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Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 5th Lesson General Principles of Metallurgy Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 5th Lesson General Principles of Metallurgy

Very Short Answer Questions

Question 1.
What is the role of depressant in froth floatation?
Answer:
By using depressants in froth floatation process, it is possible to separate a mixture of two sulphide ores.
Eg: In the ore containing ZnS and PbS, the depressant used is NaCN. It prevents ZnS from coming to the froth but allows PbS to come with the froth.

Question 2.
Between C and CO, which is a better reducing agent at 673K.
Answer:

• Out of C and CO, Carbon nonoxide (CO) is a better reducting agent at 673K.
• At 983K (or) above coke(C) is better reducing agent.
• The above observations are form Ellingham diagram.

Question 3.
Name the common elements present in the anode mud in the eletrolytic refining of copper.
Answer:

• The common elements present in the anode mud in eletrolytic refining of copper are less reactive valuable metals, silver (Ag), gold (Au) and platinum (Pt) etc….
• These elements donot lose eletrons at anode and collect under the anode as anode mud.

Question 4.
State the role of silica in the metallurgy of copper.
Answer:
The role of silica in the metallurgy of copper is to àcts as an acidic flux. Silica reacts with the impurities of iron and form slag.
FeO (Gangue) + SiO₂ (flux) → FeSiO₃ (Slag)

Question 5.
Explain “poling. [A.P. Mar.16, 15]
Answer:
When the metals are having the metal oxides as impurities this method is employed. The impure metal is melted and is then covered by carbon powder. Then it is stirred with green wood poles. The reducing gases formed from the green wood and the carbon, reduce the oxides to the metal.
Eg: Cu & Sn metals are refined by this method.

Question 6.
Decribe a method for the refining of nickel.
Answer:
Mond’s process:

• In Mond’s process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetra carbonyl.
  
• Nickel tetra carbornyl is strongly heated to decompose and gives the pure Nickel
  

Question 7.
How is cast iron different from pig iron ?
Answer:

• Cast iron is formed by melting pig iron with scrap iron and coke in presence of blast of hot air. It contains carbon content 3%(approximately). It is extremely hard and brittle.
• Pig iron is formed from blast furnace. It contains carbon 4% (approximately). It contains small quantities of impurities S, P, Si, Mu etc….

Question 8.
What is the difference between a mineral and an ore ?
Answer:
Minerals : The naturally occuring chemical compounds of metal in the earth’s crust are called minerals.
Ore : The mineral from which metal can be extracted economically is called ore.

Question 9.
Why copper matte is put in silica lined converter ?
Answer:
Copper matte conains Cu₂ sand FeS. In this mixture FeS is gangue. For removing the gangue, silica present in the lining of the Bessemer’s converter acts as acidic flux and forms slag.
2FeS (gangue) + 3O₂ → 2FeO + 2SO₂
FeO (gangue) + SiO₂ (flux) → FeSiO₃ (Slag)

Question 10.
What is the role of cryolite in the metallurgy of aluminium? [T.S. Mar. 16, 15]
Answer:
By adding the cryolite to the pure Alumina, the melting point of pure Alumina is lowered (which is very high 2324K) and eletrical conductivity of pure alumina is increased.

Question 11.
How is leaching carried out in the case of low grade copper ores?
Answer:
In case of low grade ores of copper, hydrometallurgy technique is used for extraction. Here leaching process can be done by using acids (or) bacteria, The solution containing Cu⁺² is treated with scrap iron (or) H₂.
Cu⁺²_(aq) + H_2(g) → Cu_(S) + 2H⁺_(aq)

Question 12.
Why is zinc not extracted from zinc oxide through reduction using CO?.
Answer:
Zinc is not extracted from zinc oxide through reduction by using CO.
Explanation:
2Zn + O₂ → 2ZnO, ∆G° = -650 IcI
2C0 + O₂ → 2CO, ∆G° = -450 kJ
2ZnO + 2CO → 2Zn, 2CO₂, ∆G° = 200 kJ
∆G° = Positive indicatis that the reaction is not feasible.

Question 13.
Give the composition of the following alloys. [A.P. & T.S. Mar. 19, 17] [Mar. 14]
a) Brass
b) Bronze
c) German Silver
Answa:
a) Composition of Brass : 60 – 80% Cu, 20 – 40% Zn .
b) Composition of Bronze : 75 – 90% Cu, 10 – 25% Sn
c) Composition of German silver : 50 – 60% Cu, 10 – 30% Ni, 20 – 30% Zn.

Question 14.
Explain the terhts gangue and slag.
Answer:
Gangue : The ore is contaminated with the minerals of earth crust and undesired chemical compounds. These are known as gangue (or) matrix.
Slag: The fused material obtained during the refining (or) smelting of metals by combining the flux with gangue is called slag.
Eg : FeO (gangue) + SiO₂ (flux) → FeSiO₃ (Slag)

Question 15.
How is Ag or Au obtained by leaching from the respective ores ?
Answer:
Ag and Au are leached with a dilute solution of NaCN (or) KCN in presence of 02(air). From the leached solution the metal is obtained through displacement by Zinc. .
4M_(s) + 8eN^–_(aq) + 2H₂O_(aq) + O_2(g) → 4[M(CN)₂]^–_(aq) + 4OH^–_(aq)
2[M(CN)₂ ]^–_(aq) + Zn_(s) → [Zn(CN)₄]^2-_(aq) + 2M_(s)

Question 16.
What are the limitations of Ellingham diagram ?
Answer:
Limitations of Ellingham diagram :

• Ellingham diagram is based only on the thermodynamic concepts. It does not explain the kinetics of the reduction process. The graph simply indicates whether a reacton is possible or not but not the kinetics of the reaction.
• The interpretation of ∆G° is depends on K [ ∆G° = -RTl.nK]
  It is presumed that the reactants and products are in equilibrium. .
  M_xO + A_red ⇌ xM + AO_ox
  This is not always true because the reactant or product may be solid.

Question 17.
Write any two ores with formulae of the following metals :
(a) Aluminium
(b) Zinc
(c) Iron
(d) copper
Answer:
a) Ores of Aluminium : Bauxite – Al₂ O₃. 2H₂O
Cryolite – Na₃AlF₆
b) Ores of Zinc : Zinc blende – Zns
Calamine – ZnCO₃
c) Ores of Iron : Haematite – Fe₂O₃
Magnetite – Fe₃O₄
d) Ores of copper : Copper pyrites – CuFeS₂
Copper glance – Cu₂S.

Question 18.
What is matte ? Give its composition.
Answer:
During the extraction of ‘Cu’ from copper pyrites the product of the blast furnace consists mostly of Cu₂S and a little of FeS. This product is known as “Matte”. It is collected from the outlet at the bottom of the fumance.

Question 19.
What is blister copper ? Why is it so called ? [T.S. Mar. 18]
Answer:
During the extraction of ‘Cu’ from copper pyrites when the matte is charged into a Bessemer converter then cuprous oxide combine with cuprous sulphide and forms Cu metal.
2CU₂O + Cu₂S → 6Cu + SO₂
The ‘Cu’ meted is cooled.
The obtained copper metal is impure and is known as “Blister copper” (98% pure).
The solid fied copper has blistered appearance due to evolution of SO₂ Hence it is called blister copper.

Question 20.
Explain magnetic separation of impurities from an ore.
Answer:
Electro-magnetic method : The method is used if the gangue or the ore particles are magnetic in nature. The finely powdered ore is dropped on a belt moving on two strong electromagnetic rollers. The magnetic and the non-magnetic substances form two separate heaps.

For example, let the ore particles be magnetic and then the non-magnetic material will be the gangue. The non-magnetic material forms a heap nearer to the magnetic roller.
Eg : Haematite and magnetite hae magnetic ore particles. Cassiterite or Tin stone has Wolframite as magnetic impurity.

Question 21.
What is flux ? Give an example.
Answer:
Flux : An outside substance added to ore to lower its melting point is known as flux.

• Flux combines with gangue and forms easily fusible slag.
  gangue + flux → slag
  Eg: FeO (gangue) + SiO₂ (flux) → FeSiO₃ (Slag

Question 22.
Give two uses each of the following metals :
(a) Zinc
(b) Copper
(c) Iron
(d) Aluminium
Answer:
a) Uses of Zinc :

• Zinc is used in large quantities in batteries
• Zinc is used for galvanising iron.
• Zinc is used in manufacturing alloys.
  Eg: Brass, German silver,

b) Uses of copper :

• Copper is used in manufacturing of water pipes and steam pipes.
• Copper is used in making wires used in electrical industry.
• Copper is used in manufacturing of alloys.
  Eg : Brass, Bronze. .

c) Uses of Iron:

• Cast iron is used in casting stoves, railway siéepers- gutter pipers, toys etc.
• It is used in manufacturing of wrought iron and steel.
• Wrought iron is used in making anchors, wires, bolts, chains etc.

d) Uses of Aluminium :

• Aluminium foils are used as wrappers for chacolates.
• Aluminium is fine dust used in paints and lacquers.
• Aluminium is used in photo frames.
• Aluminium is used in extraction of Cr and Mn from their oxides.

Question 23.
Between C and CO, which is a better reducing agent for ZnO ?.
Answer:
Case – I : [Coke as reducing agent]
ZnO + C → Zn + CO, ∆G° become lesser as the T is more then 1120K
Case – II : [CO as reducing agent]
ZnO + CO₂ → Zn + CO, ∆G° becomes lesser when the T is more then 1323K.

• The value of ∆G° is negative for a reaction to occur.
• In the equation (1) ∆G° becomes negative at low temperature- so equation (1) is feasible i. r. C is a better reducing agent for ZnO.

Question 24.
Give the uses of
a) Cast iron
b) Wrought iron
c) Nickel steel
d) Stainless steel
Answer:
a) Uses of Cast iron :

• Cast iron is used for casting stoves, railway sleepers, gutter pipes, toys etc.
• It is used in manufacturing of wrought iron and steel.

b) Uses of wrought iron :

• Wrought iron is used in making anchors, wires.
• Wrought iron is used in making chains and agricultural implements.

c) Uses of Nickel steel:

• Nickel steel is used for making cables, automobiles and aeroplane parts.
• Nickel steel is used for making pendulum, measuring tapes, chrome steel for cutting tools and crashing machines.

d) Uses of stainless steel:

• Stainless steel is used in manufacturing of cycles, automobiles.
• Stainless steel is used in manufacturing of utensils, pens etc.

Question 25.
How is aluminium useful in the extraction of chromium and manganese from their oxides ?
Answer:

• ’AT’ is used as reducing agent.
• 4 By Alumino thermite process Cr, Mn are extracted from their oxides.
• The reactions are highly exothermic.
  Cr₂O₃ + 2Al → 2Cr + Al₂O₃
  3Mn₃O₄ + 8Al → 4Al₂O₃ + 9Mn

Short Answer Questions

Question 1.
Copper can be extracted by hydrometallurgy but not zinc – explain.
Answer:
Copper can be extracted by hydromefallurgy but not zinc.
Explanation:

• E⁰ Value of Zn⁺²/Zn = -0.762V is less than that of E⁰ value of Cu⁺²/Cu = 0.337V .
• From the above data zinc is a stronger reducing agent and can easily displace the Cu⁺² ions present in the complex.
  [Cu(CN)₂] + Zn (Soluble complex → [Zn(CN)₂] + Cu (Precipitate)
• Zinc can be isolated by hydrometallurgy only when stronger reducing agents than zinc are present.
• So zinc cannot be extracted by hydrometallurgy.

Question 2.
Why is the extraction of copper form pyrites more difficult then that from its oxide ore through reduction?
Answer:
The extraction of copper from pyrites is more difficult than that from its oxide ore through reduction.
Explanation:
Pyrites (Cu₂S) Cannot be reduced by carbon(or) hydrogent because the standard free energy of formation (∆G°) of Cu₂S is greater then those of CS₂ and H₂S.

The ∆G° of copper oxide is less than that of CO₂.
∴ The sulphide ore is first converted to oxide by roasting and then reduced.

Question 3.
Explain zone refining.
Answer:
Zone refining:

• Zone refining is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
• A circular mobile heater is fixed at one end of a rod of impure metal.
• The molten zone moves along with the heater moves forward the pure metal crystallises out of the melt and the impurities pass into the adjacent molten zone.
• The above process is repeated several times and the heater is moved in the same direction form one end to the other end. At one end impurities get concentrated. This end is cut off.
• This method is very useful for producing semiconductor grade metals of very high purity. Eg : Ge, Si, B, Ga etc…
  

Question 4.
Write down the chemical reactions taking place in the extraction of zinc from zinc blende.
Answer:
In the extraction of zinc from zinc blende the following chemical reactions taking place.
i) Roasting : The process of strong heating the zinc blende in presence of air is roasting.

ii) Reduction : Zinc oxide obtained by roasting undergo reduction with coke.

Question 5.
Write down the chemical reactions taking place in different zones in the blast furnace during the extraction of iron.
Answer:
The chemical reactions taking place in different zones in the blast furnace during the extraction of iron are

Question 6.
How is alumina separated from silica in the bauxite ore associated with silica ? Give equations.
Answer:
Serpeck’s process is used for the purification of white Bauxite i.e., Bauxite containing silica as impurity.
In this process bauxite is mixed with coke and is then heated to 2075K in a current of nitrogen. Silica is reduced to silicon and escapes as a vapour.

Aluminium nitride formed in the above step is hydrolysed to form aluminium hydroxide.
AlN + 3H₂O → AZ(OH)₃ ↓ + NH₃ ↑
Al(OH)₃ PPt is washed, dried and then ignited to from purified bauxite.

Question 7.
Giving examples to differentiate roasting and calcination. [T.S. Mar. 19, 18, 16, 15; A.P. Mar. 16, 15]
Answer:
Roasting: Removal of the volatile components of a mineral by heating mineral either alone (or) mixed with some other substances to a high temperature in the presence of air is called Roasting.

• It is applied to the sulphide ores.
• SO₂ gas is producted along with metal oxide.
  Eg : 2ZnS + 3O₂ 2ZnO + 2SO₂ ↑

Calcination: Removal of the volatile components of a mineral by heating in the absence of air is called calcination.

•  It is applied to carbonates and bicarbonates.
• CO₂ gas is produced along with metal oxide.
  Eg: CaCO₃ CaO + CO₂
  ZnCO₃ ZnO + CO₂

Question 8.
The Value of ∆G° for the formation of Cr₂O₃ is – 540kJ mol^-1 and that of Al₂O₃ is – 827kJ mol^-1. Is the reduction of Cr₂O₃ Possible with Al?
Answer:
From the given data the following are the thermo chemical equations
\(\frac{4}{3}\) Cr_(s) + O_2(g) → \(\frac{2}{3}\) Cr₂O_2(s), ∆G° – 540 KJ ……………….. (1)
\(\frac{4}{3}\) Al_(s) + O_2(g) → \(\frac{2}{3}\) Al₂O_3(s), ∆G° – 827 KJ ……………….. (2)
Equation (1) – Equation (2)
\(\frac{2}{3}\) Cr₂O_3(s) + \(\frac{4}{3}\) Al_(s) → \(\frac{2}{3}\) Al₂O_3(s) + \(\frac{4}{3}\) Cr_(s), ∆G° – 287 KJ
∆G° = -ve, So the reaction is feasible
Hence the reduction of Cr₂O₃ is possible with ‘Al’.

Question 9.
What is the role of graphite rod in the electrometallurgy of aluminium ? [A.P. Mar. 16]
Answer:

• In the electrolytic reduction of alumina b_y Hall – Heroult process graphite rod acts as anode.
• At anode, O₂ gas is liberated which reacts with the carbon of anode to form CO₂ gas. So these graphite rods are consumed slowly and need to be replaced from time to time.
  Al₂O₃ ⇌ 2Al⁺³ + 3O^-2
  Cathode : Al⁺³ + 3e^– → Al
  Anode : 3O^-2 → 3[O] + 6e^–
  C_(s) + O_2(g) → CO_2(g)

Question 10.
Outline the principles of refining of metals by the following methods.
(a) Zone refining
(b) Electrolytic refining
(c) Poling
(d) Vapour phase refining.
Answer:
a) Zone refining :

• Zone refining is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal. .
• A circular mobile heater is fixed at one end of a rod of impure metal.
• The molten zone moves along with the heater moves forward the pure metal crystallises out of the melt and the impurities pass into the adjacent molten zone.
• The above process is repeated several times and the heater is moved in the same direction form one end to the other end. At one end impurities get concentrated. This end is cut off.
• This method is very useful for producing semiconductor grade metals of very high purity.
  Eg : Ge, Si, B, Ga etc..
  

b) Electrolytic refining : This process is used for less reactive metals like Cu, Ag, AZ, Au etc.

• In this process anode is made by impure metal and a thin strip of pure metal acts as cathode. .
• On electrolysis metal dissolves from anode and pure metal gets deposited at cathode.
  M → Mⁿ⁺ + ne^–
  Impure
  Mⁿ⁺ + ne^– → M (Cathode) Pure metal
  Impurities settle down below anode in the form of anode mud.

c) Poling : When the metals are having the metal oxides as impurities this method is employed. The impure metal is melted and is then covered by carbon powder. Then it is stirred with green wood poles. The reducing gases formed from the green wood and the carbon, reduce the oxides to the metal.
Eg : Cu & Sh metals are refined by this method.

d) Vapour phase refining: In this method the metal is converted into its volatile compound and collected. It is then decomposed to give pure metal. .

1. The metal should form a volatile compound with an available reagent.
2. The volatile compound should be easily decomposable. So the the recovery is easy.
   E.g : Mond’s process :
3. In Mond’s process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetra carbonyl.
   
4. Nickel tetra carbomyl is strongly heated to decompose and gives the pure Nickel
   

Question 11.
Predict the conditions under which Al might be expected to reduce MgO.
Answer:
The Equations for the formation of two oxides are
\(\frac{4}{3}\)Al_(S) + O_2(g) → 2Mg_(S) + \(\frac{2}{3}\) O_2(g) \(\frac{4}{3}\) 2MgO_(S)
From the Ellingham diagram the two curves of these oxides formation intersect each otherat a certain point. The corresponding value of ∆G° becomes zero for the reduction of MgO by aluminium metal,
2MgO_(S) + \(\frac{4}{3}\) Al_(S) ⇌ 2Mg_(S) + \(\frac{2}{3}\) Al₂O_3(S)

• From the above information the reduction of MgO by Al metal cannot occur below this temperature (1665 K) instead, Mg can reduce  Al₂O₃ to Al below 1665 K.
• Al-Metal can reduce MgO to Mg above 1665K because ∆G° for Al₂O₃ is less compared to that of MgO.
  

Question 12.
Explain the purification of sulphide ore by froth floatation method. [A.P. Mar.’19, 17, 15; T.S. Mar. 15]
Answer:
Froth floatation method :
This method is used to concentrate sulphide ores.

• In this process a suspension of the powdered ore is made with water.
• A rotating paddle is used to agitate the suspension and air is blown into the suspension in presence of an oil.
• Froth is formed as a result of blown of air, which carries the mineral particles.
• To the above slurry froth collectors and stabilizers are added.
• Collectors like pine oil enhance non-wettability of the mineral particles.
• Froth stabilizers like cresol stabilize the froth.
• The mineral particles wet by oil and gangue particles wet by water.
• The broth is light and is skimmed off. The ore particles are then obtained from the froth. By using depressants in froth floatation process, it is possible to separate a mixture of two suplhide ores.
  Eg : In the ore containing ZnS and PbS, the depressant used is NaCN. It prevents ZnS from coming to the froth but allows PbS to come with the froth.

Question 13.
Explain the process of leaching of alumina from bauxite. [T.S. Mar. 17]
Answer:
The principal ore of aluminium, bauxite, usually contains SiO₂, iron oxides and titanium oxide (TiO₂) as impurities, concentration is carried out by digesting the powdered ore with a concentrated solution of NaOH at 473 – 523 K and 35 – 36 bar pressure. The way. Al₂O₃ is leached out as sodium aluminate (and SiO₂ too as sodium silicate) leaving the other impurities behind :
Al₂O₃ (S) + 2NaOH(aq) + 3H₂O(l) → 2 Na[Al(OH)₄](aq)
The aluminate is alkaline in nature and is neutralised by passing CO₂ gas and hydrated Al₂O₃ is precipitated. At this stage, the solution is seeded with freshly prepared samples of hydrated Al₂O₃ which induces the precipitation of Al₂O₃ xH₂O
2 Na[Al(OH)₄](aq) + CO₂(g) → Al₂O₃ xH₂O(s) + 2NaHCO₃(aq)
The sodium silicate remains in the solution and the insoluble hydrated alumina is filtered, dried and heated to give pure Al₂O₃:

Question 14.
What is Ellingham diagram ? What information can be known from this in the reduction of oxides ?
Answer:
The graphical representation of Gibbs energy which provides a sound basis for considering the choice of reducing agent in the reduction of oxides. This graphical representation is known as Ellingham diagram.

• This diagram helps us in predicting the feasibility of thermal reduction of an ore.
• The Criterion of feasibility of a reaction is that at a given temperature, Gibbs energy of the reaction must be negative.
•  Ellingham diagram normally consists plots of ∆G° vs T for formation of oxides of elements.
• The graph indicates whether a reaction is possible or not, i.e., the tendency of reduction with a reducing agent is indicated with a reducing agent is indicated.
• The reducing agent forms its oxide when the metal oxide is reduced. The role of reducing agent is to make the sum of ∆G° values of the two reactions negative.
• Out of C and CO, Carbon monoxide (CO) is a better reducting agent at 673K.
• At 983K (or) above coke(C) is better reducing agent.
• The above observations are form Ellingham diagram.
  Zinc is not extracted from zinc oxide through reduction by using CO.

Explanation :

• 2Zn + O₂ → 2ZnO, ∆G° =-650 kJ
• 2CO + O₂ → 2CO₂, ∆G° = -450 kJ
• 2ZnO + 2CO → 2Zn, 2CO₂, ∆G° = 200 kJ
  ∆G° = Positive indicatis that the reaction is not feasible.
  The above fact is explained on the basis of Ellingham diagram.
  

Question 15.
How is copper extracted from copper pyrites?
Answer:
Extraction of copper from copper pyrites :
Copper pyrite is the main source of copper metal. Various steps involved in the extraction of copper discussed below.
Step – I:
Concentration of ore by froth floatation process :
The ore is first crushed in ball mills. The finely divided ore is suspended in water. A little pine oil is added and the mixture is vigorously agitated by a current of air. The froth formed carries the ore particles almost completely. The gangue sinks to the bottom of the tank. The froth is separated and about 95% concentrated ore is obtained.

Step -II:
Roasting : To remove the volatile impurities like As (or) Sb, the ore is roasted in a free supply of air. A mixture of sulphides of copper and iron are obtained and these are partially oxidised to respective oxides.
Cu₂S. Fe₂S₃ + O₂ → Cu2S + 2FeS + SO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2FeS + 3O₂ → 2FeO + 2SO₂

Step -III:
The roasted ore is mixed with a little coke and sand (Silica) and smelted in a blast furnace and fused. A blast of air, necessary for the combustion of coke, is blown through the tuyeres present at the base of the furnace. The oxidation of the sulphides of copper and iron will be completed further. A slag of iron silicate is formed according to the reactions given below :
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → Fe SiO₃ Ferrous Silicate (Slag)
Cu₂O + FeS → Cu₂S + FeO

Step -IV : After smelting the copper ore in blast furnace, the product of the blast furnace consists mostly of Cu₂S and a little of ferrous sulphide. This product is known as “Matte.” It is collected from the outlet at the bottom of the furnace. After then the following processes are carried out for getting the pure copper.

Bessemerization : The matte is charged into a Bessemer converter. A bessemer converter is a pear-shaped furnace. It is made of steel plates. The furnace is given a basic lining with lime or magnesium oxide (obtained from dolomite or magnesite). The converter is held in position by trunnions and can be tilted in any position. A hot blast of air and sand is blown through the tuyeres present near the bottom. Molten metal, the product in the furnace, collects at the bottom of the converter.

Reactions that took place in blast furnace go to completion. Almost all of iron is eliminated as a slag. Cuprous oxide combines with cuprous sulphide and forms Cu metal.
2Cu₂O + Cu₂S → 6Cu + SO₂
The molten metal is cooled in sand moulds. S02 escapes. The impure copper metal is known as “Blister copper” and is about 98% pure.

Step -V:
Refining : The Blister copper is purified by electrolysis. The impure copper metal is made into plates. They are suspended into lead – lined tanks containing Copper (II) Sulphate solution. Thin plates of pure copper serve as cathode. The cathode plates are coated with graphite. On electrolysis, pure copper is deposited at the cathode. The copper obtained is almost 100% pure Cu.

Question 16.
Explain the extraction of Zinc form Zinc blende.
Answer:
“Zinc blende (ZnS)” is an important source for ‘Zn’ metal.
Extraction of Zinc : Zinc blende ore is treated in the following stages.
i) Crushing : The ore is crushed to a fine powder in ball mills.

ii) Concentration of the ore : The ore is concentrated first by gravity process. The crushed ore is washed with a stream of water on a Wilfley’s table. The tables have a corrugated top and are under a rocking motion. Due to this motion, the lighter gangue particles are washed away by the steam. The heavier ore particles settle to the bottom of the table.

The partially concentrated ore is further concentrated by froth floatation process. The ore particles go with the froth.

The concentrated ore is subjected further to electromagnetic separation, if iron oxide is present in the gangue. Iron oxide is magnetic and so forms a heap nearer to the magnetic pole.

iii) Roasting: The concentrated ore thus obtained is roasted in rotary shelf burner which is provided with horizontal shelves and raking arms. The ore is added at the top and zinc oxide is collected from the bottom. The following reactions take place.
2ZnS + 3O₂ → 2ZnO + 2SO₂
ZnS + 2O₂ → ZnSO₄
2ZnSO₄ → 2ZnO + 2SO₂ + O₂.
iv) Reduction : Three methods are in practice for the reduction of the oxide to the metal., The most commonly used method is the Belgian process. In this process the roasted ore is mixed with coal or coke intimately and is taken in fireclay or earthem ware retorts. The retorts are made of fireclay which are bottle shaped tubes. These are closed at one end. The other end is connected to air cooled earthem ware condenser. A large number of retorts are placed in tiers in a large furnace and heated in 1100°C by burning the gas. ‘Prolongs’ made of sheet iron are attached to the condensers. The metal condenses in these earthem ware condensers and the prolongs. The metal powder collected is mixed with some zinc oxide and is known as “zinc dust”. Some of the zinc metal is collected in the fused state. This is solidified in moulds. This metal is called “zinc spelter”.
ZnO + C → Zn + CO
ZnO + CO → Zn + CO₂
Electrolytic refining : Very pure zinc is obtained by electrolysis. The electrolyte is zinc sulphate solution containing a little H₂SO₄. Impure zinc is made the anode and the pure zinc plates are made the cathode.

Question 17.
Explain smelting process in the extraction of copper.
Answer:
The roasted ore is mixed with a little coke and sand (Silica) and smelted in a blast furnace and fused. A blast of air, necessary for the combustion of coke, is blown through the tuyeres present at the base of the furnace. The oxidation of the sulphides of copper and iron will be completed further. A slag of iron silicate is formed according to the reactions given below :
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → Fe SiO₃
Ferrous Silicate (Slag)
Cu₂O + FeS → Cu₂S + FeO

Question 18.
Explain electrometallurgy with an example.
Answer:
Electro metallurgy : Metallurgy involving the use of electric arc furnaces, electrolysis and other electrical operations is called electrometallurgy.
In the reduction of a molten metal salt, electrolysis is used. These methods are based on electro chemical principles.
These are expained by the equation
∆G° = -n FE°
n = no. of electrons
E° = Electrode potential.
Electrolytic Reduction of Alumina : Pure Alumina (Al₂O₃) is a bad conductor of electricity and it has high melting point (2050°C). So it cannot be electrolysed. Alumina is electrolysed by dissolving in fused cryolite to increase the conductivity and small amount of Fluorspar is added to reduce its melting point. Thus the electrolyte is a fused mixture of Alumina, Cryolite and Fluorspar.
Electrolysis is carried out in an iron tank lined inside with graphite (carbon) which functions as cathode. A number of carbon rods (or) copper rods suspended in the electrolyte functions as Anode.

An electric current of 100 amperes at 6 to 7 volts is passed through the electrolyte. Heat produced by the current keeps the mass in fused state at 1175 to 1225K. The following reactions take place in the electrolytic cell under these conditions.
Na₃AlF₆ → 3NaF + AlF₃
Cryolite
4AlF₃ → 4Al³⁺ + 12F^–
At cathode : 4Al³⁺ + 12e^– → 4Al
At anode : 12F^– → 6F₂ + 12e^–
F₂ formed at the anode reacts with alumina and forms Aluminium fluoride.
2Al₂O₃ + 6F₂ → 4AlF₃ + 3O₂
Aluminium, produced at the cathode, sinks to the bottom of the cell. It is removed from time to time through topping hole.

Question 19.
Explain briefly the extraction of aluminium from bauxite.
Answer:
Extraction of Aluminium from Bauxite:
For the purpose of extraction of Al, Bauxite is the best source.
purification of Bauxite : Bauxite containing Fe₂O₃ as impurity is known as red Bauxite.
Bauxite containing SiO₂ as impurity is known as white Bauxite and can be purified by “Serpeck’s Process”. Red Bauxite is purified by Bayer’s process and Hall’s process.
Bayer’s Process : Red bauxite, is roasted and digested in concentrated NaOH at 423 K. Bauxite dissolves in NaOH to form sodium meta aluminate while impurity Fe₂O₃ does not dissolve which can be removed by filtration.
Al₂O₃.2H₂O + 2NaOH → 2NaAlO₂ + 3H₂O
The solution which contains sodium meta aluminate is diluted and crystals of AZ(OH)3, are added which serves as seeding orgent. Sodium meta aluminate undergoes hydrolysis to precipitate Al(OH)₃.
2NaAlO₂ + 4H₂O → 2NaOH + 2Al(OH)₃ ↓
Al(OH)₃ is filtered and ignited at 1200°C to get anhydrous alumina.
2Al(OH)₃ Al₂O₃ + 3H₂O
Halls’ Process: Red Bauxite is fused with sodium carbonate to form sodium meta aluminate which is extracted with water. The impurity Fe₂O₃ is filtered out.
Al₂O₃ + Na₂CO₃ → 2NaAlO₂ + CO₂
Into the solution of sodium meta aluminate, CO₂ gas is passed to precipitate Al(OH)₃.

