\(\operatorname{LiI}(s)\) has a heat of formation of \(-272 \mathrm{~kJ} / \mathrm{mol}\) and a lattice energy of \(-753 \mathrm{~kJ} / \mathrm{mol}\). The ionization energy of \(\mathrm{Li}(g)\) is 520 . \(\mathrm{kJ} / \mathrm{mol}\), the bond energy of \(\mathrm{I}_{2}(g)\) is \(151 \mathrm{~kJ} / \mathrm{mol}\), and the electron affinity of \(\mathrm{I}(g)\) is \(-295 \mathrm{~kJ} / \mathrm{mol}\). Use these data to determine the heat of sublimation of \(\operatorname{Li}(s)\).

Short Answer

Expert verified
The heat of sublimation of Li(s) can be determined using the Born-Haber cycle equation and the given data. By plugging in the known values and solving for the heat of sublimation, we find that the heat of sublimation of Li(s) is \(180.5\,\text{kJ/mol}\).

Step by step solution

01

Understand the Born-Haber cycle

The Born-Haber cycle is an enthalpy cycle that relates various energies involved in the formation of ionic compounds (like the formation of LiI from lithium and iodine). The cycle helps us determine one value if all others are known. The steps involved in this cycle are: 1. Sublimation of solid lithium (Li) to gaseous lithium. 2. Ionization of gaseous lithium (Li) to form Li+. 3. Dissociation of gaseous iodine (I2) into gaseous iodine atoms (2*I). 4. Electron affinity: the transfer of an electron from iodine atom to form an iodide ion (I-). 5. Lattice energy: the formation of the ionic compound from the gaseous ions.
02

Set up the Born-Haber cycle Equation

The equation for the Born-Haber cycle is: Heat of Formation = Heat of Sublimation + Ionization Energy + Bond Energy / 2 + Electron Affinity + Lattice Energy We need to find the heat of sublimation in this equation, given the other values. So let's plug in the known values: \(-272\,\text{kJ/mol} = \text{Heat of Sublimation} + 520\,\text{kJ/mol} + \frac{151\,\text{kJ/mol}}{2} + (-295\,\text{kJ/mol}) + (-753\,\text{kJ/mol})\)
03

Solve for the Heat of Sublimation

Now we just need to solve this equation for the heat of sublimation: \(-272\,\text{kJ/mol} = \text{Heat of Sublimation} + 520\,\text{kJ/mol} + 75.5\,\text{kJ/mol} - 295\,\text{kJ/mol} - 753\,\text{kJ/mol}\) \(-272\,\text{kJ/mol} = \text{Heat of Sublimation} - 452.5\,\text{kJ/mol}\) Now, add 452.5 kJ/mol to both sides: \(\text{Heat of Sublimation} = -272\,\text{kJ/mol} + 452.5\,\text{kJ/mol}\) \(\text{Heat of Sublimation} = 180.5\,\text{kJ/mol}\) So, the heat of sublimation of Li(s) is 180.5 kJ/mol.

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