Phase Equilibria

Question 1

What is a phase?

Answer 1

This is a homogeneous part of a system which is physically distinct from other parts of the system and separated from them by a boundary.

Each of the three states of matter i.e. solid, gas and liquid constitute a phase.

Question 2

What is a component?

Answer 2

This is any of the chemical spaces contained in a phase system.

A phase system can be a one component phase system; two component phase system or a three component phase system.

Question 3

What is a solution?

Answer 3

This is a homogeneous mixture of two or more substances i.e. a solution is a one phase system.

Question 4

What are the factors that define a given phase?

Answer 4

• Temperature
• pressure
• composition.

Question 5

What is a triple point?

Answer 5

This is the point on a phase diagram representing the temperature and pressure at which the solid, liquid and gaseous phases co-exist in equilibrium.

Question 6

What is the critical point in phase equilibrium?

Answer 6

The point on a phase diagram representing the temperature and pressure above which the liquid and gaseous phases are indistinguishable.

Question 7

What is critical temperature?

Answer 7

This is the temperature above which a gas cannot be liquefied by increase in pressure alone without further cooling no matter how much pressure is applied.

Question 8

What are the differences between phase diagram of water and carbon dioxide?

Answer 8

1) The critical pressure of carbon dioxide is above atmospheric pressure while that of water is below atmospheric pressure.
2) The melting point of solid carbon dioxide increases with increase in pressure while that of water decreases with increases in pressure.

Question 9

What are Miscible liquid mixtures?

Answer 9

These are liquids that dissolve in each other quickly at all compositions, temperature and pressure.

Mixtures of miscible liquids result into either ideal solutions or non-ideal solutions

Question 10

What are ideal solutions?

Answer 10

An ideal solution consisting of two components A and B is one in which the intermolecular attractions A—A, B—B and A—B are all equal.

OR:

An ideal solution is one that obeys Raoult’s law throughout its composition. In such a solution, the cohesive forces between the molecules of components are equal in magnitude and kind to the adhesive forces between the molecules of the components in the mixture.

Question 11

What are the characteristics of ideal solutions?

Answer 11

• It has equal intermolecular attractions between all molecules in the solution.
• There is no change in volume when it is made by mixing of its components i.e. ∆Vm = 0
• The enthalpy change on mixing of its components to form the solution is zero i.e. ∆Hm = 0

Question 12

Give examples of liquids that form ideal solutions

Answer 12

• Benzene and Methylbenzene
• Hexane and Heptane
• Octane and Nonane
• Chlorobenzene and Bromobenzene.
• Bromoethane and Chloroethane

Question 13

State Raoult’s law for ideal solutions

Answer 13

It states that the partial pressure of any volatile component of an ideal solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution provided temperature is constant.

Question 14

What are the conditions for an ideal solution to obey Raoult’s Law?

Answer 14

• Temperature must be constant.
• The ideal solution must be sufficiently dilute

Question 15

Describe fractional distillation of an ideal solution

Answer 15

• The liquid mixture is separated by fractional distillation.
• The liquid mixture of composition P is heated and its temperature rises until T1 when it boils to form a vapour of composition Q which is richer in the more volatile component A than the liquid mixture from which it is formed.
• The rising vapour is condensed by the cold glass beads forming a liquid of the same composition as the vapour at point R.
• The formed liquid flows down the fractionating column where it meets hot rising vapor causing it boil again at a lower temperature T2 forming a vapour of composition S with increased percentage composition of the more volatile component A.
• Successive boiling and condensation eventually results into pure A as the distillate and pure B as the residue in the flask.

Question 16

What is a non-ideal solution?

Answer 16

This is one in which the intermolecular attractions A—B are less or greater than the average A—A and B—B attractions

Question 17

Explain negative deviation from Raoult’s law

Answer 17

• A solution consisting of two liquids A and B exhibits negative deviation from Raoult’s law when the intermolecular attractions A—B are greater than the intermolecular forces of attraction A— A and B—B.

Question 18

What are the characteristics of solutions that show negative deviation?

Answer 18

• The intermolecular attractions between the two liquid components is greater than the intermolecular attractions of the pure components i.e. A—B > A—A or B—B
• There is decrease in volume when it is made by mixing of its components
• There is evolution of heat on mixing of its components
• There is reduced vapour pressure above the solution than for pure components
• It exhibits minimum vapour pressure and maximum boiling point

Question 19

What is an azeotropic mixture?

Answer 19

This is a liquid mixture which at constant pressure boils at a constant temperature without change in its composition forming vapour of the same composition as the liquid mixture.

The composition of an azeotropic mixture is always equal to that of the vapour above it.

