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

1 Answer
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 #"AlCl"_3(s)# from the elemental states of the atoms.

Well...

  • Chlorine naturally exists as a diatomic gas, #"Cl"_2(g)# at #25^@ "C"# and #"1 atm"#.
  • Aluminum naturally exists at #25^@ "C"# and #"1 atm"# as a solid.

Therefore, the unbalanced standard formation reaction is:

#"Cl"_2(g) + "Al"(s) -> "AlCl"_3(s)#

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

#color(blue)(barul(|stackrel(" ")(" "3/2 "Cl"_2(g) + "Al"(s) -> "AlCl"_3(s)" ")|))#

And this has an enthalpy of formation of #DeltaH_f^@ = -"705.63 kJ/mol"#.

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

#EN_(Cl) - EN_(Al) = 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^@ "C"#, showing it is a solid at room temperature and pressure.