## Lattice Energy Formula :

Equilibrium value of the lattice energy U_{L} _{(µl) }in milliliter is equal to the product of Avogadro’s constant N_{A(6.022 × 1022)} , Madelung constant ∝_{(r/s2)} in radius per second square, Cation and anion charge Z^{2 }_{(}_{cl}^{–}_{)} in ^{–} ions charge, Electron charge e^{2} is divided by Permittivity of free space ∈o(8.8541878128(13)×10^{−12} F⋅m^{−1 }) in farad per meter, Closest ion distance r ^{2}o _{(m/s) } in meter per second and multiply by Born Exponent n. Hence lattice energy formula has been written as,

U_{L} _{(µl) }= (N_{A(6.022 × 1022)} ∝_{(r/s2)}Z^{2 }_{(}_{cl}^{–}_{)} e^{2 }/

4π∈_{o(}8.8541878128(13)×10^{−12} F⋅m^{−1}) r ^{2}o_{(m/s)}) (1-1/n)

N_{A} →Avogadro’s constant (6.022 × 1022)

α → Madelung constant in r/s^{2 }

e →Electron charge in (1.6022 × 10^{-19}C)

Z^{+ }and Z^{– }→Cation and anion charge in cl^{–} and Na^{+}

ϵ_{o} → Permittivity of free space in 8.8541878128(13)×10^{−12} F⋅m^{−1}

n → Born Exponent

r_{0} →Closest ion distance in cm

U_{L }→equilibrium value of the lattice energy in (µl),

## Sample Examples :

### Example 1:

Calculate the Lattice energy of NaCl by using

Born-Landed equation

Given :

α = 1.74756

Z^{–} = -1 (the Cl^{–} ions charge)

Z^{+ }= +1 (the charge of the Na^{+} ion)

N_{A} = 6.022 × 10^{23} ion pairs mol^{-1}

C = 1.60210 × 10^{-19}C (the charge on the electron)

π = 3.14159

εo = 8.854185 × 10^{-12 }C^{2} J^{-1} m^{-1}

r_{o} = 2.82× 10^{-12} m, the sum of radii of Born-Landed equation.

Na^{+} and Cl^{–}

n = 9.1 the average of the values for Na^{+} and Cl^{–}.

### Solution :

Using the Born –Landed equation

U_{L}=(N_{A}∝Z^{2}e^{2}/4π∈or^{2}o)(1-1/n)

U_{L}=(6.022×1023/mol)(1.74756)(1.6022×10−19)2(1.747558)/

4×3.14159 (8.854×10−^{10})(2.82×10−12)(1–1/9.1)

U_{L }= –756 kJ per mol

Therefore lattice energy for NaCl is 756 KJ per mol.

### Example 2 :

The lattice energy of AgBr is 795 KJ mol^{-1}. Predict the Lattice energy of the isomorphous Agl using Born-Landed equation . the numeric of r_{c}+r_{a} is 221 pm for AgBr and 242 pm for Agl.

### Answer:

if the only variance between AgBr and Agl were in the size of the anion, one would expect the lattice energies to be relational to the inverse ratio of r_{c}+r_{a. }hence forth we expect the lattice energy of Agl to be,

=(795) (221/242)

=425 J mole^{-1}.