Overview — Module 4: Chemical Thermodynamics

Chemical reactions are accompanied by energy changes — as heat in combustion, as light in photosynthesis, as electricity in batteries. The formation of glucose by photosynthesis absorbs solar energy; burning kerosene or cooking gas releases heat and light. Even dry cells convert stored chemical energy to electrical energy through carefully controlled redox reactions. This lesson focuses on reactions where heat is evolved or absorbed, developing the language and laws of thermochemistry.

You will learn systems and surroundings, exothermic and endothermic reactions, thermochemical equation conventions, the first law of thermodynamics, internal energy and enthalpy (and their relationship for gaseous reactions), standard enthalpies of formation, combustion, neutralization, atomisation, phase transition, solution and ionization, Lavoisier–Laplace and Hess's laws, and bond enthalpy calculations with worked examples from the NIOS textbook.

Section 1: Basic Terms and Processes (9.1)

9.1.1 System and Surroundings

The system is the part of the universe under study; everything else is surroundings. A reaction mixture in a beaker is the system; the beaker and room are surroundings.

• Isolated system: exchanges neither matter nor energy (perfect thermos flask).
• Closed system: exchanges energy only, not matter (stoppered flask).
• Open system: exchanges both matter and energy (open flask, plants, animals).

9.1.2–9.1.3 State Functions and Properties

State functions depend only on initial and final state, not the path (pressure, temperature, internal energy, enthalpy). Distance travelled is path-dependent; separation between two cities is a state function. Changing a gas from (p₁, T₁) to (p₂, T₂) gives the same ΔT and Δp whether you heat first then compress, or compress first then heat — but heat and work exchanged along the way differ.

Extensive properties depend on system size (mass, volume, total internal energy, heat content). Intensive properties do not (temperature, pressure, density, viscosity, refractive index). Density is intensive because it is mass per unit volume — doubling the sample doubles both mass and volume, leaving density unchanged. This distinction matters when scaling reactions from laboratory to industrial plant size.

9.1.4 Types of Processes

• Isothermal: temperature constant (ice melting at 273 K, 1 atm) — heat supplied or removed to maintain T.
• Adiabatic: no heat exchange (q = 0); temperature changes (acid + base in thermos). All enthalpy change appears as temperature rise.
• Reversible: infinitesimally slow; system always at equilibrium; can be exactly reversed (Fig. 9.3 — piston on liquid-vapour equilibrium, infinitesimal pressure change causes slow condensation or evaporation at constant T).
• Irreversible: rapid changes; non-uniform T and P; cannot be exactly undone — all spontaneous real reactions.

Adiabatic and isothermal are ideal limits. A reaction in an open test tube is neither perfectly isothermal (some heat escapes) nor perfectly adiabatic (some heat exchanges with air) — but ΔH measured under controlled calorimetry approximates the ideal case.

Standard state: 1 bar pressure, substance in most stable form at specified temperature — used to compare enthalpies.

Reversible processes are idealised limits — real laboratory reactions are irreversible because they proceed at finite speed with friction, mixing, and temperature gradients. Nevertheless, thermodynamic state functions like ΔH remain valid because they depend only on initial and final states, not whether the path was reversible.

Fig. 9.2 illustrates that pressure and temperature differences between initial and final states are independent of whether the change occurred via path I, II, or III. This path-independence is what makes Hess's law possible later in the chapter.

Section 2: Exothermic, Endothermic and Thermochemical Equations (9.2–9.3)

9.2 Exothermic and Endothermic Reactions

Exothermic: heat evolved to surroundings (Zn + HCl, quick lime + water, fuel combustion). Test tube feels hot. Endothermic: heat absorbed from surroundings (NH₄Cl, KNO₃ in water, Ba(OH)₂ + NH₄Cl). Test tube feels cold.

9.3 Thermochemical Equations

Equations showing heat change with physical states: (g), (l), (s), (aq). Allotropes specified: C(graphite). ΔH negative = exothermic; positive = endothermic.

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔH = −891 kJ. H₂(g) + I₂(g) → 2HI(g); ΔH = +52.2 kJ. If coefficients are multiplied, ΔH is multiplied by the same factor: 2H₂ + O₂ → 2H₂O; ΔH = 2(−242) = −484 kJ.

Physical state matters enormously: burning methane produces different ΔH depending on whether water forms as liquid or steam — liquid H₂O releases more heat because condensation releases additional energy. Always include (g), (l), (s), (aq) in thermochemical equations. Fractional coefficients are allowed when ΔH is scaled accordingly — ½N₂ + ³⁄₂H₂ → NH₃ with ΔH = −46 kJ refers to exactly those stoichiometric amounts.

