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Pro 🔒~15 min

Thermochemistry & Hess's Law

Energy in chemical reactions — exothermic vs endothermic

How it works

Thermochemistry studies energy changes in chemical reactions. Enthalpy (H) is heat flow at constant pressure. Exothermic reactions release heat (ΔH < 0): combustion, neutralization, most phase changes from gas → liquid → solid. Endothermic reactions absorb heat (ΔH > 0): photosynthesis, dissolving NH₄NO₃, most decompositions. Bond energy approach: breaking bonds requires energy (endothermic), forming bonds releases energy (exothermic). ΔH = energy in (break bonds) - energy out (form bonds). Hess's Law: ΔH for a reaction equals the sum of ΔH for any series of steps with the same overall equation — enthalpy is a state function (path-independent). Standard enthalpies of formation (ΔHf°) are measured for 1 mol from elements in standard state.

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Step-by-step

  1. Select a reaction type and watch the energy diagram animate.
  2. In Combustion mode, CH₄ + O₂ → CO₂ + H₂O releases −890 kJ/mol.
  3. Drag the activation energy hill to see transition state.
  4. Switch to Hess's Law mode to build complex reactions from simpler steps — add or reverse steps and watch ΔH accumulate.

Key formulas

  • ΔHrxn=∑ΔHf(products)−∑ΔHf(reactants)\Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})Standard enthalpy of reaction from formation enthalpies
  • ΔHrxn=∑BE(bonds broken)−∑BE(bonds formed)\Delta H_{rxn} = \sum BE(\text{bonds broken}) - \sum BE(\text{bonds formed})ΔH from bond energies (energy in = bonds broken, out = bonds formed)
  • q=mcΔT(calorimetry)q = mc\Delta T \quad (\text{calorimetry})Heat measured in a calorimeter: mass × specific heat × ΔT

Frequently asked questions

Burning 1 mol of propane (C₃H₈) releases 2220 kJ. How much heat is released burning 44 g of propane? (M = 44 g/mol).
44 g / 44 g/mol = 1 mol → 2220 kJ released.
Using bond energies: H₂ + Cl₂ → 2HCl. H-H = 432, Cl-Cl = 243, H-Cl = 431 kJ/mol. Calculate ΔH.
Bonds broken: H-H (432) + Cl-Cl (243) = 675. Bonds formed: 2×H-Cl = 2×431 = 862. ΔH = 675 - 862 = -187 kJ.
Dissolving 8 g of NH₄NO₃ in water cools 100 g of water from 25°C to 19°C. Calculate ΔH in kJ/mol. (c_water = 4.18 J/g°C, M(NH₄NO₃) = 80 g/mol).
Q_water = 100×4.18×(-6) = -2508 J (water lost heat). ΔH_rxn = +2508 J for 0.1 mol → +25080 J/mol = +25.1 kJ/mol (endothermic).
Use Hess's Law: C(s)+O₂(g)→CO₂(g), ΔH₁=-393. CO(g)+½O₂(g)→CO₂(g), ΔH₂=-283. Find ΔH for C(s)+½O₂(g)→CO(g).
Target = Eq1 - Eq2. ΔH = ΔH₁ - ΔH₂ = -393 - (-283) = -110 kJ/mol.