PART 8 – Changes
Ch
iin St
States,
t
Colligative Properties,
Properties and Phase
Diagram
Reference: Chapter
p 5,, 12 in textbook
1
States of Materials
Solid
Liquid
Gas
2
Gas and Ideal Gas
z
z
Properties
ope es o
of a Gas
„
Low density, No fixed shape
„
Can be significantly expanded and compressed
„
Can be mixed by any ratios
Ideal Gas – Assumption
„
Gas molecules have no volume;
„
No intermolecular interaction
3
Ideal Gas: P, V, T
z
P vs.
s V
„
z
P1 * V1 = P2 * V2
Æ V / T = constant
or
V1 / T1 = V2 / T2
or
P1 / T1 = P2 / T2
P vs. T
„
z
or
V vs. T
„
z
Æ P * V = constant
Æ P / T = constant
t t
Note: the units of P
P, V
V, T
4
Ideal Gas Law
z
Ideal Gas Equation
q
„
P*V=n*R*T
¾
¾
¾
¾
¾
z
P： 101.3
P
101 3 kPa = 1 atm = 760 mmHg = 760 torr
V： 1 m3 = 1 × 103 dm3 ( L ) = 1 × 106 cm3 (mL)
T： K = C° + 273.15
n ： # of moles of all gases
R = 8.31 J • mol-1•K-1 = 0.082 atm • L • mol-1 • K-1
Avogadro’s Law
„
Under same P
P, T
T, 1 mole of gas occupies the same V
V.
„
At the Standard Temperature & Pressure (STP), 1
mole of any ideal gas is 22.4 liter.
5
Practice
z
Q1 N2H4 is a fuel
Q1：N
fuel, prepared by the following reaction:
2NH3 (g) + NaOCl (aq) → N2H4 (aq) + NaCl (aq) + H2O (l)
To synthesize 15.0 kg N2H4, how much volume of 10oC,
3.63 atm of NH3 (g) is needed?
z
Solution:
n (N2H4) = m (N2O4) / MW (N2O4) = 1.50
1 50 x 104 g / 32 g/mol = 469 mol
V(NH3) = n (NH3) RT/ P
= 2n (N2H4) RT/ P
= 2 x 469 x 0.082 x (273 + 10) / 3.63
= 5.99 x 103 L
6
Partial Pressure
z
Dalton’s Law
„
Under constant temperature & volume, the total
pressure off a gas mixture
i t
equals
l th
the partial
ti l pressure
of each gas component.
Ptotal = PA + PB + PC + …… + Pi
z
Gas Partial Pressure (Pi)
„
P of a gas component solely occupying whole volume
7
Molar Fraction
Each gas component is valid for the Ideal Gas Law:
PA = nART/ V,
PB = nBRT/ V
Ptotal = nART/ V + nBRT/ V +
……
……
， Pi= niRT/ V
+ niRT/ V
= ((nA + nB + … … + ni ))RT/ V
= ntotal RT/ V
Pi / Ptotal = (niRT/ V) / (ntotalRT/ V) = ni / ntotal
Æ
ni / ntotal = Xi (molar fraction)
Æ
Pi = Ptotal × Xi
8
Partial Volume
z
z
Partial Vol
Volume
me of a gas component
„
Volume of a gas component under T and Ptotal
„
Vtotal = VA + VB + VC + …… + Vi
Molar Fraction (Xi)
„
Xi = ni / ntotal = Pi / Ptotal = Vi / Vtotal
„
Pi × Vtotal= Ptotal × Vi
9
Practice
z
z
Q2: At 25°C,, 1 atm,, we collected a mixture of H2 and
saturated water vapor 152 mL. If the saturated water
vapor is 23.76 mmHg, calculate: (a) Partial pressure of
H2 ((g);
) (b) A
Amountt off substance
b t
off H2 (g);
( ) ((c)) V
Volume
l
off
this H2 (g) if it is dry.
Solution:
((1)) PH2 = Ptotal- PH2O = 760 - 23.76 = 736.2 mmHg
g
(2) PH2Vtotal = nH2RT
nH2 = PH2Vtotal / RT
= 736.2 x 0.152 / 62.4 x 298.15 = 6.2 x 10-3 mol
(3) PH2Vtotal= VH2 Ptotal
VH2 = PH2Vtotal / Ptotal = 736.2 x 152 / 760 = 147 ml
10
Practice
z
z
Q3: At 250°C, PCl5 completaly evaporates and then
partially dissociates into PCl3 and Cl2。If
If 2.98
2 98 g PCl5 is
placed in a 1.00 dm3 container and completely evaporates
into g
gas,, the total p
pressure is 113 kPa. Calculate the
partial pressure of each gas component.
