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 can utilize the provided values.

1. The enthalpy change for the melting of iodine (I₂(s) → I₂(l)) is +16 kJ mol⁻¹.
2. The enthalpy change for the sublimation of iodine (I₂(s) → I₂(g)) is +62 kJ mol⁻¹.

Using Hess's Law, we can express the vaporisation reaction as follows:

I₂(s) → I₂(l) → I₂(g)

The total enthalpy change for vaporisation can be calculated as:

$\Delta H_{vaporisation} = \Delta H_{sublimation} - \Delta H_{fusion}$

Substituting the values gives:

$\Delta H_{vaporisation} = 62 \, \text{kJ mol}^{-1} - 16 \, \text{kJ mol}^{-1} = 46 \, \text{kJ mol}^{-1}$

Therefore, the answer is C. +46 kJ mol⁻¹.