The precipitated Al(OH)₃ is ignited at 1200°C to get anhydrous alumina.
2Al(OH)₃ → Al₂O₃ + 3H₂O
Serpeck’s Process: Powdered bauxite is mixed with coke and heated to 2075 K in a current of nitrogen gas. Aluminium Nitride is formed while Si02 is reduced to SiO₂ which escapes out.

Aluminium nitride is hydrolysed to get aluminium hydroxide which on ignition gives anhydrous alumina.
AlN + 3H₂O → Al(OH)₃ ↓ + NH₃ ↑
2Al(OH)₃ 3 Al₂O₃ + 3H₂O

Electrolytic Reduction of Alumina : Pure Alumina (Al₂O₃) is a bad conductor of electricity and it has high melting point (2050°C). So it cannot be electrolysed. Alumina is electrolysed by dissolving in fused cryolite to increase the conductivity and small amount of Fluorspar is added to reduce its melt¬ing point. Thus the electrolyte is a fused mixture of Alumina, Cryolite and Fluorspar.

Electrolysis is carried out in an iron tank lined inside with graphite (carbon) which functions as cathode. A number of carbon rods (or) copper rods suspended in the electrolyte functions as Anode.
Na₃AlF₆ → 3NaF + AlF₃
Cryolite
4AlF₃ → 4Al³⁺ + 12F^–
At cathode : 4Al³⁺ + 12e^– → 4Al
At anode : 12F^– → 6F₂ + 12e^–
F₂ formed at the anode reacts with alumina and forms Aluminium fluoride.
2Al₂O₃ + 6F₂ → 4AlF₃ + 3O₂
Aluminium, produced at the cathode, sinks to the bottom of the cell. It is removed from time to time through topping hole.

Purification of Aluminium : (Hoope’s Process)
The impurities present are Si, Cu, Mn etc.,
The electrolytic cell used for refining of aluminium consists of iron tank lined inside with carbon. This acts as anode. The tank contains three layers of fused masses. The bottom layer contains impure aluminium. Middle layer contains mixture of AlF₃, NaF and BaF₂ saturated with Al₂O₃. Top layer contains pure aluminium and graphite rods kept in it act as cathode.

On passing current aluminium ions from the middle layer are discharged at the cathode as pure aluminium. Equivalent amount of aluminium from the bottom layer passes into middle layer.

Long Answer Questions

Question 1.
The choice of a reducing agent in a particular case depends on thermo-dynamic factor, •lain with two examples.
Answer:
The choice of a reducing agent in a particular case depends on thermodynamic factor. This fact can be explained by considering the following examples.

• Out of C and CO, Carbon nonoxide (CO) is a better reducting agent at 673K.
• At 983K (or) above coke(C) is better reducing agent.
• The above observations are form Ellingham diagram.
• Zinc is not extracted from zinc oxide through reduction using CO.

Explanation :
2Zn + O₂ → 2ZnO, ∆G° = -650 kJ
2CO + O₂ → 2CO₂, ∆G° = -450 kJ
2ZnO + 2CO → 2Zn + 2CO₂, ∆G° = 200 kJ
∆G° = Positive indicatis that the reaction is not feasible.
The extraction of copper from pyrites is more difficult than that from its oxide ore through reduction. ‘ .
Explanation : Pyrites (Cu₂S) Cannot be reduced by carbon (or) hydrogent because the standard free energy of formation (∆G°) of Cu₂S is greater then those of CS₂ and H₂S.

The ∆G° of copper oxide is less than that of CO₂.
∴ The sulphide ore is first converted to oxide by roasting and then reduced. Roasting

The following are the conditions that Al might be expected to reduce MgO. The Equations for the formation of two oxides are
\(\frac{4}{3}\)Al_(S) + O_2(g) → \(\frac{2}{3}\) Al₂O_3(S)
2Mg_(S) + O_2(g) → 2MgO_(S)
From the Ellingham diagram the two curves of these oxides formation intersect each other at a certain point. The corresponding value of ∆G° becomes zero for the reduction of MgO by aluminium metal,
2MgO_(S) + \(\frac{4}{3}\)Al_(S) → 2Mg_(S) + \(\frac{2}{3}\) Al₂O_3(S)

• From the above information the reduction of MgO by Al metal cannot occur below this temperature (1665 K) instead, Mg can reduce Al₂O₃ to Al below 1665 K.
• Al-Metal can reduce MgO to Mg above 1665 K because ∆G° for Al₂O₃ is less compared to that of MgO.
  

Question 2.
Discuss the extraction of Zinc from Zinc blend
Answer:
“Zinc blende (ZnS)” is an important source for ‘Zn’metal.
Extraction of’Zinc : Zinc blende ore is treated in the following states.
i) Crushing : The ore is crushed to a fine powder in ball mills.
ii) Concentration of the ore : The ore is concentrated first by gravity process. The crushed ore is washed with a stream of water on a Wilfley’s table. The tables have a corrugated top and are under a rocking motion. Due to this motion, the lighter gangue particles are washed away by the steam. The heavier ore particles settle to the bottom of the table.

The partially concentrated ore is further concentrated by froth floatation process. The ore particles go with the froth.

The concentrated ore is subjected further to electromagnetic separatioh, if iron oxide is present in the gangue. Iron oxide is magnetic and so forms a heap nearer to the magnetic pole.

iii) Roasting: The concentrated ore thus obtained is roasted in rotary shelf burner which is provided with horizontal shelves and raking arms. The ore is added at the top and zinc oxide is collected from the bottom. The following reactions take place.
2ZnS + 3O₂ → 2ZnO + 2SO₂
ZnS + 2O₂ → ZnSO₄
2ZnSO₄ → 2ZnO + 2SO₂ + O₂
iv) Reduction : Three methods are in practice for the reduction of the oxide to the metal. The most commonly used method is the Belgian process. In this process the roasted ore is mixed with coal or coke intimately and is taken in fireclay or earthern ware retorts. The retorts are made of fireclay which are bottle shaped tubes. These are closed at one end. The other end is connected to air cooled earthern ware condenser. A large number of retorts are placed in tiers in a large furnace and heated in 1I00°C by burning the gas. ‘Prolongs’ made of sheet iron are attached to the condensers. The metal condenses in these earthern ware condensers and the prolongs. The metal powder collected is mixed with some zinc oxide and is known as “zinc dust”. Some of the zinc metal is collected in the fused state. This is solidified in moulds. This metal is called “zinc spelter”.
ZnO + C → Zn + CO
ZnO + CO → Zn + CO₂
Electrolytic refining : Very pure zinc is obtained by electrolysis. The electrolyte is zinc sulphate solution containing a little H₂SO₄. Impure zinc is made the anode and the pure zinc plates are made the cathode.

Question 3.
Explain the reactions occuring in the blast furnace in the extraction of iron.
Answer:
In the Blast furnace, reduction of iron oxides takes place in different temperature ranges. Hot air is blown from the bottom of the furnace and coke is burnt to give temperature upto about 2200K in the lower portion itself. The burning of coke therefore supplies most of the heat required in the process. The CO and heat moves to upper part of the furnace. In upper part, the temperature is 1o%er and the iron oxides (Fe₂O₃ and Fe₃O₄) coming from the top are reduced in steps to FeO. Thus, he reduction reactions taking place in the lower temperature range and in the higher temperature range, depend on the points of corresponding intersections in the ∆_rG° vs T plots.

These reactions can be summarised as follows:
At 500 – 800 K (lower temperature range in the blast furnace)
3 Fe₂O₃ + CO → 2 Fe₃O₄ + CO₂
Fe₃O₄ + 4 CO → 3 Fe + 4 CO₂
Fe₂O₃ + CO → 2 FeO + CO₂
At 900 – 1500 K (higher temperature range in the blast furnace)
C + CO₂ → 2 CO
FeO + CO → Fe + CO₂
Lime stone is also decomposed to CaO which removes silicate impurity of the ore as CaSiO₃ slag. The slag is in molten state and separates out from iron.

The iron obtained from Blast furnace contains about 4% carbon and many impurities in smaller amount (e.g., S, P, Si, Mn). This is known as pig iron. Cast iron is different from and is made by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about 3%) and is extremely hard and brittle.
Fe₂O₃ 3 C → 2 Fe + 3 CO

Question 4.
Discuss the extraction of copper from copper pyrites.
Answer:
Extraction of copper from copper pyrites:
Copper pyrite is the main source of copper metal. Various steps involved in the extraction of copper discussed below.
Step – I:
Concentration of ore by froth floatation process:
The ore is first crushed in ball mills. The finely divided ore is suspended in water. A little pine oil is added and the mixture is vigorously agitated by a current of air. The froth formed carries the ore particles almost completely. The gangue sinks to the bottom of the tank. The froth is separated and about 95% concentrated ore is obtained.

Step – II: Roasting :
To remove the volatile impurities like As (or) Sb, the ore is roasted in a free supply of air. A mixture of sulphides of copper and iron are obtained and these are partially oxidised to respective oxides.
Cu₂S. Fe₂S₃ + O₂ → Cu₂S + 2FeS + SO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2FeS + 3O₂ → 2FeO + 2SO₂

Step – III :
The roasted ore is mixed with a little coke and sand (Silica) and smelted in a blast furnace and fused. A blast of air, necessary for the combustion of coke, is blown through the tuyeres present at the base of the furnace. The oxidation of the sulphides of copper and iron will be completed further. A slag of iron silicate is formed according to the reactions given below :
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → Fe SiO₃
Ferrous Silicate (Slag)
Cu₂O + FeS → Cu₂S + FeO
Step -IV:
After smelting the copper ore in blast furnace, the product of the blast furnace consists mostly of Cu2S and a little of ferrous sulphide. This product is known as “Matte.” It is collected from the outlet at the bottom of the furnace . After then the following processes are carried out for getting the pure copper.

1. Bessemerization: The matte is charged into a Bessemer converter. A bessemer converter is a pear-shaped furnace. It is made of steel plates. The furnace is given a basic lining with lime or magnesium oxide (obtained from dolomite or magnesite). The converter is held in position by trunnions and can be tilted in any position. A hot blast of air and sand is blown through the tuyeres present near the bottom. Molten metal, the product in the furnace, collects at the bottom of the converter.

Reactions that took place in blast furnace go to completion. Almost all of iron is eliminated as a slag. Cuprous oxide combines with cuprous sulphide and forms Cu metal.
2Cu₂O + Cu₂S → 6Cu + SO₂

The molten metal is cooled in sand moulds. S02 escapes. The impure copper metal is known as “Blister copper” and is about 98% pure.

Step -V:
Refining : The Blister copper is purified by electrolysis. The impure copper metal is made into plates. They are suspended into lead – lined tanks containing Copper (II) Sulphate solution. Thin plates of pure copper serve as cathode. The cathode plates are coated with graphite. On electrolysis, pure copper is deposited at the cathode. The copper obtained is almost 100% pure Cu.

Question 5.
Explain the various steps involved in the extraction of almninium from bauxite.
Answer:
Extraction of Aluminium from Bauxite:
For the purpose of extraction of Al, Bauxite is the best source.
purification of Bauxite : Bauxite containing Fe₂O₃ as impurity is known as red Bauxite.
Bauxite containing SiO₂ as impurity is known as white Bauxite and can be purified by “Serpeck’s Process”. Red Bauxite is purified by Bayer’s process and Hall’s process.
Bayer’s Process : Red bauxite, is roasted and digested in concentrated NaOH at 423 K. Bauxite dissolves in NaOH to form sodium meta aluminate while impurity Fe₂O₃ does not dissolve which can be removed by filtration.
Al₂O₃.2H₂O + 2NaOH → 2NaAlO₂ + 3H₂O
The solution which contains sodium meta aluminate is diluted and crystals of AZ(OH)3, are added which serves as seeding orgent. Sodium meta aluminate undergoes hydrolysis to precipitate Al(OH)₃.
2NaAlO₂ + 4H₂O → 2NaOH + 2Al(OH)₃ ↓
Al(OH)₃ is filtered and ignited at 1200°C to get anhydrous alumina.
2Al(OH)₃ Al₂O₃ + 3H₂O
Halls’ Process: Red Bauxite is fused with sodium carbonate to form sodium meta aluminate which is extracted with water. The impurity Fe₂O₃ is filtered out.
Al₂O₃ + Na₂CO₃ → 2NaAlO₂ + CO₂
Into the solution of sodium meta aluminate, CO₂ gas is passed to precipitate Al(OH)₃.

The precipitated Al(OH)₃ is ignited at 1200°C to get anhydrous alumina.
2Al(OH)₃ → Al₂O₃ + 3H₂O
Serpeck’s Process: Powdered bauxite is mixed with coke and heated to 2075 K in a current of nitrogen gas. Aluminium Nitride is formed while Si02 is reduced to SiO₂ which escapes out.

Aluminium nitride is hydrolysed to get aluminium hydroxide which on ignition gives anhydrous alumina.
AlN + 3H₂O → Al(OH)₃ ↓ + NH₃ ↑
2Al(OH)₃ 3 Al₂O₃ + 3H₂O

Electrolytic Reduction of Alumina : Pure Alumina (Al₂O₃) is a bad conductor of electricity and it has high melting point (2050°C). So it cannot be electrolysed. Alumina is electrolysed by dissolving in fused cryolite to increase the conductivity and small amount of Fluorspar is added to reduce its melt¬ing point. Thus the electrolyte is a fused mixture of Alumina, Cryolite and Fluorspar.

Electrolysis is carried out in an iron tank lined inside with graphite (carbon) which functions as cathode. A number of carbon rods (or) copper rods suspended in the electrolyte functions as Anode.
Na₃AlF₆ → 3NaF + AlF₃
Cryolite
4AlF₃ → 4Al³⁺ + 12F^–
At cathode : 4Al³⁺ + 12e^– → 4Al
At anode : 12F^– → 6F₂ + 12e^–
F₂ formed at the anode reacts with alumina and forms Aluminium fluoride.
2Al₂O₃ + 6F₂ → 4AlF₃ + 3O₂
Aluminium, produced at the cathode, sinks to the bottom of the cell. It is removed from time to time through topping hole.

Purification of Aluminium : (Hoope’s Process)
The impurities present are Si, Cu, Mn etc.,
The electrolytic cell used for refining of aluminium consists of iron tank lined inside with carbon. This acts as anode. The tank contains three layers of fused masses. The bottom layer contains impure aluminium. Middle layer contains mixture of AlF₃, NaF and BaF₂ saturated with Al₂O₃. Top layer contains pure aluminium and graphite rods kept in it act as cathode.

On passing current aluminium ions from the middle layer are discharged at the cathode as pure aluminium. Equivalent amount of aluminium from the bottom layer passes into middle layer.

Textual Examples

Question 1.
Suggest a condition under which magnesium could reduce alumina.
Answer:
The two equations are :
a) \(\frac{4}{3}\) Al + O₂ → \(\frac{2}{3}\) Al₂O₃
b) 2 Mg + O₂ → 2MgO
At the point of intersection of the Al₂O₃ and MgO curves (marked. “A” in Ellingham diagram), the ∆G^⊖ becomes ZERO for the reaction :
\(\frac{2}{3}\) Al₂O₃ + 2Mg → 2MgO + \(\frac{4}{3}\) Al
Below that point magnesium can reduce alumina.

Question 2.
Although thermodynamically feasible, in practice, magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why ?
Answer:
Temperatures below the point of intersection of Al₂O₃ and MgO curves, magnesium can reduce alumina. But the process will be uneconomical.

Question 3.
Why is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of reduction ?
Solution:
The entropy is higher if the metal is in liquid state than when it is in solid state. The value of entropy change (∆S) of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of ∆G^⊖becomes more on negative side and the reduction becomes easier.

Question 4.
At a site, low grade copper ores are available and zinc and iron scraps are also available. Which of the two scraps would be more suitable for reducing the leached copper ore and why ?
Solution:
Zinc being above iron in the electrochemical series (more reactive metal is zinc), the reduction will be faster in case zinc scraps are used. But zinc is costlier metal than iron. So using iron scraps will be advisable and advantageous.

Intext Questions

Question 1.
Which of the ores mentioned in Table can be concentrated by magnetic separation method ?
Answer:
Ores in which one of the components (either the impurity or the actual ore) is magnetic can be concentrated.
E.g.: ores containing iron (haematite, magnetite, siderite and iron pyrites).

Question 2.
What is the significance of leaching in the extraction of aluminium ?
Answer:
Leaching is significant as it helps in removing the impurities like SiO₂, Fe₂O₃ etc. from the bauxite ore.

Question 3.
The reaction, Cr₂O₃ + 2 Al → Al₂O₃ + 2 Cr (∆G^⊖ = -421 kJ) is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?
Answer:
Certain amount of activation energy is essential even for such reactions which are thermodynamically feasible, therefore heating is required.

Question 4.
Is it true that under certain conditions Mg can reduce Al₂O₃ and Al can reduce MgO ? What are those conditions ?
Answer:
Yes, below 1350°C Mg can reduce Al₂O₃ and above 1350°C, Al can reduce MgO. This can be inferred from ∆G^⊖ vs T plots.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 4th Lesson Surface Chemistry Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 4th Lesson Surface Chemistry

Very Short Answer Questions

Question 1.
What is an interface ? Give one example.
Answer:
The interface is normally a few molecules thick but its area depends on the size of the particles of bulk.
Interface is represented by separating the bulk phases by hyphen (-) (or) a slash (/).
Eg : Interface between a solid and a gas may be represented by solid-gas (or) solid/gas.

Question 2.
What is adsorption ? Give one example.
Answer:
Adsorption : The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg : Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.

Question 3.
What is absorption ? Give one example.
Answer:
Absorption : The uniform distribution of a substance through out the bulk of the solid substance is known as absorption.
Eg : Chalk stick dipped in ink.

Question 4.
Distinguish between adsorption and absorption. Give one example of each.
Answer:
Adsorption
1. The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg :Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.

Absorption
1) The uniform distribution of a susbstance through out the bulk of the solid substance is known as absorption.
Eg : Chalk stick dipped in ink.

Question 5.
The moist air becomes dry in the presence of silica gel. Give reason for this.
Answer: The moist air becomes dry in presence of silica gel because the water molecules present in air get adsorbed on the surface of the gel.

Question 6.
Methylene blue solution when shakes with animal charcoal gives a colourless filtrate on filteration. Give the reason.
Answer:
Methylene blue solution (organic dye) when shaken with animal charcoal gives a colourless filtrate on filtration. The molecules of the dye from the solution are adsorbed on the charcoal.

Question 7.
A small amount of silica gel and a small amount of anhydrous calcium chloride are placed separately in two corners of a vessel containing water vapour. What phenomena will occur ?
Answer:
A small amount of silica gel and a small amount of anhydrous CaCl₂ are placed separately in two comers of a vessel of water vapour. Water vapours are is absorbed by anhydrous CaCl₂ but adsorbed by silical gel.

Question 8.
What is desorption ?
Answer:
Desorption: The process of removing aa adsorbed substance from a surface on which it is adsorbed is called desorption.

Question 9.
What is sorption ?
Answer:
In case of some substances both adsorption and absorption takes place. This phenomenon is called sorption.

Question 10.
Amongst adsorption, absorption which is a surface phenomena and why ?
Answer:

1. Adsorption is a surface phenomenon.
2. In adsorption the concentration of the adsorbate increases only at the surface of the adsorbent. While in absorption the concentration is uniformly distributed throughout the bulk of the compound.

Question 11.
What is the name given to the phenomenon when both absorption and adsorption take place together ?
Answer:
The name of the phenomenon when both adsorption and absorption take place together is sorption.

Question 12.
Chalk stick dipped in an ink solution exhibits the following :
a) The surface of the stick retains the colour of the ink.
b) Breaking the chalk stick,-it is found still white from inside.Explain the above observations.
Answer:
a) Chalk stick dipped in an ink solution, the surface of the stick retains the colour of the ink due to adsorption of coloured molecules in the ink.
b) Chalk stick dipped in an ink solution, the surface of the stick retains the colour of the ink due to adsorption of coloured molecules in the ink while the solvent of the in k goes deep into the stick due to absorption so on breaking the chalk stick, it is found to be white inside.

Question 13.
What are the factors which influence the adsorption of a gas on a solid ?
Answer:
Adsorption of a gas on solids is influenced by the following factors.
a) Surface area of the adsorbent, b) Nature of the gas, c) Pressure of the adsorbate, d) Temperature.

Question 14.
Why is adsorption always exothermic ?
Answer:
During the adsorption, there is always a decrease in residual forces of the surface. There is decrease in surface energy which appears as heat.
∴ Adsorption is always Exothermic.

Question 15.
Give the signs of ∆H & ∆S, when ammonia gas gets adsorbed on charcoal.
Answer:
When NH₃ gas gets adsorbed on charcoal the sign of ∆H is negative and the sign of ∆S is also negative.

Question 16.
How many types of adsorption are known ? What are they ?
Answer:
Adsorption process is divided into two types.
1) Physisorption 2) Chemisorption

Question 17.
What types of forces are involved in physisorption of a gas on solid ?
Answer:
Weak Vander waals forces are involved in physisorption of a gas on solid.

Question 18.
What type of interaction occurring between gas molecules and a solid surface is responsible for chemisorption of the gas on solid.
Answer:
Chemical bonds occur between gas molecules and a solid surface, is responsible for chemisorption of the gas on solid.

Question 19.
Why chemisorption is called activated adsorption ?
Answer:
Chemisorption involves a high energy of activation. Hence it is referred as activated adsorption.

Question 20.
What is the difference between physisorption and chemisorption ?
Answer:
1) In physisorption the forces present between adsorbate and adsorbent are weak vander . waal’s forces.
2) In chemisorption the forces present between adsorbate and adsobent are chemical bonds (valency forces).

Question 21.
Out of physisorption and chemisorption, which can be reversed ?
Answer:
Out of physisorption and chemisorption, phyisorption is reversed (reversible).

Question 22.
How is adsorption of a gas related to its critical temperature ?
Answer:
Higher the critical temperature, greater is the case of liquefaction of a gas. Then the extent of adsorption will be high.

Question 23.
The critical temperature of SO₂ is 630 K and that of CH₄ is 190 K. Which is adsorbed easily on activated charcoal ? Why ?
Answer:
Given the critical temperature of SO₂ is 630 K and that of CH₄ is 190 K.
SO₂ gas is adsorbed easily on activated charcoal because of higher critical temperature. Higher the critical temperature of gas, extent of adsorption is high.

Question 24.
Easily liquefiable gases aye readily adsorbed on solids. Why ?
Answer:
Easily liquefiable gases are readily adsorbed on solids because vander waal’s forces are stronger near the critical temperatures.

Question 25.
Amongst SO₂, H₂ which will be adsorbed more readily on the surface of charcoal and why ?
Answer:
Among SO₂, H₂ gases, SO₂ gas adsorb more readily on the surface of charcoal because SO₂ has high critical temperature (630 K) than dihydrogen (33K).

Question 26.
Compare the enthalpy of adsorption for physisorption and chemisorption.
Answer:

1. In physical adsorption enthalpy of adsorption is low (20 – 40 KJ/mole)
2. In case of chemical adsorption enthalpy of adsorption is high (80 – 240 KJ/mole)

Question 27.
What is the magnitude of enthalpy of physical adsorption ? Give reason for this magnitude.
Answer:
The magnitude of enthalpy of physical adsorption is low 20 – 40 KJ ,/mole. This is due to the presence of weak vander waal’s forces between gas molecules and solid surface.

Question 28.
What is the magnitude of enthalpy of chemisorption ? Give reason for this magnitude.
Answer:
The magnitude of enthalpy of chemical adsorption is high 80 – 240 KJ /mole. This is due to the presence of chemical bonds between gas molecules and solid surface.

Question 29.
Give any two applications of adsorption.
Answer:
Applications of adsorption :
a) Separation of inert gases : Different noble gases adsorb at different temperatures on coconut charcoal. By this principle (adsorption) mixture of noble gas is separated by adsorption on coconut charcoal.
b) Gas masks : Gas mask is a device which consists of activated charcoal (or) mixture of adsorbents is used by coal miners to adsorb poisonous gases during breathing.

Question 30.
Why physisorption suffers from lack of specificity ?
Answer:
Physisorption suffers from lack of specificity.
Explanation: A given surface of an adsorbent doesnot show any preference of a particular gas as the vander waal’s forces are universal.

Question 31.
What is an adsorption isotherm ? Write the equation of Freundlich adsorption isotherm.
Answer:
Adsorption Isotherm : The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve known as adsorption isotherm.
Freundlich adsorption isotherm equation is \(\frac{x}{m}\) = k. P^1/n
x = mass of the gas adsorbed
m = mass of the adsorbent
P, k and n are constants.

Question 32.
In the Freundlich adsorption isotherm, mention the conditions under which, following graph will be true ?

Answer:
In the above graph extent of adsorption does not depends on the pressure.
When \(\frac{1}{n}\) = 0, \(\frac{x}{m}\) = constant, the adsorption is independent of pressure.

Question 33.
What role does adsorption play in heterogenous catalysis ?
Answer:
In heterogeneous catalysis, adsorption of reactants on the solid surface of the catalysts increases the rate of reaction.
Eg : Manufacturing of NH₃ using Fe-as catalyst [Haber’s process].

Question 34.
What is the role of MnO₂ in the preparation of O₂ from KClO₃ ?
Ans. The chemical equation for the preparation of O₂ from KClO₃ is

MnO₂ increases the rate of reaction. Hence it is a catalyst for the above reaction.

Question 35.
Define “promoters” and “poisons” in the phenamenon of catalysis ?
Answer:
Promoters : The substances with enhance the activity of catalyst are known as promoters. Poisons : The substances which decrease the activity of a catalyst are known as poisons.

Question 36.
What is homogeneous catalysis ? How is it different from heterogeneous catalysis ?
Answer:
Homogeneous Catalysis : The catalysis in which reactants and catalyst are in same phase is called Homogeneous catalysis.
In case of heterogeneous catalysis, catalyst and reactants are present in different phases where as in case of homogeneous catalysis catalyst and reactants are present in same phase.

Question 37.
Give two examples for homogeneous catalytic reactions.
Answer:
The following are the examples for homogeneous catalytic reactions.

Question 38.
Give two examples for heterogeneous catalysis.
Answer:
The following are the examples of heterogeneous catalytic reactions.

Question 39.
Give two examples which indicate the selectivity of heterogeneous catalysis.
Answer:
The following reactions indicate the selectivity of heterogeneous catalysis.
Starting with H₂ and CO, and using different catalysts, we get different products.

The action of a catalyst is highly selective in nature. A substance which acts as a catalyst in one reaction may fail to catalyse another reaction.

Question 40.
Why zeolites are treated as shape selective catalysts ?
Answer:

• The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysts.
• Zeolites are shope selective catalysts because of their honey comb-like structure. They are microporous alumino silicates with 3-dimensional net work of silicates.

Question 41.
Which zeolite catalyst is used to convert alcohols directly into gasoline ?
Answer:
Zeolite ZSM – 5 is used to convert alcohols directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.

Question 42.
What are enzymes ? What is their role in human body ?
Answer:
Enzymes are complex nitrogenous organic compounds which are produced by living plants and animals.
These act as specific catalysts in biological reactions.
These catalyse the numerous reactions that occur in the bodies of animals and plants to maintain the life process.

Question 43.
Can catalyst increase the yield of reaction ?
Answer:
Catalyst does not increase the yield of reaction. Catalyst just speed up the product formation.

Question 44.
Name any two enzyme catalyzed reactions. Give the reactions.
Answer:
1) Inversion of Cane Sugar :

2) Decomposition of urea into ammonia and CO₂:

Question 45.
Name the enzymes obtained from soyabean source.
Answer:
The enzyme obtained from soyabean source is urease.

Question 46.
Name the enzymes used in
a) Decomposition of urea into ammonia.
b) Conversion of proteins into peptides in stomach.
Answer:
a) The enzyme used in the decomposition of urea into ammonia is urease.
b) The enzyme used in conversion of proteins into peptides in stomach is pepsin.

Question 47.
What enzymes are obtained from yeast ?
Answer:
The enzymes obtained from yeast are invertase, zymase and maltase.

Question 48.
At what ranges of temperature and pH, enzymes are active ?
Answer:
The optimum temperature range for enzymatic activity is 298 – 310 K. Human body temperature being 310 K is suited for enzyme catalysed reactions. The rate of an enzyme – catalysed reaction is maximum at a particular pH called optimum pH, which lies between 5-7.

Question 49.
Represent diagrammatically the mechanism of the enzyme catalyis.
Answer:

Question 50.
Name any two industrially important heterogeneous catalytic reactions mentioning the catalysts used.
Answer:
i) Manufacturing of NH₃ by Haber’s process

ii) Hydrogenation of vegetable oils in presence of finely divided nickel as catalyst.

The above mentioned are industrially important heterogeneous catalytic reactions.

Question 51.
What is a colloidal solution ? How is it different from a true solution with respect to dispersed particle size and homogeneity ?
Answer:
A heterogeneous system in which one substance is dispersed as large particles in another substance is called colloidal solution.

• In colloidal solution the particle size of the dispersed phase is of the order of 1 mμ – 1 μ where as in true solution the particle size of the solute are the order of mp or less.
• Colloidal solution is a heterogeneous binary system where as true solution is a homogeneous binary system.

Question 52.
Name the dispersed phase and a dispersion medium in the following colloidal systems

1. fog
2. smoke
3. milk.

Answer:

1. Fog : Dispersed phase → liquid
   Dispersion medium → gas
2. Smoke : Dispersed phase → carbon particles (solid)
   Dispersion medium → Air
3. Milk : Dispersed phase → liquid fat
   Dispersion medium → water

Question 53.
What are lyophilic and lyophobic sols ? Give one example for each type.
Answer:
Lyophilic colloid : The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.

The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 54.
Explain the terms with suitable examples .

1. aerosol
2. hydrosol.

Answer:

1. Aerosol: The colloidal solution in which dispersed phase is solid and dispersion medium is gas is called an aerosol.
   Eg : Smoke, dust.
2. Hydrosol: In a colloidal solution dispersion medium is water them it is called aquasol (or) hydrosol.
   Eg : Milk, Gold sol.

