# What is the formation reaction of aluminum chloride molecule and its enthalpy of formation?

Jul 26, 2017

You may also mean the liquid or the gas, but I assume you mean the solid. Thus, I assume you mean the process that produces ${\text{AlCl}}_{3} \left(s\right)$ from the elemental states of the atoms.

Well...

• Chlorine naturally exists as a diatomic gas, ${\text{Cl}}_{2} \left(g\right)$ at ${25}^{\circ} \text{C}$ and $\text{1 atm}$.
• Aluminum naturally exists at ${25}^{\circ} \text{C}$ and $\text{1 atm}$ as a solid.

Therefore, the unbalanced standard formation reaction is:

${\text{Cl"_2(g) + "Al"(s) -> "AlCl}}_{3} \left(s\right)$

Balancing it to ensure that we form one mol of product (as it is defined for a standard formation reaction), we write:

$\textcolor{b l u e}{\overline{\underline{| \stackrel{\text{ ")(" "3/2 "Cl"_2(g) + "Al"(s) -> "AlCl"_3(s)" }}{|}}}}$

And this has an enthalpy of formation of $\Delta {H}_{f}^{\circ} = - \text{705.63 kJ/mol}$.

Keep in mind that this compound is not much of a molecule; its electronegativity difference is

$E {N}_{C l} - E {N}_{A l} = 3.16 - 1.61 = 1.55$,

which indicates at least moderate ionic character in the bonds, despite the trigonal planar symmetry making this compound nonpolar. This is reflected somewhat in its melting point of ${192.4}^{\circ} \text{C}$, showing it is a solid at room temperature and pressure.