Question 20

Give the similarity between an azeotropic mixture and a pure compound.

Answer 20

• Both the azeotrope and a pure compound boil at constant temperature without change in composition provided pressure stays constant.

Question 21

Give differences between an azeotropic mixture and a pure compound.

Answer 21

• The composition of the azeotrope changes with pressure while that of a compound is not affected by pressure changes.
• The components of the azeotropic mixture can be separated by physical mean whereas elements in a compound cannot be separated by physical means.
• The components of an azeotropic mixture are not chemically combined together whereas the elements in a compound are chemically combined together.

Question 22

Explain fractional distillation of a mixture showing negative deviation using a boiling point-composition diagram.

Answer 22

• The liquid mixture of composition X is heated and its temperature rises until point p when it boils at temperature T1 forming a vapour of composition q with increased percentage composition of component B, the more volatile component.
• The vapour is condensed by the glass beads forming a liquid of the same composition as the vapour at point r.
• The formed liquid flows down the fractionating column where it is heated again by hot rising vapour from the flask and boils at a much lower temperature T2 forming a vapour of composition s, much richer in B.
• This vapour is condensed by the glass beads forming a liquid of the same composition as the vapour at point t.
• Successive boiling and condensation results into pure B as the distillate and the residue in the flask being an azeotropic mixture of composition Y consisting of both components A and B.
• Similarly when a mixture of composition Z is fractionally distilled, the distillate will be pure A and the residue in the flask an azeotropic mixture of composition Y

Question 23

State means of separation of an azeotropic mixture.

Answer 23

• If one of the components of the azeotrope is water, the water can be removed by adding a suitable drying agent e.g. Silicon(IV) oxide.
• By solvent extraction using a suitable solvent.
• By passing the azeotrope over a suitable adsorbent (e.g. silica gel) onto which one of the components is adsorbed.
• By distillation in presence of a third component which forms a binary azeotrope with one of the components which boils at a lower temperature than the second component of the original mixture.

Question 24

Explain positive deviation from Raoult’s law.

Answer 24

A solution consisting of two liquids A and B exhibits positive deviation from Raoult’s law when the intermolecular attractions A—B are less than the A—A and B—B intermolecular attractions.

Question 25

What are characteristics of solutions that show positive deviation?

Answer 25

- The intermolecular attractions between the two liquid components is less than the intermolecular attractions of the pure components i.e. A---B < A---A and B---B. - There is increase in volume when it is made by mixing of its components. - There is absorption of heat on mixing of its components. - There is increased vapour pressure above the solution than for pure components at the same temperature - It exhibits maximum vapour pressure and minimum boiling point

Question 26

Give examples of liquid mixtures that exhibit positive deviation from Raoult’s law.

Answer 26

- Carbon disulphide and propanone - Benzene and methanol - Water and ethanol - Ethanoic acid and Methylbenzene.

Question 27

Explain fractional distillation of a mixture showing positive deviation using a boiling point-composition diagram.

Answer 27

- The liquid mixture of composition Z is heated and its temperature rises until point P when it boils at temperature T1 forming a vapour of composition Q with richer percentage composition of component B. - The vapour is condensed by the glass beads to form a liquid of the same composition as the vapour at point R. - The formed liquid flows down the fractionating column and it is heated again by hot rising from the flask and boils at a lower temperature T2 forming a vapour of composition S which is condensed by the glass beads forming a liquid of the same composition as its vapour at point T. - Repeated boiling and condensation eventually results into an azeotropic mixture as the distillate and pure A as the residue in the flask.

Question 28

Explain why a mixture of Sulphuric acid and water shows negative deviation from Raoult's law.

Answer 28

- Pure sulphuric acid has weak Van der Waal forces while pure water has strong intermolecular hydrogen bonds. - Sulphuric acid in water dissociates forming ions which become hydrated resulting into ion to dipole forces which are stronger forces than both the Van der waal forces in pure sulphuric acid and the hydrogen bonds in pure water. - Therefore, in the mixture, the escaping tendency of both water and sulphuric acid molecules into vapour state is increased leading to reduced vapour pressure and negative deviation.

Question 29

Explain why a mixture of Ethanol and water shows a positive deviation from Raoult’s law.

Answer 29

- Both water and ethanol are polar molecules and form intermolecular hydrogen bonds as pure liquids with the hydrogen bonds in water being stronger. - When the two liquids are mixed together, some of the hydrogen bonds are broken down due to presence of the non-polar hydrocarbon chain of ethanol. - Therefore, the resulting solution contains weaker intermolecular hydrogen bonds between water and ethanol molecules compared to the pure liquids. - Consequently, the escaping tendency of molecules into vapour state from the solution increases resulting into increased vapour pressure. Hence positive deviation.

