The ionization energy for rubidium is 403 kJ/mol. The electron affinity for iodine is -295 kJ/mol. Use these values and Hess's law to calculate the change in enthalpy for the following reaction per mole of reagent:
Rb(g) + I(g) > Rb+(g)+I−(g), ΔH=?

108 kJ/mol the change in enthalpy for the reaction per mole of reagent.

Explanation:

Ionization energy of the rubidium = I.E

....[1]

An electron affinity for iodine  = E.A

....[2]

Enthalpy change for the reaction : [1]+ [2] ( Hess's law)

108 kJ/mol the change in enthalpy for the reaction per mole of reagent.

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