📌 Snapshot
- Establishes thermodynamics as the study of energy transformations in macroscopic systems at equilibrium — based only on initial and final states, not on rate or mechanism.
- Builds the toolkit: system/surroundings, state functions (U, H, S, G), first law (ΔU = q + w), enthalpy (H = U + pV), heat capacities (Cp − Cv = R), Hess's law, and entropy-based criteria for spontaneity.
- Closes with the third law (S = 0 for a pure crystal at 0 K) and the link ΔG° = −RT ln K connecting thermodynamics to equilibrium.
- CUET tests this chapter heavily for sign conventions, ΔH ↔ ΔU conversion via Δng RT, Hess-cycle arithmetic, and spontaneity prediction from ΔH/ΔS/T.
📖 Detailed Notes
2.1 Core concepts
- System and surroundings. System = part of universe under observation; surroundings = remaining universe that can interact with it; the wall between them is the boundary, real or imaginary (NCERT §5.1.1, p. 137).
- Types of system. Open = exchanges both matter and energy (open beaker); closed = exchanges only energy (sealed copper vessel); isolated = exchanges neither (thermos flask) (NCERT §5.1.2, p. 137–138).
- State of a system. Specified by state variables p, V, T, n; once a minimum set is fixed, all other macroscopic properties follow. State functions depend only on present state, not on path taken (NCERT §5.1.3, p. 138). The minimum set is independent of the choice of path used to reach that state.
- Internal energy U as state function. Sum of all forms of energy (chemical, mechanical, electrical, kinetic, potential, vibrational, rotational) of the system; can change by heat flow, work, or matter transfer. Joule's adiabatic experiments (1840–50) showed wad is path-independent, so ΔU = wad in an adiabatic process — a numerically equal amount of work whether performed mechanically, electrically or by friction (NCERT §5.1.4, p. 138–139).
- Sign convention (IUPAC). q positive when heat is absorbed by the system; w positive when work is done on the system; both negative when energy leaves the system. ΔU > 0 if the system gains energy and < 0 if it loses energy (NCERT §5.1.4, p. 139).
- First law of thermodynamics. ΔU = q + w; energy of an isolated system is constant (q = 0, w = 0 ⇒ ΔU = 0). Energy can neither be created nor destroyed, only converted from one form to another (NCERT §5.1.4(c), p. 140, Eq. 5.1).
- Pressure–volume work. w = −pext ΔV for a single-step (irreversible) compression; for reversible isothermal expansion of an ideal gas, wrev = −2.303 nRT log(V₂/V₁). The reversible work is the maximum work the gas can deliver because pext is set infinitesimally lower than pin at every step (NCERT §5.2.1, p. 140–142, Eq. 5.2 & 5.5).
- Free expansion. pext = 0, so w = 0; for an isothermal ideal-gas free expansion ΔU = 0 (no temperature change) and q = 0 too (NCERT §5.2.1, p. 142).
- Special cases. Constant volume: ΔU = qV (because w = 0). Adiabatic: q = 0, so ΔU = wad. Constant pressure: ΔH = qp (NCERT §5.2.1, p. 142).
- Enthalpy. H = U + pV is a state function; at constant pressure ΔH = qp = heat absorbed. ΔH = ΔU + Δng RT relates enthalpy and internal energy for gas-phase reactions; Δng = moles of gaseous products − moles of gaseous reactants (NCERT §5.2.2(a), p. 143, Eq. 5.7 & 5.10). For solids/liquids, ΔH ≈ ΔU because pΔV is negligible.
- Extensive vs intensive. Extensive properties (mass, V, U, H, S, heat capacity) depend on amount; intensive properties (T, p, density, molar quantities, viscosity, refractive index) do not. Ratios of two extensive properties (density = m/V; molar volume V/n) are intensive (NCERT §5.2.2(b), p. 144).
- Heat capacity. q = C ΔT; molar heat capacity Cm = C/n; specific heat c = q/(m ΔT). For an ideal gas, Cp − Cv = R; the extra heat at constant pressure goes into pV expansion work. For monatomic ideal gas Cv = (3/2)R, Cp = (5/2)R, γ = 5/3 (NCERT §5.2.2(c–d), p. 144–145, Eq. 5.13).