Question 30

Explain why a mixture of Phenol and Phenylamine shows a negative deviation from Raoult’s law.

Answer 30

- Both pure phenol and pure phenylamine are polar compounds containing benzene ring and form intermolecular hydrogen bonds between their molecules. - A solution of phenol and phenylamine (aniline) forms stronger intermolecular hydrogen bonds than the pure liquids because the hydroxyl group of phenol is much more polar than the amine group of phenylamine. - Consequently, the escaping tendency of molecules into vapour state from the solution is decreased resulting into decreased vapour pressure. Hence negative deviation.

Question 31

Explain why a mixture of benzene and ethanol shows a positive deviation from Raoult’s law.

Answer 31

- Benzene is a non-polar compound whose molecules are joined by weak Van der waal forces while ethanol is a polar compound whose molecules associate via strong hydrogen bonds. - When the two liquids are mixed together, strong repulsion occurs between their molecules due to the difference in the nature of the intermolecular forces. - This increase escaping tendency of the molecules in the mixture to vapour state which increases the vapour pressure and resulting into positive deviation.

Question 32

Explain why a mixture of benzene and trichloromethane shows a negative deviation from Raoult’s law.

Answer 32

- This is because when benzene and trichloromethane are mixed together, strong hydrogen bonding takes places between the hydrogen atom trichloromethane and the pi electron cloud of the benzene ring. - The resulting hydrogen bonding is stronger than the intermolecular forces in pure benzene and pure trichloromethane. - Therefore the escaping tendency of molecules of the mixture is less that of the pure liquids. Hence reduced vapour pressure and negative deviation from deviation from Raoults law.

Question 33

What are Immiscible liquid mixtures?

Answer 33

These are liquid mixtures in which the intermolecular bonding present in the two liquids is sufficiently different e.g. water (which is a highly hydrogen-bonded compound) and tetrachloromethane (whose molecules are Van der Waal interlinked) form an immiscible liquid mixture.

Question 34

State the Partition law.

Answer 34

It states that “when two immiscible liquids are in contact with each other and a solute soluble in both is added, the solute will distribute itself in such a way that at equilibrium its concentration in the two solvents has a definite ratio at a constant temperature”.

Question 35

What is Partition co-efficient, KD?

Answer 35

This is the constant ratio of the equilibrium concentrations of a solute distributed between two immiscible solvents in contact at a constant temperature.

Question 36

State the conditions under which the distribution law is valid.

Answer 36

- Temperature should kept constant - The two solvents should be immiscible - The solute should be of the same molecular mass in both solvents i.e. the solute should not associate or dissociate in either solvent. - The two solutions in contact should be reasonably dilute.

Question 37

Describe the experiment to determine the partition co-efficient of ammonia between water and trichloromethane.

Answer 37

- To a known volume of aqueous ammonia in a separating funnel an equal volume of trichloromethane is added, the mixture vigorously shaken and left to stand for about twenty minutes for equilibrium between the aqueous and the organic layers to be established. - An equal volume of each layer is pipette and titrated with standard hydrochloric acid using methyl orange indicator. - The volume of the acid required to reach the endpoint in each case is noted. - Ammonia reacts with hydrochloric acid according to the following equation: 𝐻𝐶𝑙(𝑎𝑞) + 𝑁𝐻3(𝑎𝑞) → 𝑁𝐻4𝐶𝑙(𝑎𝑞) - The concentration of ammonia in each layer is calculated from which the KD is determined i.e. - This is the KD of ammonia between water and trichloromethane at the experimental temperature

Question 38

Describe the experiment to determine the partition co-efficient of iodine between trichloromethane and water.

Answer 38

- To a known volume of water in a separating funnel, an equal volume of trichloromethane is added followed by a known small mass of iodine. - The mixture is vigorously shaken to dissolve all the iodine and left to stand for about twenty minutes for the two layers to separate out and equilibrium to be established between the aqueous and the organic layers to be established. - An equal volume of each layer is pipetted and titrated against standard sodium thiosulphate solution using starch indicator. - The volume of sodium thiosulphate required to reach the endpoint in each is noted. - Iodine reacts with sodium thiosuplhate according to the following equation: - The concentration of iodine in each layer is calculated from which the KD is determined as follows: - This is the KD of iodine between trichloromethane and water at the experimental temperature.

Question 39

Describe the experiment to determine the partition co-efficient of ethanoic acid between tetrachloromethane and water.