Section 3: First Law, Internal Energy and Enthalpy (9.4)

First law: Energy cannot be created or destroyed. Total energy of universe or isolated system is constant.

Internal energy (U) is the sum of translational, vibrational, rotational, electronic and nuclear energies of all atoms, molecules and ions in the system — a state function. Absolute U cannot be measured; only ΔU = U₂ − U₁. Internal energy changes through heat flow into/out of the system and work done on/by the system. Heat and work are not state functions — they depend on path. Two routes between the same states can give identical ΔU but different combinations of q and w.

Work of expansion at constant external pressure p: when volume expands from V₁ to V₂, the system does work w = −pΔV on surroundings. Compression (ΔV negative) means positive work on the system. Combining with heat absorption gives the complete energy balance via the first law.

Example: q = +50 kJ, w = −30 kJ (work done by system) → ΔU = +50 + (−30) = +20 kJ. At constant volume: ΔU = q_v.

Intext 9.1: For ½N₂ + ³⁄₂H₂ → NH₃, Δn_g = 1 − 2 = −1. At 298 K, ΔU = ΔH − Δn_gRT = −46 − (−1)(8.314×298/1000) kJ ≈ −43.5 kJ. When Δn_g is negative (fewer gas moles in products), ΔH is more negative than ΔU — contraction of gas volume releases pV-type work to surroundings.

Sign convention summary: heat given to system increases U (+q); work done on system increases U (+w); work done by system decreases U (w negative in ΔU = q + w). A bomb calorimeter measures q_v at constant volume; an open beaker reaction at atmospheric pressure effectively measures ΔH.

Section 4: Standard Enthalpies (9.5)

Δ_rH° = H°_products − H°_reactants. Negative → exothermic (CH₄ combustion −890.4 kJ). Positive → endothermic (H₂ + I₂ → 2HI, +52.5 kJ).

• Combustion (Δ_combH°): complete burning of 1 mol in O₂ — C₂H₅OH: −1365.6 kJ mol⁻¹.
• Neutralization (Δ_neutH°): H⁺(aq) + OH⁻(aq) → H₂O(l); ≈ −57 kJ mol⁻¹ for strong acid + strong base.
• Atomisation (Δ_aH°): 1 mol → gaseous atoms. C(graphite) → C(g): 716.68 kJ. For liquids = enthalpy of vaporisation.
• Transition (Δ_trsH°): phase change — sublimation Δ_subH°, vaporisation Δ_vapH°, fusion Δ_fusH°, allotrope change C(graphite) → C(diamond): +1.90 kJ.
• Solution (Δ_solH°): 1 mol solute in specified moles of solvent — HCl(g) + 10H₂O: −69.5 kJ mol⁻¹.
• Ionization (Δ_ionH°): weak electrolyte fully ionizes — HCN → H⁺ + CN⁻: +43.7 kJ mol⁻¹.

Standard enthalpies are tabulated at 298 K and 1 bar. Δ_combH° values are large and negative — useful for comparing fuel energy content (ethanol −1365.6 kJ mol⁻¹ per mole of ethanol). Neutralization enthalpy is less negative when a weak acid or weak base is involved because energy is consumed ionizing the weak electrolyte before H⁺ and OH⁻ can combine.

Phase transition enthalpies form a hierarchy: sublimation (solid → gas) equals fusion (solid → liquid) plus vaporisation (liquid → gas) at the same temperature (Hess's law applied to phases). Graphite-to-diamond transition is endothermic (+1.90 kJ) — diamond is metastable at 1 bar, which explains why graphite is the standard state for carbon.

Using formation data for any reaction: identify all products and reactants, multiply each Δ_fH° by stoichiometric coefficient, sum products, subtract sum of reactants. Elements in standard state contribute zero. Example 9.2 strategy: construct formation reaction for C₂H₆ by combining combustion equations of C, H₂, and C₂H₆ itself — multiply (1) by 2, (2) by 3, add, subtract (3).

Section 5: Laws of Thermochemistry (9.6)

Lavoisier–Laplace law: reversing a reaction reverses the sign of ΔH. N₂ + O₂ → 2NO; ΔH = +180.5 kJ ↔ 2NO → N₂ + O₂; ΔH = −180.5 kJ.

Hess's law: enthalpy change is independent of path — depends only on initial and final states. Thermochemical equations add/subtract like algebraic equations.

CO from C + ½O₂ cannot be measured directly (incomplete combustion). Using: (1) C + O₂ → CO₂, ΔH₁° = −393.5; (3) CO + ½O₂ → CO₂, ΔH₃° = −283.0; then ΔH₂° = ΔH₁° − ΔH₃° = −110.5 kJ mol⁻¹.