Question 55.
Explain why lyophilic colloids are relatively more stable than lyophobic colloids ?
Answer:

• Lyophilic colloids are reversible and are quite stable. These are not coagulated.
• Lyophobic sols are irreversible and need stabilizing agents for their stabilization. These sols are readily precipitated on the addition of small amounts of electrolytes.

Question 56.
Give two examples of colloidal solutions of liquids dispersed in solid. What is the name given to the colloidal solution ?
Answer:
Cheese, butter, jellies are examples of colloidal solutions of liquids dispersed in solid. The name of the colloidal solution given to this type is Gels.

Question 57.
What is the difference between multimolecular and macromolecular colloids ? Give one example for each.
Answer:

• In multimolecular colloids, a large number of atoms (or) small molecules of the dispersed phase aggregate together to form species in the colloidal range.
• Macro molecules in suitable solvents form solutions in which the sizes of the macromolecules are in the colloidal range.

Question 58.
What are micelles ? Give one example.
Answer:
Micelles : Some substances which at low concentrations behave as normal strong electrolytes, but at high concentrations exhibit colloidal behaviour due to formation of aggregates. The aggregated particles thus formed are called micelles.
Eg : Stearate ions (C₁₇H₃₅COO^–) associate together in high concentration, in a solution of soap in water and forms a micelle.

Question 59.
How do micelles differ from a normal colloidal solutions ?
Answer:
Some substances which at low concentrations behave as normal strong electrolytes, but at high concentrations exhibit colloidal behaviour due to formation of aggregates. The aggregated particles thus formed are called micelles.

• hese are also known as associated colloids.
• These colloids have both lyophilic and lyophobic parts.
• Micelles may contain as many as 100 or more of normal molecules.
• On dilution these colloids revert back to individual electrolytes.

Question 60.
Give two examples of associated colloids.
Answer:
Surface active agents such as soaps and synthetic detergents are examples of associated colloids.

Question 61.
Can the same substance act both as colloid and crystalloid ?
Answer:
Yes, Micelles, (associated colloids) act as normal strong electrolytes at low concentrations and exhibit colloid behavior at high concentrations.

Question 62.
Give two examples of lyophobic sols.
Answer:
Metal sols and metal suiphide sols are examples of lyophobic colloids

• Gold sol is lyophobic colloid

Question 63.
Give examples of’colloidal system of

1. Liquid In solid
2. Gas in solid.

Answer:

1. Cheese, butter, jellies are examples of liquid in solid type colloidal system.
2. Foam rubber, pumice ‘stone are examples of gas in solid type of colloidal system.

Question 64.
What type of substances form lyophobic sols?
Answer:
Substances like metals, their sulphides.donot form colloidal sol simply by mixing with the dispersion medium. These form colloids by special methods, These sols are lyophobic colloids.

Question 65.
What is critical micelle concentration (CMC) and kraft temperature (T_k) ?
Answer:
The formation of micelles takes place only above a particular temperature called Kraft temperature (T_k) and above a particular concentration called Critical micelle concentration (CMC).

Question 66.
Why lyophobic colloids are called irreversible colloids ?
Answer:

• Lyophobic colloids are coagulated by the addition of small amounts of electrolytes.
• The precipitates does not give back the colloidal sol by same addition of the dispersion medium.
• Hence lyophobic sols are called irreversible sols.

Question 67.
How a colloidal sol of arsenous sulphide is prepared ?
Answer:
Colloidal sol of arsenous sulphide is prepared by the double decompostion of As₂O₃ and H₂S as follows.

Question 68.
What is Peptization ?
Answer:
Peptization : The process of converting a precipitate into colloidal sol by shaking it with the dispersion medium in the presence of a small amount of electrolyte is called Peptization.

Question 69.
What is dialysis ? How is dialysis can be made fast ?
Answer:
Dialysis : The process of removing a dissolved substance from a colloidal solution using a suitable membrane is called dialysis.
Dialysis is made faster by applying an Emf. This is known as Electrodialysis.

Question 70.
What is collodion solution ?
Answer:
Collodion solution : Collodion is a 4% solution of nitro cellulose in a mixture of alcohol and ether.

Question 71.
How an ultrafilter paper is prepared from ordinary filter paper ?
Answer:
Ultra filter paper is prepared by soaking the filter paper in collodion solution, hardening by formaldehyde and then finally dried.

Question 72.
What is Tyndall effect ?
Answer:
Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

This is an optical property exhibited by colloidal solution. This phenomenon is clearly seen with a microscope placed at right angles to the path of light. Colloidal particles become self luminous due to absorption of light. A true solution does not show Tyndall effect.

Question 73.
What conditions is tyndall effect observed ?
Answer:
Tyndall effect is observed only when

1. The diameter of the dispersed particles is not much smaller than the wave length of the light used.
2. The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

Question 74.
Can Tyndall effect be used to distinguish between a colloidal solution and a true solution ? Explain.
Answer:
Tyndall effect is used to distinguish between a colloidal solution and a true solution.

• True solution placed in dark and is observed in the direction of the light passed beam passing through it. It appears clear and it is observed from a direction at right angles to the direction of the light beam, it appears perfectly dark.
• Colloidal solution viewed in the same way may appear reasonably clear in the direction of light but they show a mild to strong opalescence when viewed at right angles to the direction of light.

Question 75.
Sky appears blue in colour. Explain.
Answer:
Dust particles along with water vapour suspended in air scatter blue light which reaches our eyes and hence the sky looks blue to us.

Question 76.
What is Brownian movement.
Answer:
Brownian movement: This is a kinetic property of colloidal solution.

When a colloidal solution is examined by ultramicroscope, the colloidal particles are seem to be moving in a rapid zig-zag motion.
This rapid motion of colloidal particles is called Brownian movement.

This motion is due to unequal bombardment of colloidal particles by molecules of dispersion medium. Smaller the colloidal particles the more rapid is the Brownian movement.

Question 77.
What is the main cause for charge on a colloidal solution ?
Answer:
The charge on the colloidal particles is due to electron capture by sol particles during electro dispersion of metals , and due to preferential adsorption ions from solution (or) due to formulation of electrical double layer.

Question 78.
What is electrokinetic potential or zeta potential ?
Answer:
In a colloidal sol the charges of opposite signs on the fixed and diffused parts of the double layer results in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charge is called electro kinetic potential (or) zeta potential.

Question 79.
Write the chemical formula of positively charged and negatively charged hydrated ferric oxide colloidal solutions.
Answer:

• Formula of positively charged hydrated ferric oxide is Fe₂O₃.xH₂O/Fe⁺³
• Formula of negatively charged hydrated ferric oxide is Fe₂O₃.xH₂0/OH^–

Question 80.
Give the order of coagulating power of Cl, SO₄^2-, PO₄^3- in the coagulation of positive sols.
Answer:
The order of coagulating ability’of given ions with positive sols is PO₄^3- > SO₄^2- > Cl^–

Question 81.
Amongst Na⁺, Ba²⁺, Al³⁺, which coagulates negative sol readily and why ?
Answer:
The order of coagulating ability of given ions with negative sols is Al⁺³ > Ba⁺² > Na⁺

Question 82.
A colloidal solution of AgI is positively charged when prepared from a solution containing excess of Ag⁺ ions and negatively charged when prepared from a solution containing excess of I^– ions Explain.
Answer:
When a dilute AgNO₃ solution is added to a dilute KI solution taken in excess, the precipitated Agl adsorbs I^– ions present in excess and a negatively charged AgI colloidal solution is formed. However when dilute KI solution is added to dilute AgNO₃ solution taken in excess, AgI precipitate adsorbs Ag⁺ ions present in excess and a positively charged AgI colloidal solution is formed. Generally the ion common to dispersed particle is adsorbed. .
AgI/I^–
Negatively charged
AgI/Ag⁺
Positively charged.

Question 83.
What is electrophoresis ?
Answer:
When electric potential is applied across two platinum electrodes dipping in a colloidal solution,-the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Question 84.
What is electro osmosis ?
Answer:
If the movement of colloidal particles is arrested by some suitable means, the dispersion medium moves in opposite direction. This phenomenon is termed “electro osmosis”.

Question 85.
What is coagulation ?
Answer:
The stability of the lyophilic sols is due to the presence of charge on the colloidal particles. If this charge is neutralised the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. This process of settling down of colloidal particles is called coagulation (or) flocculation (or) precipitation.

Question 86.
Define flocculation value.
Answer:
The minimum concentration of an electrolyte in millimoles per litre required to cause coagulation of a sol in two hours is called “coagulating value” (or) flocculation value.

Question 87.
State Hardy – Schulze rule.
Answer:
Greater the valence of the coagulating ion added, the greater is its power to cause coagulation. This is known as Hardy-Schulze rule.

Questions 88.
Coagulation takes place when sodium chloride solution is added to a colloidal solution of hydrated ferric oxide. Explain.
Answer:
When NaCl solution is added to the colloidal solution of hydrated ferric oxide coagulation takes place (precipitation formed). .
The reason is that colloidal particles interact with ions (Na⁺, Cl^–) carrying charge opposite to that present on themselves. This causes neutralisation of charges leading to their coagulation.

Question 89.
How are lyophobic solutions protected from phenomenon of coagulation.
Answer:
Lyophobic sols are protected from the phenomenon of coagulation by adding suitable lyophilic sol.

Question 90.
What is protective colloid ?
Answer:
Lyophilic colloids used for the prevention of coagulation of lyophobic colloids are called protective colloids.

• Lyophilic colloids protect the lyophobic colloids.

Question 91.
What is an emulsion ? Give two examples. [A.P. Mar. 17]
Answer:
Emulsion: The colloidal system in which a dispersion of finely divided droplets of a liquid in another liquid medium is called emulsion.
Eg : Milk, Vanishing cream, Cold cream.

Question 92.
How emulsions are classified ? Give one example for each type of emulsion. [A.P. Mar. 17]
Answer:
Emulsion : A dispersion of finely divided droplets of a liquid in another liquid medium is called’emulsion’.
Ex: Milk.
In Milk, the droplets of liquid fat are dispersed in water. This is an example for oil in water type emulsion.
Classification of emulsions : Emulsions are “classified into two classes. These are
a) oil in water (O/W) and
b) water in oil (W/O), (O = Oil; W = Water).

These emulsions are classified as such depending on which is dispersed phase and which is dispersion medium.
a) Oil in Water (O/W) type emulsions : In this type of emulsions, the dispersed phase is oil and the dispersion medium is water.
Ex : Milk, liquid, fat (oil) in water.
Vanishing cream; fat in water.

b) Water in Oil (W/O) type emulsions : In this type of emulsions, the dispersed phase is water and the dispersion medium is oil.
Ex : Stiff greases : water in lubrication oils
Cod liver oil : water in cod liver oil
Cold cream : water in fat.

Question 93.
What is an emulsifying agent ?
Answer:
Emulsifying agent: The third substance which is added in small amounts to an emulsion to stabilize the emulsion is called emulsifying agent.

Question 94.
What is demulsification ? Name two demulsifiers.
Answer:
Demulsification : The separation of an emulsion into constituent liquids is known as demulsification.

Question 95.
How is artificial rain produced ?
Answer:
Artificial rain is produced by throwing electrified sand (or) spraying a sol carrying charge opposite to the one on clouds from an aeroplane.

Question 96.
Bleeding from fresh cut can be Stopped by applying alum. Give reasons.
Answer:
Bleeding from fresh cut can be stopped by applying alum. This is due to styptic action of alum which coagulates the blood and stops further bleeding.

Question 97.
Deltas are formed at the points where river enters the sea. Why ?
Answer:
Deltas are formed at the points where river enters the sea.
Explantation : River water is a colloidal solution of clay. Sea water contains a number of electrolytes. When river water meets the sea water, the electrolytes present in sea water coagulate the colloidal solution of clay resulting in its deposition of clay with the formation of delta.

Question 98.
Name any two applications of colloidal solutions.
Answer:
Applications of colloidal solutions :
Rubber: Plant latex is a colloidal solution of rubber particles which are negatively charged. Rubber is obtained from latex by coagulation.
Industrial Products : Paints, inks, synthetic plastics, rubber, graphite, lubricants, cement, etc., are all colloidal in nature.

Question 99.
How can aerial pollution by colloidal particles of smoke be prevented ? Explain.
Answer:
Aerial pollution by colloidal particles of smoke be prevented by electrical precipitation of smoke. Smoke is a colloidal solution of solid particles such as carbon, arsenic compounds, dust, etc., in air. The smoke, before it comes out from the chimney, is led through a precipitator containing plates having a charge opposite to that carried by smoke particles. The particles on coming in contact with these plates lose their charge and get precipitated. The particles thus settle down on the floor of the chamber. The precipitator is called cottrell precipitator.

Question 100.
Alum is used to purify water obtained from natural sources. Explain.
Answer:
The water obtained from natural sources often contains suspended impurities. Alum is added to such water to coagulate the suspended impurities and make water fit for drinking purposes.

Question 101.
Why medicines are more effective in colloidal state ?
Answer:
Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated.

Question 102.
How rubber is obtained from latex ?
Answer:
Plant latex is a colloidal solution of rubber particles which are negatively charged. Rubber is obtained from latex by coagulation.

Question 103.
Name the type of emulsion to which milk belongs.
Answer:
Milk is a oil in water type of emulsion.
Dispersed phase : liquid fat.
Dispersion medium : water.

Short Answer Questions

Question 1.
What is adsorption ? Discuss the mechanism of adsorption of gases on solids.
Answer:
Adsorption : The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg : Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.
Mechanism of Adsorption :

• Adsorption arises due to the fact that the surface particles of the adsorbent are not in the same environment as the particles inside the bulk.
• Adsorbent possess unbalanced or residual attractive forces. These forces are responsible for attracting the adsorbate molecules on adsorbent surface.
• The extent of adsorption increases with increase in surface area of adsorbent at a given temperature and pressure.
• During the adsorption, there is always a decrease in residual forces of the surface. There is decrease in surface energy which appears as heat. So, Adsorption is always Exothermic.
• Adsorption is accompanied by decrease in enthalpy as well as decrease in entropy of system. .
• As the adsorption proceeds AH becomes less and less negative, ultimately, AH becomes equal to TAS and AG becomes zero. At this state equilibrium is attained.

Question 2.
What are different types of adsorption ? Give any four differences between characteristics of these different types. [T.S. Mar. 19, 15; A.P. Mar. 15]
Answer:
Adsorption process is divided into two types.
1) Physisorption
2) Chemisorption.
Distinguishing characteristics of Physisorption and Chemisorption are given in the following table:
Physisorption

1. This process is weak, due to vander Waal’s forces.
2. The process is reversible.
3. This is a quick process, i.e., takes place quickly.
4. The process decreases with increase of temperature.
5. This is a multilayered process.
6. The process depends mainly on the nature of the adsorbent.

Chemisorption

1. This process is strong, due to chemical forces.
2. The process is irreversible.
3. This is a slow process.
4. The process increases with increase of temperature.
5. This is a unilayered process.
6. The process depends both on the nature of adsorbent and adsorbate.

Question 3.
What do you understand by the terms given below (a) absorption (b) Adsorption (c) Adsorbent and Adsorbate
Answer:
Absorption : The uniform distribution of a susbstance through out the bulk of the solid .substance is known as absorption. Eg: Chalk stick dipped in ink.
Adsorption : The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg: Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.

Adsorbent: The substance on whose surface the adsorption occurs is known as adsorbent.

Adsorbate : The substance whose molecules get adsorbed on the surface of the adsorbent is known as adsorbate.
Eg: In the adsorption of acetic acid by charcoal, acetic acid is adsorbate and the charcoal is adsorbent.

Question 4.
Adsorption of a gas on the surface of solid is generally accompanied by decrease in entropy. Still it is a spontaneous process. Explain.
Answer:
Adsorption is accompanied by decrease in enthalpy as well as decrease in entropy of the system.

• For a process to be spontaneus there is a decrease in Gibbs Energy. .
• On the basis of equation ∆G = ∆H – T∆S; ∆G = -Ve
•  If AH has sufficiently high negative value and T∆S is positive.
• In an adsorption process which ¡s spontaneous ∆G becomes negative by combining above.
• As the adsorption proceeds ∆H becomes less and less negative ultimately, ∆H becomes equal to T∆S and ∆G becomes zero. At this state equilibrium is attained.

Question 5.
How can the constants k and n of the Freundlich adsorption equation be calculated?
Answer:
Freundlich adsorption isotherm equation is
\(\frac{x}{m}\) = k. P =, x/m = Extent of adsorption ⇒ P = Pressure
k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.

Applying logarithm to the above equation
log \(\frac{x}{m}\) = log k + \(\frac{1}{n}\) log P.

• A graph is plotted taking log \(\frac{x}{m}\) on y – axis and log P on x – axis. 1f the graph is a straight line then Freundlich isotherm is valid.
• The slope of the straight line gives \(\frac{1}{n}\) value.
• The intercept on the y-axis gives value of log k.
• \(\frac{1}{n}\) has values between 0 and 1
• When \(\frac{1}{n}\), = 0, \(\frac{x}{m}\) = constant, the adsorption is independent of pressure
  \(\frac{1}{n}\) = 1, \(\frac{x}{m}\) = kP i.e., \(\frac{x}{m}\) ∝ p.

Question 6.
How does the extent of adsorption depend upon
a) Increasing the surface area per unit mass of adsorbent.
b) Increasing temperature of the system.
c) Increasing pressure of the gas.
Answer:
a) The extent of adsorption increases by increasing the surface area per unit mass of adsorbent.
b) The extent of adsorption decrease with an increase in temperature.
c) The extent of adsorption increase with increasing pressure of the gas.

Question 7.
What is catalysis ? How is catalysis classified ? Give two examples for each type of catalysis. [A.P. Mar. 16] [Mar. 14]
Answer:
Catalysis : A substance which alters the rate of a chemical reaction without itself being consumed in the process, is called a catalyst.
The action of catalyst in altering the rate of a chemical reaction is called catalysis.

Types of catalysis: Catalysis is classified into two types as
a) Homogeneous catalysis and b) Heterogeneous catalysis.

a) Homogeneous catalysis : The catalytic process in which the catalyst is present in the same phase as that of reactants, is known as homogeneous catalysis.

b) Heterogeneous catalysis : The catalytic process in which the catalyst is present in a phase different from that of the reactants is known as heterogeneous catalysis.

Question 8.
Discuss the mechanism involved in adsorption of heterogeneous catalysis.
Answer:
Adsorption theory of heterogeneous catalysis :
Adsorption theory explains the mechanism of heterogeneous catalysis.

• The modern adsorption theory is the combination of the intermediate compound formation theory and the old adsorption theory.
•  The mechanism of catalysis involves five steps.
• Diffusion of the reactants to the surface of the catalyst.
• Adsorption of the reactant molecules on the surface of the catalyst.
• Occurrence of chemical reaction on the catalyst’s surface through formation of an . intermediate.
• Desorption of reaction products from the catalyst surface, and thereby making the surface available again for more reaction to occur.
• Diffusion of reaction products away from the catalyst’s surface.

Question 9.
Discuss some features of catalysis by zeolites.
Answer:
The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysts.

• Zeolites are shope selective catalysts because of their honey comb-like structure. They are microporous alumino silicates with 3-dimensional net work of silicates.
• Zeolites are microporous alumino silicates with dimensional network of silicates in which some Si-atoms are replaced by aluminium atoms giving Al-o-Si frame work.
• The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as on the pores and cavities of the zeolites.
• Zeolites are found in nature and also synthesized for catalytic selectivity.

Uses:

• Zeolites are used as catalysts in petrochemical industries for cracking.
• Zeolite ZSM – 5 is used to convert alcohol directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.

Question 10.
Give brief account of mechanism of enzyme catalysis with suitable diagrams.
Answer:
Mechanism of enzyme catalysis :

The enzyme – catalysed reactions may be considered to proceed in two steps.
Step 1 : Binding of substrate to enzyme to form an activated complex (ES^≠).
E + S → E S^≠
Step 2 : Decomposition of the activated complex to form product.
E S^≠ → E + P

Question 11.
Discuss the factors that influence the catalytic activity of enzymes.
Answer:
Factors influencing the catalytic activity of enzymes :

• Highly specific nature: One catalyst catalyses only one reaction. Each enzyme is specific ‘ for a given reaction.
  Eg : The enzyme urease catalyses the hydrolysis of urea only and not any other amide.
• Highly active under optimum temperature: The rate of an enzyme reaction is maximum at a temperature called optimum temperature.
• The optimum temperature range for enzymatic activity is 298 – 310 K.
• Highly active under optimum pH : The rate of an enzyme – catalysed reaction is maximum at a particular pH called optimum pH which lies between 5 – 7.
• Increasing activity in presence of activators and co-enzymes : The enzymatic activity is increased in presence of co-enzymes (or) in presence of activators such as Na⁺, Mn⁺² etc.,
•  Influence of inhibitors and poisons: The inhibitors (or) poisons interact with the active functional groups on the enzyme surface and reduce or completely destroy the catalytic activity.

Question 12.
Name any six enzyme catalysed reactions. [A.P. Mar. 19]
Answer:
i) Inversion of Cane sugar : Enzyme : Invertase

Question 13.
What do you mean by activity and selectivity of catalysts ?
Answer:
Activity :
The ability of a catalyst in increasing the rate of reaction is defined as activity of catalyst.

• The activity of a catalyst depends upon the strength of chemisorption to a large extent.
• The reactants must get adsorbed reasonably strongly on to the catalyst to become reactive.
  Eg : The catalystic activity increases from Group – 5 to Group – 11 for hydrogenation reactions.
  The maximum activity being shown by 7 – 9 group metals.
  

Selectivity:
The selectivity of a catalyst is its-ability to direct a reaction to form specific products. The following reactions indicate the selectivity of heterogeneous catalysis.
Starting with H₂ and CO, and using different catalysts, we get different products.

The action of a catalyst is highly selective in nature. A substance which acts as a catalyst in one reaction may fail to catalyse another reaction.

Question 14.
How are colloids classified on the basis of physical states of components ?
Answer:
On the basis of physical states of dispersed phase and dispersion medium colloidal solutions are classified into eight types.

Question 15.
How are colloids classified on the basis of nature of the dispersion medium ?
Answer:
On the basis of nature of dispersion medium colloids are classified as follows.

•  If the dispersion medium is ‘air’, the sols are called aerosols.
  Eg : Smoke.
• If the dispersion medium is ‘water’ the sols are called Hydrosol (or) aquasol.
  Eg : starch solution.
• If the dispersion medium is ‘alcohol’ then the sols are called alcosols.

Question 16.
How are colloids classified on the basis of interaction between dispersed phase and dispersion medium ?
Answer:
Lyophilic colloid : The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex: Starch solution.
The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 17.
What is the difference between a colloidal sol, gel, emulsion and a foam?
Answer:

Question 18.
What are lyophilic and lyophobic sols? Compare the two terms in terms of stability and reversibility.
Answer:
Lyophilic colloid : The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.
The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 19.
Name a substance whose molecules consist of lyophilic as well as lyophobic parts. Give its use in our daily life.
Answer:

• Associated colloids (or) micelles contains both lyophobic and lyophillic parts.
• Examples are soaps and synthetic detergents.
• The cleaning action of soap is due to the fact that soap molecule form a micelle around the oil droplet.

Question 20.
Describe Bredig’s arc method of preparation of colloids with a neat diagram.
Answer:
Electrical disintegration or Bredig’s Arc method : This process involves dispersion as well as condensation. Colloidal sols of metals such as gold, silver, platinum, etc., are prepared by this method. In this method, electric arc is struck between electrodes of the metal immersed in the dispersion medium. The intense heat produced vapourises the metal, which then condenses to form particles of colloidal size.

Question 21.
Name any four examples of preparation of colloids by chemical methods with necessary chemical equations.
Answer:
Chemical methods : Colloidal solutions are prepared by chemical reactions leading to the formation of species by double decompostion, oxidation, reduction or hydrolysis. These species then aggregate leading to the formation of sols.

Question 22.
Describe the purification of colloidal solutions by the phenomenon of dialysis with a neat diagram. [A.P. Mar. 18]
Answer:
Colloidal solutions generally contain excess amount of electrolytes and some other soluble impurities. It is necessary to reduce the concentration of soluble impurities to a requisite minimum. The impurities presence required in traces.
“The process used for reducing the amount of impurities to a requisite minimum is known as purification of colloidal solution”.

Purification of colloidal solution by Dialysis :
Dialysis : The process of removing a dissolved substance from a colloidal solution using a suitable membrane is called dialysis.

• In a true solution particles can pass through animal membrane (or) cellophane sheet (or) parchment paper but not colloidal particles.
• The apparatus used for the dialysis is called dialyser.
• A bag of suitable membrane containing the colloidal solution is suspended in a vessel containing a continuously flowing water.
• The molecules and ions diffuse through the membrane into the water and pure colloidal solution is left behind in the bag.

Question 23.
Explain the formation of micelles with a neat sketch.
Answer:
Mechanism of micelle formation: Let us take the example of soap solution. Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOC^– Na⁺ (e.g., sodium strearate CH₃ (CH₂)₁₆ COO^– Na⁺, which is a major component of many bar soaps). When dissolved in water, it dissociates into RCOO^– and Na⁺ ions. The RCOO^– ions, however, consist of two parts – a long hydrocarbon chain R (also called non-polar ‘tail’) which is hydrophobic (water repelling), and a polar group COO^–(also called polar-ionic ‘head’), which is hydrophilic (water loving).

The RCOO^– ions are, therefore, present on the surface with their COO^– groups in water and the hydrocarbon chains (R) staying away from it and remain at the surface. But at critical micelle concentration, the COO^– irons are pulled into the bulk of the solution and are aggregated to form a spherical shape with their hydrocarbon chains pointing towards the centre of the . sphere with COO^– part remaining outward on the surface of the sphere. The aggregate thus formed is known as ‘ionic micelle’. These micelles may contain as many as 100 simple molecules.

Similarly, in case of detergents, e.g., sodium laurylsulphate. CH₃ (CH₂)₁₁ SO₃^–Na⁺, the polar group is -SO₃^– along with the long hydrocarbon chain. Hence, the mechanism of micelle formation here also is same as that of soaps.

Question 24.
Action of soap is due to emulsification and micelle formation. Comment.
Answer:
Soap is sodium stearate, C₁₇H₃₅COONa, in water soap gives the ions stearate anion and sodium ion.

Cleaning action of soap : Soap anions form a micelle. The grease or dirt of the cloth are absorbed into the interior of the micelle. The tails of the anion are pegged into micelle and these micelle are washed away with the soap solution.

Soap thus, functions as an emulsifying agent for the water dirt emulsion. The emulsified grease or dirt is then washed away with soap solution.

Question 25.
Explain the phenomenon of Brownian movement giving reasons for the occurrence of this phenomena.
Answer:
Brownian movement: This is a kinetic property of colloidal solution.

When a colloidal solution is examined by ultramicroscope, the colloidal particles are seem to be moving in a rapid zig – zag motion.
This rapid motion of colloidal particles is called Brownian movement. This motion is due to unequal bombardment of colloidal particles by molecules of dispersion medium. Smaller the colloidal particles the more rapid is the Brownian movement.

Question 26.
Name the four positively charged sols. .
Answer:
The following are the positively charged sols.

• Hydroated metallic oxide sols. Eg : Al₂O₃.xH₂O, CrO₃.xH₂O etc.
• Basic dye stuffs Eg : Methylene blue sol.
• Hemoglobin (blood).
• Oxides Eg : TiO₂ sol.

Question 27.
Name the four negatively charged sols.
Answer:
The following are the negatively charged sols.

• Metal sols Eg : Ag, Au-sols.
• Metallic suphides sols Eg : ArS₃, CdS sols.
• Acid dye stuffs Eg: eosinol.
• Sols of starch, gum, clay etc.

Question 28.
Explain the terms helmholtz electrical double layer and zeta potential. What are their significancies in the colloidal solutions ?
Answer:
Helmholtz electrical double layer :
The combination of the two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer.

• In a colloidal sol the charges of opposite signs on the fixed and diffused parts of the double layer results in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charge is called electro kinetic potential (or) zeta potential.
• The above concepts applicable in colloidal solution in which the solid particles carry one kind of charge while liquid medium carries opposite charge to that of solids.

Question 29.
Explain with a neat sketch the phenomenon of electrophoresis.
Answer:
Electrophoresis : The existence of charge on colloidal particles is confirmed by electro-phoresis experiment.
When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal articles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Positive charged particles move towards the cathode while negative charged particles towards anode.

Question 30.
Explain the following terms, i) Electrophoresis ii) Coagulation iii) Tyndall effect.
Answer:
Electrophoresis: When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Coagulation : The stability of the lyophilic sols is due to the presence of charge on the colloidal particles. If this charge is neutralised the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. This process of settling down of colloidal particles is called coagulation (or) flocculation (or) precipitation.
Coagulation of lyophobic sols can be carried out by
i) Electrophoresis, ii) Boiling, iii) Adding Electrolytes, iv) Prolonged dialysis etc.

Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

Question 31.
Explain the phenomenon observed
i) When a beam of light is passed through a colloidal sol.
Answer:
Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

This is an optical property exhibited by colloidal solution. This phenomenon is clearly seen with a microscope placed at right angles to the path of light. Colloidal particles become self luminous due to absorption of light. A true solution does not show Tyndall effect.

ii) An electrolyte, NaCl is added to hydrated ferric oxide.
Answer:
When NaCl solution is added to the colloidal solution of hydrated ferric oxide coagulation takes place (precipitation formed).

The reason is that colloidal particles interact with ions (Na⁺, Cl^–) carrying charge opposite to that present on themselves. This causes neutralisation of charges leading to their coagulation.

iii) An electric current is passed through a colloidal solution.
Answer:
Electrophoresis : the existence of charge on colloidal particles is confirmed by electrophoresis experiment.

When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Positive charged particles move towards the cathode while negative charged particles towards anode.

Question 32.
Describe cottrell smoke precipitator with a neat diagram.
Answer:
Aerial pollution by colloidal particles of smoke be prevented by electrical precipitation of smoke. Smoke is a colloidal solution of solid particles such as carbon, arsenic compounds, dust, etc., in air. The smoke, before it comes out from the chimney, is led through a precipitator containing plates having a charge opposite to that carried by smoke particles. The particles on coming in contact with these plates lose their charge and get precipitated. The particles thus settle down on the. floor of the chamber. The precipitator is called conttrels precipitator.

