Thermochemistry
IV : Enthalpy Changes and
Changes of State
Review:
• KE= energy of motion
• PE= stored energy
• T = measure of average KE of
particles of system (or
surroundings)
Phase change:
• A change in matter
without any change in
chemical composition of
the system
• Just like chem rxns
changes of state involve
P.E changes in the system
*** Always involves energy changes, but not
temperature changes. There is constant
temperature during a phase change. Why?
Because…
• This is not a change in kinetic
energy
• But we know that energy is
entering the system because the
bonds that are holding the
molecules together are being
broken or altered.
• This increases PE of the
molecules
• It requires 10-100 times more
energy for a chemical reaction
to take place than a phase
change
• Why? because a chemical
reaction requires stronger ionic
and covalent bonds to be
broken, but a phase change only
requires intermolecular bonds
to be broken.
Changes of State cont’d
• In general, enthalpy changes
btw liq and gas or greater then
btw liq and solid
• Why?
Phase changes:
(surroundings)
(s) to (l)
endo
T
endo
T
(l) to (g)
Fusion
(melting)
vaporization
(g) to (l)
condensation exo
T
(l) to (s)
exo
T
(g) to (s)
Solidifying
(freezing)
sublimation
exo
T
(s) to (g)
sublimation
endo
T
Molar Enthalpy for changes
of state
• Aka: Latent Heat of Phase
Change
• You can find tables of these
values listed in data booklet as
well as pg 647
• Unit: kJ/mol
Opposites!!
∆Hvap = - ∆Hcond
∆Hmelt = - ∆Hfre
• Freezing and condensing releases
heat to surroundings
• Vaporization and melting absorb
heat from surroundings
• Same units: kJ
Definitions
• Molar Heat of Fusion (melting) = energy
that must be absorbed in order to convert
one mole of solid to liquid at its melting
point (ΔH melt )
•
Molar Heat of Freezing (solidifying) =
energy that must be removed in order to
convert one mole of liquid to solid at its
freezing point (- ΔH freez)
Cont’d
• Molar Heat of Vaporization = energy that
must be absorbed in order to convert one
mole of liquid to gas at its boiling point
(ΔH vap )
• Molar Heat of Condensation = energy that
must be removed in order to convert one
mole of gas to liquid at its condensation
point (- ΔH cond)
Can be represented just like a
chem rxn:
H2O (s) + 6.02 kJ  H2O (l)
or
H2O (s)  H2O (l)
∆H = 6.02 kJ
Example:
• The molar heat of fusion of water is 6.009 kJ/mol.
How much energy is needed to convert 60g of ice at
0  C to liquid water at 0  C?
• 60g x 1 mol H2O = 3.3 mol H2O
18.02 g
q = n ∙ ∆Hfus
= (3.3 mol) (6.009 KJ)
= 19.8 kJ
HW:
• Pg 648-649 # 24-26,28,29