PHASE EQUILIBRIUM

one of the most important sources of information concerning the behavior of elements, compounds and solutions.

They provide us with the knowledge of phase composition and phase stability as a function of temperature (T), pressure (P) and composition (C).

PHASECOMPONENTS

PHASE RULE: DEGREE OF FREEDOM

PHASE

•A chemically and structurally homogeneous

portion of material•Separated with other parts

of the system

•Determines the number of independent variables needed•Is the correlation between the

number of phase (P), components (C), and 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)

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)

OR

PHASE RULES

F = C – P +2

Degree of freedom or the

number of independent

variables

Number of component

The number of phase

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

TYPES

UNIVARIANT

BIVARIANT

EXAMPLES

CaCO3(s) CaO(s) + CO2(g) Calculate the degree of freedom (F)

F = C – P +2 = 2 – 3 + 2 = 1 (univariant)

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

ONE COMPONENT SYSTEM

Standard phase diagram for water (H2O)

Standard phase diagram for carbon dioxide (CO2)

Standard phase diagram for one component system

Critical point ???

B

A

What does it means by:1)AO curve2)OB curve3)OC curve4)AOB curve5)BOC curve6)AOC curveO

H2O

CO2

?????

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.

This unique properties of water is due to the network of hydrogen bonding in ice is

more extensive than in liquid

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

50

(a)

(d)

(c)(b)

LEARNING CHECK !!!!!

50

(a) (b) (c) (d) (e)

LEARNING CHECK !!!!!

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.

O

sublimation

Has a positive slope

So, its shows that the increases of pressure, will increased the melting point for CO2 solid

Critical point

TWO COMPONENT SYSTEM

Two completely miscible liquid – ideal,

non-ideal, positive and negative

deviation

Raoult’s law – in explaining the effect of non-volatile solute on vapour pressure of solvent and its melting

and boiling point

Composition diagram vs boiling point

composition for ideal, non-ideal, negative n

positive deviation

Fractional distillation and azeotropic

system

Eutectic system and cooling curves

VAPOUR PRESSURE

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

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.

sublimation

Vapour pressure increases with increasing temperature due to its KE

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

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:

•relatively strong, the vapor pressure will be relatively low.

•relatively weak, the vapor pressure will be relatively high.

ethyl ether (C4H10O)Pvapor (25oC) = 520 torr

ethyl alcohol (C2H6O)Pvapor (25oC) = 75 torr

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.

Low Temperature

High Temperature

1

2

COMPLETELY MISCIBLE LIQUID

RAOULT’S LAW

Liquid solution in

liquid

3 types :1)Complete Miscible liquid

2)Half miscible liquid3)Immiscible liquid

Oil and water

ether and water

Methanol and ethanol

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 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

Ideal solution- mostly involve the substance that have similar

physicochemical properties. Ex: MeOH/EtOH, benzene/toluene, n-

hexane/n-heptane

AA A B=

•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

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= 0.88Therefore,

Psolution = 0.88 x 23.76 mmHg

= 20.96 mmHg

PT = PA + P B

@Ptotal = XA Po

A + XB PoB

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 Po

A + XB PoB

Calculate the number of moles for benzene and toluene:n of benzene : PA + P B = 60/78 = 0.77 molen of toluene : PA + P B = 40/92 = 0.43 mole

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.

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.

VP BP

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 : AA B B

A B

=

WEAKER THAN

•SO,the molecules in the solution have lower tendency to escape into vapour phase.

•Therefore the process is EXOTHERMIC

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.

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.

Notice the much bigger difference this time due to the presence of the new ionic interactions

Azeotrope

USING THE DIAGRAM

Distilling dilute nitric acidStart with a dilute solution of nitric acid with a composition of C1and trace through what happens.

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.

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.

Distilling nitric acid more concentrated than 68% by massThis 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.

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.

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.

Formed when the intermolecular forces between molecules in the mixture are weaker than those in pure liquids.

AA B B

A B

=

STRONGER THAN

Vapour pressure of the solution is higher than expected

The solution has a greater tendency to evaporate or

escape into vapour

The solution has a greater tendency to evaporate or

escape into vapour

The process is endothermicThe 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.

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.

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

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.

the thermometer bulb is placed exactly at the outlet

from the fractionating column

Relating what happens in the fractionating column to the phase diagram

Boil a mixture with composition C1.

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?

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.

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 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.

The boiling points of the two liquids. The closer they are together, the longer the column has to be.

Which compound that rich at this point? What phase?

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

BOILING POINT : is the temperature at which its vapour pressure equals the external pressure

∆Tb = Tb - T ob ∆Tb = Tb - T ob

Boiling point of pure solvent

Boiling point of solution

Boiling point elevation

Molal (m) = mol solute Kg solventMolal (m) = mol solute Kg solvent

∆Tb is proportional to molality of solute in the solution, so…. ∆Tb = Kb m∆Tb = Kb m

The molal boiling point elevation constant with unit oC/m or oC kg/mol

Molar mass of solute = Kb (gram solute) (∆Tb ) (kg solvent)

Molar mass of solute = Kb (gram solute) (∆Tb ) (kg solvent)

Boiling point solvent is

higher than boiling point of

solution

Boiling point solvent is

higher than boiling point of

solution

WHY??

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.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 mol0.6693676 mol / 0.200 kg = 3.3468 m

EXAMPLES:

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

FREEZING POINT DEPRESSION : is the temperature at which solid begins to appear in liquid or solution

∆Tf = T of - Tf ∆Tf = T of - Tf

Molal (m) = mol solute Kg solventMolal (m) = mol solute Kg solvent

∆Tf is proportional to molality of solute in the solution, so…. ∆Tf = Kf m∆Tf = Kf m

Molar mass of solute = Kf (gram solute) (∆Tf ) (kg solvent)

Molar mass of solute = Kf (gram solute) (∆Tf ) (kg solvent)

Boiling point of solution

Boiling point of pure solution

T of > Tf T of > Tf

•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.

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.

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.

Learning check !!!!!!

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

It’s the BONUS TIME AND YOUR HAPPY HOUR. THIS IS YOUR

HOMEWORKS TO FIND ABOUT EUTECTIC MIXTURE AND

COOLING CURVES

GOOD LUCK GUYS

GOOD LUCK GUYS

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

YOU HAVE TO MAKE A SHORT NOTES ON THIS SUBTOPIC AND SUBMIT IT TO YOUR

‘KAWAII’ LECTURER