Question 33.
Among NaCl, Na₂SO₄, Na₃PO₄ electrolytes, which is more effective for coagulation of hydrated ferric oxide sol and why ?
Answer:
Among NaCl, Na₂SO₄, Na₃PO₄ electrolytes, Na₃PO₄ is more effective for coagulation of hydrated ferric oxide sol. This is explained by Hardy-Schulze rule.
Greater the valence of the coagulating ion added, the greater is its power to cause coagulation. This is known as Hardy-Schulze rule.

Coagulating ability order of the anions in the above salts is
PO₄^-3 > SO₄^-2 > Cl^–.

Question 34.
Discuss how a lyophilic colloids protect a lyophobic colloids.
Answer:
Lyophilic colloids are much stable than lyophobic colloids. A very small amount of electrolyte can precipitate the lyophobic colloids. But their precipitation can be prevented by the addition of small amounts of lyophilic colloids. This lyophilic colloids form a protecting layer around the lyophobic colloids and protect them from precipitation (coagulation). Lyophilic colloids used for. this purpose are called protective colloids.
Ex : Gelatin, gum, starch sols are generally used as protective colloids.

Question 35.
Discuss the use of colloids in
i) Purification of drinking water ii) Tanning iii) Medicines.
Answer:
i) Purification of drinking water: The water obtained from natural sources often contains suspended impurities. Alum is added to such water to coagulate the suspended impurities and make water fit for drinking purposes.

ii) Tanning: Animal skins are colloidal in nature. When a skin, which has positively charged particles, is soaked in tannin, which contains negatively charged colloidal particles, mutual coagulation takes place. This results in the hardening of skin (leather). This process is termed as tanning. Chromium salts are also used in place of tannin.

iii) Medicines : Most of the medicines are colloidal in nature. For example argyrol is a silver sol used as an eye lotion. Colloidal antimony is used in curing kalaazar. Colloidal gold is used as intramuscular injection. Milk of magnesia, an emulsion, is used for stomach disorders. Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated.

Question 36.
Define Gold Number.
Answer:
The capacity of the lyophilic colloid in protecting a lyophobic colloid is measured in terms of Gold Number, introduced by “Zigmondy”.

Gold Number:” It is the number of milligrams of lyophilic colloid that is to be added to 10 ml of standard gold sol to prevent its precipitation by the addition of 1ml of a 10% sodium chloride solution”.
Lesser in the gold number the greater is the protecting capacity. Gold number of a few colloids are given below.
Starch = 25 .
Gelatin = 0.005 – 0.01
Albumin =0.1 – 0.2

Question 37.
How do emulsifiers stabilize emulsion ? Name two emulsifiers.
Answer:
Emulsifying agent: The third substance which is added in small amounts to an emulsion to stabilize the emulsion is called emulsifying agent.
—> The emulsifying agent forms an interfacial film between suspended particles and the medium.
Eg : 1) Carbon black stabilizes water in oil type emulsions.
2) Casein and silica stabilize oil in water type emulsions.

Long Answer Questions

Question 1.
Explain the terms absorption, adsorption and sorption. Describe the different types of adsorption.
Answer:
Absorption : The uniform distribution of a substance through out the bulk of the solid substance is known as absorption.
Eg : Chalk stick dipped in ink.

Adsorption : The accumulation (or) concentration of a substance on the surface, rather than in the bulk of solid (or) liquid is known as adsorption.
Eg : Adsorption of gases like O₂, H₂, Cl₂, etc., on charcoal

Sorption : In case of some substances both adsorption and absorption takes place. This phenomenon is called sorption.

Types of adsorption : On the basis of forces present between adsorbate and adsorbent molecules, adsorption classified into two types :
1) Physisorption, 2) Chemisorption.

1) Physisorption: If the forces that are responsible for the adsorption of adsorbate molecules on the surface of the adsorbent are weak vander waal’s for the adsorption is known as physisorption.
Eg : Adsorption of H₂, on O₂, on charcoal.
Characteristics :

• It is a weak adsorption.
• Enthalpy of adsorption is low (20 – 40 KJ / mole).
• It is reversible and occurs rapidly.
• It decrease with increase in temperature.
• It increase with increase in pressure.
• It is multilayered and not specific.

2) Chemisorption : If the forces that are responsible for the formation of adsorbate molecules on the source of adsorbent are chemical forces, the adsorption is called chemisorption.
Eg : Adsorption of H₂ on Ni – metal surface.
Characteristics :

• It is a strong adsorption. ,
• Enthalpy of adsorption of high (80 – 240 KJ / mole).
• It is irreversible and occurs slowly.
• It increases with increase of temperature and finally decrease.
• Pressure has no effect on chemisorption.
• It is unilayered and specific.

Question 2.
Discuss the characteristics of physical adsorption.
Answer:
Characteristics of physical adsorption :

• Lack of specificity : Physisorption is not specific. In this adsorption the given surface of adsorbent doesnot show any preference for a particular gas as the forces present are variderwaal’s forces (Universal).
• Nature of adsorbate : The amount of gas adsorbed by a solid depends on the nature of gas. Easily liquefiable gases are readily adsorbed.
• The gas with high critical temperature value is easily liquefied and gets readily adsorbed.
• Reversible Nature : Physical adsorption is reversible and occurs rapidly at low temperatures.
• Surface area of adsorbent: The extent of adsorption increase with increase in surface area of adsorbent. .
• Finely divided metals and porous substances having large surface area are good adsorbents.
• Enthalpy of adsorption : Physisorption is an exothermic process but the enthalpy of adsorption is low (20 – 40 KJ/mole). This is due to the presence of Vander waal’s forces.

Question 3.
Discuss the characteristics of chemisorption. .
Answer:

1. High specificity: Chemisorption is highly specific and it will only occur if there is some possibility of chemical bonding between adsorbate and adsorbent.
2. Irreversibility : Chemisorption involves compound formation and is irreversible in nature.
   • Chemisorption is exothermic and increases with increase of temperature.
3. Surface area : Chemisorption increase with increase of surface area of the adsorbent.
4. Enthalpy of adsorption : Enthalpy of chemisorption is high (80 – 240 KJ/mole).

Question 4.
Compare and contrast the phenomenon of physisorption and chemisorption.
Answer:
Characterstics of Physical adsorption :
Physisorption: If the forces that are responsible for the adsorption of adsorbate molecules on the surface of the adsorbent are weak vander waal’s for the adsorption is known as physisorption.
Eg : Adsorption of H₂ on O₂ on charcoal.

• Lack of specificity : Physisorption is not specific. In this adsorption the given surface of adsorbent doesnot show any preference for a particular gas as the forces present are vander waal’s forces (Universal). ‘
• Nature of adsorbate : The amount of gas adsorbed by a solid depends on the nature of gas.
• Easily liquefiable gases are readily adsorbed:
  The gas with high critical temperature value is easily liquefied apd gets readily adsorbed.
• Reversible Nature : Physical adsorption is reversible and occurs rapidly at low temperatures.
• Surface area of adsorbent: The extent of adsorption increase with increase in surface area of adsorbent.
• Finely divided metals and porous substances having large surface area are good adsorbents. .
• Enthalpy of adsorption : Physisorption is an exothermic process but the enthalpy of adsorption is low (20 – 40 KJ/mole). This is due to the presence of Vander waal’s forces.

2) Chemisorption : If the forces that are responsible for the formation of adsorbate molecules on the source of adsorbent are chemical forces, the adsorption is called chemisorption.
Eg : Adsorption of H₂ on Ni – metal surface.

1. High specificity: Chemisorption is highly specific and it will only occur if there is some possibility of chemical bonding between adsorbate and adsorbent.
2.  Irreversibility : Chemisorption involves compound formation and is irreversible in nature.
   • Chemisorption is exothermic and increases with increase of temperature.. ,
3. Surface area : Chemisorption increase with increase of surface area of the adsorbent.
4. Enthalpy of adsorption : Enthalpy of chemisorption is high (80 – 240 KJ/mole).

Question 5.
What is an adsorption isotherm ? Discuss the phenomenon of adsorption of gases on solids with the help of Freundlich adsorption isotherm.
Answer:
Adsorption Isotherm : The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve known as adsorption Isotherm.

• Freundlich adsorption isotherm equation is
  \(\frac{\mathrm{x}}{\mathrm{m}}\) = k, P^1/n
  x = mass of the gas adsorbed
  m = mass of the adsorbent
  P, k and n are constants.
• k and n depend on the nature of the adsorbent and the gas at a particular temperature.
• The relation ship is generally represented in the form of a curve where x/m is plotted against pressure.
• These curves indicate that at a fixed pressure there is a decrease in physical adsorption with rise of temperature.
  
  \(\frac{\mathrm{x}}{\mathrm{m}}\) = k.P^1/4
  Applying logarithm
  Log \(\frac{\mathrm{x}}{\mathrm{m}}\) = log k + \(\frac{\mathrm{1}}{\mathrm{n}}\) log p.
• The validity of Freundlich isotherm can be verified by plotting log \(\frac{\mathrm{x}}{\mathrm{m}}\) on y-axis and log P on x-axis. If it is a straight line then the isotherm is valid.
• The slope of the straight line give the value of \(\frac{\mathrm{1}}{\mathrm{n}}\)
• The intercept on the y-axis gives the value of log k.

Question 6.
Give a detailed account of applications of adsorption.
Answer:
Applications of adsorption : The phenomenon of adsorption funds a number of applications. Important ones are listed here.

1. Production of high vacuum: The traces of air still remaining in the vessel evacuated by a vacuum pump to give high vacuum can be adsorbed by charcoal.
2. Gas masks : Gas mask, a device which consists of activated charcoal or mixture of adsorbents is usually used by coal miners to adsorb poisonous gases during breathing.
3. Control of humidity : Silica gel and alumina gel are used as adsorbents for removing mixture and controlling humidity of air in rooms.
4. Removal of colouring matter from solutions : Animal charcoal removes colours of impure coloured solutions by adsorbing impurities responsible for the colour.
5. Separation of inert gases : Due to the difference in degree of adsorption of gases by charcoal, a mixture of noble gases can be separated by adsorption on coconut charcoal at different temperatures.
6. In curing diseases : A number of drugs kill germs by getting themselves adsorbed on germs.
7. Froth floatation process : A low grade sulphide ore is concentrated by separating silica and other earthy matter by this method using pine oil and frothing agent.
8. Adsorption indicators : Surfaces of certain precipitates such as silver halides have the property of adsorbing some dyes like eosin, fluorescein, etc. and thereby producing a characteristic colour change at the end point in argentometric titrations.
9. Chromatographic analysis : Chromatographic analysis based on the phenomenon of adsorption finds a number of applications in analytical and industrial methods.

Question 7.
What is catalysis ? How is catalysis classified ?* Give four examples for each type of catalysis.
Answer:
Catalysis: A substance which alters the rate of a chemical reaction without itself being consumed in the process, is called a catalyst.
The action of catalyst in altering the rate of a chemical reaction is called catalysis.
Types of catalysis : Catalysis is classified into two types as .
a) Homogeneous catalysis and
b) Heterogeneous catalysis.
Homogeneous catalysis : The catalytic process in which the catalyst is present in the same phase as that of reactants, is known as homogeneous catalysis.

Heterogeneous catalysis: The catalytic process in which the catalyst is present in a phase different from that of reactarts is known as heterogeneous catalysis.

Question 8.
Discuss the mechanism of heterogeneous catalysis.
Answer:
Adsorption theory of heterogeneous catalysis :
Adsorption theory explains the mechanism of heterogeneous catalysis.

• The modern adsorption theory is the combination of the intermediate compound formation theory and the old adsorption theory.
• The mechanism of catalysis involves five steps.
• Diffusion of the reactants to the surface of the catalyst.
• Adsorption of the reactant molecules on the surface of the catalyst.
• Occurrence of chemical reaction on the catalyst’s surface through formation of an intermediate.
• Desorption of reaction products from the catalyst surface, and thereby making the surface available again for more reaction to occur.
• Diffusion of reaction products away from the catalyst’s surface.

Question 9.
What are enzymes ? Explain in detail the enzyme catalysis, with necessary examples.
Answer:
Enzymes are complex nitrogenous organic compounds which are produced by living plants and animals.

• These act as specific catalysts in biological reactions.
• These catalyse the numerous reactions that occur in the bodies of animals and plants to maintain the life process.
  Mechanism of enzyme catalysis :
  
  The enzyme – catalysed reactions may be considered to proceed in two steps.
  Step 1 : Binding of substrate to enzyme to form an activated complex (ES^≠)
  E + S → ES^≠
  Step 2 : Decomposition of the activated complex to form product.
  ES^≠ → E + P

Factors influencing the catalytic activity of enzymes :

• Highly specific nature: One catalyst catalyses only one reaction. Each enzyme is specific for a given reaction.
  Eg : The enzyme urease catalyses the hydrolysis of urea only and not any other amide
• Highly active under optinum temperature: The rate of an enzyme reaction is maximum at a temperature called optimum temperature.
• The optimum temperature range for enzymatic activity is 298 – 310 K.
•  Highly active under optimum pH : The rate of an enzyme – catalysed reaction is maximum at a particular pH called optimum pH which lies between 5-7
•  Increasing activity in presence of activators and co-enzymes : The enzymatic activity is increased in presence of co-enzymes (or) in presence of activators such as Na⁺, Mn⁺² etc.,
• Influence of inhibitors and poisons: The inhibitors (or) poisons interact with the active functional groups on the enzyme surface and reduce or completely destroy the catalytic activity.
  i) Inversion of Cane sugar: Enzyme : Invertase
  

Question 10 & 11.
What are colloidal solutions ? How are they classified ? Give examples.
Answer:
A heterogeneous system in which one substance is dispered as large particles in another substance is called colloidal solution.

• In colloidal solution the particle size of the dispersed phase is of the order of 1 mμ – 1 μ where as in true solution the particle size of the solute are the order of mp or less.
• Colloidal solution is a heterogeneous binary system where as true solution is a homogeneous binary system.
  On the basis of physical states of dispersed phase and dispersion medium colloidal solutions are classified into eight types.
  

On the basis of nature of dispersion medium colloids are classified as follows.
If the dispersion medium is ‘air’, the sols are called aerosols.
Eg : Smoke.

If the dispersion medium is ‘water’ the sols are called Hydrosol (or) aquasol.
Eg: starch solution.
If the dispersion medium is ‘alcohol’ then the sols are called alcosols.

The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.
The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution. Ex : Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 12.
How are colloids classified on the basis of the nature of interaction between a dispersed phase and a dispersion medium ? Describe an important characteristic of each class. Which of the sols need stabilising agents for preservation ?
Answer:
The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.

The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Lyophilic colloids are much stable than lyophobic colloids. Avery small amount of electrolyte can precipitate the lyophobic colloids. But their precipitation can be prevented by the addition of small amounts of lyophilic colloids. This lyophilic colloids form a protecting layer around the lyophobic colloids and protect them from precipitation (coagulation). Lyophilic colloids used for this purpose are called protective colloids.
Ex : Gelatin, gum, starch sols are generally used as protective colloids.

Question 13.
What are micelles ? Discuss the mechanism of micelle formation and cleaning action of soap.
Answer:
Micelles : Some substances which at low concentrations behave as normal strong electrolytes, but at high concentrations exhibit colloidal behaviour due to formation of aggregates. The aggregated particles thus formed are called micelles.
Eg : Stearate ions (C₁₇H₃₅COO^–) associate together in high concentration, in a solution of soap in water and forms a micelle.

Mechanism of micelle formation : Let us take the example of soap solution. Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOC^– Na⁺ (E.g., sodium strearate CH₃ (CH₂)₁₆ COO^– Na⁺, which is a major component of many bar soaps). When dissolved in water, it dissociates into RCOO^– and Na⁺ ions. The RCOO^– ions, however, consist of two parts – a long hydrocarbon chain R (also called non-polar ’tail) which is hydrophobic (water repelling), and a polar group COO^–(also called polar-ionic ‘head), which is hydrophilic (water loving).

The RCOO^– ions are, therefore, present on the surface with their COO^– groups in water and the hydrocarbon chains (R) staying away from it and remain at the surface. But at critical micelle concentration, the COO^– irons are pulled into the bulk of the solution and are aggregated to form a spherical shape with their hydro¬carbon chains pointing towards the centre of the sphere with COO- part remaining outward on the surface of the sphere. The aggregate thus formed is known as ‘ionic micelle’. These micelles may contain as many as 100 simple molecules.

Similarly, in case of detergents, E.g., sodium laurylsulphate. CH₃ (CH₃)₁₁ SO₃^–Na⁺, the polar group is -SO₃^–along with the long hydrocarbon chain. Hence, the mechanism of micelle formation here also is same as that of soaps.
Soap is sodium stearate, C₁₇H₃₅COONa, in water soap gives the ions stearate anion and sodium ion.
C₁₇H₃₅COONa → C₁₇H₃₅COO^– + Na⁺ (Stearate ion)
The anion of soap = C₁₇H₃₅COO^–

Cleaning action of soap : Soap anions form a micelle. The grease or dirt of the cloth are absorbed into the interior of the micelle. The tails of the anion are pegged into micelle and these micelle are washed away with the soap solution.

Soap thus, functions as an emulsifying agent for the water dirt emulsion. The emulsified grease or dirt is then washed away with soap solution.

Question 14.
Describe the properties of colloids with necessary diagrams wherever necessary.
Answ:
i) Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles.. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

This is an optical property exhibited by colloidal solution. This phenomenon is clearly seen with a microscope placed at right angles to the path of light. Colloidal particles become self luminous due to absorption of light.
A true solution does not show Tyndall effect.

ii) Brownion movement: This is a kinetic property of colloidal solution.

When a colloidal solution is examined by ultramicroscope, the colloidal particles are seem to be moving in a rapid zig-zag motion.
This rapid motion of colloidal particles is called Brownian movement.

This motion is due to unequal bombardment of colloidal particles by molecules of dispersion medium. Smaller the colloidal particles the more rapid is the Brownian movement.

iii) Charge on colloidal particles : The charge on the colloidal particles is due to electron capture by sol particles during electro dispersion of metals and due to preferrential adsorption ions from solution (of) due to formulation of electrical double layer.

When a dilute AgNO₃ solution is added to a dilute KI solution taken in excess the precipitated AgI adsorbs I^– ions present in excess and a negatively charged AgI colloidal solution is formed. However when dilute KI solution is added to dilute AgNO₃ solution taken in excess, AgI precipitate adsorbs Ag⁺ ions present in excess and a positively charged Agl colloidal solution is formed. Generally the ion common to dispersed particle is adsorbed.
AgI/I^–
Negatively charged
AgI/Ag⁺
Positively charged.

iv) Zeta Potential: In a colloidal sol the charges of opposite signs on the fixed’and diffused parts of the double layer results in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charge is called electro kinetic potential (or) zeta potential.

v) electrophoresis : When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.
If the movement of colloidal particles is arrested by some suitable means, the dispersion medium moves in opposite direction. This phenomenon is termed “electro osmosis”.

vi) Coagulation : The stability of the lyophilic sols is due to the presence of charge on the colloidal particles. If tins charge is neutralised the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. This process of settling down of colloidal particles is called coagulation (or) flocculation (or) precipitation.

Question 15.
What are emulsions ? How are they classified ? Describe the applications of emulsion. [T.S. Mar. 18, 16]
Answer:
Emulsion: The colloidal system in which a dispersion of finely divided droplets of a liquid in another liquid medium is called Emulsion.
Ex: Milk. .
In Milk, the droplets of liquid fat are dispersed in water. This is an example for oil in water type emulsion.
Classification of emulsions : Emulsions are classified, into two classes. These are
a) oil in water (O/W) and ‘
b) water in oil (W/O), (O = Oil; W = Water).
These emulsions are classified as such depending on which is dispersed phase and which is dispersion medium.

a) Oil in Water (O/W) type emulsions : In this type of emulsions, the dispersed phase is oil and the dispersion medium is water.
Ex : Milk, liquid, fat (oil) in water.
Vanishing cream; fat in water.

b) Water in Oil (W/O) type emulsions : In this type of emulsions, the dispersed phase is water and the dispersion medium is oil.
Ex : Stiff greases : water in lubrication oils
Cod liver oil : water in cod liver oil
Cold cream : water in fat.

Applications of Emulsions : Emulsions are useful

• In the digestion of fats in intestines.
•  In washing processes of clothes and crockery.
•  In the preparation of lotions, creams, ointments in pharmaceutical and cosmetics.
• In the extraction of metals (froth flotation).
• In the conversion of cream into butter by churning.
• To break oil and water emulsions in oil wells.
• In the preparation of oily type of drugs for easy adsorption to the body.

Intext Questions

Question 1.
Write any two characteristics of Chemisorption. .
Solution:

1. High specificity: Chemisorption; is highly specific and it will only occur when adsorbent and adsorbate molecules can chemically react with each other.
   Eg.: oxygen is adsorbed on metals by oxide formation.
2. Surface area : Chemisorption increases with increase in surface area of the adsorbent.

Question 2.
Why does physisorption decrease with the increase of temperature ?
Solution:
Physisorption is an exothermic process :
Solid (Adsorbent) + Gas (Adsorbate) ⇌ Gas adsorbed on solid + Heat
When temperature is increased, the equilibrium shifts towards the backward direction to neutralise the effect of the change (Le-Chatelier’s principle). So, physisorption decreases with increase in temperature.

Question 3.
Why are finely powdered substances more effective adsorbents than their non powdered crystal forms ? .
Solution:
Finally divided substances provide increased surface area for adsorption which is not available to such extent in their crystalline forms. That’s why the powdered forms are more effective than crystalline forms for the purpose of adsorbents.

Question 4.
Hydrogen used in Haber’s process is obtained by reacting methane with steam in presence of NiO as catalyst. The process is known as steam reforming. Why is it necessary to remove CO formed in steam reforming when ammonia is obtained by Haber’s process ?
Solution:
Carbon monoxide (CO) acts as a poison for catalyst iron and promoter molybdenum in Haber’s process, i.e., the efficiency of catalyst and promoter is decreased. It also combines with Fe to form iron carbonyl, Fe(CO)₅ which interfere in the production of ammonia. Hence, CO must be removed from the reaction mixture.

Question 5.
Why is ester hydrolysis slow in the beginning but is fast after sometime ?
Solution:
The chemical equation for ester hydrolysis is as follows :

Carboxylic acid produced during hydrolysis releases H+ ions in the solution which act as catalyst (auto-catalysis) for the reaction. Therefore, ester hydrolysis is slow in the beginning but becomes faster after sometime.

Question 6.
What is role of desorption in the process of adsorption catalysis.
Solution:
The reaction products formed on the catalyst surface get detached from the surface as a result of desorption, thereby making the surface available again for more reactant particles.

Question 7.
What modification can you suggest in the Hardy-Schulze law ?
Solution:
According to Hardy Schulze law, the ions carrying charge opposite to the charge on sol particles neutralise their charge and thus cause their coagulation or precipitation. But actually, the sol carrying these ions also get coagulated since the ions neutralize their charge. So, Hardy Schulze law can be modified as :

“When equimolar proportions of two oppositely charged sols are mixed, they mutually neutralize their charge and both get coagulated”.

Question 8.
Why is it essential to wash the precipitate in gravimetric chemical analysis with wash liquid before drying and weighing it quantitatively ?
Solution:
Some of the reactant ions my be adsorbed or may adhere to the surface of the particles of the precipitate formed during an ionic reaction. In order to remove these reactant ions, the precipitate should.be washed with water. If this is not done, an error may be produced during quantitative analysis.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 3(b) Chemical Kinetics Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 3(b) Chemical Kinetics

Very Short Answer Questions

Question 1.
Define the speed or rate of a reaction.
Answer:
The change in the concentration of a reactant (or) product in unit time is called the speed or rate of a reaction.
(or)
The decrease in the concentration of a reactant (or) increases in the concentration of product per unit time.

Question 2.
Assuming that the volume of the system is constant, derive the expression for the average rate of the system R → P in terms of R and R [time = ‘t’ sec] [R = reactant, P = product].
Answer:

Question 3.
What are the units of rate of reaction ?
Answer:
Units of rate of reaction – moles/Lit × sec – moles. Lit^-1 . sec^-1

Question 4.
Draw the graphs that relate the concentrations .(C) of the reactants and the reaction times (t) and the concentrations of the products (C) and the reaction times (t) in chemical reactions.
Answer:

Question 5.
Write the equation for the rate of the reaction.
\(5 \mathrm{Br}_{(\mathrm{aq})}^{-}+\mathrm{BrO}_{3(\mathrm{aq})}^{-}+6 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 3 \mathrm{Br}_{2(\mathrm{aq})}+3 \mathrm{H}_2 \mathrm{O}_{(\text {) }}\)
Answer:
Given reaction is
\(5 \mathrm{Br}_{(\mathrm{aq})}^{-}+\mathrm{BrO}_{3(\mathrm{aq})}^{-}+6 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 3 \mathrm{Br}_{2(\mathrm{aq})}+3 \mathrm{H}_2 \mathrm{O}_{(\text {) }}\)
Rate of reaction = \(=-\frac{1}{5} \frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}=\frac{-\Delta\left[\mathrm{BrO}_3^{-}\right]}{\Delta \mathrm{t}}=\frac{-1}{6} \frac{\Delta\left[\mathrm{H}^{+}\right]}{\Delta \mathrm{t}}=\frac{-1}{3} \frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta \mathrm{t}}=\frac{1}{3} \frac{\Delta\left[\mathrm{H}_2 \mathrm{O}\right]}{\Delta \mathrm{t}}\)

Question 6.
What is rate law ? Illustrate with an example.
Answer:
The equation that describes mathematically the dependence of the rate of a reaction on the concentration terms of the reactions is known as the rate equation (or) rate law.
Eg : 2A + 3B → 3C
Rate, of the given reaction ∝ [A]² [B]³

Question 7.
Mention a reaction for which the exponents of concentration terms are not the same as their stoichiometric coefficients in the rate equation.
Answer:
The following are the reactions for which the exponents of concentration terms are not the same as their stoichiometric coefficients in the rate equation.
CHCl₃ + Cl₂ → CCl₄ + HCl
rate = k[CHCl₃] [Cl₂]^1/2
CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
rate = k[CHCOOC₂H₅] [H₂O]^1/2

Question 8.
Define Order of a reaction. Illustrate your answer with an example. [T.S. Mar. 15]
Answer:
Order of a reaction : The sum of the powers of the concentration terms of the reactants present in the rate equation is called order of a reaction.
Order of a reaction can be 0, 1, 2, 3, and even a fraction
Eg. : 1) N₂O₅ → N₂O₄ + \(\frac{1}{2}\) O₂
rate ∝ [N₂O₅]
∴ It is a first order reaction.

2) 2N₂O → 2N₂ + O₂
rate ∝ [N₂O]²
∴ It is 2^nd order reaction.

Question 9.
What are elementary reactions ?
Answer:
The reactions taking place in one step are called elementary reactions.

Question 10.
What are complex reactions ? Name one Complex reaction.
Answer:
A sequence of elementary reactions, reactants give the products, the reactions are called complex reactions.
Eg: Oxidation of Ethane to CO₂ and H₂O passes through a series of intermediate steps in which alcohol, aldehyde and acid are formed.

Question 11.
Give the units of rate constants for Zero, first order and second order reactions.
Answer:

Question 12.
Define molecularity of a reaction. Illustrate with an example.
Answer:
The number of reacting species (atoms, ions or molecules) taking parts in an elementary reaction, which must colloid simultaneously to bring about a chemical reaction is called molecularity of a reaction.
NH₄NO₂ → N₂ + 2H₂O (Unimolecular)
2HI → H₂ + I₂ (Bimolecular)
2NO + O₂ → 2NO₂ (Trimolecular)

Question 13.
What is rate determining step in a complex reaction ?
Answer:
The overall rate of a reaction is controlled by the slowest step in a reaction is called the rate determining step in a complex reaction.

Question 14.
Give the mechanism for the decomposition reaction of H₂O₂ in alkaline medium catalysed by I^– ions.
Answer:
Chemical equation of decomposition of H₂O₂ in alkaline medium is

Mechanism:

• It is a first order reaction w.r.t. both H₂O₂, I^–
• This reaction takes place is two steps
  1. H₂O₂ + I^– → H₂O + IO^–
  2. H₂O₂ + IO^– → H₂O + I^– + O₂

Question 15.
Write the equation relating [R], [R]₀ and reaction time ‘t’ for a zero order reaction. [R] = concentration of reactant at time ‘t’ and [R]₀ = initial concentration of reactant.
Answer:
Zero order reaction Rate constant k = \(\frac{[R]_0-[R]}{t}\)
Eg : Decomposition of NH₃

Rate = k [NH₃]⁰ = k
∴ \(\frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}\) = k ⇒ ∆x = ∆t × k

Question 16.
Draw the graph that relates the concentration ‘R, of the reactant and ‘t’ the reaction time for a zero Order reaction.
Answer:

Question 17.
Give two examples for zero Order reactions. [A.P. Mar. 19]
Answer:
Examples for zero order reactions

Question 18.
Write the Integrated equation for a first order reaction in terms of [R], [R]₀ and ‘t’.
Answer:
[Rl = Concentration of reaction after time ‘t
[R]₀ = Initial concentrations of reactant
∴ k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)
This is the integrated Equation for a first order reaction.

Question 19.
Give two examples for gaseous first order reactions. [Mar. 14]
Answer:
The following are the examples for gaseous first order reactions
N₂O_5(g) → N₂O_4(g)g + \(\frac{1}{2}\) O_2(g)
SO₂Cl_2(g) → SO_2(g) + Cl_2(g)

Question 20.
For the reaction A(g) → B(g) + C(g), write the integrated rate equation in terms of total pressure ‘P and the partial pressures p_Ap_Bp_C.
Answer:
Given
A_(g) → B_(g) + C_(g)
p = p_A + p_B + p_C
k = (\(\frac{2.303}{t}\)) log (\(\frac{\mathrm{p}_0}{\mathrm{p}_{\mathrm{A}}}\))
k = (\(\frac{2.303}{t}\)) log (\(\frac{\mathrm{p}_0}{2 \mathrm{p}_{\mathrm{i}}-\mathrm{p}_{\mathrm{t}}}\))
p₀ = initial pressure
P_i = Total pressure
p_A, p_B, p_C are partial pressures.

