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

A. –751 kJ
B.+29 kJ
C. –1401 kJ
D.–29 kJ
E.+751 kJ

Answer 1

The lattice energy for NaBr(s) is calculated using the Born-Haber cycle and Hess's Law by adding the sublimation energy. The correct answer is option E.

Step-by-step explanation:

To calculate the lattice energy of NaBr(s) using the given thermochemical equations, we can apply Hess's law and the Born-Haber cycle. The relevant stages in the Born-Haber cycle include sublimation of sodium, dissociation of bromine, ionization of sodium, electron affinity of bromine, and finally the formation of NaBr(s).

The lattice energy of NaBr(s) can be found using the following steps: Sublimation energy of Na(s) to Na(g): +107 kJ. Ionization energy of Na(g) to Na+(g): +496 kJ. Dissociation energy of Br2(g) to 2 Br(g): (112 kJ) / 2 = +56 kJ. Electron affinity of Br(g), to form Br-(g): -325 kJ.

Heat of formation of NaBr(s), from Na(s) and 1/2 Br2(g): -361 kJ. By summing these values, we get: (+107 kJ) + (+496 kJ) + (+56 kJ) - (-325 kJ) - (-361 kJ) = +751 kJ. Therefore, the calculated lattice energy for NaBr(s) is +751 kJ, which corresponds to answer choice E.