹ for CH₃COOH (2022).
• Temperature rise: find the mol of water from the limiting H⁺ or OH⁻; heat = 57.1 × that; ΔT = heat / (total mass × 4.18).

Example (2021) · 400 mL 0.2 M H₂SO₄ (80 mmol H⁺) + 600 mL 0.1 M NaOH (60 mmol OH⁻). Water = 0.060 mol, heat = 3.426 kJ, mass 1000 g. ΔT = 3426 / (1000 × 4.18) = 0.82 K → 82 (× 10⁻²).

8 · Spontaneity, entropy and Gibbs energy

The first law puts no limit on the direction of a change. Heat flows from hot to cold, a gas fills its container, carbon burns to CO₂; none reverses by itself.

• Spontaneous means having the potential to proceed without outside help. It says nothing about the rate: H₂ and O₂ can sit mixed for years and the reaction is still spontaneous.
• A spontaneous process is irreversible and can be reversed only by an outside agency.

Is a decrease in enthalpy the criterion? No. Many spontaneous reactions are exothermic (NH₃, HCl, H₂O formation), but some are endothermic and still spontaneous: ½N₂ + O₂ → NO₂, ΔH = +33.2; C(graphite) + 2S(l) → CS₂(l), +128.5. And two gases diffuse into each other in an isolated box with ΔH = 0.

Entropy S is a measure of disorder (randomness). It is a state function.

• Solid < liquid < gas for a given substance; the crystalline solid at 0 K is the most ordered.
• Entropy rises with temperature, with more gas moles, with more particles, with dissolving or mixing.
• Heat added at a lower temperature causes more disorder than the same heat at a higher temperature:

At equilibrium entropy is at its maximum and ΔS(total) = 0. For isothermal expansion of an ideal gas ΔU = 0 both ways, but ΔS(total) ≠ 0 for the irreversible path: ΔS tells reversible from irreversible, ΔU can't.

Example (NCERT Problem 5.10) · Entropy decreases when a liquid crystallises. It increases when a solid is warmed from 0 to 115 K, when 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g), and when H₂(g) → 2H(g).

Example (2021) · Entropy decreases in: freezing water at 0 °C and at −10 °C, N₂ + 3H₂ → 2NH₃, adsorption of CO on lead. It increases when NaCl dissolves. Adsorption always lowers the entropy of the gas.

Example (NCERT Problem 5.11) · 4Fe + 3O₂ → 2Fe₂O₃: ΔS(sys) = −549.4 J K⁻¹ mol⁻¹, ΔH = −1648 kJ mol⁻¹. ΔS(surr) = 1648 × 10³ / 298 = 5530. ΔS(total) = +4980.6 J K⁻¹ mol⁻¹, so it is spontaneous: the heat released disorders the surroundings.

Example (NCERT exercise 5.22) · Forming 1 mol H₂O(l), ΔfH = −286 kJ: ΔS(surr) = 286 000 / 298 = +959.7 J K⁻¹.

Entropy changes you calculate · Beyond NCERT (NCERT gives only q(rev)/T)

Example (2019) · 1 kg ice at 273 K → steam at 383 K: 334/273 + 4.2 ln(373/273) + 2491/373 + 2.0 ln(383/373) = 1.22 + 1.32 + 6.68 + 0.05 = 9.26 kJ K⁻¹. Each step is ∫C dT/T or ΔH/T (2025 asked the formula).

Example (2023) · NaCl melts with ΔH = 30.4 kJ and ΔS = 28.4 J K⁻¹: T = 30 400 / 28.4 = 1070 K. At a transition, ΔG = 0.

Gibbs energy G (the Gibbs function), for closed and open systems where both H and S change:

G is an extensive state function. Why ΔG decides: at constant T and p, ΔS(surr) = −ΔH(sys)/T, so TΔS(total) = TΔS(sys) − ΔH(sys) = −ΔG(sys). Hence ΔS(total) > 0 is the same as ΔG < 0.

• ΔG < 0: spontaneous. ΔG > 0: non-spontaneous. ΔG = 0: equilibrium.
• TΔS is energy not available for useful work; −ΔG is the net energy available for useful (non-pV) work, the free energy.

NCERT Table 5.4, effect of temperature:

ΔH, ΔS	ΔG	Spontaneous?
−, +	always −	at all T
−, −	− at low T	below T = ΔH/ΔS
+, +	− at high T	above T = ΔH/ΔS
+, −	always +	never

"Low" and "high" are relative: for some reactions room temperature is already "high". A +, + reaction with a small ΔS needs a high T, one reason reactions are run hot.

Example (NCERT exercise 5.17, asked in 2024) · 2A + B → C, ΔH = 400 kJ mol⁻¹, ΔS = 0.2 kJ K⁻¹ mol⁻¹. Spontaneous above T = 400 / 0.2 = 2000 K.