S l ti
Solution:
MWPCl5 = 208.5
nPCl5 = 2.98 / 208.5 = 1.43 x 10-2 mol
When PCl5 completely evaporates but before dissociation:
P = nRT/V = 1.43 x 10-2 x 8.314 x (250 + 273.15) / 1.00 x 10-3 = 62.14 kPa
When partially dissociate:
PPCl5 = 62.14 – x,
PCl
C 5 (g)
( ) ↔ PCl
C 3 (g)
( ) + Cl
C 2 (g)
( )
PPCl3 = x,
PCl2 = x
Ptotal = PPCl5 + PPCl3 + PCl2 = 62.14
62 14 – x + x + x = 113 kPa
So: PPCl5 = 62.14 – x = 11.28 kPa, PPCl3 = PCl2 = 50.86 kPa
Æ x = 50.86
50 86 kPa
11
Phase
z
Phase
„
A material or material fraction with (macroscopically)
homogeneous
g
p
physical
y
& chemical p
properties
p
„
z
Phase Change
„
z
e.g. Gas, Liquid, (water vs. oil), Solid, (different solids)
A same material changes from one phase to another.
Ph
Phase
E
Equilibrium
ilib i
„
For a material,, when its all fractions of different
phases remain unchanged under a certain condition.
12
Liquid
z
z
Liquid
„
Li id iis a ki
Liquid
kind
d off "C
"Condensed
d
d Ph
Phase""
„
Short-range
g ordered;; Long-range
g
g disordered
Liquid Evaporation
Vapor：gas molecules
Vapor：
above liquid
Open
System
Liquid
Molecules with higher energy will escape from liquid phase to vapor phase first
13
Liquid Evaporation
Closed
System
v
Evaporation
Condensation
t
14
((Saturated)) Vapor Pressure
z
Vapor Pressure / Saturated Vapor Pressure
„
P of a vapor phase at equilibrium with its liquid phase
„
Vapor pressure increases with temperature
„
At a liquid’s boiling point (Tb), Vapor pressure = 1 atm
Artificial Piston
15
How Vapor Pressure Changes
g with T
Vapor
p Pressure ~ Temperature
p
lg P =
„
− ΔH vap
2.303RT
+ const
ΔHvap: Heat of vaporization (Enthalpy of vaporization)
ΔH vap ⎛ T2 − T1 ⎞
p2
⎟⎟
=
× ⎜⎜
lg
p1 2.303R ⎝ T2 × T1 ⎠
lg P (vapor of H2O)
z
1/T
16
Practice
z
Q: For phenol (C6H5OH), ΔHvap = 48.139 kJ/mol,
and its vapor pressure is 100 mmHg at 392.5 K.
What is the vapor pressure of phenol at 350 K?
z
Solution:
For phenol: ΔHvap = 48.139 kJ/mol T1 = 392.5 K
T2 = 350 K
P1 = 100 mmHg
P2 = ？mmHg
p2
48.139 × 1000J/mol ⎛ 350 − 392.5 ⎞⎛ K ⎞
=
lg
⎟⎜ 2 ⎟
⎜
100 2.303
2 303 × 8.314J/(mo
8 314J/( l ⋅ K) ⎝ 350 × 392.5
392 5 ⎠⎝ K ⎠
p2
p2
lg
= −0.778,
= 0.1668
100
100
p2 = 16.68mmHg
17
Practice
z
Q: At 40oC, compress 1.00 L air containing saturated
bezene (C6H6) vapor from 1 atm to 5 atm
atm. If the vapor
pressure of bezene is 181.7 mmHg at 40oC, how many
grams off benzene
b
is
i condensed
d
d tto liliquid
id iin thi
this process?
