Transcript PHASE EQUILIBRIUM

They provide us with the knowledge of phase
composition and phase stability as a function of
temperature (T), pressure (P) and composition (C).
PHASE
EQUILIBRIUM
one of the most important sources of information
concerning the behavior of elements, compounds
and solutions.
DEFINITION
ONE COMPONENT
SYSTEM
TWO COMPONENT
SYSTEM
COMPONENTS
PHASE
DEFINITION
PHASE RULE: DEGREE
OF FREEDOM
1) The mixture of ice and water = have two
phase which is solid and liquid
2) The mixture of oxygen gas and nitrogen gas =
have one phase which is gas phase (the
system is homogen)
•A chemically and
structurally homogeneous
portion of material
•Separated with other parts
of the system
3) The mixture of oil and water = have 2 same
phase(liquid). Oil and water are not
homogen and have the boundaries to
separate both phase
4) CaCO3(s)
CaO(s) + CO2(g)
= 3 phase (2 solid,1 gas)
•Determines the number of
independent variables needed
•Is the correlation between the
number of phase (P), components
(C), and degree of freedom
PHASE
OR
The number of chemical
species that can explained
the composition of all phase
in a system
The least number of
different substances
required to describe the
composition of all phases in
the system
1) water, CO2 = one
component
2) Aqueous solution of
potassium nitrate = 2 system
component because have
potassium nitrate salt and
water.
COMPONENT
PHASE
RULES
Also known as Gibbs phase rule
F = C – P +2
Degree of
freedom or the
number of
independent
variables
The number
of phase
Number of
component
2 variables
(temperature
and pressure)
DEGREE OF FREEDOM (F)
The number of variables that may
be changed independently without
causing the appearance of a new
phase or disappearance of an
existing phase
UNIVARIANT
TYPES
BIVARIANT
EXAMPLES
CaCO3(s)
CaO(s)
F = C – P +2
=2–3+2
= 1 (univariant)
+ CO2(g)
Calculate the degree of freedom (F)
Means: only one
variable, either
temperature or pressure
can be changed
independently
The number of components is not
always easy to determine at first
glance, and it may require careful
examination of the pyhsical
conditions of the system at
equilibrium
Standard phase diagram for water (H2O)
ONE COMPONENT
SYSTEM
Standard phase diagram for carbon dioxide
(CO2)
H2O
Standard phase diagram for one component system
B
O
CO2
Critical
point ???
What does it
means by:
1)AO curve
2)OB curve
3)OC curve
4)AOB curve
5)BOC curve
6)AOC curve
A
?????
Standard phase diagram for water (H2O)
Special case
!!!!!
TA curve = known as melting point or
freezing point
Represent the equilibrium between ice
and liquid
Has a negative slope
Water as the liquid is denser than the solid
(ice floats on water).
That means that an increase of pressure
favors the formation of liquid and that the
melting point of water falls with increasing
pressure.
According to Le Chatelier’s principle,
when pressure is applied, the reaction
shifts in the direction that can release the
stress and cause ice to melt
This unique properties of water is due to
the network of hydrogen bonding in ice is
more extensive than in liquid
LEARNING CHECK !!!!!
(d)
50
(c)
(b)
(a)
LEARNING CHECK !!!!!
50
(a)
(b)
(c)
(d)
(e)
Standard phase diagram for carbon dioxide
(CO2)
The point O is the triple point for
CO2 (at 5.1 atm, -57o C). So,
CO2 solid can’t changed to liquid
form at 1 atm.
Critical point
Has a positive slope
So, its shows that the increases
of pressure, will increased the
melting point for CO2 solid
O
sublimation
Raoult’s law – in
explaining the effect
of non-volatile solute
on vapour pressure of
solvent and its melting
and boiling point
Two completely
miscible liquid – ideal,
non-ideal, positive
and negative
deviation
TWO COMPONENT
SYSTEM
Composition diagram
vs boiling point
composition for ideal,
non-ideal, negative n
positive deviation
Eutectic system and
cooling curves
Fractional distillation
and azeotropic
system
VAPOUR PRESSURE
Vapour pressure
increases with
increasing temperature
due to its KE
When a liquid evaporates in a
closed vessel, its gaseous
molecules formed above the
liquid have high KE and exert a
vapour pressure.
The molecules collide with the
pinston and push the pinston
upward
sublimation
Microscopic equilibrium between gas and
liquid. Note that the rate of evaporation of the
liquid is equal to the rate of condensation of
the gas.
Microscopic equilibrium between gas and solid.
Note that the rate of evaporation of the solid is
equal to the rate of condensation of the gas.
