The enthalpy of sublimation of iodine is 60.2 kJ/mol, and its enthalpy of vaporization is 45.5 lz.1/mol. What is the enthalpy of fusion of iodine?

ΔH = 14,7kJ/mol

Explanation:

It is possible to make algebraic sum of several chemical process to obtain enthalpy of a determined reaction (Hess's law).

Sublimation of iodine is (Transition from solid to gas):

I₂(s) → I₂(g) ΔH = 60.2kJ/mol

Vaporization of iodine (From liquid to gas):

I₂(l) → I₂(g) ΔH = 45.5kJ/mol

Fusion of iodine (From solid to liquid can be obtained subtracting the sublimation process - Vaporization process:

I₂(s) → I₂(g) ΔH = 60.2kJ/mol

I₂(g) → I₂(l) ΔH = -45.5kJ/mol

I₂(s) → I₂(l) ΔH = 60.2kJ/mol - 45.5kJ/mol

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .

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