C_-_102_F11_Notes_new

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Unformatted text preview: C)
Liquids,
Solids
&
Solutions
 Petrucci
et
al.,
10th
Edition:
12.1‐12.4,
13.1‐13.9
 
 Some
Properties
of
Liquids
 Vapour
Pressure
 Surface
Tension
 • Energy
required
to
increase
the
surface
area
of
a
liquid
 • Substances
that
reduce
surface
tension
are
called
wetting
agents
 • Cohesive
Forces
–
forces
exerted
between
molecules
holding
them
together
in
a
 droplet
 • Adhesive
Forces
–
forces
between
the
liquid
and
the
surface
resulting
in
surface
 wetting
 [see
Petrucci
et
al.,
Figure
12­13]
 
 Enthalpy
of
Vapourization
 • Vapourization
(evaporation)
–
the
passage
of
molecules
from
the
surface
of
a
liquid
 into
the
vapour
state.

Rate
of
evaporation
increases
with:
 • • • • 
 __________________
in
temperature
 __________________
in
liquid
surface
area
 • __________________
in
intermolecular
forces
 Condensation
–
conversion
of
a
vapour
to
a
liquid
 Enthalpy
of
Vapourization
(ΔHvapourization)
–
the
quantity
of
heat
that
must
be
 absorbed
to
vapourize
a
certain
quantity
of
liquid
at
a
constant
temperature.
 
 ΔHcondensation
=
‐ΔHvapourization
 
 
 Phase
Equilibrium
 Water
in
an
open
vessel
would
evaporate
completely
 At
phase
equilibrium
(Figure
12­13
(c)):
 • Equilibrium
is
a
dynamic
process
 • No
net
conversion
from
one
phase
to
another

 • Properties
are
uniform
throughout
 [see
Petrucci
et
al.,
Figure
12­13(c)]
 
 
 
 
 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C ‐1 
 Vapour
Pressure
 Vapour
pressure
(Pvap)
is
the
pressure
measured
when
the
system
reaches
phase
 equilibrium
 • Dependent
on
temperature
and
gas
type
 • An
intensive
property
(independent
of
system
size)
 Example:

H2O
at
25°C
has
Pvap
=
23.8
mm
Hg
 For
mixtures
at
equilibrium,
the
partial
pressure
of
each
component
is
a
function
of
Pvap
 [see
Petrucci
et
al.,
Figure
12­17]
 [see
Petrucci
et
al.,
Table
12.4]
 
 Question
C1

 A
liquid
is
in
equilibrium
with
its
vapour
at
a
given
temperature.

Which
of
the
following
 conditions
are
applicable?
 I) There
is
no
transfer
of
molecules
between
liquid
and
vapour.
 II) The
vapour
pressure
has
a
unique
value.
 III) The
opposing
processes,
(liquid
to
vapour)
and
(vapour
to
liquid),
proceed
at
 equal
rates.
 IV) The
concentration
of
vapour
is
dependent
on
time.

 
 
 Vapour
Pressure
of
Selected
Liquids
 Boiling
Point
–
where
the
pressure
of
the
escaping
molecules
equals
that
of
the
molecules
 in
the
vapour
phase
above

 ­­­­­­
=
___________________
Boiling
Point

(at
P
=
760
mmHg
=
1
atm)
 
 [see
Petrucci
et
al.,
Figures
12­18,
12­20]
 
 
 Antoine’s
Equation
for
Pvap
 • Vapour
pressure
of
a
compound
is
a
strong
function
of
temperature
 • The
most
important
correlation
for
vapour
pressure
is
Antoine’s
equation:
 
 
 
 • A,
B
and
C
are
constants
that
depend
on
the
type
of
substance
in
a
given
temperature
 range
 • Antoine’s
equation
is
an
_____________________________
equation
–
it
is
based
on
the
 observation
that
plots
of
log10
Pvap
versus
1/T
are
approximately
linear
 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C ‐2 
 Energy
Changes
During
Phase
Changes
 • Energy
(heat)
is
necessary
to
accomplish
a
phase
transition
(vapourization,
 liquefaction,
or
sublimation)
 • H2O
(l)
+
44.01
kJ/mol
→
H2O
(g)