Question 21.
What is half-life of a reaction ? Illustrate your answer with an example. .
Answer:
The time required for the initial concentration of the reactants to become half of it’s value during the progress of the reaction is called half life (t_1/2) of reaction.
Eg : The radio active of C-14 is exponential with a half life of 5730 years.

Question 22.
Write the equation relating the half-life (t_1/2) of a reaction and the rate constant ‘k’ for first order reaction.
Answer:
Half life of first order reaction (t_1/2) = \(\frac{0.693}{k}\)
k = rate constant

Question 23.
Write the equation useful to calculate half-life (t_1/2) values for zero and first order reactions.
Answer:
Half life (t_1/2) of zero order reaction
At half life (t_1/2) [R] = \(\frac{[\mathrm{R}]_0}{2}\)
∴ t_1/2 = \(\frac{[\mathrm{R}]_0 / 2}{\mathrm{k}}\)
∴ t_1/2 = \(\frac{\left[\mathrm{R}_0\right]}{2 \mathrm{k}}\)
Half life of a first order reaction
t_1/2 = \(\frac{0.693}{\mathrm{k}}\)
k = rate constant

Question 24.
What are pseudo first order reactions ? Give one example.
Answer:
First order reactions whose molecularity is more than one are called pseudo first order reactions.

Order = 1
molecularity = 2

Question 25.
Write the Arrhenius equation for the rate constant (k) of a reaction.
Answer:
Arrhenius equation is
k = A × e^-Ea/RT
k = Rate constant
E_a = activation energy
R = gas constant
T = Temperature (K)

Question 26.
By how many times the rate constant increases for a rise of reaction temperature by 10°C ?
Answer:
For every 10°C rise of temperature rate constant of chemical reactions may be doubled (some times tripled).

Question 27.
Explain the term ‘activation energy’ of a reaction with a suitable diagram.
Answer:
Activation Energy : The energy required to for an intermediate called activated complex (C) during a chemical reaction is called activation energy.

Question 28.
Write the equation which relates the rate constants k₁ and k₂ at temperatures T₁ and T₂ of a reaction.
Answer:
\(\log \left(\frac{\mathrm{k}_2}{\mathrm{k}_1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]\)
E_a = activation energy; R = Universal gas constant.

Question 29.
What is collision frequency (Z) of a reaction ? How is rate related to it for the reaction A + B → Products.
Answer:
The number of collisions per second per unit volume of the reaction mixture is called collision frequency (Z). Fora bimolecular elementary reactions.
A + B → products; Rate = Z_AB.e – E_a/RT.

Question 30.
Draw the graphs between potential energy – reaction coordinates for catalysed and uncatalysed reactions.
Answer:

Question 31.
What is the effect of temperature on the rate constant ?
Answer:
Most of the chemical reactions are accelerated by increase of temperature.

For a chemical reaction with rise of temperature by 10°C the rate constant is nearly doubled k = A.e_-Ea/RT

Short Answer Questions

Question 1.
Define average rate of a reaction. How is the rate of reaction expressed in terms of change in the concentration of reactants and products for the following reactions.
1) 2HI_(g) → H_2(g) + I_2(g)
2) Hg_(l) + Cl_2(g) → HgCl_2(g)
3) 5 Br_(aq) + \(\mathrm{BrO}_{3(\mathrm{aq})}^{-}\) + \(6 \mathrm{H}_{(\mathrm{aq})}^{+}\) → 3 Br_2(aq) + 3 H₂O_(l)
Answer:
Average rate of a reaction: The change in the concentration of any one of the reactants or products per unit time is called average rate of a reaction.

Question 2.
What is rate equation ? How is it obtained ? Write the rate equations for
1) 2NO_(g) + O₂ (g) → 2NO₂ (g)
2) CHCl₃ + Cl₂ → CCl₄ + HCl
3) CH₃COOC₂H₅ (l) + H₂O(l) → CH₃COOH (aq) + C₂H₅OH(aq)
Answer:
The equation that describes mathematically the dependence of the rate of a reaction on the concentration terms of the reactions is known as the rate equation (or) rate law.
Eg, : 2A + 3B → 3C
Rate of the given reaction ∝ [A]² [B]³
How to obtain: Each concentration term is raised to some power which may or may not be same as the stoichiometric coefficient of the reacting species
i) 2NO_(g) + O_2(g) → 2NO_2(g)
\(\frac{\Delta[\mathrm{R}]}{\Delta t}\) = k [NO]² [O₂]

ii) CHCl₃ + Cl₂ → CCl₄ + HCl
\(\frac{\Delta[\mathrm{R}]}{\Delta t}\) = k [CHCl₃] [Cl₂]^1/2

iii) CH₃COOC₂H_5(l) + H₂O_(l) → CH₃COOH_(aq) + C₂H₅OH_(aq)
\(\frac{\Delta[\mathrm{R}]}{\Delta t}\) = k [CH₃COOC₂H₅].

Question 3.
Define and explain the order of a reaction. How is it obtained experimentally?
Answer:
Order of a reaction :. The sum of the powers of the concentration terms of the reactants
present in the rate equation is called order of a reaction.
Order of a reaction can be 0, 1, 2, 3, and even a fraction
Eg. : 1) N₂O₅ → N₂O₄ + \(\frac{1}{2}\) O₂
rate ∝ [N₂O₅]
∴ It is a first order reaction.

2) 2N₂O → 2N₂ + O₂
rate ∝ [N₂O]²
∴ It is 2^nd order reaction.

Order of a reaction can be determined experimentally
Half – Time (t1) method : The time required for the initial concentration (a) of the reactant to become half its value (a/2) during the progress of the reaction is called half-time (t_1/2) of the reaction.
A general expression for the half life, (t_1/2) is given by
t_1/2 ∝ \(\frac{1}{a^{n-1}}\)
Therefore, for a given reaction two half time values (t’_1/2and t”_1/2) with initial concentrations a’ and a” respectively are determined experimentally and the order is established from the equation.
\(\left(\frac{t_{1 / 2}^{\prime}}{t_{1 / 2}^n}\right)=\left(\frac{a^n}{a^{\prime}}\right)^{n-1}\)
Where ‘n’ is the order of the reaction.

Question 4.
What is “moleculartiy” of a reaction ? How is it different from the ‘order’ of a reaction ? Name one bimolecular and one trimolecular gaseous reactions. [T.S. Mar. 17] [Mar. 14]
Answer:
The number of reacting species (atoms, ions or molecules) taking parts in an elementary reaction, which must colloid simultaneously to bring about a chemical reaction is called molecularity of a reaction.
NH₄NO₂ → N₂ + 2H₂O (Unimolecular)
2HI → H₂ + I₂ (Bimolecular)
2NO + O₂ → 2NO₂ ( Trimolecular)

• Molecularity has only integer values (1, 2, 3 )
• It has nop zero, non fraction values while order has zero, 1, 2, 3 and fractional values.
• It is determined by reaction mechanism, order is determined experimentally.

Question 5.
Derive the integrated rate equation for a zero order reaction.
Answer:
Zero order reaction is the reaction in which rate of reaction does not depends on the concentration of reactants.
R → P
Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k [R]°
Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\); d[R] = – k. dt
Integrating on both sides
[RJ = -kt + I …………….. (1)
I – Integration constant
At t = 0 → R = [R]₀ initial concentration
I = [R]₀
Substituting I = [R]₀ in the above equation (1)
[R] = – kt + [R]₀
k = \(\frac{\left[R_0\right]-[R]}{t}\)
This is the integrated rate equation for a zero order reaction.

Question 6.
Derive an integrated rate equation for a first order reaction.
Answer:
In first order reactions rate depends on only one concentration term.
R → P
Rate = k [R]; -M = – k. dt
Integration on both sides
ln [R] = – kt + I
I = Integration constant
At t = 0, [R] = [R]₀ ⇒ l_n[R]₀ = I
Substituting I = ln [R] in the above equation (1)
ln [R] = – kt + ln [R]₀
ln \(\frac{[\mathrm{R}]}{\left[\mathrm{R}_0\right]}\) = -kt ………………. (2)
k = \(\frac{1}{\mathrm{t}} \ln \frac{\left[\mathrm{R}_0\right]}{[\mathrm{R}]}\)
Taking antilog on both sides of eq. (2)
R = [R]₀. e^-kt
This is first order rate equation.

Question 7.
Derive an integrated rate equation in terms of total pressure (P) and the partial pressures p_A, p_B, p_C for the gaseous reaction A(g) → B(g) + C(g).
Answer:
Given first order gas phase equations
A(g) → B(g) + C(g)
Let p_i be the initial pressure of A and pt the total pressure at time’t’. Integrated rate equation for such a reaction can be derived as
Total pressure p_t = p_A + p_B + p_C (pressure units)
p_A, p_B and p_C are the partial pressures of A, B and C, respectively.
If x atm be the decrease in pressure of A at time t and one mole each of B and C is being formed, the increase in pressure of B and C will also be x atm each.

Thus the rate expression for 1st order gaseous phase reaction derived.

Question 8.
What is half-life (t_1/2) of a reaction ? Derive the equations for the ‘half-life’ value of zero and first order reactions.
Answer:
The time required for the initial concentration of the reactants to become half of it’s value during the progress of the reaction is called half life (t_1/2) of reaction.
Eg : The radio active of C-14 is exponential with a half life of 5730 years.
Half life of zero order reaction :

Question 9.
What is Arrhenius equation ? Derive an equation which describes the effect of rise of temperature (T) on the rate constant (k) of a reaction.
Answer:
The temperature dependence of the rate of reaction can be explained by Arrhenius equation.
k = A.e^-Ea/KT
A = Arrhenius factor.
E_a = activation Energy
R = gas constant
T = Temperature. (K)
k = A.e^-Ea/KT
ln k = ln A – E_a/RT
2.303 log k = 2.303 log A – E_a/RT
T₁, T₂ are Temperatures
k₁ is rate constant at temperature T₁
k₂ is rate constant at temperature T₂
∴ 2.303 (log k₂ – log k₁) = \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_2}-\frac{1}{\mathrm{~T}_1}\right)\)
\(\log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]\)
The above equation describes the effect of temperature (T) on rate constant (k)

Question 10.
Discuss the effect of catalyst on the kinetics of a chemical reaction with a suitable diagram. [T.S. Mar. 15]
Answer:
The substance which increases the rate of reaction without being consumed in the chemical reaction is called a catalyst.
In presence of catalyst the reaction proceeds in a new path which lowers the activation energy.

Question 11.
Describe the salient features of the collision theory of reaction rates of bimolecular reactions. [T.S. Mar. 18, 16; A.P. Mar. 17]
Answer:
Collision theory of reaction rate bimolecular reactions salient features.

• The reaction molecules are assumed to be hard spheres
• The reaction is postulated to occur when molecules collide with each other.
• The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).
• For a bimolecular elementary reaction
  A + B → products
  Rate = Z_AB. e^-E_a/RT ; Z_AB = collision frequency.
• All collisions do not lead to product formation.
• The collisions with sufficient kinetic energy (Threshold energy) are responsible for product formation. These are called as effective collisions.
• To account for effective collisions a factor p called to probability factor or steric factor is introduced.
  Rate = P Z_AB. e^-E_a/RT

Question 12.
Explain the terms
a) Activation energy (E_a)
b) Collision frequency (Z) :
c) Probability factor (P) with respect to Arrhenius equation.
Answer:
a) Activation Energy: The energy required to for an intermediate called activated complex (C) during a chemical reaction is called activation energy.

Diagram showing plot of potential energy vs reaction coordinate

b) Collision frequency: The number of collisions per second per unit volume of the reaction mixture is called collision frequency (Z). For a bimolecular elementary reactions
A + B → products
Rate = Z_AB. e^-E_a/RT

c) Probability factor (P) with respect to Arrhenius equation : To account for effective collisions a factor p called to probability factor or steric factor is introduced.
Rate = P Z_AB. e^-E_a/RT

Long Answer Questions

Question 1.
Explain the following terms with suitable examples.
a) Average rate of a reaction
b) Slow and fast reactions
c) Order of a reaction
d) Molecularity of a reaction
e) Activation energy of reaction.
Answer:
a) Average rate of a reaction : The change in the concentration of any one of the reactants or products per unit time is called average rate of a reaction.

b)

1. Fast Reactions : In case pf ionic compounds reactions takes place fastly i.e., rate is high
   NaCl + AgNO₃ → NaNO₃ + AgCl
2. Slow reactions : In case of covalent compounds reactions takes place slowly i.e., rate is low.
   

c) Order of reaction : The number of reacting species (atoms, ions or molecules) taking parts in an elementary reaction, which must colloid simultaneously to bring about a chemical reaction is called molecularity of a reaction.
NH₄NO₂ → N₂ + 2H₂O (Unimolecular)
2HI → H₂ + I₂ (Bimolecular)
2NO + O₂ → 2NO₂ (Trimolecular)
Molecularity has only integer values (1, 2, 3 …………..)
It has non zero, non fraction values while order has zero, 1, 2, 3 ……….. and fractional values.
It is determined by reaction mechanism, order is determined experimentally.

d) Molecularity of a reaction: Zero order reaction is the reaction in which rate of reaction does not depends on the concentration of reactants.
R → P
Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k [R]°
Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\); d[R] = – k. dt
Integrating on both sides
[RJ = -kt + I …………….. (1)
I – Integration constant
At t = 0 → R = [R]₀ initial concentration
I = [R]₀
Substituting I = [R]₀ in the above equation (1)
[R] = – kt + [R]₀
k = \(\frac{\left[R_0\right]-[R]}{t}\)
This is the integrated rate equation for a zero order reaction.

e) Activation energy of reaction: The energy required to for an intermediate called activated complex (C) during a chemical reaction is called activation energy

Diagram showing plot of potential energy vs reaction co-ordinate

Question 2.
Give two examples for each of zero order and first order reactions. Write the equations for the rate of a reaction in terms of concentration changes of reactants and products for the following reactions.
1) A(g) + B(g) → C(g) + D(g)
2) A(g) → B(g) + C(g) 3) A(g) + B(g) – C(g)
Answer:
Examples of zero order reaction :

Examples of first order reaction :
N₂O_5(g) → N₂O_4(g) + \(\frac{1}{2}\) O_2(g)
SO₂Cl_2(g) → SO_2(g) + Cl_2(g)

Question 3.
Discuss the effect of temperature on the rate of a reaction. Derive necessary equations in this context. [T.S. Mar. 15]
Answer:
Most of the chemical reactions are accelerated by increase of temperature.
For a chemical reaction with rise of temperature by 10°C the rate constant is nearly doubled
k = A.e^-Ea/RT
The temperature dependence of the rate of reaction can be explained by Arrhenius equation.
k = A.e^-Ea/RT
A = Arrhenius factor.
E_a = activation Energy
T = Temperature (°K)
R = gas constant
k = A.e^-E_a/RT
In k = ln A – E_a/RT
2.303 log k = 2.303 log A – E_a/RT
T₁, T₂ are Temperatures
k₁ is rate constant at temperature T₁
k₂ is rate constant at temperature T₂
∴ 2.303 (log k₂ – log k₁) = \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_2}-\frac{1}{\mathrm{~T}_1}\right)\)
\(\log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]\)
The above equation describes the effect of temperature (T) on rate constant (k)

Question 4.
Give a detailed account of the collision theory of reaction rates of bimolecular gaseous reactions. [A.P. Mar. 17, 16]
Answer:
Collision theory of reaction rate bimolecular reactions salient features.

• The reaction molecules are assumed to be hard spheres
• The reaction is postulated to occur when molecules collide with each other.
• The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).
• For a bimolecular elementary reaction
  A + B → products
• Rate = Z_AB.e^-E_a/RT; Z_AB = collision frequency.
• All collisions do not lead to product formation.
• The collisions with Sufficient kinetic energy (Threshold eriergy) are responsible for product formation. These are called as effective collisions.
• To account for effective collisions a factor p called to probability factor or steric factor is introduced.
  Rate = PZ_AB.e^-E_a/RT
• The proper orientation of reactant molecular lead to bond formation where as improper orientation makes them back and no product are formed.
  

In this theory activation energy and proper orientation of the molecules to gather determine the creteria for an effective collision and hence the rate of a chemical reaction.

Problems

Numerical Data Based And Concept Oriented Questions

Question 1.
A reaction is 50% completed in 2 hours and 75% completed in 4 hours. What is the order of the reaction ? [T.S. Mar. 16]
Solution:
Given that a reaction is 50% completed in 2 hrs.
75% completed in 4 hrs.
From the data half life is independent of initial concentration so it is a first order reaction.

Question 2.
A reaction has a half – life of 10 minutes. Calculate the rate constant for the first order reaction. [T.S. Mar. 16]
Solution:
In case of fist order reaction t1/2 = \(\frac{0.693}{k}\)
∴ k = \(\frac{0.693}{t_{1 / 2}}=\frac{0.693}{10}\) = 0.0693 min^-1

Question 3.
In a first order reaction, the concentration of the reactant is reduced from 0.6 mol/L to 0.2 mol/L in 5 min. Calculate the rate constant (k).
Solution:
a = 0.6mol L^-1; a – x = 0.2 mol L^-1; t = 5 min.
Since it is a first order reaction.
k = \(\frac{2.303}{t} \log _{10} \frac{a}{(a-x)}\)
k = \(\frac{2.303}{t}\) log \(\frac{0.6}{0.2}\) = 0.2197 min^-1.

Question 4.
The rate constant for a zero order reaction in A is 0.0030 mol L^-1 s^-1. How long it will take for the initial concentration of A to fall from 0.10 M to 0.075 M.
Solution:
In case of zero order reaction
k = \(\frac{1}{t}\) [[A₀] – [A]]
[A]₀ = Initial concentration
[A] = concentration after time t
k = 0.0030 mol L^-1s^-1
[A]₀ = 0.10 M
[A] = 0.075 M
0.0030 = \(\frac{1}{t}\) [0.10 – 0.075]
t = \(\frac{0.025}{0.0030}\) = 8.33 seconds.

Question 5.
A first order decomposition reaction takes 40 min. for 30% decomposition. Calculate it’s t_1/2 value.
Solution:
Given t = 40 min, a = 100
a – x = 100 – 30 = 70
In case of first order reaction
k = \(\frac{2.303}{t} \log \frac{a}{(a-x)}\)
k = \(\frac{2.303}{40} \log \frac{100}{70}\) = 0.0576 [2.0000 – 1.8451]
= 0.0576 (0.1549) = 8.922 × 10^-3
\(\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\mathrm{k}}=\frac{0.693}{8.922 \times 10^{-3}}\) = 77.673 min.

Question 6.
Calculate the half-life of first order réaction whose rate constant is 200 s^-1.
Solution:
Half – life period for first order reaction
\(\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\mathrm{k}}\)
\(\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{200 \mathrm{~s}^{-1}}\) = 0.346 × 10^-2s = 3.46 × 10^-3s

Question 7.
The thermal decomposition of HCOOH is a first order reaction. The rate constant is 2.4 × 10^-3 s^-1 at a certain temperature. Calculate how long will it take for 3/4 of initial quantity of HCOOH to decompose. ‘
Solution:
Given reaction is a first order reaction.
\(\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\mathrm{k}}=\frac{0.693}{2.4 \times 10^{-3}}\)
= 288.75 sec.
To convert into \(\frac{3}{4}\) of the original, two half lifes are required.
Time to decompose = 2 × 288.75
= 577.5 sec.
= 5.775 × 10² sec.

Question 8.
The decomposition of a compound is found to follow first order rate law. If it takes 15 minutes for 20% of original material to react, calculate the rate constant.
Solution:
Given t = 15 min., a = 100
a – x = 100 – 20 = 80
k = \(\frac{2.303}{t} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\)
k = \(\frac{2.303}{15} \log \frac{100}{80}\) = 0.1535 [2.0000 – 1.9031] = 0.1535 × 0.0969
k = 0.0148 min^-1.

Question 9.
In a pseudo first order hydrolysis of ester in water, the following results are obtained

Calculate the average rate of reaction between the time interval 30 to 60 s.
Solution:

Question 10.
The half-life for a first order reaction is 5 × 10^-6s. What percentage of the initial reactant will react in 2 hours ?
Solution:
Given t_1/2 = 5 × 10^-6 sec.
k = \(\frac{0.693}{1_{1 / 2}}=\frac{0.693}{5 \times 10^{-6}}\) = 0.1386 × 10⁶ sec^-1
k = 0.1386 × 10⁶ sec^-1 = 1.386 × 10⁵ sec^-1
Here a = 100, t = 2 hrs. = 2 × 60 × 60 sec, (a – x) = ?

Question 11.
H₂O_2(aq) decomposes to H₂O_(l) and O_2(g) in a first order reaction w.r.t. H₂O₂. The rate constant is k = 1.06 × 10^-3 min^-1. How long it will take 15% of the sample to decompose?
Solution:
k = 1.06 × 10^-3 min^-1
a = 100 .
a – x = 100 – 15 = 85
k = \(\frac{2.303}{t} \log \frac{a}{a-x}\)
1.06 x 10 = \(\frac{2.303}{t} \log \frac{100}{85}\)
t = \(\frac{2.303}{1.06 \times 10^{-3}} \log \frac{100}{85}\) = 153.4 min

Question 12.
Show that in the case of first order reaction, the time required for 99.9% completion of the reaction is 10 times that required for 50% completion. (log 2 = 0.30 10)
Solution:
t_1/2 = \(\frac{0.693}{k}\)
For 99.9% → a = 100
a – x = 100 – 99.9 = 0.1
k = \(\frac{2.303}{\mathrm{k}} \log \frac{100}{0.1}\)
But k = \(\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{2.303}{0.693}\) × t_1/2 log 1000 = 3.33 × t_1/2 × 3 = 9.99 × t_1/2
t_99.9% is 10 times t_1/2

Question 13.
The rate constant of a reaction is doubled when the temperature is raised from 298 K to 308 K. Calculate the activation energy.
Solution:
Ea = ? k₂ = 2k₁; T₁ = 298 K; T₂ = 308 K

Question 14.
The first order rate constant ‘k’ for the reaction C₂H₅I(g) → C₂H₄ (g) + HI(g) at 600 K is 1.60 × 10^-5 s^-1. The energy of activation is 209 kJ/mol. Calculate ‘k’ at 700 K.
Solution:
k₁ = 1.60 × 10^-5 s^-1
k₂ = ?
T₁ = 600 K
T₂ = 700 K
E_a = 209 kJ mol^-1 = 209000 J mol^-1

Question 15.
The activation energy for the reaction 2HI_(g) → H_2(g) + I_2(g) at 581 K is 209.5 kJ/mol. Calculate the fraction of molecules having energy equal to or greater than activation energy. [R = 8.31 JK^-1 mol^-1] .
Solution:
Fraction of molecules [x] having energy equal to or more than activation energy may be calculated as fallows:
x – n/N = e^-E_a/RT
In x = \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\) or log x = \(\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)
or log x = –\(\frac{209.5 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times\left[8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right] \times 581 \mathrm{~K}}\) = -18.8323
x = Antilog [-18.8324] = Antilog \(\overline{19} .1677\) = 1.471 × 10^-19
Fraction of molecules = 1.471 × 10^-19

Question 16.
For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units seconds.
Solution:
R→ P
0.03M to 0.02 M in 25 minutes.
0.03M – 0.02M in 25 × 60 sec.
Average rate \(\frac{0.03-0.02}{25 \times 60}=\frac{0.01}{25 \times 60}\)
= \(\frac{0.01}{1500}\) = 6.66 × 10^-6 ms^-1

Question 17.
In a reaction 2A → Products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this interval.
Solution:
2A → products
0.5 – 0.4 moI lit rate ∝ [A]²
rate = \(\frac{-1}{2} \frac{\mathrm{d}}{\mathrm{dt}}\) [A] = \(\frac{1}{2}\left[\frac{(0.5-0.4)}{10}\right]\) = 0.5 × 10^-2
rate = 5 × 10^-3 mol Lit^-1 min^-1

Question 18.
For a reaction, A + B → Product : the rate law is given by r = k [A]^1/2 [B]² What is the order of the reaction?
Solution:
A + B → product
r = k [A]^1/2 [B]²
Rate of the reaction r = \(\frac{1}{2}\) + 2 = 2.5

Question 19.
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased by three times, how will it affect the rate of formation of Y.
Solution:
x → y It is a second order reaction
r ∝ [x]²
If x = 1 r = 1
If x = 3 r = 3² = 9
The rate of formation of y increases by 9 times.

Question 20.
A first order reaction has a rate constant 1.15 × 10^-3 s^-1. How long will 5g of this reactant take to reduce to 3 g?
Solution:
Rate constant [k] = 1.15 × 10^-3 s^-1
Initial amount [R]₀ = 5g; Final amount [R] = 3g

Question 21.
Time required to decompose SO₂Cl₂ to half of its initial amount Is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Solution:
For the first order reaction
Rate constant (k) = \(\frac{0.693}{60 \mathrm{~min}}=\frac{0.693}{(60 \times 60) \mathrm{s}}\) = 1.925 × 10^-4 s^-1
k = 1.925 × 10^-4 s^-1

Question 22.
From the rate expression for the following reactions, determine their order of reaction . and the dimensions of the rate constants.
i) 3NO(g) → N₂O(g) Rate = k[NO]²
ii) H₂O₂ (aq) + 3I^– (aq) + 2H₊ → 2H₂O(l) + I₃ Rate = k[H₂O₂] [I^–]
iii) CH₃CHO(g) → CH₄ (g) + CO(g) Rate = k[CH₃CHO]^3/2
iv) C₂H₅Cl(g) → C₂H₂ .(g) + HCl(g) Rate = k[C₂H₅Cl]
Solution:
i) Rate = k [NO]²
order of reaction = 2 .
units (dimensions) of rate constant
[k] = \(\frac{\text { Rate }}{[\mathrm{NO}]^2}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^2}\) = mol^-1 L s^-1

ii) Rate = k[H₂O₂] [I^–]
order of reaction = 1 + 1 = 2
Dimensions of k = \(\) = mol^-1 L s^-1

iii) Rate = k[CH₃CHO]^3/2
order of reaction = \(\frac{3}{2}\)
Dimensions of k = \(\frac{\text { Rate }}{\left[\mathrm{CH}_3 \mathrm{CHO}\right]^{3 / 2}}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^{3 / 2}}\) = mol^-1/2 L^1/2 s^-1

iv) Rate = k[C₂H₅Cl]
order of reaction = 1
Dimensions of k = \(\frac{\text { Rate }}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol} \mathrm{~L}^{-1}}\) = s^-1

Question 23.
For the reaction 2A + B → A, B, the rate = K[A] [B]² with k = 2.0 × 10^-6 v mol^-2 L² s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1, Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-2.
Solution:
i) Case I:
Rate = k [A] [B]²
= [2.0 × 10^-6 mol² L^-2 s^-1] × (0.1 mol L^-1) × (0.2 mol L^-1)²
= 8.0 × mol L^-1 s^-1

ii) Case II:
Concentration of A at a particular time = 0.06 mol L^-1
Amount of A reacted = (0.1 – 0.06) = 0.04 mol L^-1
Amount of B reacted = \(\frac{1}{2}\) × 0.04 mol L^-1 = 0.02 mol L^-1
Concentration of B at a particular time = [0.2 – 0.02] mol L^-1 = 0.18 mol L^-1.
Rate = k [A] [B]²
= [2.0 × 10^-6 mol^-2 L² s^-1] × [0.06 mol L^-1] × (0.18 mol L^-1)²
= 3.89 × 10^-9 mol L^-1 s^-1.

Question 24.
The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4 mol^-1 Ls^-1.
Solution:
For a zero order reaction
rate = \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = k.
dx = change in the concentration
dt = difference in time
k = rate constant

Rate of production of N₂ = k = 2.5 × 10^-14 mol L^-1 s^-1
Rate of production of H₂ = 3 × 2.5 × 10^-14 = 7.5 × 10^-14 mol L^-1 s^-1

Question 25.
The rate expression for the decomposition of dimethyl ether in terms of partial pressures is given as Rate = k (pCH₃ O CH₃)^3/2. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constant ?
Solution:
Unit of rate = bar min^-1
unit of k = \(\frac{\text { Rate }}{\left[\left(\mathrm{PCH}_3 \mathrm{OCH}_3\right]^{3 / 2}\right.}\)
= \(\frac{\text { bar } \min ^{-1}}{[\mathrm{bar}]^{3 / 2}}\) = bar^-1/2 min^-1

Question 26.
A reaction is second order with respect to a reactant. How is the rate of reaction is affected if the concentration of the reactant is
i) doubled
ii) reduced to half
Solution:
For a reaction A → products
Rate = k[A]² = ka²
i) When concentration of A is doubled
i.e., [A] = 2a
Rate = k[2a]² = 4ka²
Rate of reaction becomes 4 times

ii) When concentration of A is reduced to \(\frac{1}{2}\) i.e., [A] = \(\frac{1}{2}\) a
Rate = k [latex]\frac{a}{2}[/latex]² = \(\frac{1}{4}\) ka²
Rate of reaction becomes \(\frac{1}{2}\) times i.e reduced to one fourth.

Question 27.
A reaction is first order in A and second order in B.

1. Write the differential rate equation
2. How is the rate affected on increasing the concentrations of B there times?
3.  How is the rate affected when the concentrations of both A and B are doubled ?

Solution:

1. Rate k [A] [B]²
2. Rate = k [A] [3B]² = 9k [A] [B]² Rate of reaction becomes 9 times
3. Rate = k [2A] [2B]² = 8k [A] [B]² Rate of reaction becomes 8 times

Question 28.
In a reaction between A and B, the initial rate of reaction (r_o) was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B ?
Solution:
Rate law states that
Rate = k [A]^x [B]^y
(Rate)₁ = k[0.20]^x [0.30]^y = 5.07 × 10^-5 ………………. (i)
(Rate)₂ = k[0.20]^x [0.10]^y = 5.07 × 10^-5 …………….. (ii)
(Rate)₃ = k[0.40]^x [0.05]^y = 1.43 × 10^-4 …………… (iii)
Dividing equation (i) by equation (ii)

x log 2 = log 2.82
x = 1.4957 = 1.5
∴ order with respect to A = 1.5
order with respect to B = 0.

