Question

Consider the following: Li(s) + ½ I₂(g) --> LiI(s) ΔH = –292 kJ. LiI(s) has a lattice energy of –753 kJ/mol. The ionization energy of Li(g) is 520 kJ/mol, the bond energy of I₂(g) is 151 kJ/mol, and the electron affinity of I(g) is –295 kJ/mol. Use these data to determine the heat of sublimation of Li(s).

Answer 1

Answer:

Heat of sublimation of Li(s) = 160.5 kJ/mol

Explanation:

Given that:

$\mathtt{2Li(s) + \dfrac{1}{2} I_2(g) \to 2LiI(s)}$

ΔH = –292 kJ

The heat of formation for the above reaction =  –292 kJ × 2 =  -584 kJ/mol

$\mathtt{Li^+_{(g)} + I^-_{(g)} \to LiI_{(s)}}$

The lattice energy of LiI(s) = -753 kJ/mol

$\mathtt{Li(g)\to Li^+(g) + e^- }$

The ionization energy of LiI(s) = +520 kJ/mol

$\mathtt{I_2_{(g)} \to 2I_{(g)} }$

The Bond Energy of I₂(g) = 151 kJ/mol

$\mathtt{I_{(g)} + e^- \to I^-_{(g)}}$

The electron affinity of I(g) = -295 kJ/mol

Heat of sublimation: Sublimation is the process of changing of a solid matter into gas without passing through the liquid stage, Now, the molar heat of sublimation is the amount of energy that must be added to a mole of solid  to turn it directly into a gas without any interference through the liquid phase provided the pressure is  constant.

From  the above reactions: The heat of sublimation  of Li(s) can be calculated by the sum total of the following.

$\mathtt{Li_{(s)} + \dfrac{1}{2} I_2_{(g)} \to LiI_{(s)} \ \ -292 kJ/mol} \\ \\\mathtt{I_{(g)} \to \dfrac{1}{2} I2(g) \ \ -75.5 kJ/mol} \\ \\ \mathtt{I^-(g) \to I(g) + e^- \ \ +295 kJ/mol} \\ \\ \mathtt{LiI(s) \to Li^+_{(g)} + I^-_{(g)} \ \ +753 kJ/mol} \\ \\ \mathtt{Li^+_{(g)} + e^- \to Li(s) \ \ -520 kJ/mol}$

$\mathtt{Li(s) \to Li(g)}$ = (-292 +(-75.5)+295+753+(-520)) kJ/mol

$\mathtt{Li(s) \to Li(g)}$ =  160.5 kJ/mol

Heat of sublimation of Li(s) = 160.5 kJ/mol