The answer is 14.3 g
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Answer:
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