Final answer:
The energy change for the reaction Hg(l) → Hg+(g) + e⁻ is greater than the first ionization energy of mercury, which is 1006 kJ/mol, because it includes the energy needed to vaporize liquid mercury before ionization.
Step-by-step explanation:
Ionization energy (IE) is defined as the amount of energy required to remove an electron from a gaseous atom to form a cation. For mercury (Hg), the first ionization energy is 1006 kJ/mol. This value pertains to turning Hg in its gaseous form into Hg+ and releasing an electron. The energy change for the reaction Hg(l) → Hg+(g) + e⁻ corresponds to this ionization process, which means the energy required is equivalent to the first ionization energy specified above.
Given this information, the answer to the question is that the energy change for the reaction Hg(l) → Hg+(g) + e⁻ is equal to the first ionization energy of mercury, which is 1006 kJ/mol. Therefore, the correct answer is that the energy change for the reaction is greater than 1006 kJ/mol, considering that we need extra energy to convert liquid mercury to its gaseous form before ionization can occur.
Hence, this eliminates options A (less than 1006 kJ/mol), C (electron affinity of mercury), D (1036 kJ/mol), and E (second ionization energy of mercury). Ionization energy takes into consideration an atom in gaseous state, hence the given value doesn't immediately apply to the liquid state without accounting for the heat of vaporization