Final answer:
The enthalpy of hydration for an iodide ion in the process RbI → Rb+ + I- is c) -284.0 kJ/mol. So Option C is correct.
Step-by-step explanation:
The enthalpy of hydration for an iodide ion in the process RbI → Rb⁺ + I⁻ can be determined using the equation ΔHsol = -ΔHiat + ΔHhydr. Here, ΔHsol is the enthalpy change of solution, ΔHiat is the ionization enthalpy, and ΔHhydr is the enthalpy of hydration. The negative sign in front of ΔHiat indicates the energy absorbed during ionization.
In this case, we are dealing with the dissociation of RbI into Rb⁺ and I⁻ ions. The enthalpy change of ionization (ΔHiat) for Rb⁺ is positive since energy is required to remove an electron. However, the enthalpy change of hydration (ΔHhydr) for I⁻ is larger in magnitude, and it is negative because energy is released when water molecules surround and solvate the ion.
Plugging the given values into the equation ΔHsol = -ΔHiat + ΔHhydr, we get -ΔHiat = ΔHsol - ΔHhydr. Substituting the known values, we find -ΔHiat = -284.0 kJ/mol. Therefore, the enthalpy change of ionization for Rb⁺ is -284.0 kJ/mol, leading to the final answer of -284.0 kJ/mol for the enthalpy of hydration for an iodide ion in the given process.