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2.1
What happens when an ionic compound dissolves in water and the
energetics of why?
Section 2.1
discusses a dissolving enthalpy cycle – another application of Hess's
Law – in which the process of dissolving an ionic compound such as a
halide salt or a metal oxide, is broken down into theoretical stages to help
understand the factors involved in deciding whether of not a substance
will dissolve readily or not at all. For the moment, only enthalpy
changes will be considered but eventually entropy changes must be
discussed too!
|
This ΔH is called the
ENTHALPY of SOLUTION |
(i) ionic compound(s)
+ aq
 cations(aq)
+ anions(aq) |
|
(ii) solid ionic compound
vapourised to give gaseous ions – this
ΔH
is called the LATTICE ENTHALPY |
(endothermic) |
gaseous
cations and anions |
(exothermic) |
(iii)
hydration of the cations and anions – these
ΔHs are called the ENTHALPIES of HYDRATION of the ION |
|
In terms of enthalpy changes: (i) = (ii) + (iii)
Enthalpy
of solution ΔHsolution(compound) is defined
as the heat absorbed or released when 1 mole of compound
dissolves in water
Lattice
enthalpy ΔHLE(compound) is defined as the energy released
when 1 mole of a solid ionic is formed from its separated
constituent gaseous ions (an exothermic definition).
Lattice enthalpy can also
be defined as the energy required to completely dissociate 1
mole of an ionic compound into its gaseous constituent ions
(endothermic definition).
Enthalpy
of hydration is defined as the energy released when 1 mole of
gaseous ions interacts with water to give a solution of the hydrated
ions
Standard values of these enthalpy changes are
usually at 298K and 1 atm./101kP pressure.
All three enthalpy terms are re–defined in
more detail and explained further as the discussion proceeds. |
2.1a The changes that occur in
the dissolving process
Despite the strong
ionic bonding forces in most salts or simple binary compounds like oxide
or chloride
crystals i.e. the strong electrostatic attraction between positive ions
(cations) and negative ions (anions) many ionic compounds readily
dissolve in water. Therefore, not surprisingly, a great deal of energy
is required to separate the ions, but dissolving can still take place.
So how can we explain this?
 Water
consists of highly polar molecules due to the great electronegativity
difference between hydrogen and oxygen (O > H hence the polarity of the
bond δ–O–Hδ+).
When salts dissolve in water a process of solvation occurs in which the
ions become solvated by association with the solvent molecules. If water
is the solvent, the process is called hydration and is exothermic
because it involves particles coming together via intermolecular forces
or covalent bonds in the case of aqua–cation complex ions – so this is a
compensating source of energy.
In some cases
cations become hydrated via dative covalent bonds to form a complex ions
e.g.
Li(H2O)4]+,
Cu(H2O)4]2+, Mg(H2O)6]2+,
Al(H2O)6]3+ etc. because in the case of water,
the most electronegative part of the highly polar water molecule (>Oδ–)
will be attracted to the positive ion and, since the oxygen atom has two
lone pairs of electrons, it is also the source of the dative covalent
bond by donation of one of these pairs of electrons into a vacant metal
ion orbital.
In the case of
anions, the positive ends of the water molecules (–Hδ+)
will orientate themselves towards the negative anion and the water
molecules become weakly associated with anion, but no covalent bonds are
formed.
This solvation of
the ions means the ions are effectively bigger particles which makes the
distance between the positive and negative ion centres greater, and by
the laws of electrostatics, the attractive forces is weakened and the
hydration process is always exothermic.
PLEASE note that
dissolving a solute in this situation cannot be simply regarded as a
physical change. The ionic lattice is broken down, but NOT by melting,
and chemical bonds are formed between the cation and water molecules to
form hydrated ions or aqua–ions such as those listed above.
2.1b
Diagram illustrating the dissolving–solvation–hydration process for
sodium chloride crystals forming salt solution
-
For diagram
simplicity, each ion is surrounded by four water molecules, though in
reality, much more than this – see later.