?
z
Solution:
During the whole process, the amount of air remains same, which is:
1.00 × (750-181.7) = Vtotal (5 × 760-181.7) Æ So: Vtotal = Vfinal= 0.157 L
At this time, moles of bezene is:
V finalٛ×181.7
(40 + 273)× 62.4
1.00 × 181.7
(40 + 273)× 62.4
Initially, moles of bezene is:
So condensed bezene is:
So,
(
1.00 − V )×181.7
W=
× 78 = 0.612ٛg
finalٛ
ٛ
(40 + 273)× 62.4
18
Melting
g & Freezing
g Curves
z
Melting
g Curve and Melting
g Point ((Tm)
z
Freezing
ee g Cu
Curve
e & Freezing
ee g Ptt
„
„
e.g. Cooling curve of water at 1 atm
Supercooling phenomenon
A
BC D
19
Sublimation & Deposition
Sublimation: direct change
g from solid p
phase to
gas phase (e.g. I2 (s) Æ I2 (g), CO2 (s) dry ice )
lg P (vapor of solid)
z
ΔH sub
lg p = −
+B
2.303RT
1/T
ΔHsub: Heat of sublimation
(Enthalpy of sublimation)
Vapor pressure vs. Temperature
20
Phase Diagrams
g
z
Phase Diagram for water – Understand the
meaning
i off different
diff
t Zones,
Z
Lines,
Li
and
d points.
i t
„
Melting point(s), Boiling point(s), at different T & P.
„
Triple point (for H2O: 0.006 atm, 0.01 oC)
„
Critical point (for H2O: 218 atm, 374 oC)
21
New Slide of Practice
Textbook,
Te
tbook page 472
472, 12
12.54
54
Use the water diagram:
g
((a)) What is the
phase of water at 2 mmHg and -3C? (b)
What term describe water at 400C and
225 atm? (c) What will happen if water at
-20C
20C and 3 mmHg is heated to 20C at
same pressure? (d) Sketch part c in figure.
( )D
(e)
Describe
ib what
h will
ill h
happen if water at
200C and 1 atm is increased to 50 atm at
constant temperature?
z
22
Solution Revisited ((Chap. 13.5—13.7))
z
Different ways
y to represent
p
concentration:
„
Solubility, S (Mass percentage)
„
Molarity, C (# of solute moles per solution volume in liter)
„
Molality, m (# of solute moles per solvent mass in kg)
„
Mole fraction, X (Percentage of moles):
nsoluteٛ
,
X soluteٛ =
nsolute
l t + nsolvent
l t
nsolventٛ
,
X solventٛ =
nsoluteٛ+ nsolventٛ
X ٛsolute + X solventٛ = 1
23
Colligative
g
Properties
z
z
Colligative
g
Properties
p
of Nonvolatile Solutes
„
Do not depend on the intrinsic properties of solute itself
„
Depend ONLY on the number of solute particles
4 common Colligative Properties
„
Vapor pressure lowering
„
Boiling point elevation
„
Freezing point depression
„
Osmotic pressure
24
1. Vapor Pressure Lowering
g
z
Raoult’s Law:
„
The solvent (volatile) vapor pressure is lowered when
it di
dissolves
l
a nonvolatile
l til solute.
l t
P = P0 * X ٛsolvent
„
For a two-component
two component (i
(i.e.
e solute + solvent) system:
P = P0 * X solvent
l t = P0 * (1 − X solute
l t )
P0 − P = P0 * X solute
so : ΔP = P0 * X solute
25
A Molecular View of Raoult’s Law
z
Solvent (volatile) + Solid (nonvolatile)
P = P0 * X ٛsolvent
z
A mixture of two volatile materials
26
Vapor Pressure of 2 Volatile Molecules
z
Ptotal = P1,0 * X1 + P2,0 * X2
„
P1,0 and P2,0 : vapor pressure of pure substance
„
X1, X2 : mole fraction of each substance in the mixture
mixture,
X 1 + X2 = 1
27
Practice
z
Q: A solution is composed by hexane (C6H14) and
heptane (C7H16).
) The saturated vapor pressures for
hexane and heptane are 0.198 atm and 0.06 atm,
respectively. If the mole fraction of hexane in the solution
is 0.4.
(a) What is the vapor pressure of the solution?
(b) What are the mole fractions of hexane and heptane in the vapor
phase above the solution?
(c) If we collect all vapor phase and condense into another empty
container,
t i
and
d llett th
them re-evaporate
t to
t reach
h equilibrium.
ilib i
Wh
Whatt
are the mole fractions in the new vapor?
(d) What can you conclude from the above results?
