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# 5.8: Lattice Energy and the Born-Haber Cycle

An important enthalpy change is the Lattice Energy, which is the energy required to take one mole of a crystalline solid to ions in the gas phase. For $$\ce{NaCl(s)}$$, the lattice energy is defined as the enthalpy of the reaction

$\ce{ NaCl(s) \rightarrow Na^{+}(g) + Cl^{-}(g) }$

with $$\Delta H$$ called the lattice energy ($$\Delta H_{Lat}$$).

## The Born-Haber Cycle

A very handy construct in thermodynamics is that of the thermodynamic cycle. This can be represented graphically to help to visualize how all of the pieces of the cycle add together. A very good example of this is the Born-Haber cycle, describing the formation of an ionic solid.

Two pathways can be envisioned for the formation. Added together, the two pathways form a cycle. In one pathway, the ionic solid if formed directly from elements in their standard states.

$\ce{Na(s) + 1/2 Cl_2 \rightarrow NaCl(s)}$

with $$\Delta H_f(NaCl)$$.

The other pathway involves a series of steps that take the elements from neutral species in their standard states to ions in the gas phase.

$Na(s) \rightarrow Na(g)$

with $$\Delta H_{sub}(Na)$$

$Na(g) \rightarrow Na^+(g) + e^-$

with $$1^{st}\, IP(Na)$$

$½ Cl_2(g) \rightarrow Cl(g)$

with $$½ D(Cl-Cl)$$

$Cl(g) + e^- \rightarrow Cl^-(g)$

with $$1^{st} EA(Cl)$$

$Na^+(g) + Cl^-(g) \rightarrow NaCl(s)$

with $$\Delta H_{Lat}(NaCl)$$

It should be clear that when added (after proper manipulation if needed), the second set of reactions yield the first reaction. Because of this, the total enthalpy changes must all add.

$\Delta H_{sub}(Na) + 1^{st} IP(Na) + ½ D(Cl-Cl) + 1^{st} EA(Cl) + \Delta H_{lat}(NaCl) = \Delta H_f(NaCl)$

This can be depicted graphically, the advantage being that arrows can be used to indicate endothermic or exothermic changes. An example of the Born-Haber Cycle for NaCl is shown below. Figure 3.6.1: the Born-Haber Cycle for NaCl.

In many applications, all but one leg of the cycle is known, and the job is to determine the magnitude of the missing leg.

Exercise $$\PageIndex{1}$$: Potassium Bromide

Find $$\Delta H_f$$­ for KBr given the following data.

$\ce{K(s) \rightarrow K(g)} \nonumber$

with $$\Delta H_{sub} = 89\, kJ/mol$$

$\ce{Br_2(l) \rightarrow Br_2(g) } \nonumber$

with $$\Delta H_{vap} = 31\, kJ/mol$$

$\ce{Br_2(g) \rightarrow 2 Br(g)} \nonumber$

with $$D(Br-Br) = 193\, kJ/mol$$

$\ce{K(g) \rightarrow K^+(g) + e^- } \nonumber$

with $$1^{st} IP(K) = 419 kJ/mol$$

$\ce{Br(g) + e^- \rightarrow Br^-(g) } \nonumber$

with $$1^{st} EA(Br) = 194 kJ/mol$$

$\ce{K^+(g) + Br^-(g) \rightarrow KBr(s)} \nonumber$

with $$\Delta H_{Lat} = 672 kJ/mol$$

Answer

$$\Delta H_f = -246 \,kJ/mol$$

Note: This cycle required the extra leg of the vaporization of Br2. Many cycles involve ions with greater than unit charge and may require extra ionization steps as well!

## Contributors and Attributions

• Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)

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