Example (2021) · FeO + C → Fe + CO. ΔH = −110.5 + 266.3 = 155.8 kJ. ΔS = 27.28 + 197.6 − 57.49 − 5.74 = 161.65 J K⁻¹. T = 155 800 / 161.65 = 964 K.

Example (NCERT exercise 5.19) · 2A(g) + B(g) → 2D(g), ΔU = −10.5 kJ, ΔS = −44.1 J K⁻¹, 298 K. Δn(g) = −1, so ΔH = −10.5 − 2.48 = −12.98 kJ. ΔG = −12.98 + 298 × 0.0441 = +0.16 kJ: not spontaneous.

Example (2020) · A(l) → 2B(g), ΔU = 2.1 kcal, ΔS = 20 cal K⁻¹, 300 K, R = 2 cal. ΔH = 2.1 + 2 × 2 × 300 × 10⁻³ = 3.3 kcal. ΔG = 3.3 − 300 × 0.020 = −2.7 kcal.

Second law: in an isolated system energy is constant, so an increase in entropy is the natural direction of spontaneous change. It explains why exothermic reactions are so often spontaneous: the heat they release raises the entropy of the surroundings.

Third law: the entropy of a pure crystalline substance approaches zero as T approaches 0 K (perfect order). Not true for solutions or supercooled liquids. It lets absolute entropies be found from thermal data (adding q(rev)/T from 0 to 298 K); standard entropies then give ΔrS° by a Hess-type sum: ΔrS° = ΣS°(products) − ΣS°(reactants). Elements do not have S° = 0 at 298 K, unlike ΔfH°.

Example (2025) · S° of X₂ 70, Y₂ 50, XY₅ 110 J K⁻¹ mol⁻¹. ½X₂ + 5/2 Y₂ → XY₅: ΔS = 110 − 35 − 125 = −50 J K⁻¹. With ΔH = −35 kJ, equilibrium at T = 35 000 / 50 = 700 K.

Partial derivatives · Beyond NCERT (2025) · (∂G/∂T)ₚ = −S; (∂G/∂p)ₜ = V; (∂H/∂T)ₚ = Cₚ; (∂U/∂T)ᵥ = Cᵥ.

9 · Gibbs energy and equilibrium

A reversible reaction can go either way only if the system sits at the minimum of G. So at equilibrium ΔrG = 0. With all species in standard states:

• ΔG° < 0 ⇔ K > 1; ΔG° = 0 ⇔ K = 1; ΔG° > 0 ⇔ K < 1 (2019 asked for the wrong match).
• Strongly endothermic: K ≪ 1, little product. Strongly exothermic: ΔG° large and negative, K ≫ 1, near completion. ΔS shifts this either way.
• From ΔH° and ΔS° you get ΔG° and so K at any temperature; from a measured K you get ΔG°.
• R = 8.314 J K⁻¹ mol⁻¹, so ΔG° comes out in J. Convert kJ to J before using it.

Example (NCERT Problem 5.12, asked in 2024) · 3/2 O₂ → O₃, Kp = 2.47 × 10⁻²⁹ at 298 K. ΔG° = −2.303 × 8.314 × 298 × log(2.47 × 10⁻²⁹) = +163 kJ mol⁻¹.

Example (NCERT Problem 5.13) · ΔG° = −13.6 kJ mol⁻¹ at 298 K: log K = 13 600 / (2.303 × 8.314 × 298) = 2.38, K = 2.4 × 10².

Example (NCERT exercise 5.20, 2024) · K = 10 at 300 K: ΔG° = −2.303 × 8.314 × 300 = −5744 J = −5.74 kJ mol⁻¹ (2024 answer 57 × 10⁻¹).

Example (2022) · 2O₃ ⇌ 3O₂, 50% dissociated at 300 K, 1 atm. From 1 mol O₃: 0.5 O₃, 0.75 O₂, total 1.25. p(O₃) = 0.4, p(O₂) = 0.6. Kp = 0.216/0.16 = 1.35. ΔG° = −8.3 × 300 × ln 1.35 = −747 J mol⁻¹.

NCERT's Problem 5.14 (N₂O₄, 50% dissociated at 60 °C, Kp = 1.33) prints ΔG° = −763.8 kJ mol⁻¹. The working gives −8.314 × 333 × 2.303 × 0.1239 = −790 J mol⁻¹, under 1 kJ. Trust the method, not that printed answer.

log K against 1/T · Beyond NCERT (2020, 2026)

Slope −ΔH°/2.303R, intercept ΔS°/2.303R. For an exothermic reaction K falls as T rises.