-
The strong crystal
lattice (giant ionic structure) is broken down by the solvation process
so the salt dissolves and the ions are free to move around in the water
solvent.
-
Its worth noting that this involves an increase in entropy
–
the solution is more disordered (more possible arrangements of the
particles) compared to the
pure liquid solvent and the highly ordered crystal lattice of very
limited possible arrangements.
-
However, when the
ions become hydrated there is a decrease in entropy, which
counts against a substance dissolving. This decrease in entropy is
due to the orientation of the water molecules when they associate
with the cations and anions i.e. a decrease in the possible ways the
water molecules can be arranged.
-
In other words the
hydrated ions produce a more ordered arrangement of some of the
water molecules.
-
This decrease in
entropy does not favour dissolving, (and neither does a very
endothermic enthalpy of solution)
-
Therefore you need to consider entropy changes to
fully explain why substances do/do not dissolve – especially when the
enthalpy of solution is positive or only slightly negative.
In order to try to
understand processes you can use a Hess's Law cycle to break the process
down into theoretical steps, each of which can measured experimentally
or theoretically calculated. The relative magnitude of each energy
change can help understand why a substance will dissolve or be
insoluble. However, this cycle only uses enthalpy values and excludes
entropy changes – which I'll deal with later.
TOP OF PAGE
2.1c The connection
between lattice enthalpy, enthalpies of ion hydration and enthalpy of
solution
(i) The energy change for
a substance dissolving in a solvent is called the enthalpy of solution.
The enthalpy
of solution ΔHsolution(compound) is defined as the heat
absorbed or released when 1 mole of compound (the solute) dissolves
in a solvent to form an 'infinitely' dilute solution where no
further heat change takes place at 298K e.g.
NaCl(s)
+ aq ===> Na+(aq) + Cl–(aq)
ΔHsolution(sodium chloride) = +4 kJ mol–1
Al2(SO4)3(s)
+ aq ===> 2Al3+(aq) + 3SO42–(aq)
ΔHsolution(aluminium sulfate) = –318 kJ mol–1
Examples
of enthalpy of solution (kJ mol–1) are tabulated below and refer to an
infinitely dilute solution
|
cation\anion |
OH– |
F– |
Cl– |
Br– |
I– |
CO32– |
NO3– |
SO42– |
|
Li+ |
–21.2 |
+4.5 |
–37.2 |
–49.1 |
–63.3 |
–17.6 |
–2.7 |
–30.2 |
|
Na+ |
–42.7 |
+0.3 |
+3.9 |
–0.6 |
–7.6 |
–24.6 |
+20.5 |
–2.3 |
|
K+ |
–55.2 |
–17.7 |
+17.2 |
+20.0 |
+20.5 |
–32.6 |
+34.9 |
+23.8 |
|
NH4+ |
– |
+5.0 |
+15.2 |
+16.2 |
+13.4 |
– |
+25.8 |
+6.2 |
|
Mg2+ |
+2.8 |
–17.7 |
–155 |
–186 |
–214 |
–25.3 |
–85.5 |
–91.2 |
|
Ca2+ |
–16.2 |
+13.4 |
–82.9 |
–110 |
–120 |
–12.3 |
–18.9 |
–17.8 |
|
Sr2+ |
–46.0 |
+10.9 |
–52.0 |
–71.6 |
–90.4 |
–3.4 |
+17.7 |
–8.7 |
|
Al3+ |
– |
–209 |
–332 |
–360 |
–378 |
– |
– |
–318 |
Comments on the enthalpy of solution
values
-
Quite a few of
the values are relatively endothermic and perhaps at first sight
dissolving seems
unfavourable e.g. potassium nitrate (+34.9) and ammonium nitrate
(+25.8), but both these salts are very soluble in water – you need
to consider entropy too! (in Part 3)
-
The value
depends on the structure and strength of the ionic lattice and the
hydration enthalpies of the constituent ions.
-
You can get
simple trends, but they are not easy to explain at times e.g.