28
Practice
z
Answer:
(a) VP of hexane: P0,, hexane* Xhexane = 0.198 * 0.4 = 0.0792 atm
VP of heptane: P0, heptane* Xheptane = 0.06 * 0.6 = 0.036 atm
Ptotal = P0, hexane* Xhexane + P0, heptane* Xheptane = 0.1152 atm
(b) Mole fraction in the vapor phase – using Y to represent:
Yhexane = 0.0792
0 0792 / 0
0.1152
1152 = 0
0.688,
688 Yheptane = 0.036
0 036 / 0
0.1152
1152 = 0
0.312
312
(c) Liquid phase in the 2nd container = Vapor phase in the 1st container
VP of hexane: P0, hexane* Xhexane = 0.198 * 0.688 = 0.136 atm
VP of heptane: P0, heptane* Xheptane = 0.06 * 0.312 = 0.0187 atm
Ptotal = P0, hexane* Xhexane + P0, heptane* Xheptane = 0.155
0 155 atm
Yhexane = 0.136 / 0.155 = 0.879, Yheptane = 0.0187 / 0.155 = 0.121
(d) The one with higher vapor pressure (hexane) is enriched in the
vapor phase – a phenomenon known as Distillation.
29
Fractional Distillation
Mixture of A & B
(if: VP: P0,A > P0,B; or
More A
than B
Boiling pt: Tb, A < Tb,B)
30
2. Boiling
g Point Elevation
P/atm
m
Solvent
ΔTb = Tb′ − Tb = K b × m
Solution
T/K
Recall: @ Boiling point (Tb), the vapor pressure = External pressure (1 atm);
So when the solvent dissolves some solute (with molality m)
m), the vapor
pressure drops, thus leading to an elevation of its boiling point.
31
Practice
z
z
Q: Dissolve 17.6 g of an unknown solute inside 250 g
benzene this benzene solution’s
benzene,
solution s boiling point increases
1.00 oC. If Kb of benzene is 2.53 oC · kg / mol, what is
the molecular weight
g of this unknown solute?
Answer:
wsoluteٛ
×1000
M soluteٛ
wsoluteٛ×1000
ΔTb = K b × m = K b ×
= Kb ×
wsolventٛ
M soluteٛ× wsolventٛ
K b × wsoluteٛ×1000
so : M soluteٛ =
wsolventٛ× ΔTb
=
2.53 ×17.60 ×1000
= 178 g / mol
250 ×1.00
32
3. Freezing
g Point Depression
ΔT f = T f′ − T f = K f × m
33
4. Osmotic Pressure
34
Osmotic Pressure
z
Osmotic Pressure: the p
pressure needed to stop
p
the net movement of solvent molecules through
a semi-permeable
p
membrane
nsolute
l t
Π = C solute * R * T =
* R *T
Vsolution
35
Practice
z
z
Q: Dissolve 2
2.00
00 g of a protein in 0
0.100
100 L water
water, and
measure the osmotic pressure of this solution at 25 oC is
0.021 atm. What is the molar mass of this protein?
p
Answer:
C = Π / RT
= 0.021 / 0.082 x 298 = 8.6 x 10-4 mol/L
That is: C = 8
8.6
6 x 10-44 mol/L= (2
(2.00
00 g / Mprotein ) / 0.100
0 100 L
So: Mpprotein = 2.00 / (8.6 x 10-4 x 0.1) = 2.3 x 104 g/mol
36
Practice
z
z
Electron microscopy
py image
g of
cells in high osmotic pressure
solution
Fact: A method to estimate the
osmotic pressure of a cell, is to place
the cells into a series of NaCl
solutions with different
concentrations, and to find out the
concentration in which the cells do
not expand or shrink.
Q: If at 37oC,
C cardiac muscle cells in a unknown
concentration of NaCl solution do not expand or shrink.
The equilibrium
q
vapor
p p
pressure of water over this NaCl
solution is 6.214 kPa. The pure water’s vapor pressure is
6.276 kPa. What is the osmotic pressure of this kind of
cardiac
di muscle
l cells?
ll ?
37
Practice
z
Ans er
Answer:
ΔP = P0,H2O
* XNaCl = P0,H2O
* nNaCl / (nNaCl + nH2O)
,
,
ΔP = P0,H2O – Psolution = 6.276 – 6.214 = 0.062 kPa
Æ (nNaCl + nH2O) / nNaCl = P0,H2O / ΔP = 6.276 / 0.062 = 101
Æ nH2O / nNaCl = 100
Æ Assume 1000 g H2O: nH2O = 1000 g / 18 g/mol = 55.5 mol,
Æ nNaCl = 0.555 mol Æ CNaCl = 0.555 mol/L
pressure: Π = C R T = 0.555 x 0.082 x ((273.15+37)) = 14.11 atm
Æ Osmotic p
38