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