Question 29.
The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D

Determine the rate law and rate constant for the reaction.
Solution:
Rate law may be expressed. as
Rate = k [A]^x [B]^y
[Rate]1 = 6.0 × 10^-3 = k (0.1)^x (0.1)^y …………….. (i)
[Rate]2 = 7.2 × 10^-2 = k (0.3)^x (0.2)^x …………….. (ii)
[Rate]3 = 2.88 × 10^-1 = k (0.3)^x (0.4)^x ……………. (iii)
[Rate]4 = 2.40 × 10^-2 = k (0.4)^x (0.l)^x …………….. (iv)

∴ Rate law expression is given by Rate = k [A] [B]²
Rate constant k, can be determined by placing the values of A,B and rate of formation of D.
By taking the values from experimant II.
Rate k[A] [B]²
k = \(\frac{\text { Rate }}{[\mathrm{A}][\mathrm{B}]^2}=\frac{7.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.3 \mathrm{molL}^{-1}\right)\left(0.2 \mathrm{molL}^{-1}\right)^2}\) = 6.0 mol^-2 L² min^-1
∴ k = 6.0 mol^-2 L² min^-1

Question 30.
The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce. the initial concentration of the reactant to its 1/ 16^th value?
Solution:
For 1st order reaction
t = \(\frac{2.303}{k} \log \frac{a}{(a-x)}\) …………….. (i)
Given (a – x) \(\frac{a}{16}\), k = 60 s^-1
Placing the values in eq. (i)

Question 31.
For a first order reaction, show that the time required for 99% completion is twice the time required for completion of 90% of reaction.
Solution:
Case I:
If a = 100; (a – x) = (100 – 99) = 1
For 99% completion of the reaction

It means that time required for 99% completion of the reaction is twice the time required to complete 90% of the reaction.

Question 32.
Here the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Calculate the rate constant.
Solution:

Total pressure after time t,
i e., [p_t] = [p_i – p] + p + p = p_i + p
Or p = p_t – p_i
a = p_i [a – x] = p_i – p on substituting the values of pi
[a – x] = p_i – [p_t – p_i]
i. e., [a – x] = 2p_i – p_i
The decomposition reaction is of gaseous nature and the rate constant k can be calculate as :

Average rate constant k = \(\frac{(2.17+2.24) \times 10^{-3} \mathrm{~s}^{-1}}{2}\)
k = 2.21 × 10^-3 s^-1

Question 33.
The following data were obtained during the first order thermal decomposition of SO₂C₂ at a constant volume.

Calculate the rate of reaction when total pressure is 0.65 atm.
Solution:

Total presstce after time t
i.e., p_t = p_i – p + p + p = p_i + p
So, a = p_i
a – x = p_i – (p_t – p_i) p_i – p_t + p_i = 2p_i – p_t
Substitutions of the value of a and (a – x) gives

b. Calculation of reaction rate when total pressure is 0.65 atm.
P_SO2Cl2 = 0.5 – (0.65 – 0.50) = (1 – 0.65) = 0.35atm
k = 2.23 × 10^-3 s^-1
Rate = k × P_SO2Cl2 = (2.23 × 10^-3 s^-1) × (0.35 atm)
Rate = 7.3 × 10^-4 atm s^-1

Question 34.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol. What will be the value of pre-exponential factor ?
Solution:
According to Arrhenius equation,
log k = log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)
k = 2.418 × 10^-5 s^-1
E_a = 179.9 KJ mol^-1 or 179900 J mol^-1
R = 8.314 JK^-1 mol^-1
T = 546K
log A = log k + \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)
= log (2.418 × 10^-5 s^-1) + \(\frac{179900 \mathrm{Jmol}^{-1}}{2.303 \times\left(8.314 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right) \times 546 \mathrm{k}}\)
= -4.6184 + 17.21 = 12.5916
A = Antilog 12.5916 = 3.9 × 10¹²s^-1
A = 3.9 × 10¹² s^-1

Question 35.
Consider a certain reaction A → Products with k = 2.0 × 10^-2s^-1. Calculate the concentration of A remaining after 100 s If the initial concentration of A is 1.0 mol L^-1.
Solution:
For the first order reaction

Question 36.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_{\(\frac{1}{2}\)} = 3.00 hours. What fraction of sample of sucrose remains 8 after 8 hours?
Solution:

Question 37.
The decomposition of hydrocarbon follows the equation K = (4.5 × 10¹¹ s^-1) e^-28000K/T. Calculate E_a.
Solution:
According to Arrhenius equation
k = Ae^-E_a/RT ………………. (i)
According to the available data
k = (4.5 × 10 s^-1)e^-28000k/T ………………….. (ii)
on comparing both equations
\(-\frac{E_{\mathrm{a}}}{\mathrm{RT}}=\frac{-28000 \mathrm{k}}{\mathrm{T}}\)
E_a = (28000 k) × R
= (28000 k) × (8.314 K^-1 J mol^-1)
= 232792 Jk mol^-1
E_a = 232.792 kJ^-1 mol^-1

Question 38.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation: log k = 14.34 – 1.25 × 10⁴ K/T. Calculate E_a for this reaction and at what temperature will its half-life period be 256 minutes?
Solution:
a) Calculation of activation energy E_a
According to Arrhenius equation; k = Ae^-E_a/RT
log k = log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\) …………………. (i)
log K = 14.34 – \(\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}\) …………………. (ii)
on comparing both equations.
\(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}=\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}\)
E_a = 1.25 × 10⁴ K × 2.303 × 8.314 (JK^-1 mol^-1)
= 23.93 × 10⁴ J mol^-1 = 239.3 kJ mol^-1

b) Calculation of required temperature
If t_1/2 = 256 min. for 1^st order reaction;

Question 39.
The decomposition of A into product has value of k as 4.5 × 10³ s^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 × 10⁴ s^-1?
Solution:
According to Arrhenius equation
\(\log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}} \times \frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\)
k₁ = 4.5 × 10³ s^-1
k₂ = 1.5 × 10⁴ s^-1
T₁ = 10°C = 283 K

T₂ = 297.19 K = (297.19 – 273.0) = 24.19°C
Temperature = 24.19°C

Question 40.
The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s^-1. calculate k at 318K andE .
Solution:
Calculation of activation energy (E_a)
For 1^st order reaction :

According to Arrhenius theory

b) Calculation of rate constant (k)
According to Arrhenius equation
log k = log A \(\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)
log k = log (4 × 10¹⁰) – \(\frac{76640 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times\left(8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right) \times(318 \mathrm{~K})}\)
log k = 10.6021 – 12.5870 = – 1.9849
k = Antilog (-1.9849) = Antilog \((\overline{2} .0151)\) = 1.035 × 10^-2 s^-1
E_a = 76.640 kJ mol^-1
k = 1.035 × 10^-2 s^-1

Question 41.
The rate of a reaction quadruples when the temperature changes from 293 K to 313K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution:
According to Arrhenius equation

Textual Examples

Question 1.
From the concentrations of C₄H₉Cl (butyl chloride) at different times given below, calculate the average rate of the reaction :
C₄H₉Cl + H₂O → C₄H₉OH + HCl
during different intervals of time.

Solution:
We can determine the difference in concentration over different intervals of time and thus determine the average rate by dividing ∆ [R] by ∆t
Average rates of hydrolysis of butyl chloride

Question 2.
The decomposition of N₂O₅ in CCl₄ at 318K has been studied by monitoring the concentration of N₂O₅ in the solution. Initially the concentration of N₂O₅ is 2.33 mol L^-1 and after 184 minutes, it is reduced to 2.08 mol L^-1. The reaction takes place according to the equation
2 N₂O₅ (g) → 4 NO₂ (g) + O₂ (g)
Calculate the average rate of this reaction in terms of hours, minutres and seconds. What is the rate of production of NO₂ during this period?
Solution:
Average Rate = \(\frac{1}{2}\left\{-\frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}\right\}=-\frac{1}{2}\left[\frac{(2.08-2.33) \mathrm{mol} \mathrm{L}^{-1}}{184 \mathrm{~min}}\right]\)
= 6.79 × 10^-4 mol L^-1/min
= (6.79 × 10^-4 mol L^-1 min^-1) × (60 min / 1h)
=4.07 × 10^-2 mol L^-1/h .
= 6.79 × 10^-4 mol L^-1 × 1mm / 60s = 1.13 × 10^-5 mol L^-1s^-1
It may be remembered that
Rate = \(\frac{1}{4}\left\{\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}\right\}\)
\(\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}\) = 6.79 × 10^-4 × 4 mol L^-1 min^-1
= 2.72 × 10^-3 mol L^-1 min^-1

Question 3.
Calculate the overall order of a reaction which has the rate expression
a) Rate = k [A]^1/2 [B]^3/2
b) Rate = k [A]^3/2 [B]^-1
Solution:
a) Rate = k [A]^x [B]^y
order = x + y
So order = 1/2 + 3/2 = 2, i.e., second order

b) order = 3/2 + (- 1) = 1/2, i.e., half order.
A balanced chemical equation never gives us a true picture of how a reaction takes place since rarely a reaction gets completed in one step. The reactions taking place in one step are called elementary reactions. If a sequence of elementary reactions (called mechanism), reactants give in the products, the reactions are called complex reactions. These may be consecutive reactions (e.g., oxidation of ethane to CO₂ and H₂O passes through a series of intermediate steps in which alcohol, aldehyde and acid are formed), reversible reactions and parallel reactions (e.g., nitration of phenol yields o-nitrophenol and p-nitrophenol).

Question 4.
Identify the reaction order from each of the following rate constants.

1. k = 2.3 × 10^-5 L mol^-1 s^-1
2. k = 3 × 10^-4 s^-1

Solution:

1. The unit of second order rate constant is L mol^-1 s^-1, therefore k = 2.3 × 10^-5 L mol^-1 s^-1represents a second order reaction.
2. The unit of a first order rate constant is s^-1 therefore k = 3 × 10^-4 s^-1 represents a first order reaction.

Question 5.
The initial concentration of N₂O₅ in the following first order reaction
N₂O₅(g) → 2 NO₂(g) + \(\frac{1}{2 \mathrm{O}_2}\) (g) was 124 × 10^-2 mol L^-1 at 318 K. The concentration of N₂O₅ after 60 minutes was 0.20 × 10^-2 mol L^-1 . Calculate the rate constant of the reaction at 318 K.
Solution:
For a first order reaction
\(\log \frac{[\mathrm{R}]_1}{[\mathrm{R}]_2}=\frac{\mathrm{k}\left(\mathrm{t}_2-\mathrm{t}_1\right)}{2.303}\)

k = 0.0304 min^-1.

Question 6.
The following data were obtained during the first order thermal decomposition of N₂O₅ (g) at constant volume:
2N₂O₅ (g) → 2₂ O₄ (g) + O₂ (g)

Calculate the rate constant.
Solution:
Let the pressure of N₂O₅ (g) decrease by 2x atm. As two moles of N₂O₅ decompose to give two moles of N₂O₄ (g) and one mole of O₂ (g), the pressure of N₂O₄ (g) increases by 2x atm and that of O₂ (g) increases by x atm.

Question 7.
A first order reaction is found to have a rate constant, k = 5.5 × 10^-14 s^-1. Find the half¬life of the reaction.
Solution:
Half-life for a first order reaction is
t_1/2 = \(\frac{0.693}{k}\)
t_1/2 = \(\frac{0.693}{5.5 \times 10^{-14} \mathrm{~S}^{-1}}\) = 1.26 × 10¹³ s

Question 8.
Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t_1/2) of the reaction.
Solution:
When reaction is completed 99.9%, [R]_n = [R]₀ – 0.999 [R]₀

Question 9.
Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below.

Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant (55 mol L^-1), during the course of the reaction. What is the value of k’ in this equation ?
Rate = k’ [CH₃ COOCH₃] [H₂O]
Solution:
For pseudo first order reaction, the reaction should be first order with respect to ester when [H₂O] is constant. The rate constant k for pseudo first order reaction is
k = \(\frac{2.303}{t} \log \frac{C_0}{C}\) where k = k’ [H₂O]
From the above data we note

It can be seen that k’ [H₂O] is constant and equal to 2.004 × 10^-3 min^-1 and hence, it is pseudo first order reaction. We can now determine k from
k’ [H₂O] = 2.004 × 10^-3 min^-1
k'[55 mol L^-1] = 2.004 × 10^-3 min^-1
k’ = 3.64 × 10^-5 mol^-1 L min^-1

Question 10.
The rate constants of a reaction at 500 K and 700 K are 0.02s^-1 and 0.07s^-1 respectively. Calculate the values of E_a and A.
Solution:

Question 11.
The first order rate constant for the decomposition of ethyl iodide by the reaction.
C₂H₅I(g) → C₂H₄ (g) + HI(g) at 600K is 1.60 × 10^-5 s^-1. It is energy of activation is 209 kj/mol. Calculate the rate constant of the reaction at 700 K.
Solution:
We know that

Intext Questions

Question 1.
For the reaction, R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 min. Calculate the average rate of reaction using units of time both in minutes and seconds.
Solution:
For the reaction R → P

Question 2.
In a reaction, 2A → products, concentration of A decreases from 0.5 mol L^-1 to 0.4 to mol L^-1 in 10 min. Calculate the rate during this interval.
Solution:
Rate of reaction = rate of disappearance of A
= \(-\frac{1}{2} \frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}\)
= \(-\frac{1}{2} \frac{(0.4-0.5) \mathrm{mol} \mathrm{L}^{-1}}{10 \mathrm{~min}}\)
= 0.005 mol L^-1 m^-1

Question 3.
For a reaction, A + B → product; the rate law is given by, r = k[A]^1/2 [B]². What is the order of the reaction ?
Solution:
Order of reaction = \(\frac{1}{2}\) + 2 = \(\frac{1}{2}\)
Order = \(\frac{5}{2}\)

Question 4.
The conversion of molecules X to Y follows second order kinetics. If concentration of x is increased to three times how will it affect the rate of formation of y ?
Solution:
For the reaction x → y
Reaction rate (r) = k[x]² …………….. (i)
If the concentration of x is increased three times, then
Reaction rate (r’) = k[3x]² = kx [9x]² ……….. (ii)
Dividing Eq. (ii) by Eq. (i) .
\(\frac{\mathrm{r}^{\prime}}{\mathrm{r}}=\frac{\mathrm{k} \times[9 \mathrm{x}]^2}{\mathrm{k} \times[\mathrm{x}]^2}=9\)
It means that the rate of formation of y will increase by nine times.

Question 5.
What will be the effect of temperature on rate constant ?
Solution:
The rate constant for a reaction is nearly doubled with about 10° rise in temperature. The reason is that the number of effective collision becomes almost double. The exact dependence of the rate of reaction can be given by Arrhenius equation; k = Ae^-E_a/RT. Where, A is called the frequency factor and E_a is the activation energy of the reaction.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 3(a) Electro Chemistry Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 3(a) Electro Chemistry

Very Short Answer Questions

Question 1.
What is a galvanic cell or a voltaic cell? Give one example. [T.S. Mar. 19]
Answer:
Galvanic cell: A device that converts chemical energy into electrical energy by the use of spontaneous redox reaction is called a Galvanic cell (or) voltaic cell.
Eg.: Daniell cell.

Question 2.
Write the Chemical reaction used in the construction of the Daniell cell together with the half-cell reactions.
Answer:
The chemical reactions used in the construction of the Daniell cell.

Question 3.
Name the two half-cell reactions that are taking place in the Daniell cell.
Answer:
The two half cell reactions that are takes place in the Daniell cell are oxidation and reduction.
Zn → Zn⁺² + 2e^– (Oxidation)
Cu⁺² + 2e^– → Cu (Reduction)

Question 4.
How is a galvanic cell represented on paper as per IUPAC convention ? Give one example.
Answer:
Representation of a Galvanic cell:
As per IUPAC convention

• Oxidation half cell represented on the left side.
• Reduction half cell represented on the right side.
• Oxidation half cell and reduction half cell are connected by a salt bridge indicated by two vertical parallel lines.
  Eg.: Cu_(s) | \(\mathrm{Cu}_{(\mathrm{aq})}^{+2}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag_(s)

Question 5.
Write the cell reaction taking place in the cell
Cu_(s) | \(\mathrm{Cu}_{(\mathrm{aq})}^{+2}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag_(s)
Answer:
Given cell is Cu_(s) | \(\mathrm{Cu}_{(\mathrm{aq})}^{+2}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag_(s)
Cu → Cu⁺² + 2e^– (Oxidation)
2Ag⁺ + 2e^– → 2Ag (Reduction)

Question 6.
What is standard hydrogen electrode ?
Answer:
The electrode whose potential is known as standard electrode (or) standard hydrogen electrode.

To determine the potential of a single electrode experimentally it combine with standard hydrogen electrode and the EMF of cell so constructed is measured with potentiometer.

Question 7.
Give a neat sketch of standard hydrogen electrode.
Answer:

Standard Hydrogen Electrode (SHE)

Question 8.
What is Nernst equation ? Write the equation for an electrode with electrode reaction Mⁿ⁺(aq) + ne^– ⇌ M(s).
Answer:
The electrode potential at any concentration measured with respect to standard hydrogen electrode is represented by Nernst equation.
Nernst equation is
E = E⁰ + \(\frac{\mathrm{RT}}{\mathrm{nF}}\) ln[Mⁿ⁺] [Metal Electrodes]
Given electrode reaction is \(\mathrm{M}_{(\mathrm{aq})}^{\mathrm{n}+}\) + ne^– ⇌ M_(s)
For the above electrode reaction Nemst equation is
E_(Mⁿ⁺/M) = \(\mathrm{E}_{\left(\mathrm{M}^{\mathrm{n}+} / \mathrm{M}\right)}^0+\frac{\mathrm{RT}}{\mathrm{nF}} \ln \left[\mathrm{M}^{\mathrm{n}+}\right]\)
Here E_(Mⁿ⁺/M) = Electrode potential
E⁰E_(Mⁿ⁺/M) = Standard Electrode potential
R = gas constant = 8.314 J/k.mole
F = Faraday = 96487 c/mole
T = temperature
[Mⁿ⁺] = concentration of species Mⁿ⁺

Question 9.
A negative E⁰ indicates that the redox couple is _______________ reducing couple than H⁺/H₂, couple, (powerful or weak)
Answer:
A negative E⁰ value indicates that the redox couple is powerful reducing couple than H⁺/H₂ couple.

Question 10.
A positive E⁰ indicates that the redox couple is a weaker _____________ couple than H⁺/H₂ couple. (oxidising or reducing)
Answer:
A positive E⁰ value indicates that the redox couple is as weaker reducing couple than H⁺/H₂ couple.

Question 11.
Write the Nernst equation for the EMF of the cell .
Ni_(s) |\(\mathrm{Ni}_{(\mathrm{aq})}^{2+}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag
Answer:
Given cell is Ni_(s) |\(\mathrm{Ni}_{(\mathrm{aq})}^{2+}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag
Nernst equation for the cell is
E_cell = \(\mathrm{E}_{\text {cell }}^0+\frac{\mathrm{RT}}{\mathrm{nF}} \ln \frac{\left[\mathrm{Ag}^{+}\right]}{[\mathrm{Ni}]^{+2}}\)

Question 12.
Write the cell reaction for which E_cell = \(\mathrm{E}_{\text {cell }}^0-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\)
Answer:
Given E_cell = \(\mathrm{E}_{\text {cell }}^0-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\)
The cell reaction is Mg_(s)/\(\mathrm{Mg}_{(\mathrm{aq})}^{2+}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag_(s)

Question 13.
How is E⁰ cell related mathematically to the equilibrium constant K_c of the cell reaction?
Answer:
Relation between E₀ cell and equilibrium constant K_C of the cell reaction
\(\mathrm{E}_{\text {cell }}^0=\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\) log K_C
n = number of electrons involved
F = Faraday 96500 C mol^-1,
T = Temperature
R = gas constant

Question 14.
How is Gibbs energy (G) related to the cell emf (E) mathematically ?
Answer:
Relation between Gibb’s energy (G) and emf (E) mathematically
∆G⁰ = -nFE_(cell)
∆G = change in Gibb’s energy
n = number of electrons involved
F = Faraday = 96500 C mol^-1

Question 15.
Define conductivity of a material. Give its SI units.
Answer:
The reciprocal of specific resistance (or) resistivity is called conductivity.
It is represented by (κ)
(Or)
The conductance of one unit cube of a conductor is also called conductivity.
SI units : ohm^-1 m^-1 (or) Sm^-1 S = Siemen

Question 16.
What is cell constant of a conductivity cell ?
Answer:
Resistance R = ρ . \(\frac{l}{\mathrm{~A}}\) = \(\frac{l}{\kappa \cdot \mathrm{A}}\)
\(\frac{l}{\mathrm{~A}}\) = cell constant
Cell constant G = \(\frac{l}{\mathrm{~A}}\) = R × κ
The cell constant of a conductivity cell is the product of resistance and specific conductance.

Question 17.
Define molar conductivity ∧_m and how is it related to conductivity (κ) ?
Answer:
Molar conductivity: The conductivity of a volume of solution containing one gram molecular weight of the electrolyte placed between two parallel electrodes separated by a distance of unit length of 1 meter is called molar conductivity (∧_m).
Relation between conductivity and molar conductivity:
∧_m = \(\frac{\kappa}{\mathrm{c}}\);
∴ c = constant

Question 18.
Give the mathematical equation which gives the variation of molar conductivity with ∧_m the molarity (c) of the solution ?
Answer:
The mathematical equation which gives the variation of molar conductivity ∧_m with the molarity (c) of the solution is
∧_m = \(\frac{\kappa}{1000\left(\text { lit } / \mathrm{cm}^3\right) \times \text { molarity (moles/lit) }}=\frac{\kappa \times 1000\left(\mathrm{~cm}^3 / \mathrm{lit}\right)}{\text { molarity (mole/lit) }}\)

Question 19.
State Kohlrausch’s law of independent migration of ions.
Answer:
Kohlrausch’s law of independent migration of ions: The limiting molar conductivity of an electrolytes can be represented as the sum of the individual contributions of the anion and the cation of the electrolytes.
\(\lambda_{\mathrm{m}(\mathrm{AB})}^0=\lambda_{\mathrm{A}^{+}}^0+\lambda_{\mathrm{B}^{-}}^0\)
\(\lambda_m^0\) = Limiting molar conductivity
\(\lambda_A^0\) = Limiting molar conductivity of cation
\(\lambda_B^0\) = Limiting molar conductivity of anion

Question 20.
State Faraday’s first law of electrolysis. [A.P. Mar. 18, 16] [Mar. 14]
Answer:
The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of current passing through the electrolyte.
(Or)
The mass of the substance deposited at an electrode during the electrolysis of electrolyte is directly proportional to quantity of electricity passed through it.
m ∝ Q; m ∝ c × t
m = ect; m = \(\frac{\text { Ect }}{96,500}\)
e = electrochemical equivalent
c = Current in amperes
t = time in seconds
E = Chemical equivalent

Question 21.
State Faraday’s second law of electrolysis. [T.S. Mar. 18] [Mar. 14]
Answer:
The amounts of different substances liberated when the same quantity of electricity is passed through the electrolytic solution are proportional to their chemical equivalent weights.
m ∝ E

Qustion 22.
What are the products obtained at the platinum anode and the platinum cathode respectively in the electrolysis of fused or molten NaCl ?
Answer:
In the electrolysis of fused (or) molten NaCl with platinum anode and platinum cathode, sodium metal obtained at cathode and chlorine gas at anode
2 NaCl → 2 Na⁺ + 2Cl^–
2 Cl^– → Cl₂ + 2e^– (anode)
2 Na⁺ + 2e^– → 2 Na (cathode)

Question 23.
Give the products obtained at the platinum electrodes (cathode and anode) when aqueous solution of K₂SO₄ is electrolysed.
Answer:
During the electrolysis of aq.K₂SO₄ using pt-anode and pt-cathode hydrogen gas liberated at cathode and oxygen gas at anode.
K₂SO₄ → 2K⁺ + \(\mathrm{SO}_4^{-2}\)

Question 24.
Write the chemical equation corresponding to the oxidation of H₂O_(l) at the platinum anode.
Answer:
The chemical equation corresponding to the oxidation of H₂O_(l) at the platinum anode is
2H₂O_(l) → O_2(g) + 4H⁺_(aq) + 4e^–

Question 25.
Give the chemical equation that represents the reduction of liquid water H₂O_(l)at the platinum cathode.
Answer:
The chemical equation that represents the reduction of liquid water H₂O_(l) at the platinum cathode is H₂O_(l) + e^– → \(\frac{1}{2}\)H_2(g) + OH^–

Question 26.
What is a primary battery ? Give one example. [A.P. Mar. 17]
Answer:
The batteries which after their use over a period of time, becomes dead and the cell reaction is completed and this cannot be reused again are called primary batteries.
Eg: Leclanche cell, dry cell.

Question 27.
Give one example for a secondary battery. Give the cell reaction.
Answer:
Lead storage battery is an example of secondary battery.
The cell reactions when the battery is in use are
Pb_(s) + \(\mathrm{SO}_4^{-2} \text { (aq) }\) → PbSO_4(s) + 2e^– (Anode)
PbO_2(s) + \(\mathrm{SO}_4^{-2} \text { (aq) }\) → \(4 \mathrm{H}_{(\mathrm{aq})}^{+}\) + 2e^– → PbSO_4(s) + 2 H₂O_(l) (Cathode)
Overall cell reaction is .
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(s) + H₂O_(l)

Question 28.
Give the cell reaction of nickel-cadmium secondary battery.
Answer:
The cell reaction of nickel-cadmium secondary battery is
Cd_(s) + 2Ni(OH)_3(s) → CdO_(s) + 2 Ni(OH)_2(s) + H₂O_(l)

Question 29.
What is a fuel cell ? How is it different from a conventional galvanic cell ?
Answer:
A fuel cell is a galvanic cell in which the chemical energy of fuel-oxidant system is converted directly into electrical energy.

• Conventional Galvanic cell converts chemical energy into electrical energy by spontaneous redox reactions.
• Fuel cell convert energy of combustion of fuels like hydrogen, methane etc., into electrical energy. These cause less pollution.

Question 30.
Give the electrode reactions occurijig at the anode and at the cathode in H₂, O₂, fuel cell.
Answer:
At cathode : O_2(g) + 2 H₂, O₂ + 4e^– → 4OH^–_(aq)
At anode : 2H_2(g) + 4 OH^–_(aq) → 4H₂O_(l) + 4e^–
Overall reaction : 2H_2(g) + O_2(g) → 2H₂O_(l)

Question 31.
What is metallic corrosion ? Give one example. [T.S. Mar. 17; A.P. Mar. 15]
Answer:
Metallic corrosion: The natural tendency of conversion of a metal into its mineral compound form i n interaction with the environment is known as metallic corrosion.
Eg : 1) Iron converts itself into its oxide [Rusting] (Fe₂O₃)
2) Silver converts itself into it’s sulphate
[tarnishing] [Ag₂S]

Question 32.
Give the electro-chemical reaction that represents the corrosion or rusting of iron.
Answer:
Rusting of iron: Iron converts itself into its oxixe Fe₂O₃
The electrochemical reaction that represents the corrosion or rusting of iron is
Anode : 2 Fe_(s) → 2 Fe²⁺ + 4e^– \(\mathrm{E}_{\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)}^0\) = -0.44V
Cathode : O_2(g) + 4H⁺_(aq) + 4e^– → 2H₂O_(l) \(\mathrm{E}_{\mathrm{H}^{+} / \mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\Theta}\) = 1.23V
Overall reaction: 2 Fe_(s) + O_2(g) + 4H⁺_(aq) → 2Fe⁺²_aq + 2 H₂O_(l) \(\mathrm{E}_{\text {(cell) }}^{\Theta}\) = 1.67 V

Short Answer Questions

Question 1.
What are galvanic cells ? Explain the working of a galvanic cell with a neat sketch taking Daniell cell as example. [T.S. & A.P. Mar. 18]
Answer:
Galvanic cell: A device which converts chemical energy into electrical energy by the use of spontaneous redox reaction is called Galvanic cell (or) voltaic cell.
Eg : Daniell cell.

Daniell cell: It is a special type of galvanic cell. It contains two half cells in the same vessel. The vessel is devided into two chambers. Left chamber is filled with ZnSO₄ (aq) solution and Zn – rod is dipped into it. Right chamber is filled with aq. CuSO₄ solution and a copper rod is dipped into it. Process diaphragm acts as Salt bridge. The two half cell’s are connected to external battery.

Cell reactions :
Ion Zn/ZnSO₄ half cell, oxidation, reaction occurs
Zn → Zn²⁺ + 2e^–
Ion Cu/CuSO₄ half cell, reduction reaction occurs.
The net cell reaction is
Zn + Cu⁺² ⇌ Zn⁺² + Cu
Cell is represented as Zn / Zn⁺² || Cu⁺² / Cu

Question 2.
Give the construction and working of a standard hydrogen electrode with a neat diagram.
Answer:
To determine the potential of a single electrode experimentally, it is combined with a standard hydrogen electrode (electrode whose potential is known) and the EMF of the cell. So constructed is measured with a potentiometer. Standard hydrogen electrode is constructed and is used as standard electrode or reference electrode. Standard Hydrogen Electrode (SHE) or Normal Hydrogen Electrode (NHE).

Pure hydrogen gas is bubbled into a solution of 1M HCl along a platinum electrode coated with platinum block. A platinum block electrode placed in the solution at atmospheric pressure.

