Thermochemistry: Hess's Law & Bond Enthalpies Why thermochemistry matters (NEET view) Every chemical reaction exchanges energy with surroundings. If heat is released, it is exothermic; if absorbed, endothermic. Thermochemistry lets us quantify these heats (enthalpy changes, H ). This helps predict fuel energy, the heat of acid–base neutralization, and how much heat a refrigerant absorbs. You will meet Hess’s law problems and bond-enthalpy estimates regularly in NEET. Exothermic vs endothermic: energy profile and heat flow. Products lower than reactants (ΔH < 0): exothermic; higher (ΔH > 0): endothermic. Heat is energy in transit due to temperature difference; temperature measures average kinetic energy. Enthalpy H and internal energy U are different state functions: H = U + PV. Their changes match only under specific conditions (e.g., ΔH = Q at constant pressure). Heat (Q) and temperature (T) are the same; enthalpy (H) and internal energy (U) are interchangeable. Foundations: U, H, and Q at constant pressure First Law Internal energy change equals heat added plus work done on the system. Enthalpy is convenient for processes at constant pressure (lab/atmospheric conditions). Definition of enthalpy Enthalpy at constant pressure At constant external pressure, the heat exchanged equals the enthalpy change. For reactions with gases only: n g = moles of gaseous products − moles of gaseous reactants. Gas-phase link between ΔH and ΔU neet-alert Sign convention: Heat released by system (exothermic) ⇒ H < 0 . Heat absorbed (endothermic) ⇒ H > 0 . NEET traps often flip this. It is the opposite: exothermic ⇒ H < 0 ; endothermic ⇒ H > 0 . Keep a mental picture of the energy profile (products lower vs higher). Exothermic reactions have positive H , and endothermic have negative H . Standard state and standard enthalpy of formation Reference state at 1 bar pressure; temperature usually taken as 298.15 K (25 C) for tabulated H 298 values. Standard state Enthalpy change when 1 mol of a compound forms from its elements in their standard states at 1 bar and 298.15 K. By convention, H f = 0 for elements in their standard states (e.g., O2(g), N2(g), graphite C). Allotropes that are not the standard state (e.g., diamond, ozone, white phosphorus) have nonzero H f . Standard enthalpy of formation ( H f ) Using standard enthalpies of formation Hess’s law in formation-enthalpy form ( are stoichiometric coefficients). Only the element in its standard state has H f = 0 . Examples: C(graphite) = 0 but C(diamond) 0 ; O 2(g) = 0 but ozone O 3(g) 0 ; red P is standard, white P is not. neet-alert H2O(l) −285.8 CO2(g) −393.5 NH3(g) −46.1 CH4(g) −74.8 NaCl(s) −411.2 Glucose, C6H12O6(s) −1273 Selected standard enthalpies of formation at 298 K (kJ mol⁻¹) Substance (state) ΔH f° Species Common types of reaction enthalpies Know the names, signs, and examples. Combustion is always exothermic; atomization is always endothermic. Neutralization for strong acid + strong base is nearly constant. Type Definition (at 1 bar, 298 K unless stated) Typical sign Example (balanced) ΔH type Key enthalpy changes: definition, sign, and example Δ H f ° (formation) From elements (standard states) to 1 mol compound varies C(graphite) + O2(g) → CO2(g) Δ H c ° (combustion) Complete burning in O2 to CO2 + H2O negative CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Δ H neutralization Heat change per mol of H2O formed in acid–base reaction ≈ −57.1 kJ mol−1 (strong–strong) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Δ H solution Dissolving 1 mol of solute in large solvent NaCl(s) → Na+(aq) + Cl−(aq) Δ H hydration Hydration of 1 mol of gaseous ions to aqueous ions negative Na+(g) + aq → Na+(aq) Δ H atomization Form 1 mol of gaseous atoms from element positive 1/2 Cl2(g) → Cl(g) Δ H sublimation Solid → gas positive I2(s) → I2(g) Δ H fusion Solid → liquid (melting) positive H2O(s) → H2O(l) Δ H vaporization Liquid → gas (boiling) positive H2O(l) → H2O(g) Enthalpy of neutralization: nearly constant for strong–strong pairs due to full ionic dissociation. Acid–base reaction forming water and a salt; strong acid + strong base gives ≈ −57.1 kJ per mol H2O. ≈ −57.1 kJ mol−1 holds for strong acid + strong base. If acid or base is weak, part of the released heat is used to ionize it, so magnitude is smaller (less exothermic). Enthalpy of neutralization is always −57 kJ mol−1. Measuring heats: calorimetry essentials Calorimetry Heat change from temperature rise/fall of a known mass or a known heat capacity. Bomb calorimeter schematic: a constant-volume setup for accurate combustion heats; the water bath temperature rise gives Q. remember Food labels (kcal) and fuel ratings come from calorimetry. Cement curing releases heat (important in mass concrete). Refrigerants absorb large enthalpy of vaporization to cool. Rocket propellants are chosen for high combustion enthalpy per kg. Hess’s Law (1840): add steps, add enthalpies Hess’s Law says: if a reaction can be written as a sum of steps, its enthalpy change equals the sum of the steps’ enthalpies. Reason: enthalpy is a state function — only initial and final states matter. Hess cycle: two different pathways from reactants to products with arrows labelled ΔH; both paths show the same net ΔH. Hess's law cycle diagram: energy-level style. Left: Reactants at top, right: Products at bottom. Path A: direct arrow labeled ΔH(reaction). Path B: multi-step arrows with ΔH1, ΔH2, ΔH3 summing to same level as products. Clean vector, red arrows, labels for state function idea, white background, no text inside figure beyond labels. 2026-05-26T17:04:23.098Z gpt-image-2 Worked Hess example: ΔHf° of CO(g) Given data (at 298 K): (1) C(graphite) + O2(g) → CO2(g); ΔH° = −393.5 kJ mol−1. (2) CO(g) + 1/2 O2(g) → CO2(g); ΔH° = −283.0 kJ mol−1. Target: C(graphite) + 1/2 O2(g) → CO(g); ΔHf°(CO) = ? Reverse reaction (2): CO2(g) → CO(g) + 1/2 O2(g); ΔH° = +283.0 kJ mol−1. Add (1) and reversed (2): Net: C(graphite) + 1/2 O2(g) → CO(g). Sum enthalpies: ΔH° = (−393.5) + (+283.0) = −110.5 kJ mol−1. Therefore, ΔHf°[CO(g)] = −110.5 kJ mol−1. Bond enthalpies: estimating ΔH from bonds broken and formed Bond enthalpy (bond dissociation energy) Heat required to break 1 mol of a specific bond in the gas phase into gaseous atoms or radicals. For polyatomic molecules, average over similar bonds gives the mean bond enthalpy. Break (input, +) minus make (release, −) gives approximate ΔH. Use gaseous species and average values. Bond-enthalpy estimate C–C 348 C=C 614 C≡C 839 C–H 413 O–H 463 O=O 498 N≡N 945 C=O (average) 745 Bond Bond E (kJ mol−1) Selected average bond enthalpies (kJ mol⁻¹, gas phase) Example: combustion of methane (gas phase estimate). Reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g). Bonds broken: 4×C–H = 4×413 = 1652; 2×O=O = 2×498 = 996; total broken = 2648 kJ. Bonds formed: CO2 has 2×C=O = 2×745 = 1490; 2H2O has 4×O–H = 4×463 = 1852; total formed = 3342 kJ. ΔH ≈ 2648 − 3342 = −694 kJ mol−1. Actual ΔH°c(CH4) is more exothermic (≈ −890 kJ mol−1 to H2O(l), ≈ −802 kJ mol−1 to H2O(g)) because bond enthalpies are averages and phases matter. Bond-enthalpy walkthrough for CH4 combustion: count bonds broken (CH, O=O) and bonds formed (C=O, O–H), then apply ΔH ≈ Σbroken − Σformed. gpt-image-2 2026-05-26T17:04:23.399Z Four-panel vector diagram on white background. Panel 1: CH4(g) molecule with 4 C–H bonds highlighted (4×413). Panel 2: 2 O2(g) with O=O bonds (2×498). Panel 3: CO2(g) with two C=O (2×745). Panel 4: 2 H2O(g) with four O–H (4×463). Red arrows for 'broken', blue for 'formed'; show arithmetic totals beside panels. Clean