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