Question

Calculate the lattice energy of NaBr(s), given the following thermochemical equations, where ΔIE and ΔEA are ionization energy and electron affinity, respectively.

Na(s) → Na(g) ΔHf° = +107 kJ
Na(g) → Na+(g) + e– ΔIE = +496 kJ
1/2 Br₂(g) → Br(g) ΔHf° = +112 kJ
Br(g) + e– → Br–(g) ΔEA = –325 kJ
Na(s) + 1/2 Br₂(g) → NaBr(s) ΔHf° = -361 kJ

Answer 1

The lattice energy of NaBr(s) is calculated as 715 kJ/mol by summing sublimation energy, ionization energy, and bond dissociation energy then subtracting electron affinity and the enthalpy of formation according to the Born-Haber cycle.

Step-by-step explanation:

The lattice energy of NaBr(s) can be calculated using the Born-Haber cycle, taking into account the enthalpy changes for various steps in the formation of NaBr. Applying the values given in the question:

• Sublimation Energy for Na(s) to Na(g): +107 kJ
• Ionization Energy (IE) for Na(g): +496 kJ
• Bond Dissociation Energy for 1/2 Br₂(g): +112 kJ
• Electron Affinity (EA) for Br(g): –325 kJ
• Enthalpy of Formation (ΔHf°) for NaBr(s): -361 kJ

The lattice energy (U) is then the sum of these values but with the enthalpy of formation and electron affinity being subtracted:

U = Sublimation Energy + IE + Bond Dissociation Energy - EA - ΔHf°

Calculating this using the provided values gives:

U = 107 kJ + 496 kJ + 112 kJ - (-325 kJ) - (-361 kJ)

U = 715 kJ/mol, which is the lattice energy of NaBr(s).