Answer 39

- A known volume of tetrachloromethane is transferred into a separating funnel followed by an equal volume of water. - To the mixture a known volume of standard ethanoic acid is added, the funnel stoppered and vigorously shaken. - The mixture is allowed to stand for about 30 minutes for the two layers to separate out and the equilibrium of ethanoic acid to be established between the aqueous and the organic layers. - An equal volume of each layer is pipetted and titrated separately with standard aqueous sodium hydroxide using phenolphthalein indicator. - Ethanoic acid reacts with sodium hydroxide according to the equation: 𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝐶𝐻3𝐶𝑂𝑂𝑁𝑎(𝑎𝑞) + 𝐻2𝑂(𝑙) - The volume of sodium hydroxide required to reach endpoint for each layer is recorded and used to determine the molar concentration of ethanoic acid in each layer. - The partition coefficient of ethanoic acid between tetrachloromethane and water at the experimental temperature is calculated according to the following expression: 𝐾𝐷 = [𝐶𝐻3𝐶𝑂𝑂𝐻] 𝑖𝑛 𝑡𝑒𝑡𝑟𝑎𝑐h𝑟𝑜𝑚𝑒𝑡h𝑎𝑛𝑒 [𝐶𝐻3𝐶𝑂𝑂𝐻] 𝑖𝑛 𝑤𝑎𝑡𝑒𝑟

Question 40

What are the applications of the partition law?

Answer 40

- Solvent extraction - Determination of the formula of a complex ion - Chromatography

Question 41

Define Solvent extraction.

Answer 41

This is the separation of a solute from one solvent in which it is less soluble by shaking with an immiscible solvent in which the solute is more soluble e.g. separation of iodine from water by using trichloromethane in which it is more soluble.

Question 42

Describe the experiment for determination the formula of a complex ion formed between copper(II) ions and ammonia molecules.

Answer 42

- A known volume of excess ammonia solution is added to a glass flask containing an equal volume of solution of copper(II) ions, Cu2+ of known concentration and the mixture well stirred until a uniform deep-blue solution containg soluble complex, [Cu(NH3)n]2+. - The deep-blue solution is shaken with trichloromethane and the mixture is allowed to settle so as to reach equilibrium. - A known volume of the organic layer is pipetted into a conical flask, titrated with standard hydrochloric acid using methyl orange indicator. - The volume of hydrochloric acid required to neutralize all the free ammonia in the trichloromethane layer is noted. - The concentration of free ammonia in the trichloromethane layer is calculated using the stoichiometric ratio of the reaction between ammonia and hydrochloric acid and the volume of hydrochloric acid reacted. - A known volume of the deep-blue solution of the aqueous layer is pipetted into a conical, titrated with standard hydrochloric acid using methyl orange indicator and the volume of hydrochloric acid needed to completely neutralize both free and complexed ammonia in the aqueous layer is noted. - The concentration of total ammonia in the aqueous layer is calculated using the stoichiometric ratio of the reaction between ammonia and hydrochloric acid and the volume the acid reacted. - The concertation of free ammonia in the aqueous layer is calculated using the expression: - The concentration of complexed ammonia is calculate using the expression: [NH3] complexed = [NH3] total in aqueous layer − [NH3] free in aqueous - Ammonia complexes with copper(II) ions according to the following equation: 𝐶𝑢2+(𝑎𝑞) + 𝑛NH3(𝑎𝑞) ⇌ [𝐶𝑢(𝑁𝐻3)𝑛]2+(𝑎𝑞) - The value of n is calculated as the ratio of the concentration of complexed ammonia to the initial concentration of copper(II) ions i.e. 𝒏 = [𝑵𝑯𝟑] 𝒄𝒐𝒎𝒑𝒍𝒆𝒙𝒆𝒅 [𝑪𝒖𝟐+ ]

Question 43

Define Steam distillation.

Answer 43

This is a technique of separating a volatile substance which is immiscible with water from its non-volatile impurities by passing steam through the mixture so that the mixture boils at less than 100oC. OR It is a technique of separating a volatile liquid/substance that is immiscible with water from its nonvolatile components at temperature below its boiling point by bubbling steam through the mixture.

Question 44

What are the Principles/Conditions for steam distillation?

Answer 44

- The substance to be steam distilled must be immiscible with water – so that at a given temperature, each component will contribute its own vapour pressure to the total vapour pressure of the mixture. Since boiling takes place when the vapour pressure is equal to the atmospheric pressure, the mixture boils at a temperature lower than that of the two pure liquids. - The substance to be steam distilled should have a high molecular mass - to ensure high vapour pressure of the mixture so that it can boil at a much lower temperature. - The substance should be volatile while the impurities in it are non-volatile – so that the distillate collected contains only that substance and water. - The substance should have high vapour pressure near the boiling point of water – to ensure that the mixture boils below 100oC and much of the substance is collected in the distillate.

