Answer:
By definition of enthalpy of
formation,\[{{I}_{2}}(s)\xrightarrow{\,}\,{{I}_{2}}\ (g)\,,\,{{\Delta
}_{f}}H=62.5\,kJ\,mo{{l}^{-1}}\]
This is also the enthalpy of sublimation by definition.
Hence, \[{{\Delta }_{f}}H\,({{I}_{2}},\,g)={{\Delta
}_{sub}}\,H({{I}_{2}},\,s).\]
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