- Calorimetry. ΔU measured in a constant-volume bomb calorimeter (no work because ΔV = 0; heat evolved heats the steel bomb + water bath); ΔH measured in a constant-pressure (coffee-cup) calorimeter, where qp = ΔrH. Calorimeter constant must be known from a calibration run (NCERT §5.3, p. 145–146).
- Reaction enthalpy and standard states. ΔrH = Σai Hproducts − Σbi Hreactants. Standard state of a substance = its pure form at 1 bar (and usually 298 K); denoted by superscript °. For solutes, standard state is hypothetical 1 mol kg⁻¹ ideal solution (NCERT §5.4(a), p. 146–147, Eq. 5.14).
- Phase-change enthalpies. ΔfusH° (always positive, e.g., ice → water +6.00 kJ/mol at 273 K), ΔvapH° (water → steam +40.79 kJ/mol at 373 K), ΔsubH° (always positive) — all measured per mole at constant T and p (NCERT §5.4(b), p. 147–148).
- Standard enthalpy of formation, ΔfH°. Enthalpy change when 1 mol of a compound forms from its elements in their reference (most stable) states at 1 bar and 298 K. ΔfH° of an element in its reference state = 0 by convention (NCERT §5.4(c), p. 149–150, Eq. 5.15).
- Thermochemical equations. Must specify physical state; coefficients refer to moles; reversing the equation reverses the sign of ΔrH; multiplying by a factor multiplies ΔrH by the same factor (NCERT §5.4(d), p. 150–151).
- Hess's law of constant heat summation. Total ΔrH for a reaction is the sum of ΔrH of the steps into which it can be divided; allows indirect calculation of unmeasurable enthalpies, e.g. C + ½O₂ → CO via C → CO₂ and CO → CO₂ (NCERT §5.4(e), p. 151–152, Eq. 5.16). Direct measurement of C + ½O₂ → CO is impossible because some CO₂ always forms.
- Enthalpies of various reactions. Combustion ΔcH° (always exothermic, e.g. butane −2658, glucose −2802 kJ/mol); atomization ΔaH° (= bond dissociation enthalpy for diatomics); bond enthalpy ΔbondH°; mean bond enthalpy (e.g. ΔC–HH° = ¼·1665 = 416 kJ/mol in CH₄); ΔrH° = Σ bond enthalpies(reactants) − Σ bond enthalpies(products) for gas-phase reactions (NCERT §5.5(a–c), p. 152–155, Eq. 5.17).
- Lattice enthalpy and Born–Haber cycle. ΔlatticeH° = enthalpy change when 1 mol of ionic solid dissociates into gaseous ions; computed indirectly via Born–Haber cycle using sublimation + ionization + dissociation + electron-gain + lattice steps. ΔsolH° = ΔlatticeH° + ΔhydH° — most ionic solids dissolve in water because hydration enthalpy is exothermic and large enough to overcome lattice enthalpy (NCERT §5.5(d–e), p. 155–156).
- Spontaneity. A spontaneous process has the potential to proceed without external help; it is irreversible. Decrease in enthalpy is not a sufficient criterion — endothermic reactions like ½N₂ + O₂ → NO₂ (ΔrH = +33.2 kJ/mol) are also spontaneous. Mixing of two gases is spontaneous with ΔH ≈ 0 (NCERT §5.6 & §5.6(a), p. 157–158).
- Entropy S. A state function and measure of disorder/randomness; ΔS = qrev/T. Gaseous state has highest S, crystalline solid lowest. For a spontaneous process in an isolated system, ΔStotal = ΔSsys + ΔSsurr > 0; ΔStotal = 0 at equilibrium (NCERT §5.6(b), p. 158–159, Eq. 5.18 & 5.19).
- Gibbs energy. G = H − TS; at constant T, ΔG = ΔH − TΔS. ΔG < 0 ⇒ spontaneous; ΔG > 0 ⇒ non-spontaneous; ΔG = 0 ⇒ equilibrium. ΔG is the "free" (useful work) energy of the reaction; ΔG also equals the maximum non-pV work available from the system at constant T, p (NCERT §5.6(c), p. 160–161, Eq. 5.20 & 5.21).
