(A Level Chem) Born-Haber Cycles

The following shows two Born-Haber Cycles. Both contain the lattice energy LE.





IE = ionisation energy, EA = electron affinity

Lattice energy LE cannot be directly measured. It's found using the Born-Haber Cycle. Let's use Born-Haber Cycle 1 and the following standard values to calculate the size of LE of NaCl and that of MgO.

Sodium chloride	Magnesium oxide
Standard values: ∆Hformationº(NaCl) = -411 kJ/mol ∆Hatomisationº(Na) = +107 kJ/mol standard 1st IE(Na --> Na+) = +502 kJ/mol ∆Hatomisationº(½Cl2 --> Cl) = +121 kJ/mol standard 1st EA(Cl --> Cl-) = -355 kJ/mol	Standard values: ∆Hformationº(MgO) = -602 kJ/mol ∆Hatomisationº(Mg) = +148 kJ/mol standard 1st IE(Mg --> Mg+) = +738 kJ/mol standard 2nd IE(Mg+ --> Mg2+) = +1451 kJ/mol ∆Hatomisationº(½O2 --> O) = +294 kJ/mol standard 1st EA(O --> O-) = -141 kJ/mol standard 2nd EA(O- --> O2-) = +798 kJ/mol
Calculation: |LE| = -(-411) + 107 + 502 + 121 + (-355) = 786 kJ/mol	Calculation: |LE| = -(-602) + 148 + 738 + 1451 + 294 + (-141) + (798) = 3845 kJ/mol

We find that MgO has much greater LE than NaCl. Reason: Mg²⁺ and O^2- are smaller and have greater charge than Na⁺ and Cl^-. The ions in MgO have higher charge density and hence higher LE than those in MgO. 

Lattice energy depends on the charge density (i.e. charge per unit surface area Q/4πr²) of ions...

Note: Some A-level Chemistry sources list LE to be positive and some negative. It depends on how LE is defined. If LE is defined to be the dissociation of an ionic solid into its gaseous ions, then LE is +ve. If it's defined in the opposite way (i.e. the formation of an ionic solid from its gaseous ions), then LE is -ve.

If lattice energy is defined by bM^a+(g) + aX^b-(g) --> (M^a+)_b(X^b-)_a(s), then it's negative...

The LE affects the enthalpy of solution ∆H_solv which can be used to predict whether a solid readily dissolves in a solvent. Now, let's use Born-Haber Cycle 2 and the following standard values to calculate the enthalpy of solution ∆H_solv of NaCl and that of AgCl to predict whether they're soluble in water.

Sodium chloride	Silver chloride
Standard values: LE = -786 kJ/mol (from above) ∆Hhyd(NaCl) = -784 kJ/mol	Standard values: LE = -820 kJ/mol ∆Hhyd(AgCl) = -797.4 kJ/mol
Calculation: ∆Hsolv = ∆Hhyd - LE = -784 kJ/mol - (-788 kJ/mol)  = +4 kJ/mol NaCl has endothermic (+ve) ∆Hsolv and yet it is able to readily dissolve in water.	Calculation: ∆Hsolv = ∆Hhyd - LE = -797.4 kJ/mol - (-820 kJ/mol)  = +22.6 kJ/mol Since ∆Hsolv is +ve, we predict AgCl is insoluble or sparingly soluble. (Note: AgCl has more covalent character than ionic character. Like most covalent compounds, AgCl does not readily dissolve in water.)
Comment: The ∆H consideration is only part of the story. Another factor is ∆S in ∆G = ∆H – T∆S where ∆G is change in Gibbs free energy and ∆S is change in entropy. Because the dissolving of NaCl result in a great +ve ∆S (i.e. A large increase in disorder), the ∆G is -ve and so the dissolving of NaCl is spontaneous.	Comment: The ∆Hsolv for AgCl(s) dissolving in aqueous ammonia is -ve and so we predict AgCl is soluble in aqueous ammonia.

The ∆G of the dissolving of NaCl is -ve and so it is soluble...

Try these questions before you check with the answers below.

1. Why does the thermal stability of the oxides of Group II elements decrease down the Group? Hint: The greater the magnitude of LE of an ionic solid, the greater its stability.
   
2. The decomposition of the carbonate of Group II elements can be written as MCO₃(s) --> MO + CO₂ which involves the ability of the cation to distort the carbonate ion. Why does the temperature at which MgCO₃ decomposes is much lower than that of BaCO₃? Hint: The greater the magnitude of the LE, the higher the temperature at which the solid decomposes.
   
3. Calculate the lattice energy of copper(II) oxide. [∆H_formation^º(CuO) = -155 kJ/mol, ∆H_atomisationº(Cu) = +339 kJ/mol, ∆H_atomisationº(½O₂ --> 2O) = +294 kJ/mol, 1^st IE(Cu --> Cu⁺) = +745 kJ/mol, 2^nd IE(Cu⁺ --> Cu²⁺) = +1960 kJ/mol, 1^st EA(O --> O^-) = -141 kJ/mol, 2^nd EA(O^- --> O^2-) = +798 kJ/mol]
   
4. Calculate the enthalpy of solution of sodium hydroxide if its lattice energy is -877 kJ/mol and enthalpy of hydration is -932 kJ/mol. How does the value tell whether sodium hydroxide will dissolve in water? 

Answers: 1. The charges of the ions remain the same down the Group. The radius of the cations increases down the Group. As a result, the charge density and hence LE decreases down the Group. 2. Barium is below magnesium in Group II. The Mg²⁺ ions are smaller and have higher charge density than Ba²⁺. This makes Mg²⁺ more capable of distorting the shape of CO₃^2- than Ba²⁺. Therefore, MgCO₃ is less thermally stable and decomposes at lower temperature than BaCO₃. 3. LE(CuO) = -(-155) + 339 + 294 + 745 + 1960 + (-141) + 798 = +4150 kJ/mol 4. ∆H_solv = ∆H_hyd + LE = -932 + (-877) = -1809 kJ/mol. The enthalpy of solution of sodium hydroxide is very exothermic telling that it very soluble in water.