Example 9.1: Glucose combustion −2840 kJ/mol (180 g) → photosynthesis reverse needs 2840×(1.08/180) = 17.04 kJ for 1.08 g. Example 9.2: Ethane Δ_fH° = −96 kJ mol⁻¹ from Hess combination.

Intext 9.2 checks: (a) enthalpy of formation is per mole not per gram; (b) neutralization releases heat (exothermic, not absorbed); (c) Δ_fH° in CO₂ equation is indeed formation enthalpy. Butane combustion: 29 g = 0.5 mol of C₄H₁₀ → 0.5 × 2658 = 1329 kJ released. For 2H₂S + SO₂ → 3S + 2H₂O: Δ_rH° = [0 + 2(−289.9)] − [2(−20.6) + (−296.9)] = −62.3 kJ (exothermic).

Hess's law is the computational backbone of thermochemistry — without it, enthalpies of formation from combustion data, lattice energies from Born–Haber cycles (Lesson context), and industrial energy balances would be impossible to construct from measurable steps.

Section 6: Bond Enthalpies (9.7)

Energy changes arise from breaking bonds in reactants and forming bonds in products. H₂ → 2H; ΔH = +435 kJ mol⁻¹ is bond dissociation enthalpy (specific bond in specific molecule). Bond enthalpy is the average over equivalent bonds (O–H in H₂O: average of 502 and 427 = 464.5 kJ mol⁻¹).

Table 9.1 lists average bond enthalpies: H–H 435, C–H 415, C–C 356, C=C 598, C≡C 832, H–Cl 431, C=O 723 kJ mol⁻¹. Method (a) formation data and method (b) bond data should agree approximately (Example 9.4: +353 vs +351 kJ).

Procedure for bond enthalpy problems (Example 9.3): (1) draw structural formula, (2) list bonds broken in reactants and bonds formed in products, (3) look up B.E. values, (4) apply Δ_rH = ΣB.E.(reactants) − ΣB.E.(products). For CH₃ + Cl₂ → CH₃Cl + HCl: broken = 3 C–H + Cl–Cl; formed = 2 C–H + C–Cl + H–Cl (net: break 1 C–H and Cl–Cl, form C–Cl and H–Cl).

Intext 9.3: bond enthalpy ≠ bond dissociation energy for polyatomic molecules (average vs specific); Hess's law says overall ΔH equals sum of step enthalpies, not just the last step. H₂ + Cl₂ → 2HCl from bond data: (435 + 242) − 2(431) = −185 kJ (exothermic).

Limitations of bond enthalpy method: averages ignore environment (O–H in H₂O vs CH₃OH differ); only rigorous for gas-phase reactions where all species are gaseous; formation enthalpy method is more accurate for general stoichiometry but requires tabulated Δ_fH° values.

Calculating N–H bond enthalpy in NH₃ (Intext 9.3): use atomisation energies of N₂ and H₂ plus Δ_fH° of NH₃ — the energy to form three N–H bonds from atoms equals the enthalpy of the reverse atomisation process divided by three. This links formation enthalpies, atomisation enthalpies, and individual bond strengths into one coherent calculation framework that NIOS expects students to master.

Exam Connections and Chapter Summary

Terminal exercise topics include: heat evolved burning 1 g ethanol given Δ_combH° per mole; calculating Δ_fH° of C₂H₂ from combustion and formation data of CO₂ and H₂O; propane combustion enthalpy from given formation values; Hess's law multi-step problems. Intext 9.1 false statement: H₂ + Cl₂ → 2HCl + 185 kJ is exothermic (heat released), not endothermic — always check ΔH sign convention.

Key skills: identify system type; assign signs to q and w; convert between ΔH and ΔU using Δn_gRT; calculate Δ_rH° from Δ_fH° table; apply Hess's law by equation manipulation; compute Δ_rH from bond enthalpies for gas-phase reactions. Remember Lavoisier–Laplace when reversing photosynthesis/combustion pairs. Strong acid–strong base neutralization is always ~−57 kJ — a favourite constant.

Energy changes in chemistry power our world — from the −890 kJ mol⁻¹ released burning methane to the +2840 kJ mol⁻¹ required to synthesise glucose in photosynthesis. Batteries convert chemical ΔG (related to ΔH at constant T) to electrical work; explosives release stored bond energy in microseconds. Thermochemistry is the quantitative foundation for all these applications.

Thermodynamics provides the energy bookkeeping for chemistry: the first law tracks internal energy, enthalpy is the practical heat measure at constant pressure, standard enthalpies tabulate formation and related processes, Hess's law solves inaccessible reactions, and bond enthalpies explain energy changes at the molecular level. Mastering sections 9.4–9.7 equips you for all NIOS numerical problems on heat of reaction, formation, and combustion.