What is the lattice enthalpy of #"MgCl"_2# and how do you determine it using the Born-Haber cycle?

The exothermic lattice-forming reaction is:

#"Mg"^(2+)(g) + 2"Cl"^(-)(g) -> "MgCl"_2(s)#

The standard formation reaction is:

#"Mg"(s) + "Cl"_2(g) -> "MgCl"_2(s)#

To get to this we start from the formation reaction and use Hess's Law to plan it out in order to get the lattice-forming reaction.

#cancel("MgCl"_2(s)) -> cancel("Mg"(s)) + cancel("Cl"_2(g))#, #-DeltaH_f^@ > 0#
#cancel("Mg"(s)) -> cancel("Mg"(g))#, #DeltaH_"sub"^@ > 0#
#cancel("Mg"(g)) -> cancel("Mg"^(2+)(g)) + cancel(2e^(-))#, #DeltaH_(IE)^@("Mg"(g)) > 0#
#ul(cancel("Cl"_2(g)) + cancel(2e^(-)) -> cancel(2"Cl"^(-)(g)))#, #DeltaH_(EA)^@("Cl"_2(g)) < 0#
#cancel("Mg"^(2+)(g)) + cancel(2"Cl"^(-)(g)) -> cancel("MgCl"_2(s))#, #DeltaH_L < 0#

We wanted all of that to cancel out, because the change in enthalpy for a closed loop is zero.

Therefore, we need all of this data, and we obtain most of them from NIST. You should of course use your own textbook...

• #DeltaH_(IE)^@("Mg"(g)) = +"7.64624 eV" = +"737.65 kJ/mol"#
• #DeltaH_(EA)^@("Cl"_2(g)) = -"2.38 eV" = -"229.60 kJ/mol"#
• #DeltaH_f^@("MgCl"_2(s) = -"641.62 kJ/mol"#

This I couldn't find... and so this source is not good.

• #DeltaH_"sub"("Mg"(s)) = DeltaH_"fus" + DeltaH_"vap"#

#~~ "8.4 kJ/mol" + ~~ "140 kJ/mol"#

#~~# #"148.4 kJ/mol"#

By Hess's Law, the sum of these steps gives a closed loop.

#-(-|DeltaH_f^@|) + |DeltaH_("sub")^@| + |DeltaH_(IE)^@("Mg"(g))| - |DeltaH_(EA)^@("Cl"_2(g))| - |DeltaH_L| = 0#

Solving for the lattice energy, which is negative.

#color(blue)(-|DeltaH_L|) = -{|DeltaH_f^@| + |DeltaH_("sub")^@| + |DeltaH_(IE)^@("Mg"(g))| - |DeltaH_(EA)^@("Cl"_2(g))|}#

#= -{"641.62 kJ/mol" + "148.4 kJ/mol" + "737.65 kJ/mol" - "229.60 kJ/mol"}#

#~~ color(blue)(-"1298.07 kJ/mol")#

This is large and negative, which makes sense for an ionic compound with a #+2# metal oxidation state (which makes #DeltaH_L# larger).

You should take the time to draw out the Born-Haber Cycle that you have just performed, and then redo this calculation with your own numbers.