-
For Group 1
hydroxides and fluorides delta H solution gets more exothermic/less
endothermic BUT the chlorides, bromides and iodides get less
exothermic/more endothermic on dissolving.
-
Group 2
compounds show a mixture of trends i.e. down the group more
endothermic or more exothermic.
-
Aluminium
compounds show the most exothermicity on dissolving, perhaps because
of the high charge on the ion giving a large ion hydration enthalpy
(see section below).
You can derive an
alternative route to form the solution employing Hess's Law
...
(ii) The ionic
crystal lattice is vapourised into its gaseous positive ion (cation) and gaseous
negative ion
(anion) constituents.
This is always a
very endothermic process because you are separating strongly
attracted oppositely charged ions.
This process is
defined as happening in the reverse exothermic direction and is called the
lattice enthalpy (lattice energy).
The lattice
enthalpy ΔHLE(compound) is defined as the energy released
when 1 mole of a solid is formed from its separated gaseous ions at
298K
and is very exothermic e.g.
Na+(g)
+ Cl–(g) ==> NaCl(s)
ΔHLE(sodium chloride) = –774 kJ mol–1
2Al3+(g)
+ 3O2–(g) ==> Al2O3(s)
ΔHLE(aluminium oxide) = –15916 kJ mol–1
Factors
affecting the lattice energy:
The greater
the force of attraction the greater the energy needed to
vapourise the lattice OR the greater the energy released on
forming the ionic lattice.
The size of
the ion radius and the ionic charges are the most important
factors to consider.
(i) The smaller
the ion radius or the greater the ion charge the greater the
lattice enthalpy i.e. more exothermic. This effectively
means the ions are closer together, so the centres of charge are
closer together, hence a greater attractive force.
(ii) The greater
the charge on the ion, the greater the attractive force between
it and neighbouring ions.
The law of
attraction of electric charges states that the force of
attraction is proportional to the product of the charges divided
by the distance between them
The
attractive force is proportional to C+ x C–
/ d2
d = the distance between the ion centres
d is
effectively the sum of the cation and anion radius for
simple lattice structures
C+ and C–
= the
numerical charges on the cation and anion respectively
The greater the force of
attraction, the greater the energy needed to separate the ions
in the crystal.
This would be indicated
by increase in melting point and enthalpy of fusion, boiling
point and enthalpy of vaporisation, and of course, the
lattice enthalpy.
The data table below
illustrates the general rules outlined above.
|
Series and comments on
examples of lattice enthalpies ΔHLE |
LE quoted as –ΔHLE(compound) in kJ mol–1 |
|
Comparing Group 1 halide salts - varying the
halide Decrease from F to I as the anion radius increases
down group 7 (same trend
for Na, K, Rb and Cs halides) |
LiF, 1022 |
LiCl, 846 |
LiBr, 800 |
LiI, 744 |
|
Comparing Group 1 halide salts - varying the
cation Decrease from Li to Cs as the cation radius increases
down group 1 (same
trend with group 1 chlorides, bromides or iodides) |
LiF, 1022 |
NaF, 902 |
KF, 801 |
RbF, 767 |
|
Comparing sodium chloride
and magnesium oxide. Neglecting radii differences, MgO has over
4x the lattice energy of sodium chloride because the ion charges
are doubled i.e. 1+ x 1– compared to 2+ x 2–, a ratio of 1 : 4. |
NaCl 774 |
MgO 3889 |
|
|
|
The first three Period 3
oxides From left to right, increasing cation charge AND
decreasing cation radius both lead to increase in lattice
enthalpy.
Emphasised by considering the LE per mole
of cation. |
Na2O 2488
1244 per mole of cation |
MgO 3889
3889 per mole of cation |
Al2O3
15916
7958 per mole of cation |
|
(iii) The gaseous
ions interact with water to give the hydrated ions in aqueous
solution.
This is always a
very exothermic process.
This energy
change is called enthalpy of hydration.
The enthalpy
of hydration is defined as the energy released when 1 mole of
gaseous ions interacts with water to give a solution of the hydrated
ions at 298K e.g.
