2.  When an ionic solid dissolves in water, two processes occur  Firstly the ions are separated (endothermic)  Secondly the ions are surrounded by water (exothermic)  There are two definitions for the lattice enthalpy, ΔH ϴ L :  Lattice dissociation enthalpy, ΔH ϴ L (diss)  Lattice formation enthalpy, ΔH ϴ L (form)

3.  Definition: The lattice dissociation enthalpy, ΔH L (diss) is the enthalpy change when 1 mole of an ionic solid is separated into its gaseous ions  Example: Sodium chloride (NaCl)  NaCl (s) → Na + (g) + Cl - (g) ΔH L (diss) = + 787 kJ mol -1  Energy must be put in to break the strong ionic bonds in the lattice; therefore it is an endothermic process

4.  Definition: The lattice formation enthalpy, ΔH L (form) is the enthalpy change when 1 mole of an ionic solid is formed from its gaseous ions  Example: Sodium chloride (NaCl)  Na + (g) + Cl - (g) → NaCl (s) ΔH L (form) = - 787 kJ mol -1  Energy is released when ionic bonds form; therefore it is an exothermic process  Lattice enthalpy of formation has the same value as the lattice enthalpy of dissociation but with the reverse sign

5.  Definition: The enthalpy of hydration, ΔH Hyd is the enthalpy change when one mole of separated gaseous ions is dissolved completely in water to form one mole of aqueous ions  Example:  Na + (g) → Na + (aq) ΔH Hyd = - 406 kJ mol -1  Cl - (g) → Cl - (aq) ΔH Hyd = - 377 kJ mol -1  Hydration is exothermic

6.  Hydration means the ion is surrounded by water molecules  Water molecules interact with ions  Water (H 2 O) is a polar molecule with a δ- oxygen atom and a δ+ hydrogen atom  A negatively charged ion becomes hydrated as it attracts the δ+ hydrogen of the water molecules δ+δ+ δ-δ- δ+δ+

7.  Water (H 2 O) is a polar molecule with a δ- oxygen atom and a δ+ hydrogen atom  A positively charged ion becomes hydrated as it attracts the δ- oxygen of the water molecules δ+δ+ δ-δ- δ+δ+

8.  Definition: The enthalpy of solution, ΔH Soln is the enthalpy change when one mole of an ionic substance dissolves in enough water to ensure the ions are well separated and do not interact with one another  Example:  NaCl (s) → Na + (aq) + Cl - (aq) ΔH Soln = + 4 kJ mol -1  Enthalpy of solution, ΔH Soln can be negative or positive

9. Lattice dissociation enthalpy, ΔH ϴ L (diss) Enthalpy of hydration, ΔH ϴ Hyd SolidGaseous Aqueous Enthalpy of solution, ΔH ϴ Soln

10.  The enthalpy of solution, ΔH Soln can be calculated using: ΔH Soln = ΔH L (diss) + ΔH Hyd  Example 1: Sodium chloride, NaCl  ΔH Soln = ΔH L (diss) (NaCl) + ΔH Hyd (Na + ) + ΔH Hyd (Cl - ) = + 787 + - 406 + - 377 = + 4 kJ mol -1

11.  Example 2: Magnesium chloride, MgCl 2  Use the data below to calculate the enthalpy of solution  ΔH ϴ Soln = ΔH ϴ L (diss) (MgCl 2 ) + ΔH ϴ Hyd (Mg 2+ ) + [2 x ΔH ϴ Hyd (Cl - )] = + 2493 + - 1920 + (2 x -364) = - 155 kJ mol -1 EnthalpyValue (kJ mol -1 ) Lattice formation-2493 Hydration (Mg 2+ )-1920 Hydration (Cl - )-364 Lattice dissociation enthalpy, ΔH ϴ L (diss) is an endothermic process

12.  Enthalpy level diagrams are another way to represent the processes involved in dissolving  In these diagrams:  Endothermic process = Arrow pointing upwards  Exothermic process = Arrow pointing downwards  The diagrams should be drawn to scale, where the length of the arrow represents the enthalpy change  (i.e. bigger arrow = large enthalpy change)

13.  Enthalpy level diagram for dissolving  Example: Sodium chloride (NaCl) NaCl (s) Starting line = Starting ionic compound Na + (g) + Cl - (g) ΔH ϴ L (diss) (NaCl) + 787 Lattice diss = endothermic, arrow up Na + (aq) + Cl - (g) ΔH ϴ Hyd (Na + ) - 406 ΔH ϴ Hyd (Cl - ) - 377 Na + (aq) + Cl - (aq) ΔH ϴ Soln (NaCl) + 4 Larger enthalpy changes at top

14.  You must be able to:  Name the enthalpy change represented by each arrow  Write equations to illustrate the enthalpy changes  Calculate the value of any of the enthalpy changes given data for the others

15.  Example 1: Calculate the lattice dissociation enthalpy, ΔH ϴ L  ΔH ϴ Soln (NaCl) = ΔH ϴ L (diss) (NaCl) + ΔH ϴ Hyd (Na + ) + ΔH ϴ Hyd (Cl - )  + 4 = ? + - 406 + - 377 + 787

16.  Example 2: Ammonium nitrate (NH 4 NO 3 )  Draw an enthalpy level diagram for dissolving of ammonium nitrate in water  Use the data to calculate the enthalpy of solution of ammonium nitrate QuantitykJ mol -1 ΔH ϴ L (NH 4 NO 3 )+ 647 ΔH ϴ Hyd (NH 4 + )- 307 ΔH ϴ Hyd (NO 3 - )- 314

17.  Enthalpy level diagram for dissolving  Example: Ammonium nitrate (NH 4 NO 3 ) NH 4 NO 3 (s) NH 4 + (g) + NO 3 - (g) ΔH ϴ L (diss) (NH 4 NO 3 ) + 647 NH 4 + (aq) + NO 3 - (aq) ΔH ϴ Hyd (NH 4 + ) - 307 ΔH ϴ Hyd (NO 3 - ) - 314 NH 4 + (g) + NO 3 - (aq) ΔH ϴ Soln (NH 4 NO 3 )

18.  Example: Calculate the enthalpy of solution, ΔH Soln of ammonium nitrate  ΔH ϴ Soln = ΔH ϴ L (diss) (NH 4 NO 3 ) + ΔH ϴ Hyd (NH 4 + ) + ΔH ϴ Hyd (NO 3 - ) = + 647 + - 307 + - 314 = + 26 kJ mol -1

19.  Instant cold packs use endothermic reactions to achieve a low temperature quickly  The dissolving of ammonium nitrate in water is used in cold packs  Ammonium nitrate and water are held in two separate compartments  We activate the pack by breaking the compartments allowing both substances to mix together