Enthalpy and Ionic Compounds
Sodium and chlorine react
vigorously and spontaneously
to form sodium chloride. The
salt can be seen as the cloud
rising from the flask.
7-1
Lattice Energy: energy required to break up 1 mole of solid into gas phase ions
The energy of attraction for ions to condense to form a solid can be
rationalized by Colulombs Law
-
+
Electrostatic attraction
7-2
So: Lattice Enthalpy:
MX(s)
M+(g) + X-(g),
)H = Lattice Enthalpy
The magnitudes of Lattice Enthalpies depend on the charge of the
ions and their separation:
The Coulomb potential energy V is given by:V = k.Q1.Q2 / r
k is a constant,
Q's are the charges ,r is the separation.
If charges are of opposite sign, energy is released (exothermic); if
they have the same sign, then energy is absorbed (endothermic).
The smaller the value of r, the bigger is V.
7-3
Born Haber Cycle: just as before we can set up two equivalent paths to
produce the ionic solid : example sodium chloride
Na+ (g) + e- + Cl (g)
enthalpy H
+502
∆H
ionization energy
∆H
electron affinity
Na (g) + Cl (g)
+121 ∆H
-354
Na+ (g) + Cl- (g)
at(Cl(g))
Na (g) + ½ Cl2 (g)
∆Hlattice energy
+108 ∆Hat(Na(g))
Na (s) + ½ Cl2 (g)
-411 ∆Hf
NaCl (s)
Let’s say we wish to
determine this
*
Energies in kJmol-1 are given in green
7-4
If we know all the energies but one we can use the cycle to get it
if we go completely round the cycle then we expend no energy:
so if we start at * and go clockwise:
-∆
∆Hf + ∆Hat(Na(g)) + ∆H at(Cl(g)) + IE + EA + LE = 0
411 + 108 + 502 + 121 -354 - LE = 0
LE = 788 kJmol-1
7-5
Another way of looking at exactly the same problem:NaCl (s)
Na(s) + 0.5Cl 2(g)
)H0 = +411 kJ
Na(s)
Na(g)
)H = +108 kJ
) H = +502 kJ
Na(g)
Na+(g) + e0.5Cl 2(g)
Cl (g)
) H = +121 kJ
Cl (g) + eCl- (g)
)H = -354 kJ
ADD
NaCl (s)
Na+(g) + Cl -(g)
) H = LE kJ
Apply Hess's Law - add them all up and )H reaction has to be ZERO - why?
411 + 108 + 502 + 121 - 354 + X = 0
LE = 788kJ
Lattice Enthalpy:NaCl (s)
Na+(g) + Cl -(g) )H = +788 kJmol-1
it is mechanically easy to get the lattice energy: write all the values out as
defined (e.g., )H (at), IE) and reverse the )Hf of the ionic solid.
Then add them all up. Result is the lattice energy.
7-6
Lattice Enthalpies in kJ/mol
LiF 1046 .... LiI 759
NaF 929 .... NaI 700
KF 826 .... KI 643
MgF2 2957 .... MgI2 2327
MgO 3850 .... BaO 3114
MgS 3406 .... BaS 2832
(Increased charges on ions leads to increased lattice energies.)
7-7
Example: given the following determine the electron affinity of chlorine
MgCl2(s) )Hfo = –641.3 kJ/mol
Mg(s) + Cl2(g)
Mg(s)
)H = 148 kJ/mol
Mg(g)
Mg(g)
Mg+(g)
)H = 737 kJ/mol
Mg+(g)
Mg2+(g)
)H = 1450 kJ/mol
Cl2(g)
2Cl(g)
)H = 244 kJ/mol
Mg2+(g) + 2Cl–(g)
MgCl2(s) )H = - 2510kJmol-1
2Cl(g)
2Cl–(g) = 2EA =2(EA)
So: 641.3 + 148 + 737 +1450 +244 +2EA - 2510 = 0
EA = - 355kJmol-1
7-8
Born-Haber Cycles
magnesium chloride
Mg2+(g) + 2e- + 2Cl
enthalpy H
∆H
bond energy of chlorine
Mg2+(g)
∆H
+
2e-
+ Cl2
(g)
(g)
2 x ∆H
Mg2+
first electron affinity
(g) + 2Cl (g)
second ionization energy
Mg+(g) + e- + Cl2
(g)
∆H
first ionization energy
Mg
(g)
+ Cl2
formation
enthalpy
(g)
∆H atomization
Mg (s) + Cl2
∆H
∆H lattice
(g)
MgCl2
(s)
7-9
Problems
CaO BaO Which one has the greater lattice enthalpy?
Order of Lattice Enthalpies?
LiCl , NaCl , BeCl2
MgO, MgF2, MgBr2
MgCl2, BeO, BeCl2
And now we start a slightly new
topic……
•2 "Spontaneous” chemical
chapter 19
reactions
7-10
Introduction: Order and Disorder
•2 "Spontaneous” chemical
chapter 19
reactions
13.1/ Spontaneity
13.2/ Entropy: The Measure of Disorder
13.3/ Absolute Entropies
13.4/ Spontaneity and Free Energy
13.5/ Some Applications of Thermodynamics
13.6/ Bioenergetics
In thermodynamics spontaneity means : will occur
if left alone long enough
7-11
Schematic view of the
spontaneous process for a
water-and ice mixture on a
table top, The energyabsorbing process, melting,
is spontaneous under these
conditions.
Schematic view of the
spontaneous process for a
water-and-ice mixture in a
freezer. The energyreleasing process, freezing,
is spontaneous under
these conditions.
7-12
It is tempting to think that chemical reactions that give off
heat are spontaneous. This is close to the truth but not quite right.
Recall:
See page 4-10
Hot and cold packs:
MgSO4(s)
Mg2+(aq) + SO42-(aq) )Ho = -91.3 kJ
NH4NO3(s)
NH4+ (aq) + NO3-(aq) )Ho = +26.3 kJ
both are spontaneous :
It turns out that how disordered things become also plays a role: it will be a combination of
enthalpy and entropy that will eventually tell us if a reaction will occur
We need to learn about entropy: next lectures
7-13