Example (2020) · K = 10 at 25 °C, 100 at 100 °C. 1 = (ΔH°/19.15) × (75 / (298 × 373)), so ΔH° = 28.4 kJ mol⁻¹. ΔG°(298) = −2.303 × 8.314 × 298 × 1 = −5.71 kJ; ΔG°(373) = −2.303 × 8.314 × 373 × 2 = −14.29 kJ.

Other links · Beyond NCERT · ΔG = ΔG° + RT ln Q (Equilibrium); ΔG° = −nFE°cell (Electrochemistry, Class 12).

10 · Traps

Trap	What goes wrong	Do this
Sign of w	Using the physics sign (work by gas +)	Chemistry: w on the system is +; expansion w < 0
Δn(g)	Counting liquid water or solids	Gas moles only; check H₂O(l) or (g)
Units of RT	ΔH in kJ, RT in J (8.314 × 298 = 2478 J)	Convert to one unit before adding
ΔG = ΔH − TΔS	ΔH in kJ, ΔS in J K⁻¹	Convert ΔS to kJ K⁻¹ (or ΔH to J)
Bond enthalpy order	Products − reactants, as for ΔfH	Reactants − products; all species gaseous
Combustion to formation	Writing compound − elements	Elements' combustion − compound's combustion
Halving equations	ΔfH(HBr) = −72.8	Per 1 mol formed: −36.4
Born–Haber	Using the full Cl₂ bond enthalpy	½ ΔbondH per Cl atom; lattice H is positive
Bomb calorimeter	Taking qᵥ as ΔH	qᵥ = ΔU; then ΔH = ΔU + Δn(g) RT
Calorimeter sign	q of reaction = +CΔT	q of reaction = −CΔT (heat goes to the calorimeter)
Neutralisation heat	Per mol of H₂SO₄	Per mol of water; use the limiting H⁺ or OH⁻
Free expansion	Using −pΔV with p of the gas	pₑₓ = 0, w = 0; ideal gas: q = 0, ΔU = 0
Irreversible isothermal	Using nRT ln(V₂/V₁)	−pₑₓΔV with the external pressure
log vs ln	Forgetting 2.303 with log	ln x = 2.303 log x
"Spontaneous above T"	Reporting T = ΔS/ΔH	T = ΔH/ΔS, units matched
S° of elements	Setting S°(O₂) = 0 like ΔfH°	Only ΔfH° of elements is 0; S° is not

11 · Numbers worth knowing

• R = 8.314 J K⁻¹ mol⁻¹ = 0.08206 L atm K⁻¹ mol⁻¹ = 0.0831 L bar = 2 cal K⁻¹ mol⁻¹ (≈ 1.987).
• 1 L atm = 101.3 J; 1 L bar = 100 J; 1 cal = 4.184 J.
• RT at 298 K = 2.48 kJ mol⁻¹; at 300 K, 2.49; at 373 K, 3.10.
• 2.303 RT at 298 K = 5.71 kJ mol⁻¹; at 300 K, 5.74.
• Neutralisation (strong/strong) −57.1 kJ mol⁻¹; water: ΔfusH 6.0, ΔvapH 40.8 (373 K) and 44.0 (298 K) kJ mol⁻¹; c(water) 4.18 J g⁻¹ K⁻¹ (75.3 J mol⁻¹ K⁻¹).

Substance	ΔfH° (kJ mol⁻¹)
H₂O(l)	−285.8
H₂O(g)	−241.8
CO₂(g)	−393.5
CO(g)	−110.5
CH₄(g)	−74.8
C₆H₆(l)	+49.0 (Problem 5.9 works out +48.5; the page prints −48.51, a sign slip)
C₂H₅OH(l)	−277.7
CaCO₃(s)	−1206.9
CaO(s)	−635.1
Fe₂O₃(s)	−824.2
NaCl(s)	−411.2
HBr(g)	−36.4
NH₃(g)	−46.1
NO(g)	+90.3

NO(g) has a positive ΔfH (NCERT exercise 5.21): it is thermodynamically unstable with respect to N₂ and O₂. NO₂ (+16 from 90 − 74, NCERT's numbers) is less unstable.

Bond	kJ mol⁻¹
H–H	435.8
C–H	414
C–C	347
C=C	611
C≡C	837
C=O	741
O–H	464
O=O	498
N–H	389
N≡N	946
Cl–Cl	243
H–Cl	431

12 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• Einstein on thermodynamics: the only physical theory of universal content he was convinced would never be overthrown.
• Joule's paddle and immersion-heater experiments (1840–50) showed that adiabatic work changes the state by the same amount whatever way it is done.
• Enthalpy comes from the Greek enthalpien, "to warm" (heat content).
• Why ΔrH matters: planning the heating or cooling to keep an industrial reaction at constant temperature, and finding how K depends on temperature.
• Combustion matters in industry, rocketry and everyday life; the body releases food energy by a series of enzyme-controlled steps with the same overall result as combustion.