A-level Chemistry/WJEC/Module 3/Enthalpy

Recall:

The enthalpy of a reaction which removes 1 electron from each atom in 1 mole of gaseous atoms.

Example:

The first ionisation energy of chlorine:

Cl(g) → Cl⁺(g) + e^- E_i = +1251 kJ mol^-1

The enthalpy of a reaction which breaks 1 mole of a certain type of covalent bond by homolytic fission, under standard conditions.

Example:

Breaking 1 mole of Cl-Cl bonds:

Cl₂(g) → 2 Cl(g) E_d = 243 kJ mol^-1

Except for bonds which form diatomic molecules, multiple molecules can contain the same type of bond. For example, only Cl₂ has a Cl-Cl bond, but the C-Cl bond is present in a very wide range of compounds (CH₃Cl, CCl₄, CH₃CCl₃, etc). For most covalent bonds, we can look up the average bond enthalpy to find an estimate of the strength of the bond in a specific compound.

New definitions:

The enthalpy of a reaction which adds 1 electron to each atom in 1 mole of gaseous atoms.

Example:

The first electron affinity of oxygen:

O(g) + e^- → O^-(g) E_ea(1) = -141 kJ mol^-1

The second electron affinity of oxygen:

O^-(g) + e^- → O^2-(g) E_ea(2) = +744 kJ mol^-1

A point about E_ea: Adding an electron to a cloud of electrons is not expected to be exothermic, but the first electron affinity of an element is often exothermic as the electron completes some part of the outer shell configuration. Each following electron affinity is increasingly endothermic.

The enthalpy of a reaction which forms 1 mole of gaseous atoms of an element, under standard conditions.

For the diatomic gases, Δ_atH^⦵ is half the bond enthalpy, E_d.

Example:

Breaking 1 mole of Cl-Cl bonds:

Cl₂(g) → 2 Cl(g) E_d = 243 kJ mol^-1

Making 1 mole of Cl atoms:

½ Cl₂(g) → Cl(g) Δ_atH^⦵ = 121.5 kJ mol^-1

The enthalpy of a reaction in which 1 mole of a substance forms a dilute aqueous solution, under standard conditions.

NaCl(s) → NaCl(aq) Δ_solH^⦵(NaCl) = +17 kJ mol^-1

The enthalpy of a reaction in which 1 mole of gaseous ions/atoms/molecules forms a dilute aqueous solution, under standard conditions.

Na⁺(g) → Na⁺(aq) Δ_hydH^⦵(Na⁺) = -406 kJ mol^-1

The enthalpy of a reaction in which 1 mole of an ionic solid is formed from its constituent ions, which were in a gaseous state, under standard conditions.

Na⁺(g) + Cl^-(g) → NaCl(s) Δ_lattH^⦵(NaCl) = -787 kJ mol^-1

For an ionic compound to dissolve, the lattice must break and the ions become hydrated.

Δ_solH^⦵ = Σ (Δ_hydH^⦵) - Δ_lattH^⦵

In English: the solution enthalpy is the sum of the hydration enthalpies of the ions, minus the lattice enthalpy of the compound.

For CaCl₂:

Δ_hydH^⦵(Ca²⁺) = -1650 kJ mol^-1

Δ_hydH^⦵(Cl^-) = -364 kJ mol^-1

Δ_lattH^⦵(CaCl₂) = -2258 kJ mol^-1

Δ_solH^⦵(CaCl₂) = -1650 - 2 x 364 - (-2258) = -120 kJ mol^-1

We can safely say that CaCl₂ dissolves in water, with an exothermic release of heat.

However, for NaCl:

Δ_hydH^⦵(Na⁺) = -406 kJ mol^-1

Δ_hydH^⦵(Cl^-) = -364 kJ mol^-1

Δ_lattH^⦵(NaCl) = -787 kJ mol^-1

Δ_solH^⦵(NaCl) = -406 - 364 - (-787) = +17 kJ mol^-1

This is a low, and also endothermic Δ_solH^⦵ value. We know NaCl dissolves in water, but it does not seem to be very enthusiastic about the process! What tips the balance here is the increase in entropy caused by the regular crystal lattice dispersing into the solution. Many ionic substances dissolve endothermically like this.