Volatile liquid is a liquid that can easily
evaporate at one atmospheric pressure and
room temperature
Molecules of volatile liq escape the liquid
phase into gaseous phase.(KE)
A volatile liquid has a strong tendency to
vapourize or evaporate into vapour, creating
high vapour pressure.
On contrary a less volatile liquid has low
vapour pressure because of lower tendency
to vapourize
1
Types of Molecules: the types of molecules
that make up a solid or liquid determine its
vapor pressure. If the intermolecular forces
between molecules are:
ethyl ether (C4H10O)
Pvapor (25oC) = 520 torr
ethyl alcohol (C2H6O)
Pvapor (25oC) = 75 torr
•relatively strong, the vapor pressure will
be relatively low.
•relatively weak, the vapor pressure will be
relatively high.
Low
Temperature
2
Temperature:
at
a
higher
temperature, more molecules have
enough energy to escape from the
liquid or solid. At a lower temperature,
fewer molecules have sufficient
energy to escape from the liquid or
solid.
High
Temperature
COMPLETELY MISCIBLE
LIQUID
RAOULT’S LAW
Methanol and ethanol
Liquid
solution in
liquid
ether and
water
Oil and
water
3 types :
1) Complete Miscible liquid
2) Half miscible liquid
3) Immiscible liquid
Ideal solution- mostly involve the
substance that have similar
physicochemical properties. Ex:
MeOH/EtOH, benzene/toluene, nhexane/n-heptane
Complete Miscible liquid
• 2 types of complete miscible liquid which is ideal and non-ideal solution
• An ideal solution is a solution that obeys Raoult’s law and non-ideal solution
disobey.
A
A
B
= A
•A solution is a ideal solution when:
•The intermolecular attractions between the mixture of same molecule with the
the mixture of different molecule are equal.
•The volume of the mixture are the total volume of both liquid (volume of liquid
A add with volume of liquid B)
•No heat changes (no endo-exothermic process)
•Obeys Raoult’s law
•Relationship between vapour pressure of a solvent and its mole fraction
•States the vapour pressure of the solute containing solution (PA) is equal to the
mole fraction of the solvent (XA) times the vapour pressure of the pure solvent
(Po A)
PA = XA Po A
Psolution = Xsolvent Po solvent
PT = PA + P B
@
Ptotal = XA PoA + XB PoB
EXAMPLE !!!!!
A solution is prepared by adding 2.0 mole of glucose
in 15.0 mole of water at 25 oC. The vapour pressure
of pure water at 25 oC is 23.76 mmHg. Calculate the
vapour pressure of the solution at 25 oC
Psolution = Xsolvent Po solvent
The mole fraction:
XA = nA / nt
= 15.0 /17.0= 0.88
Therefore,
Psolution = 0.88 x 23.76 mmHg
= 20.96 mmHg
EXAMPLE !!!!!
At 25 oC the vapour pressure of pure benzene and toluene are 93.4 mmHg and
26.9 mmHg. If the mixture contains 60.0g of benzene and 40.0 g of toluene,
calculate the vapour pressure of this solution
PT = PA + P B
@
Ptotal = XA PoA + XB PoB
Calculate the number of moles for benzene and toluene:
n of benzene : PA + P B = 60/78 = 0.77 mole
n of toluene : PA + P B = 40/92 = 0.43 mole
Ptotal = XA PoA + XB PoB
= 0.77
x 93.4 mmHg
0.77 + 0.43
= 69.54 mmHg
+
0.43
x 26.9 mmHg
0.77 + 0.43
IDEAL SOLUTION
DIAGRAMS (RAOULT’S
LAW)
VAPOUR PRESSURE/
COMPOSITION
DIAGRAM
BOILING POINT/
COMPOSITION
DIAGRAM
There is actually no such thing as an ideal mixture! However, some liquid mixtures
get fairly close to being ideal. These are mixtures of two very closely similar
substances. Commonly quoted examples include:
•hexane and heptane
•benzene and methylbenzene
•propan-1-ol and propan-2-ol
Pure vapour pressure
Pure mixture vapour pressure
Notice that the vapour pressure of pure B is higher than that of
pure A. That means that molecules B must break away more
easily than of A. B is the more volatile liquid.
Total vapour pressure of the
mixture
We'll start with the boiling points of pure A and B.B has
the higher vapour pressure. That means that it will have
the lower boiling point.
VP
BP
The diagram just shows what happens if
you boil a particular mixture of A and B.
Notice that the vapour over the top of the
boiling liquid has a composition which is
much richer in B - the more volatile
component.
EXAMPLE !!!!!
