Thermodynamics

Chemistry · Class 11

Simulation · Chemistry · Class 11

Bonds broken, bonds formed

From the lesson Enthalpies of combustion, atomisation, bonds, lattices, solution and dilution in Thermodynamics. Change the values and watch what happens.

Bonds broken, bonds formedChemistry · Class 11

The idea behind it

NCERT § "Enthalpies for Different Types of Reactions"

  • Standard enthalpy of combustion ΔcH° is the enthalpy change per mole of a substance burnt completely in oxygen, all species in standard states; combustion is always exothermic.
  • Enthalpy of atomisation is the enthalpy change when all the bonds in one mole of a substance are broken to give separate gaseous atoms; for H₂ it is 435.0 kJ mol⁻¹, and for CH₄ it is 1665 kJ mol⁻¹.
  • Bond dissociation enthalpy applies to one specific bond in a diatomic or particular molecule; for polyatomic molecules a mean bond enthalpy is used (C–H in CH₄: 1665/4 = 416 kJ mol⁻¹).
  • Estimating reaction enthalpy for gas-phase reactions: ΔᵣH° = Σ bond enthalpies of reactants − Σ bond enthalpies of products (bonds broken minus bonds formed).
  • Mean bond enthalpies give only approximate reaction enthalpies because a given bond's strength varies from molecule to molecule.
  • Lattice enthalpy is the enthalpy needed to pull one mole of an ionic solid fully apart into its gaseous ions; for NaCl it is +788 kJ mol⁻¹.
  • Lattice enthalpy cannot be measured directly; it is found from a Born-Haber cycle, which applies Hess's law to steps such as sublimation, ionisation, dissociation and electron gain.
  • Enthalpy of solution ΔsolH is the enthalpy change for dissolving one mole of a substance in a stated amount of solvent; for an ionic solid, ΔsolH = ΔlatticeH + ΔhydH. For NaCl, +788 − 784 = +4 kJ mol⁻¹, so dissolving it barely changes the temperature.
  • ΔsolH is positive for most ionic salts, so their solubility in water rises with temperature; a very high lattice enthalpy can stop a salt dissolving at all.
  • Enthalpy of dilution is the enthalpy change when extra solvent is added to a solution; it depends on the starting concentration and on how much solvent is added. For HCl, going from HCl·25 aq to HCl·40 aq gives −72.79 − (−72.03) = −0.76 kJ mol⁻¹.