# Lattice energy for binary salts are find out by born haber cycle but how we will find lattice energy for other salts which are not binary like MgCl2???

##### 1 Answer

Well, you do it in the same way, but with different coefficients, and one more step...

*As with binary salts, construct a Born-Haber cycle by using Hess' law to make a closed loop. Reversed reactions have a negative sign in front of #DeltaH#, and scaled reactions scale #DeltaH# by that constant.*

The **lattice energy** is the energy released when the solid forms from the gaseous ions, so you must form that reaction by piecing together data... you'll have an easier time than me, because I actually had to look up the data.

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

#DeltaH_f^@ = -"641.62 kJ/mol"#

#cancel("Mg"(g)) -> cancel("Mg"(s))# ,

#-DeltaH_(su b) = -"146 kJ/mol"#

#"Mg"^(2+)(g) + cancel(2e^(-)) -> cancel("Mg"(g))# ,

#-("IE"_1 + "IE"_2) = -("737.750 kJ/mol" + "1450.683 kJ/mol")#

#cancel(2"Cl"(g)) -> cancel("Cl"_2(g))# ,

#-2DeltaH_"bond" = -2("122 kJ/mol")#

#2"Cl"^(-)(g) -> cancel(2"Cl"(g)) + cancel(2e^(-))#

#ul(-2("EA"_1) = -2(-"349 kJ/mol)"" "" "" "" "" "" "" "#

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

From here (

What you have to do in addition is to include the *bond dissociation energy* of

#1/2"Cl"_2(g) -> "Cl"(g)# .

Therefore, since we have fulfilled the **lattice formation reaction** by using Hess's law:

#color(blue)(DeltaH_"Lattice") = DeltaH_f^@ - DeltaH_(su b) - ("IE"_1 + "IE"_2) - 2DeltaH_"bond" - 2("EA"_1)#

#= -"641.62 kJ/mol" - "146 kJ/mol" - ("737.750 kJ/mol" + "1450.683 kJ/mol") - 2("122 kJ/mol") - 2(-"349 kJ/mol")#

#= color(blue)(-"2522.05 kJ/mol")#

**When the lattice forms, the energy change must be negative.**