Na+(g)
+ aq ==> Na+(aq) ΔHhyd(Na+(g)) =
–364 kJ mol–1
Cl–(g)
+ aq ==> Cl–(aq) ΔHhyd(Cl–(g)) =
–406 kJ mol–1
Factors
affecting the ion hydration energy:
The size of
the ion radius and the ion charge are the most important factors
to consider here.
The smaller
the ion radius or the greater the ion charge, the greater the
ion hydration enthalpy i.e. more exothermic. The smaller the
radius OR the higher the charge, the greater is the
intensity–potential of the electric field so interaction with an
oppositely charged particle (ion or polar molecule) is stronger
and therefore more
'energetic' i.e. more exothermic.
|
Series
and comments for the process Mn±(g)
+ aq ==> Mn±(aq)
(always exothermic) – ion hydration
enthalpies ΔHhyd(cation/anion) |
quoted as ΔHhyd(ion(g)) kJ mol–1 |
| The
first three Period 3 cations – decreasing radius and increasing
charge – double trend effect, so more dramatic increase from
left to right |
Na+(aq) –406 |
Mg2+(aq) –1920 |
Al3+(aq)
–4690 |
|
| Group
1 cation trend, shows the effect of increasing radius down the group
with decreasing enthalpy of ion hydration |
Li+(aq)
–519 |
Na+(aq)
–406 |
K+(aq)
–322 |
Rb+(aq)
–301 |
|
Group 7 halide anions, shows
a similar effect of increasing radius |
F–(aq)
–506 |
Cl–(aq)
–364 |
Br–(aq)
–335 |
I–(aq)
–293 |
|
Comparing the iron(II) ion
with the smaller and more highly charged iron(III) ion giving a
considerable increase in hydration enthalpy |
Fe2+(aq),
–1950 |
Fe3+(aq),
–4430 |
Energetics-Thermochemistry-Thermodynamics Notes INDEX
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QUICK INDEX for
Energetics:
GCSE Notes on the basics of chemical energy changes
– important to study and know before tackling any of the three Advanced Level
Chemistry pages Parts 1–3 here
* Part 1a–b
ΔH Enthalpy Changes
1.1 Advanced Introduction to enthalpy changes
of reaction,
formation, combustion etc. : 1.2a & 1.2b(i)–(iii)
Thermochemistry – Hess's Law and Enthalpy
Calculations – reaction, combustion, formation etc. : 1.2b(iv)
Enthalpy of reaction from bond enthalpy
calculations : 1.3a–b
Experimental methods
for determining enthalpy changes and treatment of results and
calculations :
1.4
Some enthalpy data patterns : 1.4a
The combustion of linear alkanes and linear
aliphatic alcohols
:
1.4b Some patterns in Bond
Enthalpies and Bond Length : 1.4c
Enthalpies of
Neutralisation : 1.4d Enthalpies of
Hydrogenation of unsaturated hydrocarbons and evidence of aromatic
ring structure in benzene
:
Extra Q page
A set of practice enthalpy
calculations with worked out answers **
Part 2 ΔH Enthalpies of
ion hydration, solution, atomisation, lattice energy, electron affinity
and the Born–Haber cycle : 2.1a–c What happens when a
salt dissolves in water and why? :
2.1d–e Enthalpy
cycles involving a salt dissolving : 2.2a–c
The
Born–Haber Cycle *** Part 3
ΔS Entropy and ΔG Free Energy Changes
: 3.1a–g Introduction to Entropy
: 3.2
Examples of
entropy values and comments * 3.3a ΔS, Entropy
and change of state : 3.3b ΔS, Entropy changes and the
feasibility of a chemical change : 3.4a–d
More on ΔG,
free energy changes, feasibility and
applications : 3.5
Calculating Equilibrium
Constants from ΔG the free energy change : 3.6
Kinetic stability versus thermodynamic
feasibility - can a chemical reaction happen? and will it happen?
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