Advanced Higher Chemistry Unit 2 Applications of Hess’s Law 1 Born – Haber Cycles

Born - Haber Cycles The enthalpy of formation is the energy change when 1 mole of a substance is formed from its constituent elements at standard conditions. e.g. Mg(s) + Cl2(g)  MgCl2(s) H = X ? These cycles are a means of showing clearly all of the processes going on in the formation of 1 mole of an ionic compound. It is nothing more than the application of Hess’s Law to find the enthalpy change of a particular step that is difficult to measure directly.

There are several component steps in making an ionic compound Enthalpy of atomisation - breaking up the metallic lattice - this is ENDOTHERMIC Enthalpy of atomisation - breaking up the non-metal molecules - this is ENDOTHERMIC First/ Second etc ionisation enthalpy - turning the metal atoms into positively charged ions -this is ENDOTHERMIC Electron gain enthalpy - turning the non-metal atoms into negatively charged ions - this can be EXO or ENDOTHERMIC Lattice enthalpy - forming bond between the ions to form the lattice - this EXOTHERMIC

The Enthalpy changes needed to make ionic compounds What it means The metallic lattice is broken up into individual atoms 1. Enthalpy of sublimation (or atomisation) The non-metal molecules or structure is broken up into atoms 2. Enthalpy of dissociation(or atomisation) The requisite number of outer electron(s) are removed 3. The appropriate number of ionisation energies of the metal The non-metal atoms gain electron(s) 4. The electron gain enthalpy 5. Enthalpy of lattice-forming of the ionic substance The ionic bonds are formed The overall enthalpy change for the reaction 6. Enthalpy of formation of the substance  H

The net result of all on these energy steps is the enthalpy of formation for the substance H. Let us look at an example, the formation of 1 mole of magnesium chloride MgCl2(s)

Electron gain enthalpy = 2 x - 348.7kJ mol-1 Mg2+(g) + 2Cl(g) Electron gain enthalpy = 2 x - 348.7kJ mol-1 Second ionisation enthalpy of Mg = + 1460 kJmol-1 Mg+(g) + 2Cl(g) Mg2+(g) + 2Cl-(g) First ionisation enthalpy of Mg = + 744 kJmol-1 Mg(g) + 2Cl(g) Lattice enthalpy = - 2326 kJmol-1 Enthalpy of atomistaion of chlorine = + 243 kJmol-1 Mg(g) + Cl2(g) Enthalpy of atomistaion of Mg = + 147kJmol-1 Mg(s) + Cl2(g) H Enthalpy of formation Mg2+(Cl-)2(s)

H0r = Ho for all of the stages involved H (kJ) Enthalpy Change + 147 1. Enthalpy of atomisation of Mg + 243 2. Enthalpy of atomisation of Cl 3. First ionisation energy of Mg + 744 4. Second ionisation energy of Mg + 1460 5. Electron affinity of Cl 2 x – 348.7 6. Lattice enthalpy of MgCl2 - 2326 - 697.4 7. Enthalpy of formation of MgCl2

The Enthalpy of solution of substance is the energy change when 1 mole completely dissolves in water When an ionic compound dissolves in water, two processes occur: 1. The lattice breaks down releasing ions 2. The ions then become hydrated by forming bonds with water molecules The enthalpy of solution of a salt, just like the Born – Haber cycle for the formation of an ionic compound, can be regarded as an energy balance between the endothermic process of lattice breaking and the exothermic hydration of ions.

Enthalpy change What it means 1. Enthalpy of lattice breaking The ionic lattice is broken up into gaseous ions 2. Enthalpy of hydration of metal ion Bonds being formed between the metal ions and the water molecules 3. Enthalpy of hydration of non-metal ion Bonds being formed between the non-metal ions and the water molecules 4. Enthalpy of solution Overall enthalpy change

_______________________________ Na+ (g) + Cl(g) _______________________________ Enthalpy of hydration of sodium ions Lattice enthalpy 405(H2) +771(H1) Na+Cl(s) ______________ Na+ (g) + (aq)  Na+ (aq) __________________ Enthalpy of solution of sodium chloride Enthalpy of hydration of chloride ions  362(H3) + 4(H4) Cl(g) + (aq)  Cl(aq) Na+Cl(s) +(aq)  Na+ (aq) + Cl(aq) ___________________ ___________________________

Enthalpy change described Equation H kJmol-1 Enthalpy of lattice-breaking of sodium chloride Na+Cl(s)  Na+(g) + Cl(g) +771(H1) Enthalpy of hydration of sodium ions 405(H2) Na+(g) + (aq)  Na+(aq) Enthalpy of hydration of chloride ions Cl(g) + (aq)  Cl(aq) 362(H3) Na+Cl(s) + (aq)  Na+(aq) + Cl(aq) Enthalpy of solution of sodium chloride + 4(H4)

H4 = H1 + H2 + H3 = 771 + (- 404) + (-362) = + 4kJ mol-1 On applying Hess’s law , the enthalpy of solution can be calculated thus, H4 = H1 + H2 + H3 = 771 + (- 404) + (-362) = + 4kJ mol-1