How to Calculate the Lattice Energy of NaBr Using the Born-Haber Cycle?

The lattice energy of NaBr can be calculated using the Born-Haber cycle by considering the different energy changes associated with forming NaBr from its constituent elements. We start with the following data:

• Enthalpy of sublimation of Na (ΔH_subl) = +109 kJ/mol
• Bond energy of Br₂ (D_Br-Br) = +192 kJ/mol
• Ionization energy of Na (IE_Na) = +496 kJ/mol
• Electron affinity of Br (EA_Br) = -324 kJ/mol

To find the lattice energy (U_lattice), we can follow this sequence:

1. Sublimation of solid sodium: +109 kJ/mol
2. Dissociation of Br₂ into two Br atoms: +192 kJ/mol (but since we need one Br, we take half of this value: +96 kJ/mol)
3. Ionization of Na to Na⁺: +496 kJ/mol
4. Gaining an electron by Br to form Br^–: -324 kJ/mol

The overall formation of NaBr from its gaseous ions can be summed up as follows:

U_lattice = ΔH_subl + (0.5 × D_Br-Br) + IE_Na + EA_Br + U_lattice

To rearrange for U_lattice, we can set up the equation:

U_lattice = ΔH_f – (ΔH_subl + 0.5 × D_Br-Br + IE_Na + EA_Br)

Since the standard enthalpy of formation (ΔH_f) of NaBr can be considered as -ΔH_lattice in this case, we can substitute the known values:

U_lattice = 0 – (109 + 96 + 496 – 324)

Calculating this gives:

U_lattice = – (109 + 96 + 496 – 324) = – (377) kJ/mol

Thus, the lattice energy of NaBr is approximately +377 kJ/mol (taking into account that lattice energy is conventionally expressed as a positive value).

This value represents the energy released when gaseous ions combine to form an ionic solid, illustrating the strength of the ionic bonds in NaBr.