 @
25°C
and
1
atm:
ΔHvap
=
44.01
kJ/mol
 [see
Petrucci
et
al.,
Table
12.3]
 
 Clausius­Clapeyron
Equation
for
Pvap
 • Another
useful
equation
correlating
Pvap
to
temperature
is
the
Clausius‐Clapeyron
 equation:

 
 
 
 
 • T
is
in
K

 • ΔHvap
is
____________________
not
to
vary
with
changes
in
temperature

 • Clausius‐Clapeyron
equation
is
a
theoretical
equation
that
can
be
applied
to
other
 phase
transitions
if
the
proper
heat
of
phase
transition
(ΔHvap,
ΔHsub,
ΔHmelt)
is
used.

 
 Question
C2
 Liquid
ammonia
(NH3)
has
a
normal
boiling
temperature
of
–33.6°C
and
at
–68.5°C
has
a
 vapour
pressure
of
100
mmHg.

At
which
temperature
will
liquid
ammonia
have
a
vapour
 pressure
of
10
atm?

 
 Question
C3
 The
vapour
pressure
of
benzene
(C6H6)
at
25°C
is
94.7
mm
Hg.

1.00
g
of
benzene
is
injected
 into
a
1
L
container
which
is
held
at
25°C.

After
the
system
reaches
equilibrium,
what
is
the
 partial
pressure
of
benzene
in
the
container?
 
 
 Some
Properties
of
Solids
 • As
a
crystalline
solid
is
heated,
the
atoms
/
ions
/
molecules
vibrate
more
vigorously,
 eventually
disrupting
the
crystal
structure
 • Melting
Point
–
the
temperature
where
the
solid
loses
its
shape
and
is
converted
to
a
 liquid
 • Freezing
(solidification)
Point
–
the
temperature
where
a
liquid
is
converted
to
a
 solid
 • Melting
point
=
Freezing
point
for
a
_________________________________
 • Sublimation:
 
 Solid
→
Vapour
 
 ΔHsub
=
ΔHmelt
+
ΔHvap ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C ‐3 
 Phase
Diagrams
 Phase
diagrams
show
the
regions
where
the
several
phases
of
a
given
substance
are
in
 equilibrium
 Pressure
(not
to
scale)
 [see
Petrucci
et
al.,
Figure
12­28
for
Carbon
Dioxide]
 0
 0
 Temperature
(not
to
scale)
 
 
 
 
 [see
also
Petrucci
et
al.,
Figures
12­27
and
12­29
for
Iodine]
 
 
 • • Once
the
critical
temperature
(Tc)
and
critical
pressure
(Pc)
have
been
reached,
 the
two
distinct
phases
of
liquid
and
gas
are
no
longer
visible.
 One
homogenous
"supercritical
fluid"
phase
occurs
which
has
properties
of
both
 liquids
and
gases.

 
 
 [see
also
Petrucci
et
al.,
Figures
12­30
for
Water]
 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C ‐4 
 Question
C4
 Consider
the
following
 statements
regarding
the
phase
 diagram
for
sulfur.

Note
that
 sulfur
has
two
solid
phases:
 monoclinic
and
rhombic.

How
 many
of
the
statements
are
 correct?
 i. At
STP
the
sulfur
will
be
 rhombic
solid
 ii. There
is
only
one
triple
 point
on
the
diagram
 iii. There
are
two
triple
points
 where
the
gas
phase
is
in
 equilibrium
with
two
other
phases
 iv. Sulfur
boils
only
at
a
temperature
of
444.6°C
 v. The
normal
melting
point
for
sulfur
is
for
a
transition
from
rhombic
sulfur
to
liquid
 vi. Rhombic
sulfur
cannot
be
melted
at
pressures
below
1420
atm
 
 
 Solution
Terminology
 • Solution
‐
a
homogeneous
system
that
contains
two
or
more
substances
 • Solvent
‐
major
component
of
the
solution
 • Aqueous
is
used
when
water
is
the
solvent
BUT
it
doesn’t
tell
us
how
much
water
 is
in
the
solution
 • Solute(s)
‐
the
minor
component(s)
of
the
solution
 • Saturated
solution
–
equilibrium
state
where
no
more
solute
can
be
dissolved
into
 the
solution
 
 
 Solution
Concentration
 • Concentration,
or
molarity
(M),
is
defined
as
the
number
of
moles
of
solute
per
liter
of
 solution:

