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Chemistry · Ch 3 — Thermodynamics

Lattice Enthalpy

3.5(d)

Lattice Enthalpy

Lattice Enthalpy

Definition: The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in the gaseous state.

Na+Cl−(s)→Na+(g)+Cl−(g);ΔlatticeH⊖=+788 kJ mol−1\text{Na}^+\text{Cl}^-(s) \rightarrow \text{Na}^+(g) + \text{Cl}^-(g); \quad \Delta_{\text{lattice}} H^\ominus = +788\ \text{kJ mol}^{-1}

Since it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber Cycle.

Born-Haber Cycle for NaCl

Let us calculate the lattice enthalpy of Na⁺Cl⁻(s) by following these steps:

Step 1: Sublimation of sodium metal

Na(s)→Na(g);ΔsubH⊖=108.4 kJ mol−1\text{Na}(s) \rightarrow \text{Na}(g); \quad \Delta_{\text{sub}} H^\ominus = 108.4\ \text{kJ mol}^{-1}

Step 2: Ionization of sodium atoms

Na(g)→Na+(g)+e−;ΔiH⊖=496 kJ mol−1\text{Na}(g) \rightarrow \text{Na}^+(g) + e^-; \quad \Delta_i H^\ominus = 496\ \text{kJ mol}^{-1}

Step 3: Dissociation of chlorine (half the bond dissociation enthalpy)

12Cl2(g)→Cl(g);12ΔbondH⊖=121 kJ mol−1\frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g); \quad \frac{1}{2}\Delta_{\text{bond}} H^\ominus = 121\ \text{kJ mol}^{-1}

Step 4: Electron gain by chlorine atoms

Cl(g)+e−→Cl−(g);ΔegH⊖=−348.6 kJ mol−1\text{Cl}(g) + e^- \rightarrow \text{Cl}^-(g); \quad \Delta_{\text{eg}} H^\ominus = -348.6\ \text{kJ mol}^{-1}

Step 5: Formation of NaCl(s) from gaseous ions

Na+(g)+Cl−(g)→Na+Cl−(s);ΔlatticeH⊖=?\text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{Na}^+\text{Cl}^-(s); \quad \Delta_{\text{lattice}} H^\ominus = ?

The sequence of steps is shown in the enthalpy diagram (Fig. 5.9) and is known as a Born-Haber cycle.

Important

The importance of the cycle is that the sum of the enthalpy changes round a cycle is zero (Hess's law).

Applying Hess's law:

ΔlatticeH⊖=411.2+108.4+121+496−348.6\Delta_{\text{lattice}} H^\ominus = 411.2 + 108.4 + 121 + 496 - 348.6

ΔlatticeH⊖=+788 kJ mol−1 for NaCl(s)\Delta_{\text{lattice}} H^\ominus = +788\ \text{kJ mol}^{-1} \text{ for NaCl(s)}

Note

The internal energy change is smaller by 2RT2RT (because Δng=2\Delta n_g = 2) and is equal to +783 kJ mol−1+783\ \text{kJ mol}^{-1}.

Ionization Energy and Electron Affinity

Ionization energy and electron affinity are defined at absolute zero. At any other temperature, heat capacities for the reactants and the products have to be taken into account.

For the reactions:

  • M(g)→M+(g)+e−\text{M}(g) \rightarrow \text{M}^+(g) + e^- (ionization)
  • M(g)+e−→M−(g)\text{M}(g) + e^- \rightarrow \text{M}^-(g) (electron gain)

At temperature TT:

ΔrH⊖(T)=ΔrH⊖(0)+∫0TΔrCP⊖ dT\Delta_r H^\ominus(T) = \Delta_r H^\ominus(0) + \int_0^T \Delta_r C_P^\ominus \, dT …

Figure 5.9Enthalpy (Born–Haber) cycle for the formation of NaCl and its lattice enthalpy.
Fig. 5.9 — Enthalpy (Born–Haber) cycle for the formation of NaCl and its lattice enthalpy.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

The Born–Haber cycle is a visual accounting of the energy changes linking an ionic solid, its elements, and its gaseous ions. Fig. 5.9 is an enthalpy ladder: each horizontal rung is a state, each vertical arrow a step with its enthalpy change, drawn with the figure's own printed values.

Reading the ladder upward from the elements: sublimation of solid sodium, Na(s) →\rightarrow Na(g), costs ΔsubH⊖=+108.4\Delta_{sub}H^\ominus = +108.4 kJ mol−1^{-1}; ionisation Na(g) →\rightarrow Na+^+(g) + e−^- costs +495.6+495.6 kJ mol−1^{-1} (the running text rounds this to 496); dissociating half a mole of Cl2_2 into Cl(g) costs 12ΔbondH⊖=+121\frac{1}{2}\Delta_{bond}H^\ominus = +121 kJ mol−1^{-1}; and electron gain Cl(g) + e−^- →\rightarrow Cl−^-(g) releases ΔegH⊖=−348.6\Delta_{eg}H^\ominus = -348.6 kJ mol−1^{-1}. Separately, the direct formation Na(s) + 12\frac{1}{2}Cl2_2(g) →\rightarrow NaCl(s) has ΔfH⊖=−411.2\Delta_f H^\ominus = -411.2 kJ mol−1^{-1}, and the lattice enthalpy step is the dissociation of the solid into its gaseous ions, NaCl(s) →\rightarrow Na+^+(g) + Cl−^-(g), drawn as the tall upward leg ΔlatticeH⊖\Delta_{lattice}H^\ominus.

Because enthalpy is a state function, the changes around the closed cycle sum to zero, so:

ΔlatticeH⊖=−ΔfH⊖+ΔsubH⊖+ΔiH⊖+12ΔbondH⊖+ΔegH⊖\Delta_{lattice}H^\ominus = -\Delta_f H^\ominus + \Delta_{sub}H^\ominus + \Delta_i H^\ominus + \tfrac{1}{2}\Delta_{bond}H^\ominus + \Delta_{eg}H^\ominus

=411.2+108.4+495.6+121−348.6≈+788 kJ mol−1= 411.2 + 108.4 + 495.6 + 121 - 348.6 \approx +788\ \text{kJ mol}^{-1}

Important

In this textbook's convention the lattice enthalpy is defined for breaking the lattice apart — NaCl(s) →\rightarrow Na+^+(g) + Cl−^-(g) — so ΔlatticeH⊖\Delta_{lattice}H^\ominus is positive (+788+788 kJ mol−1^{-1} for NaCl). Some books define it for the reverse (formation of the lattice from gaseous ions), which flips the sign; always check the convention before comparing values. …

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