# Born-Haber Cycle

## What is the Born–Haber Cycle (BH)?

It is a series of steps (chemical processes) used to calculate the lattice energy of ionic
solids, which is difficult to determine experimentally. You can think of BH cycle as a
special case of Hesse’s law, which
states that the overall energy change in a chemical process can be calculated by breaking
down the process into several steps and adding the energy change from each step.

To understand the BH cycle fully let us define the meaning of lattice energy first.

It is the energy released (exothermic,
energy will have a negative value!) when a metal ion in its gaseous state combines with a nonmetal anion in its
gaseous state to form an ionic solid. The magnitude of the lattice energy relates to how stable the ionic solid is.
A larger value indicates a more stable ionic compound.

The energy released in the following processes is called is called lattice energy:

 Li+(g) + F–(g) -> LiF(s) LE = -1047kJ Mg+(g) + O2-(g) -> MgO(s) LE = -3916kJ where k is a constant, Q1 and Q2 are the charges
on the cation and anion respectively, and r is the inter-nuclear distance.

Test yourself: (answers found at end of page)

1. Explain why the LE for LiF is much less than for MgO ?
2. Explain why LE of Na2S is -2203 kJ and Cs2S is -1850 kJ?

## Lattice Energy Calculations via Born-Haber cycle

Example 1: Li(s) + 1/2F2(g) -> LiF(s)           -617kJ

This overall chemical process can be broken down to many intermediate steps:

 1 Li(s) -> Li(g) energy of sublimation +161 kJ endothermic 2 Li(g) -> Li+(g) ionization energy +520 kJ endothermic 3 1/2F2(g) -> F(g) energy of dissociation +77 kJ endothermic 4 F(g) -> F–(g) electron affinity -328 kJ exothermic 5 Li+(g) + F–(g) -> LiF(s) Lattice Energy (LE) ??

If we add 1through 5, after cancelling the intermediates,
Li+(s) + 1/2F2(g) -> LiF(s)
[+161+520+77+(-328)+LE] kJ = -617 kJ
So, LE = -1047 kJ

As you can see, the lattice energy is hugely exothermic so offsets the other endothermic processes
including energy of sublimation, ionization energy of Li(g) and the bond dissociation energy of F2(g).
The final overall process of formation of LiF from lithium (s) and F2(g) is exothermic: energetically favorable.

Test yourself: (answers found at end of page)

3. From your knowledge of ionic radii and the charge on the cations and anions in the following compounds,
arrange the compounds in the order of increasing lattice energy.
MgSe,       MgTe,       K2Se,       K2Te

Example 2: Lets calculate the lattice energy for KCl from the following literature values via the Born-Haber cycle.

 Ionization energy for K 419.0 kJ/mole Electron affinity for Cl -349 kJ/mole Bond energy for Cl2 239 kJ/mole Heat of sublimation for K 64 kJ/mole ΔH0f for KCl 437 kJ/mole LE ? Test yourself: (answers found at bottom of page)

4. Through the BH cycle calculate its lattice energy from the following energy data.

 Energy of sublimation for K(s): 77.08 kJ/mole Ionization energy for K(g): 418.6 kJ/mole Bond dissociation energy for Br2(g) 193 kJ/mole Electron affinity for bromine: -324.6 kJ/mole Heat of vaporization of Br2: 29.96 kJ/mole Heat of formation for KBr(s) -393.8 kJ/mole

### Answers to Test Yourself questions:

1. , For LiF the product of Q1 and Q2 is
-1(+1 x -1) whereas for MgO the product of Q1 and Q2 is -4(+2 x -2), this makes the numerator
much larger for LE for MgO compared to LiF.
2. IN this case the product of Q1 and Q2 is the same for Na2S and Cs2S,
so the numerator for LE is the same for both, but the Cs being much larger than Na means the inter-nuclear distance, r,
for Cs2S is much larger, making the magnitude for LE for Cs2S much smaller. (LE is inversely
proportional to the r and directly proportional to the Q1 and Q2).
3. Based on the formula for LE and considering the charges (Q1 and Q2) and inter-nuclear distance
(r), the trend for LE for the compounds will be:

CaSe >> CaTe >> Na2Se >> Na2Te
4. BH cycle for the formation of KBr:

 1 K(s) + 1/2Br2(l) -> K(g) + 1/2Br2(l) 77.08 kJ 2 K(g) + 1/2Br2(l) -> K(g) + 1/2Br2(g) 29.96/2 kJ/(1/2 mole) 3 K(g) + 1/2Br2(g) -> K+(g) + 1/2Br2(g) 418.6 kJ 4 K+(g) + 1/2Br2(g) -> K+ + Br(g) 193/2 kJ 5 K+ + Br(g) -> K+(g) + 1/2Br–(g) -324.6 kJ 6 K+(g) + 1/2Br–(g) -> KBr(s) ??