# 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???

Jul 18, 2018

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 $\Delta H$, and scaled reactions scale $\Delta H$ 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),

$\Delta {H}_{f}^{\circ} = - \text{641.62 kJ/mol}$

$\cancel{\text{Mg"(g)) -> cancel("Mg} \left(s\right)}$,

$- \Delta {H}_{s u b} = - \text{146 kJ/mol}$

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

$- \left(\text{IE"_1 + "IE"_2) = -("737.750 kJ/mol" + "1450.683 kJ/mol}\right)$

$\cancel{2 {\text{Cl"(g)) -> cancel("Cl}}_{2} \left(g\right)}$,

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

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

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

$\textcolor{g r e e n}{{\text{Mg"^(2+)(g) + 2"Cl"^(-)(g) -> "MgCl}}_{2} \left(s\right)}$

From here ($\Delta {H}_{s u b}$), and here (${\text{IE"_1 + "IE}}_{2}$) and here ($\Delta {H}_{f}^{\circ}$) and here ($\Delta {H}_{\text{bond}}$), the above data was obtained. The electron affinity of $\text{Cl}$ was easy enough to google search.

What you have to do in addition is to include the bond dissociation energy of $\text{Cl"_2(g) -> 2"Cl} \left(g\right)$, i.e. the breaking of the $\text{Cl"-"Cl}$ bond in the gas phase, AND you must make sure the coefficients line up, because the value you would be given is for

$\frac{1}{2} \text{Cl"_2(g) -> "Cl} \left(g\right)$.

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

$\textcolor{b l u e}{\Delta {H}_{\text{Lattice") = DeltaH_f^@ - DeltaH_(su b) - ("IE"_1 + "IE"_2) - 2DeltaH_"bond" - 2("EA}} _ 1}$

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

$= \textcolor{b l u e}{- \text{2522.05 kJ/mol}}$

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