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, ΔHϴL:

Lattice dissociation enthalpy, ΔHϴL (diss)

Lattice formation enthalpy, ΔHϴL (form)

Definition: The lattice dissociation enthalpy, ΔHϴL (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) ΔHϴL (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

Definition: The lattice formation enthalpy, ΔHϴL (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) ΔHϴL (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

Definition: The enthalpy of hydration, ΔHϴHyd 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) ΔHϴHyd = - 406 kJ mol-1

Cl- (g) → Cl- (aq) ΔHϴHyd = - 377 kJ mol-1

Hydration is exothermic

Hydration means the ion is surrounded by water molecules

Water molecules interact with ions

Water (H2O) 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

OHH

δ+

δ-

δ+

Water (H2O) 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

OHH

δ+

δ-

δ+

Definition: The enthalpy of solution, ΔHϴSoln 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) ΔHϴSoln = + 4 kJ

mol-1

Enthalpy of solution, ΔHϴSoln can be negative or positive

Lattice dissociation enthalpy,

ΔHϴL (diss)

Enthalpy of hydration,

ΔHϴHyd

Solid Gaseous

Aqueous

Enthalpy of solution, ΔHϴ

Soln

The enthalpy of solution, ΔHϴSoln can be calculated using:

ΔHϴSoln = ΔHϴ

L (diss) + ΔHϴHyd

Example 1: Sodium chloride, NaCl

ΔHϴSoln = ΔHϴ

L (diss) (NaCl) + ΔHϴHyd (Na+) + ΔHϴ

Hyd (Cl-)

= + 787 + - 406 + - 377 = + 4 kJ mol-1

Example 2: Magnesium chloride, MgCl2 Use the data below to calculate the enthalpy of solution

ΔHϴSoln = ΔHϴ

L (diss) (MgCl2) + ΔHϴHyd (Mg2+) + [2 x ΔHϴ

Hyd (Cl-)]

= + 2493 + - 1920 + (2 x -364) = - 155 kJ mol-1

Enthalpy Value (kJ mol-1)Lattice formation -2493Hydration (Mg2+) -1920

Hydration (Cl-) -364

Lattice dissociation enthalpy, ΔHϴL (diss)

is an endothermic process

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)

Enthalpy level diagram for dissolving Example: Sodium chloride (NaCl)

NaCl (s)Starting line = Starting ionic

compound

Na+ (g) + Cl- (g)

ΔHϴL (diss) (NaCl)+ 787

Lattice diss = endothermic,

arrow upNa+ (aq) + Cl- (g)

ΔHϴHyd (Na+)- 406

ΔHϴHyd (Cl-)

- 377Na+ (aq) + Cl- (aq)

ΔHϴSoln (NaCl)+ 4

Larger enthalpy changes at top

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

Example 1: Calculate the lattice dissociation enthalpy, ΔHϴL

ΔHϴSoln (NaCl) = ΔHϴ

L (diss) (NaCl) + ΔHϴHyd (Na+) + ΔHϴ

Hyd (Cl-) + 4 = ? + - 406 + - 377

+ 787

Example 2: Ammonium nitrate (NH4NO3)

Draw an enthalpy level diagram for dissolving of ammonium nitrate in water

Use the data to calculate the enthalpy of solution of ammonium nitrate

Quantity kJ mol-1

ΔHϴL (NH4NO3) + 647

ΔHϴHyd (NH4

+) - 307

ΔHϴHyd (NO3

-) - 314

Enthalpy level diagram for dissolving Example: Ammonium nitrate (NH4NO3)

NH4NO3 (s)

NH4+ (g) + NO3

- (g)

ΔHϴL (diss) (NH4NO3)

+ 647

NH4+ (aq) + NO3

- (aq)ΔHϴ

Hyd (NH4+)

- 307

ΔHϴHyd (NO3

-)- 314

NH4+ (g) + NO3

- (aq)

ΔHϴSoln (NH4NO3)

Example: Calculate the enthalpy of solution, ΔHϴSoln of

ammonium nitrate

ΔHϴSoln = ΔHϴ

L (diss) (NH4NO3) + ΔHϴHyd (NH4

+) + ΔHϴHyd (NO3

-)

= + 647 + - 307 + - 314

= + 26 kJ mol-1

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