textbook style. Average bond enthalpies ignore molecular environment and phases. They give approximations; accurate ΔH° needs standard enthalpies of formation under specified states. Bond enthalpy estimates give the exact ΔH of a reaction. Used to define fuels’ calorific values and to test bond-enthalpy estimates. Oxidation in O2 producing CO2 and H2O with large negative ΔH. Born–Haber cycle (qualitative): lattice energy via Hess’s Law For an ionic solid like NaCl(s), lattice energy ( U latt ) is the enthalpy change when 1 mol of solid forms from its gaseous ions. We cannot measure it directly, but we can construct a Hess cycle combining known steps to reach the same final state, then solve for U latt . Born–Haber cycle of NaCl(s): vertical energy-level diagram with steps for sublimation of Na(s), ionization of Na(g), 1/2Cl2 bond dissociation, electron affinity of Cl(g), and lattice formation from Na+(g) + Cl−(g). Energy-level diagram for NaCl. Top to bottom arrows: Na(s) → Na(g) (sublimation), Na(g) → Na+(g) + e− (ionization), 1/2 Cl2(g) → Cl(g) (bond dissoc), Cl(g) + e− → Cl−(g) (electron affinity, downward), finally Na+(g)+Cl−(g) → NaCl(s) (large downward lattice energy). Clear labels, red arrows up, blue arrows down, vector style. 2026-05-26T17:04:23.581Z gpt-image-2 NaCl(s) Born–Haber steps (sign conventions) Na(s) → Na(g): Δ H sub (endothermic, +). Na(g) → Na+(g) + e−: Ionization enthalpy (endothermic, +). 1/2 Cl2(g) → Cl(g): 1/2 bond dissociation of Cl2 (endothermic, +). Cl(g) + e− → Cl−(g): Electron affinity (exothermic, −). Na+(g) + Cl−(g) → NaCl(s): Lattice enthalpy of formation (exothermic, large −). Sum of all steps equals Δ H f °[NaCl(s)] (Hess’s Law), so U latt can be obtained. Real-life and bio link Metabolism: exothermic pathways release energy stored in food, powering ATP formation and body heat. Your body is a thermochemical machine: cellular respiration of glucose is exothermic, producing heat (keeps you warm) and chemical energy (ATP) that drives muscle and nerve function. remember Quick NEET traps and techniques Always specify states: H2O(l) vs H2O(g) changes ΔH by vaporization/condensation enthalpy. Use ΔH°f tables for exact answers; use bond enthalpies for estimates (gas phase, averages). For gases, check Δng in ΔH = ΔU + ΔngRT if both ΔH and ΔU appear. Strong acid–strong base neutralization ≈ −57.1 kJ per mol H2O; weaker pairs give smaller magnitude. Only standard-state elements have ΔH°f = 0 (graphite, O2, N2, red P). Allotropes like diamond/ozone carry nonzero values. Glossary: Thermochemistry essentials Reference at 1 bar; data usually at 298.15 K. standard conditions STP (note: pressure 1 bar, not necessarily 1 atm) Standard state Standard enthalpy of formation of 1 mol from elements in standard states. Δ H f ° Standard enthalpy of complete combustion (to CO2 and H2O). Δ H c ° Enthalpy change per mol H2O formed in acid–base reaction; ≈ −57.1 kJ mol−1 for strong–strong. Δ H neutralization Heat change on dissolving a solute in large excess solvent. Δ H solution Heat change when gaseous ions become hydrated (aqueous). Δ H hydration Heat to form gaseous atoms from element (always positive). Δ H atomization Total ΔH is the sum of step ΔH values for any path from reactants to products. Hess’s Law Energy to break a specific bond in the gas phase; average values used for estimates. Bond enthalpy Average over similar bonds in a polyatomic molecule (e.g., 4 C–H in CH4). Mean bond enthalpy Hess cycle for ionic solids combining atomization, ionization, electron affinity, and lattice energy. Born–Haber cycle Enthalpy change for forming 1 mol of ionic solid from gaseous ions (exothermic as defined here). Lattice energy SMILES (helpful references): CH4 (methane) = C; O2 = O=O; CO2 = O=C=O; H2O = O. Use these to visualize bonds when counting.