 
 
 • Molality
(m),
is
defined
as
the
number
of
moles
of
solute
per
kilogram
of
solvent,
and
 is
independent
of
temperature:
 
 
 
 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C ‐5 
 Mass
and
Mol
Fractions
 • Mass
fraction:
(mass
percent
if
multiplied
by
100%):

 
 
 
 • Mole
fraction:
(mol
percent
if
multiplied
by
100%):
 
 • Percentage:
 • 25%
(v/v):
25mL
solute
per
100mL
solution
 • 25%
(w/v):
25g
solute
per
100mL
solution
 • 25%
(w/w):
25g
solute
per
100g
solution
 
 
 
 
 Question
C5
 A
chemist
titrates
25
mL
of
a
H3PO4
(aq)
solution
with
35
mL
of
a
0.20M
NaOH
(aq)
 solution.

If
only
two
hydrogen
atoms
in
the
H3PO4
react
with
the
added
OH–
ions
in
this
 titration,
what
is
the
molality
of
the
H3PO4
solution?

The
density
of
the
H3PO4
solution
is
 approximately
1.05
g/mL.

 
 
 Intermolecular
Forces
and
the
Solution
Process
 • Summarize
as
like‐dissolves‐like
 [see
Petrucci
et
al.,
Figure
13­7]
 
 
 Solubility
of
Gases
­
Henry’s
Law
 • Generally
applied
to
gas­liquid
solutions

 • Relates
the
partial
pressure
of
a
solute
gas
(Pi)
to
the
solute
mole
fraction
(xi)
in
the
 liquid
(solvent):
 Pi = ki xi 
 • ki
(Henry’s
law
constant)
depends
on
the
solute/solvent
pair
and
the
temperature
 
 
 Effect
of
Pressure
on
Gas
Solubility
 € What
happens
as
the
gas
pressure
is
increased?
 
 Increasing
the
pressure
___________________
the
solubility
of
gases
in
liquids
 
 [see
Petrucci
et
al.,
Figure
13­11]
 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C ‐6 
 Alternate
version
of
Henry’s
Law
 • Another
way
to
write
Henry’s
Law
(see
Petrucci
et
al.,
Section
13‐5):
 
 

 
 • Ci
is
the
solubility
of
the
gas
in
the
solvent
(mL
solute
gas
/
L
solution)
 • Pi
is
the
partial
pressure
of
the
gas
above
the
solution
 • ki
is
the
Henry’s
Law
constant
–
check
the
units!
 
 
 Temperature
Effect
on
Gas
Solubility
 • Increasing
the
temperature
____________________
the
solubility
of
gases
in
liquids,
in
MOST
 cases
 [see
Petrucci
et
al.,
Figure
13­10]
 
 
 Question
C6
 You
purchase
a
machine
for
your
kitchen
that
carbonates
water.

It
operates
by
bringing
 carbon
dioxide
(CO2)
at
10
atm
into
equilibrium
with
water.

What
is
the
percentage
by
 mass
of
carbon
dioxide
in
the
soda
water
it
produces?

 
 Data:
CO2
in
water
at
25°C:
k
=
1.25
×
106
mmHg/mol
fraction
 
 
 Vapour
Pressures
of
Solutions
–
Raoult’s
Law
 Raoult’s
law
relates
the
partial
pressure
of
a
gas
to
its
vapour
pressure
in
liquid‐liquid
 and
liquid‐solid
solutions:

 Pi

=

xi
Pivap
 Pivap
is
the
vapour
pressure
of
pure
substance
i

 xi
is
the
mol
fraction
of
“i”
in
the
liquid
phase
 
 For
binary
liquid‐liquid
solutions:

 
 PT

=

PA

+

PB
 
 =

xAPAvap

+

xBPBvap

 
 =

xAPAvap

+

(1
–
xA)PBvap

 
 For
binary,
solid‐liquid
solutions:

 PT
=

Psolid
+
Pliquid

 
 
=

xsolid
Pvapsolid

+

xliquid
Pvapliquid

 
 
=

xliquid
Pvapliquid

 
 Because,
in
general:

________________
>>
_______________

 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C ‐7 
 Distillation
&
Liquid­Vapour
Equilibrium
 
 [see
Petrucci
et
al.,
Figure
13­14]
 
 
 Question
C7
 Ethanol
(C2H5OH)
and
methanol
(CH3OH)
form
a
nearly
ideal
solution.