- Second law of thermodynamics. Entropy of an isolated system increases for a spontaneous process; explains why exothermic reactions tend to be spontaneous (heat released raises Ssurr, ΔSsurr = −ΔHsys/T) (NCERT §5.6(d), p. 161).
- Third law of thermodynamics. Entropy of a pure crystalline substance approaches zero as T → 0 K; lets us calculate absolute entropies by integrating qrev/T from 0 K to T. Allows tabulation of standard absolute entropies S° at 298 K (NCERT §5.6(e), p. 161–162).
- ΔG° and equilibrium constant. At equilibrium ΔrG = 0; ΔrG° = −RT ln K = −2.303 RT log K. Large negative ΔrG° ⇒ K >> 1 (reaction near completion); large positive ⇒ K << 1; ΔrG° ≈ 0 ⇒ K ≈ 1 (NCERT §5.7, p. 162, Eq. 5.23).
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| System | Part of the universe chosen for study | 137 |
| Surroundings | Remainder of the universe in contact with the system | 137 |
| Open system | Exchanges both matter and energy with surroundings | 137 |
| Closed system | Exchanges only energy (not matter) with surroundings | 137 |
| Isolated system | Exchanges neither matter nor energy with surroundings | 138 |
| State function | Property whose value depends only on the present state, not on path (e.g. U, H, S, G, p, V, T) | 138–139 |
| Path function | Property dependent on the route taken between states (q, w) | 139 |
| Adiabatic process | Process with no heat exchange between system and surroundings (q = 0) | 138 |
| Isothermal process | Process at constant temperature (ΔT = 0) | 141 |
| Isobaric process | Process at constant pressure | 142 |
| Isochoric process | Process at constant volume (w = 0) | 142 |
| Internal energy U | Sum of all forms of energy of the system; a state function | 138 |
| First law | ΔU = q + w; energy of an isolated system is constant | 140 |
| Enthalpy H | H = U + pV; ΔH = qp at constant pressure | 143 |
| Extensive property | Depends on amount/size of matter (m, V, U, H, C) | 144 |
| Intensive property | Independent of amount (T, p, density, molar quantities) | 144 |
| Heat capacity C | q = C ΔT; Cp − Cv = R for an ideal gas | 144–145 |
| Reversible process | Process carried out infinitely slowly through equilibrium states; pext = pin ± dp | 141 |
| Irreversible process | Sudden, finite-step process; usually pext ≠ pin | 141 |
| Hess's law | ΔrH for a reaction is independent of the path — equals sum of ΔrH of the steps | 151 |
| Standard enthalpy of formation ΔfH° | ΔH when 1 mol of a compound forms from its elements in their reference states at 1 bar | 149 |
| Bond dissociation enthalpy | ΔH to break 1 mol of covalent bonds in a gaseous compound to give gaseous products | 153 |
| Lattice enthalpy ΔlatticeH° | ΔH when 1 mol of ionic solid dissociates into its gaseous ions | 155 |
| Entropy S | Measure of disorder; ΔS = qrev/T; state function | 158–159 |
| Second law | Entropy of an isolated system increases in a spontaneous process | 161 |
| Gibbs energy G | G = H − TS; ΔG < 0 ⇒ spontaneous at constant T, p | 160 |
| Third law | Entropy of a pure crystalline substance is zero at 0 K | 161 |
2.3 Diagrams / processes to remember
- Fig. 5.2 (p. 137): Open, closed and isolated systems — three labelled containers showing matter/energy exchange. Open beaker shows arrows for both matter (vapour) and heat; closed system shows only heat arrows through a sealed lid; isolated system (thermos) has neither.
- Fig. 5.5(a–c) (p. 140–141): pV-plots of work done — single-step (rectangle of area pext·ΔV), finite-step (staircase whose total area is the sum of rectangles), and infinite-step reversible (smooth hyperbola whose shaded area equals nRT ln(V₂/V₁), the maximum possible work). The diagrams visually illustrate why wrev > wirrev for expansion.
