Chemistry 12
KE= energy of motion
PE= stored energy
T = measure of average KE of particles of
system (or surroundings)
A change in matter without
any change in chemical
composition of the system
Just like chemical reactions,
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?
1
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.
In general, enthalpy changes btw liquid and
gas are greater than btw liquid and solid.
Why?
2
(s) to (l)
(l) to (g)
Fusion
(melting)
vaporization
endo
T
endo
T
(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
Aka: Latent Heat (of Phase Change)
You can find tables of these values listed in
data booklet as well as pg. 647
Units: kJ/mol
3
∆Hvap = - ∆Hcond
∆Hmelt = - ∆Hfreeze (or solidification)
Freezing (solidification) and condensing
releases heat to surroundings
Vaporization and melting absorb heat from
surroundings
Same units
Molar Heat of Fusion (melting) = energy that must
be absorbed in order to convert one mole of solid
to liquid at its melting point (∆H 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 (- ∆H freeze)
Molar Heat of Vaporization = energy that must be
absorbed in order to convert one mole of liquid to
gas at its boiling point (∆H vap )
Molar Heat of Condensation = energy that must be
removed in order to convert one mole of gas to
liquid at its condensation point (- ∆H cond)
4
H2O (s) + 6.02 kJ H2O (l)
(thermochemical)
or
H2O (s) H2O (l)
∆H0fus = 6.02 kJ
Calculate 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 B ∆H0fus
= (3.3 mol) (6.009 KJ)
= 19.8 kJ
Pg 648-649 # 24-26,28,29
5