Answer:
108 kJ/mol the change in enthalpy for the reaction per mole of reagent.
Explanation:
Ionization energy of the rubidium = I.E
[tex]Rb(g)\rightarrow Rb^+(g)+e^-,I.E=403 kJ/mol[/tex]....[1]
An electron affinity for iodine = E.A
[tex]I(g)+e^-(g)\rightarrow I^-(g),E.A=-295 kJ/mol[/tex]....[2]
[tex]Rb(g) + I(g)\rightarrow Rb^+(g)+I^-(g), \Delta H=?[/tex]
Enthalpy change for the reaction : [1]+ [2] ( Hess's law)
[tex]\Delta H=I.E+E.A[/tex]
[tex]=403 kJ/mol+(-295 kJ/mol)=108 kJ/mol[/tex]
108 kJ/mol the change in enthalpy for the reaction per mole of reagent.