- Fig. 5.7 (p. 145): Bomb calorimeter — steel bomb in a water bath; constant-volume ΔU measurement. The bomb is filled with O₂ at high pressure, electrical ignition burns the sample, and the temperature rise of the surrounding water gives qV which equals ΔU directly. Used for combustion enthalpies of food, fuels.
- Fig. 5.8 (p. 146): Coffee-cup / constant-pressure calorimeter for ΔH measurement of dissolution and neutralisation reactions in aqueous solution. Polystyrene cup acts as adiabatic shield; measured qp = ΔH directly.
- Fig. 5.9 (p. 155): Born–Haber enthalpy diagram for NaCl — sublimation (+108.4 kJ/mol) + ionization (+496) + ½ bond dissociation (+121) + electron-gain (−348.6) + lattice (−788) = −411.2 kJ/mol formation enthalpy. The vertical-arrow construction shows how an indirect cycle gives the otherwise unmeasurable ΔlatticeH°.
- Fig. 5.10(a) & (b) (p. 158): Enthalpy diagrams for exothermic (products below reactants) and endothermic (products above reactants) reactions; the vertical arrow downwards/upwards represents −ΔH or +ΔH respectively.
- Fig. 5.11 (p. 158): Diffusion of two gases A and B before and after partition removal — visual for entropy increase. The number of accessible microstates rises sharply on mixing, so ΔS_mix > 0 even though ΔH_mix ≈ 0 for ideal gases.
- Table 5.4 (p. 162): Effect of ΔH, ΔS and T on the sign of ΔG (spontaneity matrix): ΔH<0, ΔS>0 → spontaneous at all T; ΔH>0, ΔS<0 → non-spontaneous at all T; ΔH<0, ΔS<0 → spontaneous at low T; ΔH>0, ΔS>0 → spontaneous at high T. A four-row table that summarises every spontaneity case.
- PE diagram of catalysed vs uncatalysed reaction (implicit, p. 158): identical reactants and products, but the activation barrier lowered by the catalyst; the overall ΔH (and ΔG) is unchanged because state functions depend only on initial/final states.
2.4 Common confusions / NTA trap points
- Sign convention. IUPAC: w is positive when work is done on the system. Older physics books use the opposite sign — NTA distractors exploit this. If the gas expands (V₂ > V₁), w < 0 (work done by gas).
- ΔH vs ΔU. They are equal only when Δng = 0 (no change in moles of gas) or for solids/liquids; otherwise ΔH = ΔU + Δng RT. For H₂(g) + ½O₂(g) → H₂O(l), Δng = −1.5, so ΔH < ΔU.
- q is path-dependent, but qV = ΔU and qp = ΔH make them appear like state functions only under those specific constraints.
- Free expansion of an ideal gas. w = 0 (because pext = 0) and q = 0 and ΔU = 0 — students often write w ≠ 0 because volume changes. No work because external pressure is zero, no heat because temperature does not change.
- ΔfH° of an element in its reference state is zero, not "unity" or "different for each element" — a frequent NCERT exercise trap (Ex. 5.3). But ΔfH° of an element in a non-reference state (e.g., O₃ from O₂, or C(diamond) instead of C(graphite)) is NOT zero.
- Spontaneity ≠ fast. A reaction can be spontaneous (ΔG < 0) yet extremely slow (H₂ + O₂ mixture at room T). Thermodynamics is silent on rate; kinetics handles speed.
- Entropy of surroundings. ΔSsurr = −ΔHsys/T (heat lost by system at temperature T is gained by surroundings). Negative ΔSsys can still give ΔStotal > 0 if ΔHsys is sufficiently negative.
- Bond enthalpy method gives ΔrH for gas-phase reactions only. If any reactant/product is in solid or liquid state, you must add appropriate phase-change enthalpies; otherwise the calculated ΔrH is wrong.
- Reversible work is the maximum the system can deliver; irreversible work is always smaller in magnitude. Students sometimes use the same w for both — wrong.
- ΔG° is a standard-state quantity at 1 bar (and the chosen T). ΔrG itself depends on actual concentrations: ΔrG = ΔrG° + RT ln Q; equilibrium reached when Q = K and ΔrG = 0, giving ΔrG° = −RT ln K.
- Heat capacity is extensive; specific heat and molar heat capacity are intensive. A common slip in MCQ definitions.
