Born-Haber Cycle Lab Report

However in example 6, the H2O molecule given in the question is in the liquid state. Thus liquid water has to be converted into steam first. To convert 2 moles of liquid water into 2 moles of steam, an additional 82 kJ of energy is required.
By using Hess’s Law,
ΔH + 2(799) + 2(41) + 4(460) = 4(410) + (496)
ΔH = -888 kJ mol-1

Alternative: energy absorbed during bond breaking = 4(410) + 2(496) = 2632 kJ energy released during bond forming = 2(799) + 4(460) + 2(41) = 3520 kJ enthalpy change of reaction = +2632 + (-3520) = –888 kJ mol-1

Born-Haber Cycle &
The lattice energy value obtained from the Born-Haber cycle is derived from experimental data. It is also possible to calculate a theoretical value for lattice energy by assuming the compound is fully ionic. For ionic compounds with high covalent character, the difference between the experimental and theoretical values would be quite significant.

Unlike the Born- Haber cycle, the energy level diagram must have an energy axis and a datum line. The datum line is taken as the arbitrary zero enthalpy content of any pure element. Positive enthalpy changes are represented by arrows pointing upwards while negative ones are represented by arrows pointing downwards.

A Born-Haber cycle comprises 6 pieces of information. They
However, most A Level questions only provide 3 of them. The ionization energy values can be found in the Data Booklet. The standard enthalpy change of atomization of a non-metal can be obtained by using the relationship: ½ bond energy = ΔHatom

e.g. ½ Cl2 (g)  Cl (g), ΔHatom of chlorine = ½ BE (Cl-Cl) = ½ x 244 = 122 kJ mol-1

Change in Gibbs Free Energy, ΔG
G is the maximum useful work that can be obtained from a reaction at constant pressure. Under standard conditions, ΔG⦵ = ΔH⦵  TΔS⦵ sign of G, at a given temperature spontaneity
-ve spontaneous
0 system is at equilibrium e.g. reversible chemical reaction or change in phase (at the melting or boiling point)
+ve non-spontaneous / spontaneous in the reverse direction

 Comment: The unit for ΔG and ΔH is usually kJ mol-1, whereas that for ΔS is usually J mol-1 K-1. Be careful with the units in your calculations.

 Comment: ΔH and ΔS are essentially constant with temperature unless a phase change occurs.

Example 7
When water freezes, 6.0 kJ mol-1 of heat enthalpy is evolved. What is the entropy change when 54 g of water freezes at