At
20°C,
the
 vapour
pressure
of
ethanol
is
44.5
mm
Hg
and
that
of
methanol
is
88.7
mm
Hg.

Calculate
 the
mole
fraction
of
ethanol
in
the
vapour
if
the
liquid
contains
one
mol
ethanol
and
one
 mol
methanol.
 
 
 Colligative
Properties
 Consider
a
solution
formed
by
dissolving
a
non‐volatile
solute
(such
as
sugar)
in
a
liquid
 solvent
(such
as
water).

 
 Ideal
Solution
‐
if
this
solution
obeys
both
Raoult’s
and
Henry’s
laws
 • Ideal
solution
assumption
holds
for
________________
solutions

 
 4
properties
of
ideal
solutions
(colligative
properties):
 1. Vapour‐pressure
lowering
 2. Boiling‐point
elevation
 3. Freezing‐point
depression
 4. Osmotic
pressure

 
 
 Vapour
Pressure
Lowering
(Colligative
Property
#1)
 Raoult’s
law
for
a
binary
solid/liquid
solution:

 P

=

Pliquid
+
Psolid

 
 
 
=

x1P1vap
+
x2P2vap
 
 If
the
solid
is
non‐volatile,

P2vap
=
0:
 
 P
=
x1P1vap
 Since
x1
<
1,
P
<
P1vap
 
 This
vapour
pressure
lowering
(∆P1)
is
calculated
with
the
equation:
 
 ∆P1

 =
x1
P1vap
–
P1vap

 

 =
–
(1
–
x1)
P1vap

 

 =
–
x2
P1vap
 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C ‐8 
 Estimating
Molar
Mass
with
∆P1
 Vapour
pressure
depression
can
be
related
to
the
molar
mass
of
the
solute
with
the
 equation:

 
 
 
 Can
be
used
to
estimate
the
molar
mass
of
an
unknown
solid
dissolved
in
a
known
liquid

 
 
 Question
C8
 Twenty
grams
of
a
non‐volatile
solute
are
added
to
100g
of
water
at
25°C.

The
vapour
 pressure
of
pure
water
is
23.76
mm
Hg;
the
vapour
pressure
of
the
solution
is
22.41
mm
 Hg.

Calculate
the
molar
mass
of
the
solute.
 
 
 Boiling
Point
Elevation
(Colligative
Property
#2:
∆Tb)
 • Normal
boiling
point
of
a
pure
liquid
is
the
temperature
where
the
vapour
pressure
is
 1
atm
 • According
to
Raoult’s
law,
the
vapour
pressure
of
a
liquid/solid
solution
is
less
than
 that
of
the
pure
liquid

 • Therefore,
the
temperature
of
the
solution
needs
to
be
increased
to
make
it
boil
(to
 make
the
vapour
pressure
reach
1
atm)

 
 At
the
normal
boiling
point:

 slope of the curve = S = − € Petrucci
et
al.,
Figure
13‐19
 
 
 
 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C ‐9 
 • 
 
 • 
 
 • For
very
dilute
solutions
(where
Raoult’s
law
is
more
applicable),
n1
>>
n2
and
the
 previous
equation
can
be
simplified
to:

 Because
S
and
M1
are
properties
of
the
pure
solvent,
they
can
be
combined
in
a
single
 constant
Kb:

 Substitute
into
the
equation
for
∆Tb
:

 
 
 
 • • • Because
m1
is
measured
in
g,
m1/1000
is
the
mass
of
solvent
in
kg
 m2/M2
is
the
number
of
moles
of
solute
 Therefore,
the
expression
in
brackets
in
the
previous
equation
is
the
molality
(m)
of
 the
solution:
 ∆ T b 

= 

K b 
 m 
 • Notice
that
Kb
depends
only
on
the
solvent
type
 [see
Petrucci
et
al.,
Table
13.2]
 
 Question
C9
 When
39.8
g
of
a
non‐dissociating,
non‐volatile
sugar
is
dissolved
in
200
g
of
water,
the
 boiling
point
of
water
is
raised
by
0.30°C.

