Question

\(\operatorname{LiI}(s)\) 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

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

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.

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: \(-272\,\text{kJ/mol} = \text{Heat of Sublimation} + 520\,\text{kJ/mol} + \frac{151\,\text{kJ/mol}}{2} + (-295\,\text{kJ/mol}) + (-753\,\text{kJ/mol})\)

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.

Key concepts

Heat of Formation

Heat of formation is an essential concept in chemistry. It refers to the energy change when one mole of a compound forms from its elements in their standard states. For example, when lithium iodide, \(\text{LiI(s)}\), forms from lithium \(\text{Li(s)}\) and iodine \(\text{I}_2(g)\), a certain amount of energy is either absorbed or released.
The heat of formation is a vital component of the Born-Haber cycle in calculating the heat of sublimation or other energy measurements required to form an ionic compound from its elements. For \(\text{LiI(s)}\), its heat of formation is given as \(\text{-272 kJ/mol}\). Such values help calculate and understand the entire cycle, providing insights into the reaction energetics.

Lattice Energy

Lattice energy describes the energy needed to form an ionic solid from its constituent gaseous ions. It is an indication of the strength of the forces holding the ions together. The greater the lattice energy, the more energy is released when the ions come together to form a solid.
In the context of lithium iodide \(\text{LiI}\), its lattice energy is given as \(\text{-753 kJ/mol}\). This significant negative value indicates a highly exothermic process, meaning a lot of energy is released, enhancing the stability of the ionic solid. Lattice energy also depends on the sizes and charges of the ions; smaller, more highly charged ions often produce higher lattice energies.
Understanding lattice energy is crucial for predicting the stability of ionic compounds and their formation processes in the Born-Haber cycle.

Ionization Energy

Ionization energy is the energy required to remove an electron from a neutral atom in its gaseous state. It reflects how strongly electrons are held in an atom.
For lithium \(\text{Li(g)}\), the ionization energy is specified as \(\text{520 kJ/mol}\). This value plays an important role in the Born-Haber cycle, as it impacts the formation of cations needed to produce ionic compounds like \(\text{LiI}\).
Higher ionization energies usually indicate that electrons are more difficult to remove. The understanding of ionization energy helps chemists predict the reactivity of elements and the energy required for converting neutral atoms into ions, which is crucial in compound formation.

Electron Affinity

Electron affinity refers to the energy change when an electron is added to a neutral atom in the gaseous state to form an anion. It measures the tendency of an atom to accept electrons.
In the case of iodine \(\text{I(g)}\), its electron affinity is \(\text{-295 kJ/mol}\). This negative value indicates that the process of gaining an electron is exothermic, signaling that energy is released when an electron is added to form iodide ions \(\text{I}^-\).
Electron affinity is a fundamental concept for understanding an atom's electron gaining tendency, influencing the types of ions that form and their energy implications in reactions, particularly in the Born-Haber cycle to calculate the energy dynamics in ionic compound formation.