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Physical Chemistry and Transition Elements

Enthalpy of solution and hydration, and the effect of ionic charge and size

  • 3 things to know
  • Components 01 and 03

Overview

Recall questions on the three linked terms, why dissolving can be endothermic, and how ionic charge and radius affect both lattice and hydration enthalpies.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Numbers

How is lattice enthalpy calculated from a Born-Haber cycle, and what are the enthalpy changes of solution and of hydration?

Process

Explain why smaller ions produce more exothermic lattice enthalpies.

Numbers

Why do smaller ions have more exothermic enthalpies of hydration?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Define enthalpy change of solution.

    Model answer

    The enthalpy change when one mole of a solute dissolves completely in a solvent to form a solution.

  2. 2

    What is the enthalpy change of hydration?

    Model answer

    The enthalpy change when one mole of gaseous ions dissolves in water to form an aqueous solution.

  3. 3

    Is enthalpy of hydration exothermic or endothermic, and why?

    Model answer

    Always exothermic because energy is released when ion-dipole bonds form between ions and water molecules.

  4. 4

    Which three enthalpy terms are linked in an enthalpy cycle for dissolving an ionic compound?

    Model answer

    Lattice enthalpy, enthalpy of solution, and enthalpies of hydration of the ions.

  5. 5

    State the relationship between enthalpy of solution, enthalpies of hydration and lattice enthalpy.

    Model answer

    Enthalpy of solution = sum of enthalpies of hydration of all ions − lattice enthalpy.

  6. 6

    Why is the lattice enthalpy term sign-adjusted in the enthalpy of solution equation?

    Model answer

    Because lattice enthalpy is defined for lattice formation (exothermic), but dissolving requires breaking the lattice (endothermic), so the sign must be reversed.

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AThe first step is Cl• + O₃ → ClO• + O₂.
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CThe overall equation is O₃ + O → 2O₂.

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In ClO⁻, chlorine has oxidation number +1.
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