Answer:
Heat of sublimation of Li(s) = 160.5 kJ/mol
Explanation:
Given that:
[tex]\mathtt{2Li(s) + \dfrac{1}{2} I_2(g) \to 2LiI(s)}[/tex]
ΔH = –292 kJ
The heat of formation for the above reaction = –292 kJ × 2 = -584 kJ/mol
[tex]\mathtt{Li^+_{(g)} + I^-_{(g)} \to LiI_{(s)}}[/tex]
The lattice energy of LiI(s) = -753 kJ/mol
[tex]\mathtt{Li(g)\to Li^+(g) + e^- }[/tex]
The ionization energy of LiI(s) = +520 kJ/mol
[tex]\mathtt{I_2_{(g)} \to 2I_{(g)} }[/tex]
The Bond Energy of I₂(g) = 151 kJ/mol
[tex]\mathtt{I_{(g)} + e^- \to I^-_{(g)}}[/tex]
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.
[tex]\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} \\ \\[/tex]
[tex]\mathtt{Li(s) \to Li(g)}[/tex] = (-292 +(-75.5)+295+753+(-520)) kJ/mol
[tex]\mathtt{Li(s) \to Li(g)}[/tex] = 160.5 kJ/mol
Heat of sublimation of Li(s) = 160.5 kJ/mol
