Enthalpy changes for the melting of iodine, I2, and for the sublimation of iodine are provided below - VCE - SSCE Chemistry - Question 4 - 2012 - Paper 1

To find the enthalpy change for the vaporisation of iodine from liquid to gas, we need to calculate the enthalpy transition between the phases.

1. We have the enthalpy change for sublimation from solid to gas: I2(s) → I2(g) = +62 kJ mol⁻¹.

2. We have the enthalpy change for melting from solid to liquid: I2(s) → I2(l) = +16 kJ mol⁻¹.

3. To find the vaporisation, we can utilize the following relationship:

   $\Delta H_{vap} = \Delta H_{sub} - \Delta H_{melt}$

   Substituting the values:

   $\Delta H_{vap} = +62 kJ mol^{-1} - (+16 kJ mol^{-1})$ $\Delta H_{vap} = +62 kJ mol^{-1} - 16 kJ mol^{-1} = +46 kJ mol^{-1}$

The answer is therefore +46 kJ mol⁻¹, which corresponds to option C.