Question

Find the electron gain enthalpy of iodine with the help of the following data: Heat of formation of NaI = -287.9 kJ/mol Heat of Sublimation of Na = 108.4 kJ/mol Ionisation energy of Na = 106.7 kJ/mol Heat of atomisation of iodine = 495.4 kJ/mol Lattice energy of NaI = -692 kJ/mol

• A. -306.4 kJ/mol
• B. -204.9 kJ/mol
• C. -406.4 kJ/mol
• D. -300 kJ/mol

Answer 1

Answer: (A)

-306.4 kJ/mol

Answer 2

The problem can be solved using the Born-Haber cycle, which relates the enthalpy of formation of an ionic compound to various energy terms. The cycle for the formation of NaI is as follows: $Na(s) + \frac{1}{2} I_2(s) \rightarrow NaI(s)$ ; $\Delta H_f = -287.9$ kJ/mol

The individual steps in the Born-Haber cycle are:

1. Sublimation of Sodium: $Na(s) \rightarrow Na(g)$ ; $\Delta H_{sub} = +108.4$ kJ/mol
2. Ionisation of Sodium: $Na(g) \rightarrow Na^+(g) + e^-$ ; $\Delta H_{IE} = +106.7$ kJ/mol
3. Atomisation of Iodine: $\frac{1}{2} I_2(s) \rightarrow I(g)$ ; $\Delta H_{atom\_I} = 495.4$ kJ/mol
4. Electron Gain by Iodine: $I(g) + e^- \rightarrow I^-(g)$ ; $\Delta H_{electron\ gain}$ (This is what we need to find)
5. Lattice Formation: $Na^+(g) + I^-(g) \rightarrow NaI(s)$ ; $\Delta H_{lattice} = -692$ kJ/mol

According to Hess's Law, the sum of the enthalpy changes of these steps equals the enthalpy of formation of NaI: $\Delta H_f = \Delta H_{sub} + \Delta H_{IE} + \Delta H_{atom\_I} + \Delta H_{electron\ gain} + \Delta H_{lattice}$

Substituting the given values into the equation: $-287.9 = 108.4 + 106.7 + 495.4 + \Delta H_{electron\ gain} + (-692)$

Rearranging to solve for $\Delta H_{electron\ gain}$: $\Delta H_{electron\ gain} = \Delta H_f - \Delta H_{sub} - \Delta H_{IE} - \Delta H_{atom\_I} - \Delta H_{lattice}$ $\Delta H_{electron\ gain} = (-287.9) - (108.4) - (106.7) - (495.4) - (-692)$ $\Delta H_{electron\ gain} = -287.9 - 108.4 - 106.7 - 495.4 + 692$ $\Delta H_{electron\ gain} = -998.4 + 692$ $\Delta H_{electron\ gain} = -306.4$ kJ/mol