Which vapour
sample rich at
this point?
composition
NON-IDEAL DIAGRAMS
• NEGATIVE DEVIATION
•POSITIVE DEVIATION
•Involves the intermolecular forces between molecules in solution are stronger
than those in pure liquid
•Therefore, vapour pressure of the solution is lower than vapour pressure of its
components or pure liquid.
•Example :
A
A
=
B
B
WEAKER THAN
A
B
•SO,the molecules in the solution have lower tendency to escape into vapour
phase.
•Therefore the process is EXOTHERMIC
Azeotrope
Nitric acid and water form mixtures
in which particles break away to
form the vapour with much more
difficulty than in either of the pure
liquids.
That means that mixtures of nitric acid
and water can have boiling points higher
than either of the pure liquids because it
needs extra heat to break the stronger
attractions in the mixture.
In the case of mixtures of nitric acid and
water, there is a maximum boiling point of
120.5°C when the mixture contains 68% by
mass of nitric acid. That compares with the
boiling point of pure nitric acid at 86°C, and
water at 100°C.
Notice the much bigger difference this time
due to the presence of the new ionic
interactions
USING THE DIAGRAM
Distilling dilute nitric acid
Start with a dilute solution of nitric acid
with a composition of C1and trace
through what happens.
As the acid loses water, it becomes more
concentrated. Its concentration gradually
increases until it gets to 68% by mass of nitric
acid. At that point, the vapour produced has
exactly the same concentration as the liquid,
because the two curves meet.
You produce a constant boiling mixture (or
azeotropic mixture or azeotrope). If you distil
dilute nitric acid, that's what you will eventually
be left with in the distillation flask. You can't
produce pure nitric acid from the dilute acid by
distilling it.
The vapour produced is richer in water than the
original acid. If you condense the vapour and
reboil it, the new vapour is even richer in water.
Fractional distillation of dilute nitric acid will
enable you to collect pure water from the top of
the fractionating column.
Distilling nitric acid more concentrated than 68% by mass
This time you are starting with a concentration C2 to the right of
the azeotropic mixture.
The vapour formed is richer in nitric acid. If
you condense and reboil this, you will get a
still richer vapour. If you continue to do this
all the way up the fractionating column, you
can get pure nitric acid out of the top.
As far as the liquid in the distillation flask is
concerned, it is gradually losing nitric acid. Its
concentration drifts down towards the
azeotropic composition. Once it reaches that,
there can't be any further change, because it
then boils to give a vapour with the same
composition as the liquid.
Distilling a nitric acid / water mixture containing
more than 68% by mass of nitric acid gives you
pure nitric acid from the top of the fractionating
column and the azeotropic mixture left in the
distillation flask.
Formed when the intermolecular forces between molecules in the mixture are
weaker than those in pure liquids.
A
A
=
B
B
STRONGER THAN
A
Vapour pressure of the solution is higher
than expected
B
The solution has a greater
tendency to evaporate or
escape into vapour
The process is endothermic
A large positive deviation from Raoult's
Law produces a vapour pressure curve
with a maximum value at some
composition other than pure A or B.
If a mixture has a high vapour pressure it
means that it will have a low boiling point
The molecules are escaping easily and you
won't have to heat the mixture much to
overcome the intermolecular attractions
completely.
The implication of this is that the boiling
point / composition curve will have a
minimum value lower than the boiling
points of either A or B.
USING THE DIAGRAM
Suppose you are going to distil a mixture of
ethanol and water with composition C1 as
shown on the next diagram. It will boil at a
temperature given by the liquid curve and
produce a vapour with composition C2.
When that vapour condenses it will, of
course, still have the composition C2. If you
reboil that, it will produce a new vapour
with composition C3.
AZEOTROPE
This particular mixture of ethanol and water
boils as if it were a pure liquid. It has a
constant boiling point, and the vapour
composition is exactly the same as the
liquid.
It is known as a constant boiling
mixture or an azeotropic mixture or
an azeotrope.
SUMMARISE
Distilling a mixture of ethanol containing
less than 95.6% of ethanol by mass lets
you collect:
• A distillate containing 95.6% of ethanol in
the collecting flask (provided you are
careful with the temperature control, and
the fractionating column is long enough
• Pure water in the boiling flask.
FRACTIONAL
DISTILLATION
Typical fractional distillation in the lab
to give the maximum
possible surface area for
vapour to condense on
the thermometer bulb is
placed exactly at the outlet
from the fractionating
column
Some fractionating columns
have spikes of glass sticking
out from the sides which
serve the same purpose
In some cases, where you are collecting
a liquid with a very low boiling point,
you may need to surround the
collecting flask with a beaker of cold
water or ice.
Relating what happens in the fractionating column to the
phase diagram
Boil a mixture with composition C1.