Generally the electrode is fitted into the tube. The tube will have two circular small holes. This tube is immersed in the acid solution such that one haf of the circular hole is exposed to air and another half in the solution.
The following equilibrium exists at the electrode.
\(\frac{1}{2}\) H_2(g) (1_atm) ⇌ \(\mathrm{H}_{(\mathrm{aq})}^{+}\) (1M) + e^–

Question 3.
State and explain Nernst equation with the help of a metallic electrode and a non-metallic electrode.
Answer:
The electrode potential at any concentration measured with respect to standard hydrogen electro represented by Nernst equation.
Nernstst equation is
E = E⁰ + \(\frac{\mathrm{RT}}{\mathrm{nF}}\) In [Mⁿ⁺] [Metal Electrodes]
Given electrode reaction is
\(\mathrm{M}_{(\mathrm{aq})}^{\mathrm{n}+}\) + ne^– ⇌ M_(s)
For the above electrode reaction Nernst equation is
E_{(Mⁿ⁺ / M)} = \(\mathrm{E}_{\left(\mathrm{M}^{\mathrm{n}+} / \mathrm{M}\right)}^0+\frac{\mathrm{RT}}{\mathrm{nF}} \ln \left[\mathrm{M}^{\mathrm{n}+}\right]\)
Here E_{(Mⁿ⁺ / M)} = Electrode potential
E⁰_{(Mⁿ⁺ / M)} = Standard Electrode potential
R = gas constant = 8.314 J/k.mole
F = Faraday = 96487 c/mole
T = temperature
[Mⁿ⁺] = concentration of species Mⁿ⁺
For non-metal electrodes :
E = E⁰ – \(\frac{\mathrm{RT}}{\mathrm{nF}}\) ln C ; C = concentration
Example for metal electrode:
Given cell is Ni_(s) | \(\mathrm{Ni}_{(a q)}^{2+}\) || \(\) | Ag
Nernst equation for the cell is
E_cell = \(\mathrm{E}_{\text {cell }}^0+\frac{\mathrm{RT}}{\mathrm{nF}} \ln \frac{\left[\mathrm{Ag}^{+}\right]}{[\mathrm{Ni}]^{+2}}\)
Example for non-metal electrode: Pt, Cl₂/Cl^–
E = E⁰ – \(\frac{\mathrm{RT}}{\mathrm{nF}}\) log C
E = E⁰ – \(\frac{\mathrm{RT}}{\mathrm{F}}\) log C

Question 4.
Explain with a suitable example the relation between the Gibbs energy of chemical reaction (G) and the functioning of the electrochemical cell.
Answer:
Electrical work done in one second in an electrochemical cell is equal to the electrical potential multiplied by the total charge passing. If we want to obtain maximum work from a galvanic cell then the charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy and therefore, if the emf is E and nF is the amount of charge passing and ∆_rG = – nFE_[cell]
It may be remembered that E_[cell] is an intensive parameter but ∆_rG is an extensive thermodynamic property and the value depends on n. Thus, if we write the reaction
Zn_[s] + \(\mathrm{Cu}_{[\mathrm{aq}]}^{2+}\) → \(\mathrm{Zn}_{\text {[aq] }}^{2+}\) + Cu_[s]
∆_rG = -2FE_[cell]
but when we write the reaction
2 Zn_[S] + 2 \(\mathrm{Cu}_{[\mathrm{aq}]}^{2+}\) → \(\mathrm{Zn}_{[\text {aq] }}^{2+}\) + 2 Cu_[s]
∆_rG = -4FE_[cell]
If the concentration of each of the reacting species is unity, then E_[cell] = \(\mathrm{E}_{\text {[cell] }}^{\Theta}\) and we have
∆_rG^⊖ -nF\(\mathrm{E}_{\text {[cell] }}^{\Theta}\)
We can calculate equilibrium constant using the equation
∆_rG^⊖ = – RTln K_c

Question 5.
On what factors the electrical conductance of an aqueous solution of electrolyte depends?
Answer:
The electrical conductance of an aqueous solution of electrolyte depends on

1. The nature of the electrolyte
2. Size and solvation of the ions formed in the dissociation of the electrolyte.
3. The nature and viscosity of the solvent.
4. Concentration of the electrolyte
5. Temperature

Question 6.
How is molar conductivity of an aqueous electrolyte solution measured experimentally ?
Answer:
Molar conductivity: The conductivity of a volume of solution containing one gram molecular weight of the electrolyte placed between two parallel electrodes separated by a distance of unit length of 1 meter is called molar conductivity (∧_m)
Relation between conductivity and molar conductivity:
∧_m = \(\frac{\kappa}{\mathrm{c}}\);
∴ c = constant .
The conductance of a solution measured in a conductivity cell.
By using the conductivity cells
Resistance (R) = \(\frac{l}{\kappa \times A}\)
l = distance between electrodes; A = Area of cross section; K = conductivity
G* = \(\frac{l}{\mathrm{~A}}\) = cell constant
-4 Cell constant is measured by measuring the resistance of the cell containing a solution whose conductivity is known.
-4 Cell constant determined is used for measuring the resistance (or) conductivity.
∴ molar conductivity ∧_(m) = \(\frac{\kappa}{\mathrm{c}}\)
∧_(m) = \(\frac{\kappa\left(\mathrm{Scm}^{-1}\right)}{1000\left(\mathrm{~L}^{-3}\right) \text { molarity (moles/lit) }}\)

Question 7.
Explain the variation of molar conductivity with the change in the concentration of the electrolyte. Give reasons.
Answer:
Molar conductivity of a solution at a given concentration is the conductance of the volume of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length.

Molar conductivity ∧_m = \(\frac{\mathrm{\kappa A}}{l}\)
∧_m = κ × V (∵ l = 1, A = V)
V = Volume containing 1 mole of electrolyte
Molar conductivity increase with decrease in concentration of electrolyte (or) increase in dilution. This is due to total volume V of solution containing one mole of electrolyte also increases.

When concentration approaches to zero, the molar conductivity is known as limiting molar conductivity and it is represented by
For strong electrolytes molar conductivity is given by \(\wedge_m^0\)
∧_m = \(\wedge_m^0\) – Ac^1/2
For weak electrolytes limiting’ molar conductivity is given by
λ_AB⁺ = λ_A⁺ + λ_B^– (Kohlrausch law)

Question 8.
State and explain Kohlrausch’s law of independent migration of ions. [T.S. Mar. 16; A.P. Mar. 15]
Answer:
Kohlrausch’s law of independent migration of ions: The limiting molar conductivity of an electrolytes can be represented as the sum of the individual contributions of the anion and the cation of the electrolytes.
\(\lambda_{\mathrm{m}(\mathrm{AB})}^0=\lambda_{\mathrm{A}^{+}}^0+\lambda_{\mathrm{B}^{-}}^0\)
\(\lambda_{\mathrm{m}}^0\) = Limiting molar conductivity
\(\lambda_{\mathrm{A}^{+}}^0\) = Limiting molar conductivity of cation
\(\lambda_{\mathrm{B}^{-}}^0\) = Limiting molar conductivity of anion

Applications:

1. Kohlrausch’s law is used in the calculation of the limiting molar conductivity of weak electrolytes.
   Eg : λ_CH3COOH = λ_CH3COONa + λ_HCl – λ_NaCl
   = λ_CH3COO^– + λ_Na⁺ + λ_H⁺
2. This law is used in the calculation of degree of dissociation of a weak electrolyte.
3. This law is used in the calculation of solubility of sparingly soluble salts like AgCl, BaSO₄ etc.

Question 9.
What is electrolysis ? Give Faraday’s first law of electrolysis. [A.P. & T.S. Mar. 15]
Answer:
Electrolysis : The decomposition of a chemical compound in the molten state or in the solution state into its constituent elements under the influence of an applied EMF is called electrolysis.
The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of current passing through the electrolyte.
(Or)
The mass of the substance deposited at an electrode during the electrolysis of electrolyte is directly proportional to quantity of electricity passed through it.
m ∝ Q; m ∝ c × t
m = ect; m = \(\frac{\text { Ect }}{96,500}\)
e = electrochemical equivalent
t = time in seconds
c = Current in amperes
E = Chemical equivalent

Question 10.
What are the products obtained at the cathode and anode during the electrolysis of the following when platinum electrodes are used in the electrolysis
a) Molten KCl
b) Aq. CuSO₄ solution
c) Aq. K₂SO₄ solution
Answer:
a) During the electrolysis of molten KCl using platinum electrodes. Potassium is obtained at cathode and chlorine at anode.
2 KCl → 2K⁺ + 2Cl^–
2 Cl^– → Cl₂ + 2e^– (anode)
2K⁺ + 2e^– → 2K (cathode)

b) During the electrolysis of aq.CuS04 solution using platinum electrodes, O₂ gas liberated at anode and Cu deposited at cathode.
2 CuSO₄ → 2 Cu⁺² + 2 SO^-2₄
2 Cu⁺² + 4e^– → 2 Cu (cathode)
2H₂O – 4e^– → O₂ + 4H⁺ (anode)

c) During the electrolysis of aq.k₂SO₄ using pt-anode and pt-cathode hydrogen gas liberated at cathode and oxygen gas at anode.

Question 11.
What are primary and secondary batteries ? Give one example for each.
Answer:
The batteries which after their use over a period of time’ becomes dead and the cell reaction is completed and this cannot be reused again are called primary batteries.
Eg : Leclanche cell, dry cell.

Secondary battery: A secondary battery is the battery in which after it’s use can be recharged and can be used again.
A good secondary battery undergoes large no. of discharging and charging cycles.
Lead storage battery is an example of secondary battery. ,
The cell reactions when the battery is in use are
Pb_(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → PbSO₄(s) + 2e^– (Anode)
PbO_2(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → \(4 \mathrm{H}_{\text {(aq) }}^{+}\) + 2e → PbSO_4(s) + 2 H₂O_(l) (Cathode)
Overall cell reaction is
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(aq) + 2H₂O_(l)

Question 12.
What are the fuel cells ? How are they different from galvanic cells ? Give the construction of H₂, O₂ fuel cell? [T.S. Mar. 17]
Answer:
A fuel cell is a galvanic cell in which the chemical energy of fuel-oxidant system is converted directly into electrical energy.

• Conventional Galvanic cell converts chemical energy into electrical energy by spontaneous redox reactions.
• Fuel cell convert energy of combustion of fuels like hydrogen, methane etc., into electrical energy. These cause less pollution.

H₂ – O₂ fuel cell: In this cell, hydrogen and oxygen are bubbled through porous carbon electrodes into Cone. NaOH solution. Electrodes are embedded with suitable catalysts. The electrode reactions are:

Overall reaction : 2H_2(g) + O_2(g) → 2 H₂O_(l)
The cell functions as long as the reacting gases are in supply. The heat of combustion is directly converted into electrical energy.

Question 13.
What is metallic corrosion ? Explain it with respect to iron corrosion.
Answer:
Metallic corrosion: The natural tendency of conversion of a metal into its mineral compound form on interaction with the environment is known as metallic corrosion.
Eg.: 1) Iron converts itself into its oxide [Rusting] (Fe₂O₃)
2) Silver converts itself into it’s sulphate [tarnishing] [Ag₂S]
Rusting of iron : Iron converts it self into its oxixe Fe₂O₃
The electrochemical reaction that represents the corrosion or rusting-of iron is
Anode : 2 Fe_(s) → 2 Fe²⁺ + 4e^– \(\mathrm{E}_{\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)}^0\) = -0.44 V
Cathode : O_2(g) + \(4 \mathrm{H}_{\text {(aq) }}^{+}\) + 4e^– → 2₂O_(l) \(\) = 1.23 V
Overall reaction : 2 Fe_(s) + O_2(g) + \(4 \mathrm{H}_{\text {(aq) }}^{+}\) → \(2 \mathrm{Fe}_{\mathrm{aq}}^{+2}\) + 2 H₂O_(l) \(\mathrm{E}_{(\text {cell) }}^{\Theta}\) = 1.67 V

Long Answer Questions

Question 1.
What are electro chemical cells ? How are they constructed ? Explain the working of the different types of galvanic cells ?
Answer:
Electrochemical cell is a device which makes use of a spontaneous redox reaction for the generation of electrical energy.
Eg : Galvanic cell, Daniell cell etc.
Galvanic cell: A device which converts chemical energy into electrical energy by the use of spontaneous redox reaction is called Galvanic cell (or) voltaic cell.
Eg : Daniell cell.

Daniell cell: It is a special type of galvanic cell. It contains two half cells in the same vessel.
The vessel is devided into two chambers. Left chamber is filled with ZnSO₄ (aq) solution and Zn – rod is dipped into it, Right chamber is filled with aq. CuSO₄ solution and a copper rod is dipped into it. Process diaphragm acts as Salt bridge. The two half cell’s are connected to external battery.

Cell reactions :
Ion Zn/ZnSO₄ half cell, oxidation, reaction occurs
Zn → Zn²⁺ + 2e^–
Ion Cu/CuSO₄ half cell, reduction reaction occurs.
Cu⁺² + 2e^– → Cu
The net cell reaction is
Zn + Cu⁺² ⇌ Zn⁺² + Cu
Cell is represented as Zn / Zn⁺² // Cu²⁺ / Cu

• The two vertical lines (||) represents a salt bridge.
• It is U shaped tube filled with KCl and agar agar gel salt bridge maintains the electrical neutrality and flow of current.

The batteries which after their use over a period of time, becomes dead and the cell reaction is completed and this cannot be reused again are called primary batteries.
Eg : Leclanche cell, dry cell.

Secondary battery: A secondary battery is the battery in which after it’s use can be recharged and can be used again.
A good secondary battery undergoes large no. of discharging and charging cycles.
Lead storage battery is an example of secondaiy battery.
The cell reactions when the battery is in use are
Pb_(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → PbSO₄(s) + 2e^– (Anode)
PbO_2(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → \(4 \mathrm{H}_{\text {(aq) }}^{+}\) + 2e → PbSO_4(s) + 2 H₂O_(l) (Cathode)
Overall cell reaction is
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(aq) + 2H₂O_(l)
In batteries also the principle involved same as in the galvanic cell.

Question 2.
What is electrical conductace of a solution? How ¡s it measured experimentally?
Answer:
The reciprocal of resistance (R) is known as conductance (c)
\(\frac{l}{\mathrm{R}}=\kappa \times \frac{1}{l / \mathrm{A}}\)
c = \(\kappa \times \frac{1}{l / \mathrm{A}}\)
The conductance or the current conducting capacity of an electrolytic solution can be expressed as .

1. Specific conductance (κ),
2. Molar conductivity (∧_m)

1. Specific conductance : The conductance of the solution enclosed between two parallel electrodes of unit area of cross section separated by unit distance is called specific conductance (κ).

2. Molar conductivity (∧_m) : The conductivity of a volume of solution containing one gram molecular weight of the electrolyte placed between two parallel electrodes separated by a distance of unit length of 1 meter is called molar conductivity (∧_m)
Relation between conductivity and molar conductivity :
∧_m = \(\frac{\kappa}{\mathrm{c}}\); ∴ c = constant
Measurement of electrical conductance :

• The resistance of a metallic wire can be measured with a Wheatstone bridge.
•  In measuring the resistance of an electrolytic solution two problems are identified.
  1. On passing direct current (DC) through the solution changes the composition of the solution due to electrolysis.
  2. A solution cannot be connected to the measuring bridge like a metallic wire.
• First problem solved by using AC instead of DC. .
• Second problem solved by using specially designed vessel called conductivity cell.
  
  By using above cells,
  Resistance R = \(\frac{l}{\kappa \times \mathrm{A}}\)
  l = distance between electrodes; A = Area of cross section; κ = conductivity
  \(\frac{l}{\mathrm{~A}}\) = cell constant = G*
  ∴ G* = \(\frac{l}{\mathrm{~A}}\) = R × κ
  The cell constant (G*) is determined by measuring the resistance of the cell containing a solution whose conductivity is already known
• Cell constant once determined, used for measuring the resistance (or) conductivity.
• The set up for the measurement of the resistance is shown in following figure.

Question 3.
Give the applications of Kohlrausch’s law of independent migration of ions. [T.S. Mar. 16]
Answer:
Kohlrausch’s law of independent migration of ions: The limiting molar conductivity of an electrolytes can be represented as the sum of the individual contributions of the anion and the cation of the electrolytes.
\(\lambda_{\mathrm{m}(\mathrm{AB})}^0=\lambda_{\mathrm{A}^{+}}^0+\lambda_{\mathrm{B}^{-}}^0\)
\(\lambda_{\mathrm{m}}^0\) = Limiting molar conductivity
\(\lambda_{\mathrm{A}^{+}}^0\) = Limiting molar conductivity of cation
\(\lambda_{\mathrm{B}^{-}}^0\) = Limiting molar conductivity of anion

Applications:
1) Calculation of limiting molar conductance at infinite dilution (∧_∞) of weak electrolytes :
Limiting molar conductance of weak electrolytes at infinite dilution- cannot be obtained graphically by extrapolation method. However the application of Kohlrausch’s law enables indirect evaluations in such cases.

Eg: Limiting molar conductance of acetic acid at infinite dilution can be calculated from the limiting molar conductances at infinite dilution of hydrochloric acid, sodium chloride and sodium acetate, as illustrated below:

2) Calculation of degree of dissociation of a weak electrolyte :
Degree of dissociation (α) = \(\frac{\lambda_{\mathrm{v}}}{\lambda_{\infty}}\)
Where λ_v = Limiting molar conductivity at a given concentration or dilution.
λ_∞ = Limiting molar conductivity of given electrolyte at infinite dilution.

3) Calculation of the solubility (s) of sparingly soluble salts like AgCl, BaSO₄ etc.
S = \(\frac{\kappa \times 1000}{\wedge_{\infty}}\)
where K = specific conductance
∧_∞ = Limiting molar conductance at infinite dilution.

Question 4.
Give the different types of batteries and explain the construction and working of each type of battery. [A.P. Mar. 19]
Answer:
The batteries which after their use over a period of time, becomes dead and the cell reaction is completed and this cannot be reused again are called primary batteries.
Eg.: Leclanche cell, dry cell.

Secondary battery:
A secondary battery is the battery in which after it’s use can be recharged and can be used again.
A good secondary battery undergoes large no. of discharging and charging cycles.
Lead storage battery is an example of secondary battery.
The cell reactions when the battery is in use are
Pb_(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → PbSO₄(s) + 2e^– (Anode)
PbO_2(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → \(4 \mathrm{H}_{\text {(aq) }}^{+}\) + 2e → PbSO_4(s) + 2 H₂O_(l) (Cathode)
Overall cell reaction is
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(aq) + 2H₂O_(l)

Construction and working of primary batteries :

Dry cell:

1. This is a modification of Leclanche cell. The liquid state electrolytes (in Leclanche cell) are replaced by paste electrolytes.
2. A cylindrical ‘Zn’ vessel is covered with a cardboard. This is sealed with pitch. Zn vessel acts as negative electrode. A carbon rod is introduced at the centre of the Zn vessel. This carbon rod acts as positive electrode.
3. Carbon rod is surrounded by a paste of (C + MnO₂). The remaining space is filled with (NH₄Cl + ZnCl₂ paste. The two pastes are separated by a porous sheet.
4. These cells are easy to handle and used in radios, watches, torch lights etc. The cell potential is approximately 1.5 V
   
5. Electrode reactions:
   Cathode : MnO₂ + \(\mathrm{NH}_4^{+}\) + e^Θ → MnO (OH) + NH₃
   Anode : Zn + 2MnO₂ + 2H₂O → Zn²⁺ + 2OH^Θ + 2MnO (OH)
   Some of the secondary reactions are
   2 NH₄Cl + 2OH^Θ → 2NH₃ + 2Cl^– + 2H₂O
   Zn²⁺ + 2NH₃ + 2Cl^– → [Zn (NH₃)₂] Cl₂

Secondary batteries – construction and working : A secondary cell after its use can be recharged and can be used again. A good secondary cell undergoes a large number of discharging and charging cycles. The most important secondary cell in use is the lead storage battery (fig.(b)). This is commonly used in automobiles and invertors. It consists of a lead anode and a grid of lead packed with lead dioxide (PbO₂) as cathode. A 38% solution of sulphuric acid is used as electrolyte.
The cell reactions when the battery is in use (discharging) are :
Anode : Pb_(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → PbSO_4(s) + 2e^–
Cathode: PbO_2(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → \(4 \mathrm{H}_{\text {(aq) }}^{+}\) + 2e → PbSO_4(s) + 2 H₂O_(l)
i.e., overall cell reaction consisting of cathode and anode reactions is
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(aq) + 2H₂O_(l)
These reactions occur during discharge i.e., during use of the battery.
On charging the discharged battery the above reaction is reversed and PbSO₄ (s) on anode and cathode is converted into Pb and PbO₂, respectively.

Problems

Question 1.
The standard potentials of some electrodes are as follows. Arrange the metals in an increasing order of their reducing power.
1) K⁺/K = -2.93 V
2) Ag⁺/Ag = 0.8OEV
3) Cu²⁺/Cu = 0.34 V
4) Mg²⁺/Mg = -2.37 V
5) Cr³⁺/Cr = – 0.74V
6) Fe²⁺/Fe = —0.44 V
Solution:
Given
E₀ of cells K⁺/K = —2.93
Mg⁺²/Mg = -2.37 V
Ag⁺/Ag = 0.80 V
Cr⁺³/Cr = -0.74 V
Cu⁺²/Cu = 0.34 V
Fe⁺²/Fe = -0.44 V
Lower the reduction potential values indicates high reducing power and high value indicates the high oxidising power.
So, increasing order of the given cells reducing power is
Ag < Cu < Fe < Cr < Mg < K.

Question 2.
Calculate the emf of the cell at 25°C
Cr | Cr³⁺ (0.1 M) || Fe²⁺ (0.01M) | Fe, given that \(\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^0\) = – 0.74V and \(\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^0\) = 0.44 V.
Solution:
Given cell is
Cr | \(\mathrm{Cr}_{(0.1 \mathrm{M})}^{+3}\) || \(\mathrm{Fe}_{(0.01 \mathrm{M})}^{+2}\) | Fe
E⁰ of Cr⁺³/Cr = -0.74 V
E⁰ of Fe⁺²/Fe = -0.44 V
E_Cr⁺³/Cr = E⁰ + \(\frac{0.059}{3}\) log10 [Cr⁺³]
= -0.74 + \(\frac{0.059}{3}\) log 0.1
= -0.76 V
E_Fe⁺²/Fe = -0.44 + \(\frac{0.059}{2}\) log 0.01
= -0.44 – 0.059 = – 0.499 V
EMF of cell = E_RHS – E_LHS
= (- 0.499) – (-0.76) = 0.26 1 V .

Question 3.
Calculate the potential of a Zn – Zn²⁺ electrode in which the molarity of Zn²⁺ is 0.001 M. Given that \(\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^0\) = – 0.76 V
R = 8.314 JK^-1 mol^-1; F = 96500 C mol^-1.
Solution:
Given electrode Zn | Zn⁺²(0.001m) \(\mathrm{E}_{\mathrm{Zn}^{+2} / \mathrm{Zn}}^0\) = -0.76V
Nemst equation is
E = E⁰ + \(\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\) log [Mⁿ⁺]
Given R = 8.314 J/K. mole
F = 96500 c/mole
E = E⁰ + \(\frac{0.059}{\mathrm{n}}\) log C
= -0.76 + \(\frac{0.059}{2}\) log 0.001
= -0.76 – \(\frac{0.059}{2}\) × 3
= – 0.76 – 0.0295 × 3
= -0.76 – 0.0885
= -0.8485 V

Question 4.
Determine ∆G⁰ for the button cell used in the watches. The cell reaction is
Zn_(s) + Ag₂O_(s) + H₂O_(l)→ \(\mathrm{Zn}_{(a q)}^{2+}\) + 2 Ag_(s) + 2 \(\mathrm{OH}_{(a q)}^{-}\)
\(\mathrm{E}_{\mathrm{Ag}^+ / \mathrm{Ag}}^0\) = + 0.80 V; \(\mathbf{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^0\) = – 0.76 V.
Solution:
Given E⁰ of Ag⁺/Ag = 0.80 V
E⁰ of Zn⁺²/Zn = – 0.76 V
Cell representation Zn/Zn⁺²|| Ag⁺ / Ag
EMF = E_RHS – E_LHS = 0.80 – (-0.76) = 1.56 V
∆G = -nFE⁰ = -2 × 96500 × 1.56 = -301.08 kJ/mole

Question 5.
Calculate the emf of the cell consisting the following half cells
Al/Al³⁺(0.001 M), Ni/Ni²⁺ (0.50 M). Given that \(\mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^0\) = – 0.25 V \(\mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^0\) = – 1.66 v (log 8 × 10^-6 = – 5.0969).
Solution:
Given E⁰ of Al⁺³/Al =-1.66 V
E⁰ of Ni⁺²/Ni = – 0.25 V
Cell representation Al/\(\mathrm{Al}_{(0.001 \mathrm{M})}^{+3}\) || \(\mathrm{Ni}_{(0.50 \mathrm{M})}^{+2}\) | Ni
Nemst equation is
E_Al⁺³/Al = \(\mathrm{E}_{\mathrm{Al}^{+3} / \mathrm{A} l}^0\) + \(\frac{0.059}{n}\) log [Al⁺³]
= -1.66 + \(\frac{0.059}{3}\) log 0.001 = – 1.66 – 0.059 = – 1.719V
E_Ni⁺²/Ni = \(\mathrm{E}_{\mathrm{Ni}^{+2} / \mathrm{Ni}}^0+\frac{0.059}{\mathrm{n}}\) log [Ni⁺²]
= -0.25 + \(\frac{0.059}{2}\) log [0.5]
= -0.25 – 0.0295 × 0.3010 = -0.25 – 0.0088795 = -0.25888
EMF of cell = E_RHS – E_LHS
= -0.25888 – (-1.719) = 1.46012 V

Question 6.
Determine the values of Kc for the following reaction
Ni_(s) + 2 \(\mathrm{Ag}_{\text {(aq) }}^{+}\) → \(\mathrm{Ni}_{\text {(aq) }}^{2+}\) + 2 Ag_(s) E⁰ = 1.05 V.
Solution:
Given
Ni_(s) + 2 \(\mathrm{Ag}_{\text {(aq) }}^{+}\) → \(\mathrm{Ni}_{\text {(aq) }}^{2+}\) + 2 Ag_(s)
E⁰ = 1.05 V.
Nernst Equation is E = E⁰ – \(\frac{0.059}{\mathrm{n}}\) log K_c
at equilibrium E = 0
E⁰ = \(\frac{0.059}{\mathrm{n}}\) log K_c
1.05 = \(\frac{0.059}{2}\) log K_c
log K_c = \(\frac{2.1}{0.059}\) = 35.59
K_c = 3.42 × 10³⁵

Question 7.
Calculate the potential of the half-cell containing 0.1 M
K₂Cr₂O_7(aq), 0.2 M \(\mathrm{Cr}_{\mathrm{aq}}^{3+}\) and 1.0 × 10^-4 M \(\mathrm{H}_{(\mathrm{aq})}^{+}\). The half-reaction \(\mathrm{Cr}_2 \mathrm{O}_{7(\mathrm{aq})}^{2-}+14 \mathrm{H}_{\text {(aq) }}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+7 \mathrm{H}_2 \mathrm{O}_{(l)}\) (E⁰ of \(\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}\) = 1.33 V)
Solution:
Given half reaction

Question 8.
Calculate K for the reaction at 298 K
\(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Zn}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\)
\(\mathrm{E}_{\mathrm{Zn}^{2+/ Z n}}^0\) = -0.76 V; \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0\) = + 0.34 V.
Solution:
Given \(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Zn}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\)
E⁰ of Zn⁺²/Zn = -0.76 V
E⁰ of Cu⁺²/Cu = -0.34 V
E⁰ of cell = E_RHS – E_LHS = 0.34 – (- 0.76) = 1.1 V
∆G⁰ = – RTlnK_c = – 2.303 RT log K_c
-212300 = -2.303 × 8.314 × 298 × log K_c
log K_c = \(\frac{212300}{2.303 \times 8.314 \times 298}\) = 37.207
K_c = 1.6 × 10³⁷

Question 9.
Calculate the emf of the cell at 298 K
Sn_(s) | Sn²⁺ (0.05 M) || \(\mathrm{H}_{(\mathrm{aq})}^{+}\) (0.02 M) | H₂ 1 atm. Pt
Given that \(\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^0\) = -0.144 V
Solution:

Question 10.
Calculate the concentration of silver ions in the cell constructed by using 0.1 M concentration of Cu²⁺ and Ag⁺ ions. Cu and Ag metals are used as electrodes. The cell potential is 0.422 V.
[\(\mathrm{E}_{\mathrm{Ag}^{2+} / \mathrm{Ag}}^0\) = 0.80 V; \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0\) = +0.34 V]
Solution:
Given E⁰ of Ag⁺/Ag = 0.80 V .
E⁰ of Cu⁺²/Cu = 0.34 V
E⁰ of cell = E_RHS – E_LHS = 0.80 – 0.34 = 0.46 V

Question 11.
Calculate the emf of the cell with the cell reaction
Ni_(s) + 2 Ag⁺ (0.002M) → Ni²⁺ (0.160 M) + 2 Ag_(s) ; \(\mathrm{E}_{\text {cell }}^0\) = 1.05 V.
Solution:
From the given cell reaction and Nernst equation

Question 12.
Cu²⁺ + 2e^– ⇌ Cu; E⁰ = +0.34 V
Ag⁺ + e^– ⇌ Ag; E⁰ = + 0.80 V
For what concentration of Ag⁺ ions will the emf of the cell be zero at 25°C. The concentration of Cu²⁺ is 0.1 M. (log 3.919 = 0.593).
Solution:
Cell is Cu|Cu⁺² | |Ag⁺|Ag
E⁰ of cell = E_RHS – E_LHS = 0.80 – 0.34 = 0.46 V
Nernst equation is E = E⁰ – \(\frac{0.059}{2} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\)
EMF of cell = 0
E⁰ = \(\frac{0.059}{2} \log \frac{0.1}{\left[\mathrm{Ag}^{+}\right]^2}\)

Question 13.
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate molar conductance.
Solution:
Conductivity (K) = 0.0248 S cm^-1 = 0.0248 ohm^-1 cm^-1
Molar concentration [c] = 0.20 mol L^-1
= \(\frac{[0.2 \mathrm{~mol}]}{\left[1000 \mathrm{~cm}^3\right]}\) = 2.0 × 10^-4 mol cm^-3
Molar conductivity [∧_m] = \(\frac{\kappa}{\mathrm{c}}=\frac{\left[0.0248 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\right]}{\left[2.0 \times 10^{-4} \mathrm{~mol} \mathrm{~cm}^{-3}\right]}\)
= 124 ohm^-1 mol^-1 cm² or 1245 mol^-1 cm²
Molar conductivity = 124 mol^-1 cm²