Question 45

Describe the process of steam distillation.

Answer 45

- The flask containing the volatile substance immiscible water and it’s non-volatile impurities is gently heated and steam is passed through the mixture - The steam heats up the mixture, causing boiling of both water and the volatile organic compound to occur at a lower temperature, and forming vapor - The hot vapor containing both water, and the volatile compound is condensed by the Liebig condenser, forming an immiscible distillate of water and the volatile organic compound - The distillate is collected in the receiver, and the volatile organic compound immiscible with water, separated from it using a separating funnel. It is then dried using a suitable drying agent.

Question 46

What are the Uses/Applications of steam distillation?

Answer 46

a) Extraction of vegetable oils from plant material. b) Purification/isolation of organic compounds such as essential oils. c) Determination of relative molecular mass of organic compounds.

Question 47

State the advantage of steam distillation.

Answer 47

- The distillation occurs at a temperature lower than the boiling point of the pure substance and pure water. - This helps to avoid thermal decomposition in case the substance being isolated from the mixture decomposes near its boiling point.

Question 48

State the disadvantage of steam distillation.

Answer 48

- Its distillate is an immiscible mixture of the substance being isolated and water which requires another technique such as use of a separate funnel to complete the separation process.

Question 49

Give examples of compounds that can be steam distilled.

Answer 49

- Chlorobenzene, - Bromobenzene, - Phenylamine (Aniline) e.t.c.

Question 50

What is Eutectic point?

Answer 50

The point on a phase diagram showing the constant temperature at which a liquid mixture solidifies without change in composition.

Question 51

What is Eutectic temperature?

Answer 51

This is the constant temperature at which a liquid mixture of two components is in equilibrium with its solid components at a given pressure.

Question 52

Describe the cooling curve for a liquid mixture of composition X.

Answer 52

- When a liquid mixture of composition X is cooled its temperature decreases from T2 without change in phase until temperature T1 when it starts to crystallize forming solid bismuth. - Crystallization of bismuth decreases the freezing point of the remaining solution due to the increased percentage composition of cadmium. - Successive crystallization of bismuth lowers the freezing point of the remaining solution further along line AB until the eutectic point is reached, at point B, when both bismuth and cadmium simultaneously start to crystallize out of solution at constant temperature, T0, without change in composition the solution. - T0 is the Eutectic point/Eutectic temperature. - Further cooling beyond the eutectic point will not cause any change in composition of the solid mixture.

Question 53

State the similarity between a eutectic mixture and a pure compound.

Answer 53

- At a constant pressure, both the Eutectic mixture and pure compound freeze at constant temperature without change in composition.

Question 54

State the differences between a Eutectic mixture and a pure compound.

Answer 54

- It can be separated by physical means unlike a compound - Its X-ray diffraction pattern does not conform to that of a pure compound. - Its composition is affected by change in pressure while that of a compound is not. - Microscopic examination shows that it is a heterogeneous mixture of individual crystals of its components. - When its composition is determined, it does not correspond to any known compound.

Question 55

What conditions necessary for formation of eutectic mixtures?

Answer 55

- Pure crystals of the components should separate from mixture on cooling. - The two substance involved must be completely miscible when melted

Question 56

Give the tests that can be carried out on a eutectic mixture to show that it is not a pure compound.

Answer 56

- Microscopy shows a heterogenous mixture of separate distinct crystals that make up the mixture. - X-ray diffraction pattern does not conform to a pure compound. - Melting point and composition of eutectic mixture change with pressure. - Composition of eutectic does not correspond to any known compound.

Question 57

State the Uses/Applications of Eutectic mixtures.

Answer 57

- Used in the formation of alloys such as solder which melt at a lower temperature than the pure metals. - Used to make safety alliances such as plugs. - Used in the extraction of metals to lower the melting point of the ores e.g. in the extraction of aluminium, cryolite is added to the purified bauxite to lower its melting point.

Question 58

Give two component systems that exhibit the similar behavior as bismuth-cadmium mixture.

Answer 58

(i) Tin and Lead (ii) Gold and Thallium (iii) Benzene and Chlorobenzene (iv) Water and potassium iodide

Question 59

Give examples of liquid mixtures that exhibit negative deviation from Raoult’s law

Answer 59

- Trichloromethane and propanone - Nitric acid and water - Hydrochloric acid and water - Phenol and Phenylamine.

Question 60

Give examples of immiscible liquids

Answer 60

- Ethoxyethane and water - Tetrachloromethane and water