Which compound
that rich at this
point? What phase?
Which compound
that rich at this
point? What phase
The vapour over the top of the boiling liquid
will be richer in the more volatile
component, and will have the composition
C2.
Which compound
that rich at this
point? What phase?
Each time the vapour condenses to a liquid,
this liquid will start to trickle back down the
column where it will be reboiled by up-coming
hot vapour. Each time this happens the new
vapour will be richer in the more volatile
component.
The aim is to balance the temperature of the
column so that by the time vapour reaches the
top after huge numbers of condensing and
reboiling operations, it consists only of the
more volatile component - in this case, B.
The boiling points of the two liquids. The
closer they are together, the longer the column
has to be.
what is the point of the packing in the column?
•To make the boiling-condensing-reboiling process as effective as
possible, it has to happen over and over again.
•By having a lot of surface area inside the column, you aim to have the
maximum possible contact between the liquid trickling down and the
hot vapour rising.
•If you didn't have the packing, the liquid would all be on the sides of the
condenser, while most of the vapour would be going up the middle and
never come into contact with it.
•BOILING POINT ELEVATION
•FREEZING POINT DEPRESSION
BOILING POINT : is the temperature at which its vapour pressure equals the
external pressure
∆Tb = Tb - T ob
Boiling point
elevation
Molal (m) = mol solute
Kg solvent
WHY??
Boiling point of
pure solvent
Boiling point
solvent is higher
than boiling
point of solution
Boiling
point of
solution
∆Tb is proportional to molality
of solute in the solution, so….
Molar mass of solute = Kb (gram solute)
(∆Tb ) (kg solvent)
∆Tb = Kb m
The molal boiling point
elevation constant with unit
oC/m or oC kg/mol
EXAMPLES:
What is the boiling point elevation when 11.4 g of ammonia (NH3) is dissolved in
200. g of water? Kb for water is 0.52 °C/m.
1) Determine molality of 11.4 g of ammonia in 200. g of water:
11.4 g / 17.031 g/mol = 0.6693676 mol
0.6693676 mol / 0.200 kg = 3.3468 m
2) Determine bp elevation:
Δt = Kb m
Δt = (0.52 °C/m) (3.3468 m)
Δt = 1.74 °C
EXAMPLES:
Calculating Molecular Mass (Formula Weight) of Solute
1.15g of an unknown, nonvolatile compound raises the boiling point of 75.0g
benzene (C6H6) by 0.275oC.
Calculate the molecular mass (formula weight) of the unknown compound.
Calculate the molality of solute particles:
m = ΔTb ÷ Kb
ΔTb = 0.275oC
Kb = 2.53oCm-1 (from table above)
m = 0.275 ÷ 2.53 = 0.109m
Calculate the moles of solute present:
molality = moles solute ÷ kg solvent
n(solute) = m x kg solvent = 0.109 x 75.0 x 10-3 = 8.175 x 10-3 mol
Calculate the molecular mass (formula weight) of the solute:
n(solute) = mass(solute) ÷ MM(solute)
MM(solute) = mass(solute) ÷ n(solute) = 1.15 ÷ 8.175 x 10-3 = 141 g/mol
FREEZING POINT DEPRESSION : is the temperature at which solid begins to appear in
liquid or solution
Boiling point of
solution
∆Tf = T of - Tf
T of > Tf
Boiling point of
pure solution
Molal (m) = mol solute
Kg solvent
∆Tf is proportional to molality of
solute in the solution, so….
Molar mass of solute = Kf (gram solute)
(∆Tf ) (kg solvent)
∆Tf = Kf m
•the boiling point of the solvent in a solution
is higher than that of the pure solvent;
•the freezing point (melting point) of the
solvent in a solution is lower than that of
the pure solvent.
Learning check !!!!!!
Automobile antifreeze is ethylene glycol, C2H6O2. It is a non-electrolyte. If a radiator
contains 40.0% antifreeze and 60.0% water, by mass, what is the freezing point of the
solution in the radiator? The normal freezing point for water is 0.0 °C and Kf is 1.86 °C
mol/kg.
Find the molality of the solution. The mass of the solvent is 0.0600 kg and the
formula weight of the solute is 62.066 g/mol.
Use the freezing point depression formula
It’s the BONUS TIME AND YOUR
HAPPY HOUR. THIS IS YOUR
HOMEWORKS TO FIND ABOUT
EUTECTIC MIXTURE AND
COOLING CURVES
EUTECTIC MIXTURE AND
COOLING CURVES
YOU HAVE TO MAKE A SHORT
NOTES ON THIS SUBTOPIC
AND SUBMIT IT TO YOUR
‘KAWAII’ LECTURER