question_answer

The enthalpy of formation of gaseous iodine is \[62.5\,kJ\,mo{{l}^{-1}}\] at 25°C. What will be the enthalpy of sublimation of iodine at 25°C?

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, \[{{\Delta }_{f}}H\,({{I}_{2}},\,g)={{\Delta }_{sub}}\,H({{I}_{2}},\,s).\]