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).

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

Explanation:

Given that:

ΔH = –292 kJ

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

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

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

The Bond Energy of I₂(g) = 151 kJ/mol

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.

= (-292 +(-75.5)+295+753+(-520)) kJ/mol

=  160.5 kJ/mol

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

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a. energy is absorbed in the reaction.

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.