🎯 Practice MCQs
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Q1. A thermos flask containing hot coffee, sealed with its lid, is best classified as which kind of thermodynamic system?
▸ Show answer & explanation
Answer: C
A thermos flask permits no exchange of either matter or energy with the surroundings — the textbook explicitly cites it as an example of an isolated system. A closed system would allow heat exchange.
Q2. Which one of the following sets contains only state functions?
▸ Show answer & explanation
Answer: C
U, H and S are all state functions (path-independent). Heat (q) and work (w) are path-dependent quantities and therefore not state functions.
Q3. For the isothermal reversible expansion of one mole of an ideal gas, the work done is given by:
▸ Show answer & explanation
Answer: B
For an isothermal reversible expansion of an ideal gas, wrev = −2.303 nRT log(V₂/V₁). Option (A) is for an irreversible single-step process with constant pext.
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Q4. The enthalpy change ΔH and internal energy change ΔU for a reaction are related by ΔH = ΔU + ΔngRT. For the reaction N₂(g) + 3H₂(g) → 2NH₃(g) at 298 K, the relation between ΔH and ΔU is:
▸ Show answer & explanation
Answer: C
Δng = (moles of gaseous products) − (moles of gaseous reactants) = 2 − 4 = −2. Hence ΔH = ΔU + (−2)RT = ΔU − 2RT, so ΔH < ΔU.
Q5. For one mole of an ideal gas, the relation between molar heat capacities at constant pressure (Cp) and at constant volume (Cv) is:
▸ Show answer & explanation
Answer: A
Starting from ΔH = ΔU + Δ(pV) and pV = RT for one mole, the textbook derives Cp − Cv = R. Cp is always larger than Cv because at constant pressure some heat does pV work.
Q6. Given the following reactions at 298 K: (i) C(graphite) + O₂(g) → CO₂(g); ΔrH° = −393.5 kJ mol⁻¹ (ii) CO(g) + ½O₂(g) → CO₂(g); ΔrH° = −283.0 kJ mol⁻¹ Using Hess's law, the enthalpy change for C(graphite) + ½O₂(g) → CO(g) is:
▸ Show answer & explanation
Answer: B
Reverse (ii) and add to (i): ΔrH° = −393.5 + (+283.0) = −110.5 kJ mol⁻¹.
Q7. Standard enthalpy of formation, ΔfH°, of an element in its reference state is, by convention:
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Answer: C
NCERT fixes ΔfH° of every element in its most stable state of aggregation at 1 bar and 298 K to be zero by convention.
Q8. In which of the following processes does the entropy of the system decrease?
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Answer: B
When a liquid crystallizes, molecules attain a more ordered state, so entropy decreases.
Q9. For a chemical reaction at constant T and p, ΔH > 0 and ΔS > 0. The reaction will be:
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Answer: C
ΔG = ΔH − TΔS. With ΔH positive and ΔS positive, the TΔS term grows with T; for sufficiently large T, TΔS outweighs ΔH and ΔG becomes negative.
Q10. The standard Gibbs energy change ΔrG° of a reaction at 300 K is −5.74 kJ mol⁻¹. Taking R = 8.314 J K⁻¹ mol⁻¹, the value of the equilibrium constant K (use ΔrG° = −2.303 RT log K) is approximately:
▸ Show answer & explanation
Answer: B
log K = −ΔrG°/(2.303 RT) = 5740/(2.303 × 8.314 × 300) ≈ 1.00, so K ≈ 10.
Q11. The expansion of an ideal gas into vacuum is called free expansion. For an isothermal free expansion:
▸ Show answer & explanation
Answer: A
Because pext = 0, no work is done (w = 0); because the temperature does not change for an isothermal process and ΔU depends only on T for an ideal gas, ΔU = 0; from ΔU = q + w with both ΔU and w zero, q must also be zero.
Q12. Which of the following statements about a catalyst is INCORRECT in a thermodynamic sense?
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Answer: C
A catalyst cannot make a non-spontaneous reaction (ΔG > 0) become spontaneous because it does not alter ΔG. It only lowers the activation energy of both directions equally, leaving K and ΔG unchanged.
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