Estimate
the
molar
mass
of
the
sugar.


 (Kb
=
0.512
°C
kg
/
mol
for
water)
 
 
 Solutions
of
Electrolytes
 • If
a
solute
dissociates,
the
effective
number
of
solute
particles
increases

 • Because
colligative
properties
depend
on
the
number
of
dissolved
particles,
the
 equations
describing
them
must
be
adjusted
to
take
dissociation
into
account

 • The
required
change
is
the
insertion
of
an
additional
factor,
the
van’t
Hoff
i:
 
 

 • • ∆Tb
=
i
Kbm

 i
is
the
number
of
particles
released
into
solution
per
formula
unit
of
solute
 • minimum
value:
i
=
1
(no
dissociation)
 • maximum
value:
i
=
number
of
particles
by
complete
dissociation
of
a
formula
unit
 For
example,
NaCl
when
it
dissolves
in
water
to
form
Na+
and
Cl–
ions

 • The
maximum
value
of
i
for
NaCl
is
2,
but
it
might
be
less
and
can
only
be
 determined
experimentally
 [see
Petrucci
et
al.,
Table
13.3]
 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C‐10
 Question
C10
 Lanthanum
(III)
trichloride
(LaCl3)
dissociates
into
ions
as
it
dissolves
in
water:
 
 LaCl3(s)
→
La3+(aq)
+
3
Cl–(aq)
 Suppose
2.453
g
of
LaCl3
is
dissolved
in
100
g
of
H2O.

What
is
the
boiling
point
of
the
 solution
at
atmospheric
pressure,
assuming
no
association
among
ions
and
that
the
 solution
behaves
ideally?
 
 
 Freezing
Point
Depression
(Colligative
Property
#3:
ΔTf)
 • An
equation
similar
to
the
one
used
to
calculate
boiling
point
elevation
(ΔTb)
is
used
 for
freezing
point
depression
(ΔTf):
 ΔTf
=
–i
Kf
m
 • Notice
that
Kf
depends
only
on
the
_________________.

 
 
 Question
C11
 The
freezing
point
depression
constant
for
HgCl2
is
34.3
K/m.

For
a
solution
of
0.849
g
of
 mercurous
chloride
(empirical
formula
HgCl)
in
50
g
of
HgCl2,
the
freezing
point
 depression
is
1.24°C.

What
is
the
molecular
formula
of
mercurous
chloride?
 
 
 Osmotic
Pressure
(Colligative
Property
#4)
 [see
Petrucci
et
al.,
Figure
13­17]
 π
=
ρ
g
h
 h Sugar solution Sugar solution pure water pure water Time → infinity (Equilibrium) Time = 0 The
semi‐permeable
membrane
allows
the
flow
of
H2O
________
________
the
flow
of
sugar
 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C‐11
 
 Van’t
Hoff’s
Equation
 For
ideal
solutions:
 
 
 c
is
the
molar
concentration
of
solute
in
solution
 
 
 Question
C12
 Maple
sap
is
approximately
a
3%
solution
of
sucrose
(C12H22O11)
in
water
(on
a
mass
 basis).

At
20°C,
the
density
of
the
solution
is
1.010
g/cm3.

Considering
only
osmotic
 pressure,
determine
the
height
(in
metres)
above
ground
to
which
the
sap
should
rise
 when
the
temperature
is
20°C.

Assume
the
ground
water
is
pure
water.


 
 
 Reverse
Osmosis
 [see
Petrucci
et
al.,
Figure
13­18]
 Desalination
of
saltwater
by
reverse
osmosis.
 • The
membrane
is
permeable
to
water
but
not
to
ions.

The
normal
flow
of
water
is
from
 side
A
to
side
B.


 • If
we
exert
a
pressure
on
side
B
that
exceeds
the
osmotic
pressure
of
the
saltwater,
a
 net
flow
of
water
occurs
in
the
reverse
direction‐from
the
saltwater
to
the
pure
water.


 • The
lengths
of
the
arrows
suggest
the
magnitudes
of
the
flow
of
water
molecules
in
 each
direction.
 ChE102
Fall
2011
Class
Notes
‐
Liquids,
Solids
&
Solutions
 
 
 
 
 
 
 C‐12
 ...
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