Question 14.
Calculate the degree of dissociation (α) of CH₃COOH at 298 K.
Given that \(\wedge_{\mathrm{CH}_3 \mathrm{COOH}}^{\infty}\) = 11.75 cm² mol^-1
\(\wedge_{\mathrm{CH}_3 \mathrm{COO}^{-}}^0\) = 40.95 cm² mol^-1
\(\wedge_{\mathrm{H}^{+}}^0\) = 349.15 cm² mol^-1
Solution:
Given that \(\wedge_{\mathrm{CH}_3 \mathrm{COOH}}^{\infty}\) = 11.75 cm² mol^-1
\(\wedge_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\infty}\) = 40.95 cm² mol^-1
\(\wedge_{\mathrm{H}^{+}}^{\infty}\) = 349.15 cm² mol^-1
\(\wedge_{\mathrm{CH}_3 \mathrm{COOH}}^{\infty}\) = \(\wedge_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\infty}\) + \(\wedge_{\mathrm{H}^{+}}^{\infty}\) = 40.95 + 349.15 = 390.1
Degree of dissociation (α) = \(\frac{\lambda^\alpha}{\lambda^{\infty}}=\frac{11.75}{390: 1}\) = 0.03 = 3 × 10^-2
∴ α = 3× 10^-2

Textual Examples

Question 1.
Represent the cell in which the following reaction takes place Mg(s) + 2Ag⁺ (0.0001M) → Mg²⁺ (0.130M) + 2 Ag(s)
Calculate its E_(cell) if \(\mathrm{E}_{(\text {cell) }}^{\Theta}\) = 3.17 V
Solution:
The cell is represented as Mg |Mg²⁺ (0.130M) || Ag⁺(0.0001M) | Ag

Question 2.
Calculate the equilibrium constant of the reaction:
Cu_(s) + 2 \({Ag}_{\text {(aq) }}^{+}\) → \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) + 2 Ag_(s) ; \(\mathrm{E}_{(\text {cell) }}^{\Theta}\) = 0.46 V
Solution:
\(\mathrm{E}_{(\text {cell) }}^{\Theta}\) = \(\frac{0.059V}{2}\) log K_c = 0.46 V or
l0g K_c = \(\frac{0.46 \mathrm{~V} \times 2}{0.059 \mathrm{~V}}\) = 156
K_c = 3.92 × 10¹⁵

Question 3.
The standard emf of Daniell cell is 1.1V. Calculate the standard Gibbs energy for the cell reactions:
Zn_(s) + \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) → \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + Cu_(s)
Solution:
∆_rG^Θ = -nF\(\mathrm{E}_{(\text {cell) }}^{\Theta}\)
n in the above equation is 2, F = 96487 C mol^-1 and \(\mathrm{E}_{(\text {cell) }}^{\Theta}\) = 1.1 V
Therefore ∆_rG^Θ = -2 × 1.1V × 96487 C mol^-1
= – 21227 J mol^-1 =-212.27 kJ mol^-1

Question 4.
Resistance of a conductivity cell filled with 0.1 mol L^-1 KCl solution is 100 Ω. If .the resistance of the same cell when filled with 0.02 mol L^-1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L^-1 KCl solution. The conductivity of 0.1 mol L^-1 KCl solution is 1.29 S/m.
Solution:
The cell constant is given by the equation :
Cell constant = G* = conductivity × resistance
= 1.29 S/m 100 Ω = 129 m^-1 = 1.29 cm^-1
Conductivity of 0.02 mol L^-1 KCl solution = cell constant / resistance
= \(\frac{\mathrm{G}^*}{\mathrm{R}}=\frac{129 \mathrm{~m}^{-1}}{520 \Omega}\) = 0.248 S m^-1
Concentration = 0.02 mol L^-1
= 1000 × 0.02 mol m^-3 = 20 mol m^-3
Molar conductivity = ∧_m = \(\frac{\kappa}{\mathrm{c}}\)
= \(\frac{248 \times 10^{-3} \mathrm{Sm}^{-1}}{20 \mathrm{~mol} \mathrm{~m}^{-3}}\) = 124 × 10^-4 Sm² mol^-1
Alternatively, κ = \(\frac{1.29 \mathrm{~cm}^{-1}}{520 \Omega}\) = 0.248 × 10^-2 S cm^-1
and ∧_m = κ × 1000 cm³L^-1 molarity^-1
= \(\frac{0.248 \times 10^{-2} \mathrm{~S} \mathrm{~cm}^{-1} \times 1000 \mathrm{~cm}^3 \mathrm{~L}^{-1}}{0.02 \mathrm{~mol} \mathrm{~L}^{-1}}\)
= 124 S cm² mol^-1

Question 5.
The electrical resistance of a column 0.05 mol L^-1 NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 10³ ohm. Calculate its resistivity, conductivity and molar conductivity.
Answer:
A = πr² = 31.4 × 0.5² cm² = 0.785 cm2 = 0.785 × 10^-4 m²
l = 50 cm = 0.5 m

Question 6.
Calculate \(\wedge_m^0\) for CaCl₂ and MgSO₄ from the data given in Table 3.4.
Solution:
We know from Kohlrausch law that
\(\wedge_{\mathrm{m}\left(\mathrm{CaCl}_2\right)}^0=\wedge_{\mathrm{Ca}^{2+}}^0+2 \lambda_{\mathrm{Cl}}^0\) = 119.0 S cm² mol ^-1 + 2(76.3) S cm² mol^-1
= (119.0 + 152.6) Scm² mol^-1
= 271.6 S cm² mol^-1
\(\wedge_{\mathrm{m}\left(\mathrm{MgSO}_4\right)}^0=\lambda_{\mathrm{Mg}^{2+}}^0+\lambda_{\mathrm{SO}_4^{2-}}^0\) =106.0 S cm² mol^-1 + 160.0 S cm² mol^-1
= 266 S cm² mol^-1

Question 7.
\(\wedge_m^0\) for NaCl, HCl and NaAc are 126.4,425.9 and 91.0 S cm² mol^-1 respectively. Calculate ∧⁰ for HAc.
Solution:
\(\wedge_{\mathrm{m}(\mathrm{HAc})}^0=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{Ac}^0}^0=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{Cl}^{-}}^0+\lambda_{\mathrm{Na}^{+}}^0-\lambda_{\mathrm{Cl}^{-}}^0-\lambda_{\mathrm{Na}^{+}}^0\)
= \(\wedge_{\mathrm{m}(\mathrm{HCl})}^0+\wedge_{\mathrm{m}(\mathrm{NaAc})}^0-\wedge_{\mathrm{m}(\mathrm{NaCl})}^0\)
= (425.9 + 91.0 – 126.4) S cm² mol^-1
= 390.5 S cm² mol^-1

Question 8.
The conductivity of 0.001028 mol L^-1 acetic acid is 4.95 × 10^-5 S cm^-1. Calculate its dissociation constant if \(\wedge_m^0\) for acetic acid is 390.5 S cm² mol^-1.
Solution:

Question 9.
A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of eopper deposited at the cathode ? [A.P. & T.S. Mar. 15] [Mar. 15]
Solution:
t = 600 s charge = current × time = 1.5A × 600 s = 900 C
According to the reaction :
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) + 2e^– = Cu_(s)
We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu.
For 900 C, the mass of Cu deposited
= (63 g mol^-1 × 900 C) / (2 × 96487 C mol^-1) = 0.2938 g

Intext Questions

Question 1.
How would you determine the standard electrode potential of the system Mg²⁺/Mg ?
Solution:
E° value of Mg²⁺/Mg electrode is determined by setting up an electrochemical cell. For this purpose, a Mg electrode is dipped in 1 M MgSO₄ solution, which acts as one half cell i.e., oxidation half-cell.

In the same way, the standard hydrogen electrode acts as the other half-cell i.e., reduction half cell. The deflection of voltmeter placed in the cell circuit is towards the Mg electrode which indicates the direction of flow of current.
The cell may be represented as : Mg | Mg²⁺ (1M) || H⁺ (M) | H₂ (1 atm). Pt The reading as given by voltmeter gives \(\mathrm{E}_{\text {cell }}^0\)

Question 2.
Can you store copper sulphate solutions In a zinc pot?
Solution:
No, zinc pot cannot store copper sulphate solutions because the standard electrode potential (E⁰) value of zinc is less than that of copper. So, zinc is stronger reducing agent then copper.
\(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + 2e^– → Zn(s); E⁰ = -0.76 V
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) + 2e^– → Cu(s); E⁰ = – 0.34 V
So, zinc will loss electrons to Cu²⁺ ions and redox reaction will occur as follows.
Zn(s) + \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) → \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + Cu(s)

Question 3.
Consult the table on standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Solution:
Fe²⁺ (ferrous) ions gets oxidised to Fe³⁺ (ferric) ion as follows
Fe²⁺ → Fe³⁺ + e^– \(\mathrm{E}_{0 \mathrm{x}}^0\) = -0.77V
Only those substances can oxidise Fe²⁺ ions to Fe³⁺ ions which can accept electrons released during oxidation or are placed above iron in electrochemical series. Three such substances are: Cl₂(g), Br₂(g) and F₂(g).

Question 4.
Calculate the potential of hydrogen electrode placed in a solution of pH 10.
Solution:

Question 5.
The cell in which the following cell reaction occurs,
\(2 \mathrm{Fe}_{(\mathrm{aq})}^{3+}+2 \mathrm{I}_{(\mathrm{aq})}^{-} \longrightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+\mathrm{I}_{2(\mathrm{~s})}\) has \(\mathrm{E}_{\text {cell }}^0\) = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Solution:
Two half reactions for the given redox reaction may be written as :
\(2 \mathrm{Fe}_{(\mathrm{aq})}^{3+}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+} ; 2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 \mathrm{e}^{-}\)
2 moles of electrons are involved in the reaction, so n = 2
∆_rG⁰ = – nF\(\mathrm{E}_{\text {cell }}^0\) = (-2 mol) × (96500 mol^-1) (0.236 V)
= -45548 CV = -45548 J
∆_rG⁰ = – 45.55 kJ
log K_c = \(\frac{\Delta \mathrm{G}^0}{2.303 \mathrm{RT}}\)
= \(\frac{(-45.55 \mathrm{~kJ})}{2.303 \times\left(8.314 \times 10^{-3} \mathrm{kJK}^{-1}\right) \times(298 \mathrm{~K})}\) = 7.983
K_c = Antilog (7.983) = 9.616 × 10⁷
K_c = 9.616 × 10⁷

Question 7.
Why does the conductivity of a solution decrease with dilution ?
Solution:
The conductivity of a solution is related with the number of ions present per unit volume of the solution. When the solution is diluted, the number of ions decreases. Hence, conductivity or specific conductance of the solution also decreases.

Question 8.
Suggest a way to determine the \(\wedge_{\mathrm{m}}^0\) value of water.
Solution:
The molar conductance of water at infinite dilution can be obtained from the knowledge of molar conductances at inifite dilution of sodium hydroxide, hydrochloric acid and sodium chloride (all strong electrolytes). This is in accordance with Kohlrausch’s law.
\(\wedge_{\mathrm{m}\left(\mathrm{H}_2 \mathrm{O}\right)}^0=\wedge_{\mathrm{m}(\mathrm{NaOH})}^0+\wedge_{\mathrm{m}(\mathrm{HCl})}^0-\wedge_{\mathrm{m}(\mathrm{NaCl})}^0\)

Question 9.
The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1. Calculate its degree of dissociation and dissociation constant.
Given, λ⁰(H⁺) = 349.6 S cm^-1 mol^-1 and λ⁰ (HCOO^–) = 54.6 S cm² mol^-1
Solution:
Step I: Calculation of degree of dissociation (a) of HCOOH

= \(\frac{\left(3.249 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\right)}{(0.886)}\)
= 3.67 × 10^-4 mol L^-1
α = 0.114
K_a = 3.67 × 10^-4 mol L^-1

Question 10.
If a current of 0.5 ampere flows through a metallic wire for 2h, then how many electrons would flow through the wire ?
Solution:
Quantity of charge (Q) passed = Current (C) × Time (t)
= (0.5 A) × (2 × 60 × 60 s)
= (3600) Ampere sec = 3600 C ,
Number of electrons flowing through the wire on passing charge of one Faraday (96500 C) = 6.022 × 10²³
Number of electrons flowing through the wire on passing a charge of 3600 C
= \(\frac{6.022 \times 10^{23} \times(3600 \mathrm{C})}{(96500 \mathrm{C})}\)
= 2.246 × 10²²
Number of electrons = 2.246 × 10²²

Question 11.
Suggest a list to metals that are extracted electrolytically.
Solution:
Those metals which are highly reactive and have large negative E⁰ values can be exerted electrolytically. This type of metals are powerful reducing agents.
Eg: Sodium, Potassium, Calcium, Magnesium etc.

Question 12.
Consider the reaction,
\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O
What is the quantity of electricity in coulombs nefeded to reduce 1 mole of \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) ?
Solution:
1 mole \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) require 6 Faraday electricity (6 F)
= 6 × 96500 C = 5.79 × 10⁵ C
Quantity of electricity = 5.79 × 10⁵ Coulomb

Question 13.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Solution:
When lead storage battery is recharged, electrical energy is supplied to the cell from the external source : i.e., the cell operates as an electrolytic cell during recharging. All the chemical reactions, which take place during the use of battery, are reversed at this time.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Solution:
Methane (CH₄) and methanol (CH₃OH)

Question 15.
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Solution:
In corrosion, a metal is oxidised by loss of electrons to oxygen with the formation of oxides. So, an electrochemical cell is set up.
Eg: Rusting of iron involves the following steps :
i) The water layer present on the surface of iron dissolves acidic oxides from air like CO₂ and forms acid to produce H⁺ ions.
H₂O + CO₂ H₂CO₃ ⇌ 2H⁺ + \(\mathrm{CO}_3^{2-}\)

ii) In the presence of H⁺ ions, iron starts losing electrons at some spot to form ferrous ions. This spot behaves as anode.

iii) The electrons released at anode move to another spot, where H⁺ ions and the dissolved oxygen gain these electrons. This spot becomes a cathode.

iv) Overall reaction, i.e., redox reaction is :

v) Ferrous ions are further oxidised by the atmospheric oxygen to ferric ions which combine with water molecules to form hydrated ferric oxide, Fe₂O₃. xH₂O. (Rust).

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 2nd Lesson Solutions Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 2nd Lesson Solutions

Very Short Answer Questions

Question 1.
Define the term solution.
Answer:
Solution : Solution is a homogeneous mixture of two (or) more components whose composition may be varied with certain limits.

Question 2.
Define molarity. [T.S. Mar.17] [Mar. 11]
Answer:
Molarity : The number of moles of solute dissolved in one litre of solution is called molarity.
Molarity (M) = \(\frac{\text { Moles of solute }}{\text { Volume (in litres) }}\)
M = \(\frac{\text { Weight }}{\text { Gram mol wt }} \times \frac{1000}{V(\mathrm{ml} l)}\)
Units: moles / litre.

Question 3.
Define molality [A.P. Mar. 15] [May 11]
Answer:
Molality: The number of moles of solute present in one kilogram of solvent is called molality.
Molality (M) = \(\frac{\text { Moles of solute }}{\text { Mass of solvent (kgs) }}\)
m = \(\frac{\text { Weight }}{\text { Gram mol wt }} \times \frac{1000}{\mathrm{G}(\mathrm{gms})}\)

Question 4.
Give an example of a solid solution in which the solute is solid.
Answer:
Example of a solid solution in which the solute is solid, is copper dissolved in gold.

Question 5.
Define mole fraction. . [T.S. Mar. 18] [Mar. 14]
Answer:
Mole fraction : The ratio of number of moles of the one component of the solution to the total number of moles of all the components of the solution is called mole fraction.
Mole fraction of solute X_s = \(\frac{\mathbf{n}_{\mathrm{S}}}{\mathbf{n}_{0}+\mathbf{n}_{\mathrm{s}}}\)
Mole fraction of solvent X₀ = \(\frac{\mathbf{n}_{\mathrm{S}}}{\mathbf{n}_{0}+\mathbf{n}_{\mathrm{s}}}\)
[n_s =number of moles of solute]
[n₀ = number of moles of solvent]
It has no units.

Question 6.
Define mass percentage solution.
Answer:
The mass percentage of a component of a solution is defined as the
mass % = × 100

Question 7.
What is ppm of a solution ?
Answer:
ppm – parts per million : It is a convenient method of expressing concentration when a solute is present in trace quantities. Parts per million is defined as the
ppm = × 10⁶

Question 8.
What role do the molecular interactions play in a solution of alcohol and water ?
Answer:
In a solution of alcohol and water the molecular interactions present are hydrogen bondings. When these components are mixed then new hydrogen bonds develop between alcohol and water. The formed forces are weaker. Due to the decrease of magnitude of attractive forces, the solution shows positive deviation from Raoult’s law. This leads to increase of vapour pressure of solution and decrease in its boiling point.

Question 9.
State Raoult’s law. [A.P. & T.S. Mar. 18; 16; A.P. Mar. 17] [Mar. 14]
Answer:
Raoult’s law for volatile solute: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

Raoult’s law for non-volatile solute : The relative lowering of vapour pressure of dilute solution containing non-volatile solute is equal to the mole fraction of solute.

Question 10.
State Henry’s law.
Answer:
At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid. .
(Or)
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.
P = K_H × x K_H = Henry’s constant
P = Partial pressure
x = Mole fraction of gas

Question 11.
What is Ebullioscopic constant ?
Answer:
Ebullioscopic constant : The elevation of boiling point observed in one molal solution containing non-volatile solute is called Ebullioscopic constant (or) molal elevation constant.

Question 12.
What is Cryoscopic constant ?
Answer:
Cryoscopic constant : The depression in freezing point observed in one molal solution containing non-volatile solute is called cryoscopic constant (or) molal depression constant.

Question 13.
Define osmotic pressure. [A.P. Mar. 18; 16]
Answer:
Osmotic pressure : The pressure required to prevent the in flow of solvent molecules into the solution when these are. separated by semi-permeable membrane.

Question 14.
What are isotonic solutions ? [T.S. Mar. 19; 16; A.P. Mar. 17, 15]
Answer:
Isotonic solutions : Solutions having same osmotic pressure at a given temperature are called isotonic solutions.
e.g.: Blood is isotonic with 0.9% \(\left(\frac{\mathrm{W}}{\mathrm{V}}\right)\) NaCl [Saline]

Question 15.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water, (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol.
Answer:

1. Phenol is partially soluble in water.
2. Toluene is insoluble in water.
3. Formic acid is highly soluble in water.
4. Ethylene glycol is highly soluble in water.
5. Chloroform is insoluble in water.
6. Pentanol is partially soluble in water.

Question 16.
Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 gm of C₉H₈O₄ is dissolved in 450g of CH₃CN.
Answer:
Given
The Mass of aspirin = 6.5 gms
Mass of acetonitrile = 450 gms
Mass of solution = 6.5 + 450
= 456.5 gms
Mass % = \(\frac{6.5}{456.5}\) × 100 = 1.424 %

Question 17.
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 ml of 0.15 M solution in methanol.
Answer:
Given
Molarity = 0.15 M
Volume V = 250 ml
Molecular weight of benzoic acid (C₆H₅COOH) = 122
Molarity (M) = \(\frac{\text { Weight }}{\text { GMw }} \times \frac{1000}{V(\mathrm{~m} l)}\)
0.15 = \(\frac{\mathrm{W}}{122} \times \frac{1000}{250}\)
W = \(\frac{122 \times 0.15}{4}\) = 4.575 gms

Question 18.
The depression in freezing point of water observed for the same amount of acetic acid, dichloro-acetic acid and trichloro acetic acid increases in the order given above. Explain briefly.
Answer:
Given acids are CH₃COOH, CHCl₂COOH, CCl₃ COOH.
The depression in freezing point of a solute in water depends upon the number of ions (or) particles in aqueous solution.
The given three acids are arranged in the order of their acidic strengths (increasing order)
CH₃COOH < CHCl₂COOH < CCl₃COOH
Due to the presence of three chlorine atoms CCl₃COOH is more acidic than CHCl₂ COOH and followed by CH₃COOH.
∴ The order of depression in freezing point is
CH₃COOH < CHCl₂COOH < CCl₃COOH

Question 19.
What is Van’t Hoffs factor ‘i’ and how is it related to ‘a’ in the case of a binary electrolyte (1:1)?
Answer:
Van’t Hoffs factor (i) : “It is defined as the ratio of the observed value of colligative property to the theoretical value of colligative property”.
i =
Solute dissociation or ionization process : If a solute on ionization gives ‘n’ ions and ‘a’ is degree of ionization at the given concentration, we will have [1 + (n – 1) α] particles.
(1 – α) ⇌ nα
Total 1 – α + nα = [1 + (n – 1) α]
∴ i = \(\frac{[1+(n-1) \alpha]}{1}\)
α = \(\frac{\mathrm{i}-1}{\mathrm{n}-1}\)
α_ionization = \(\frac{\mathrm{i}-1}{\mathrm{n}-1}\)
Solute association process :
If ‘n’A molecules combine to give A_n, we have
nA ⇌ A_n
If ‘α’ is degree of association at the given concentration.
1 – α = \(\frac{\alpha}{n}\)
= 1 – α + \(\frac{\alpha}{n}\)
i = \(\frac{1-\alpha+\frac{\alpha}{n}}{1}\)
(or) α = \(\frac{1-\mathrm{i}}{1-\frac{1}{n}}\)
α _dissociation = \(\frac{1-\mathrm{i}}{1-\frac{1}{n}}\)

Question 20.
What is relative lowering of vapour pressure ? [A.P. Mar. 19]
Answer:
Relating lowering of vapour pressure: The ratio of lowering of vapour pressure of a solution containing non-volatile solute to the vapour pressure of pure solvent is called relative lowering of vapour pressure.
R.L.V.P. = \(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{0}}\)
P₀ – P_S = lowering of vapour pressure
P₀ = Vapour pressure of pure solvent

Question 21.
Calculate the mole fraction of H₂SO₄ in a solution containing 98% H₂SO₄ by mass. [A.P. Mar. 17]
Answer:
Given a Solution containing – 98% H₂SO₄ by mass.
It means 98 gms of H₂SO₄ and 2 gms of H₂O mixed to form a solution.

Short Answer Questions

Question 1.
How many types of solutions are formed? Give an example for each type of solution.
Answer:
Solutions are classified into three types on the basis of the solvent present in that solution.

Question 2.
Define mass percentage, volume percentage and mass to volume percentage solutions.
Answer:
i) Mass percentage : The mass percentage of a component of a solution is defined as the
mass % = × 10
ii) Volume percentage \(\left(\frac{\mathbf{v}}{\mathbf{V}}\right)\) : It is defined as the volume % of a component
= × 100
iii) Mass to volume percentage \(\left(\frac{\mathbf{w}}{\mathbf{V}}\right)\) : It is defined as the mass of solute dissolved in 100 ml of the solution.
It is commonly used in medicine and pharmacy.

Question 3.
Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504g mL^-1?
Answer:
Given 68% HNO₃ by mass that means
68gms mass of HNO₃ present in 100 gms of solution.
Molecular weight of HNO₃ = 63
Number of moles of HNO₃ = \(\frac{\text { weight }}{\text { GMW }}=\frac{68}{63}\) = 1.079
Given density of the solution = 1.504 gm/ml

Question 4.
A solution of glucose in water is labelled as 10% w/w. What would be the molarity of the solution?
Answer:
Given 10% \(\left(\frac{\mathrm{W}}{\mathrm{w}}\right)\) glucose in water solution
Weight of glucose = 10 gms
GMW of glucose (C₆H₁₂O₆) = 180
Weight of water = 100 – 10 = 90 gms.
Molarity (m) = \(\frac{\text { Weight }}{\text { GMW }} \times \frac{1000}{\text { W (gms) }}\)
= \(\frac{10}{180} \times \frac{1000}{90}\) = \(\frac{100}{18 \times 9}\)
= 0.617 moles / kg
Molarity = \(\frac{\text { Weight }}{\text { GMW }} \times \frac{1000}{V(\mathrm{~m} l)}\)
Weight of solution = 100 gms
Assuming that Density of solution = 1.2 gm/ml
Volume = \(\frac{100}{1.2}\) = 83.33 ml
Molarity (M) = \(\frac{10}{180} \times \frac{1000}{83.33}\) ml
= \(\frac{1000}{18 \times 83.33}\) = 0.67 M

Question 5.
A solution of sucrose in water is labelled as 20% w/w. What would be the mole fraction of each component in the solution ?
Answer:
20% \(\left(\frac{\mathrm{W}}{\mathrm{w}}\right)\) Sucrose in water solution means
20 gms of Sucrose and 80 gms of water.
Number of moles of sucrose (n_s) = \(\frac{20}{342}\) = 0.0584
Number of moles of water (n₀) = \(\frac{80}{18}\) = 4.444
Mole fraction of sucrose X_s = \(\frac{n_{s}}{n_{0}+n_{s}}\)
= \(\frac{0.0584}{4.444+0.0584}\) + \(\frac{0.0584}{4.50284}\) = 0.01296
Mole fraction of water X₀ = \(\frac{n_{s}}{n_{0}+n_{s}}\) = \(\frac{4.444}{4.50284}\) = 0.9869
(Or)
X_s + X₀ = 1
X₀ = 1 – X_s = 1 – 0.01296 = 0987

Question 6.
How many ml of 0.1 M HCl is required to react completely with 1.0 g mixture of Na₂CO₃ and NaHCO₃ containing equal-molar amounts of both?
Answer:
Given 1gm mixture of Na₂CO₃ and NaHCO₃
Let the mass of Na₂CO₃ = a gms
Mass of NaHCO₃ = (1 – a) gms
Number of moles of Na₂CO₃ = \(\frac{\mathrm{wt}}{\text { GMW }}=\frac{\mathrm{a}}{106}\)
Number of moles of NaHCO₃ = \(\frac{\mathrm{wt}}{\mathrm{GMW}}=\frac{1-\mathrm{a}}{84}\)
Given that the mixture contains Equi molar amounts of Na₂CO₃ and NaHCO₂
∴ \(\frac{a}{106}\) = \(\frac{1-a}{84}\)
84 a =106 – 106 a
190 a = 106
a = 0.558gms
∴ Weight of Na₂CO₃ = 0.558 gms
Weight of NaHCO₃ = 1 – 0.558 = 0.442gms
Na₂CO₃ + 2 HCl → 2 NaCl + H₂O + CO₂
106 gms – 73 gms
0.558 g – ? \(\frac{73 \times 0.558}{106}\) = 0.384 gms
NaHCO₃ + HCl → NaCl + H₂O + CO₂
84 gms — 36.5 gins
0.442gms — ?
= \(\frac{36.5 \times 0.442}{84}\) = 0.1928
∴ The weight of HC1 required = 0.384 + 0.192 = 0.576 gms
Molarity (M) = \(\frac{\mathrm{Wt}}{\text { GMWt }} \times \frac{1000}{\mathrm{~V}(\mathrm{~m} l)}\)
0.1 = \(\frac{0.576}{36.5} \times \frac{1000}{V}\)
V = \(\frac{0.576 \times 1000}{36.5 \times 0.1}=\frac{576}{3.65}\) = 157.80 ml

Question 7.
A solution is obtained by mixing 300g of 25% solution and 400g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer:
Given a solution is obtained by mixing 300 gms of 25% solution and 400 gms of 40% solution by mass.
Weight of solute in 1st solution = 300 × \(\frac{25}{100}\) = 75 gms
Weight of solute in 2nd solution = 400 × \(\frac{40}{100}\) = 160 gms
Total weight of solute = 75 + 160 = 235 gms
Total weight of solution = 300 + 400 = 700 gms
Mass % of solute in resulting solution = \(\frac{235}{700}\) × 100 = 33.5%

Question 8.
An antifreeze solution is prepared from 222.6g of ethylene glycol [(C₂H₆O₂)] and 200g of water (solvent). Calculate the molality of the solution.
Solution:
Weight of Ethylene glycol = 222.6 gms .
G.mol wt =62
Weight of solvent = 200 gms
Molality (m) = \(\frac{\mathrm{Wt}}{\mathrm{GMWt}} \times \frac{1000}{\mathrm{a}(\mathrm{gms})}\)
= \(\frac{222.6}{62} \times \frac{1000}{200}=\frac{222.6}{62 \times 0.2}\) = 17.95 m
Molarity (M) = \(\frac{\mathrm{Wt}}{\mathrm{GMWt}} \times \frac{1000}{\mathrm{~V}(\mathrm{~m} l)}\)
= \(\frac{222.6}{62} \times \frac{1000}{394.2}\) = 9.1 M
Mass of solution = 200 + 222.6
= 422.6 gms
Volume = \(\frac{\text { Weight }}{\text { Density }}\) = \(\frac{422.6}{1.072}\) = 394.2 ml

Question 9.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Gas dissolved in a liquid is an exothermic reaction (H <O).
According to Lechatelier’s principle,

• If the reaction is exothermic the solubility of gas, decrease with increases in temperature.
• So gases always tend to be less soluble in liquids to as the temperature is raised.

Question 10.
What is meant by positive deviations from Raoults law and how is the sign of ∆_mix H related to positive deviation from Raoult’s law?
Answer:

• When the vapour pressure of a solution is higher than the predicted value by Raoult’s law, it is called positive deviation.
  
• In this case inter molecular interactions between solute and solvent particles (1 and 2) are weaker than those between solute and solvent particles (1 and 2) are weaker than those between solute and solute (1 – 1) and solvent – solvent (2 – 2)
• Hence the molecules of 1 or 2 will escape more easily from the surface of solution than in their pure state.
  ∴ The vapour pressure of the solution will be higher.
• The vapour pressure diagram showing positive deviation as follows.
  Eg. : Ethyl alcohol and water, Acetone and benzene.

Question 11.
What is meant by negative deviation from Raoult’s law and how is the sign of ∆_mixH related to negative deviation from Raoult’s law?
Answer:

• When the vapour pressure of a solution is lower than the predicted value by Raoult’s law, it is called negative deviation.
  
• In case of negative deviation the inter molecular attractive forces between 1 – 1 and 2 – 2 are weaker than those between 1 -2.
• It leads to decrease in vapour pressure resulting in negative deviation.
• The vapour pressure diagram showing negative deviation as follows.
• Eg.: HNO₃ and water, HCl and water.

Question 12.
The vapour pressure of water is 12.3 k P_a at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer:
Given molality of solution = 1 m
Vapour pressure of water P₀ = 12.3 kP_a
Number of moles of water = \(\frac{100}{18}\)
= 55.55 (n₀)
Mole fraction of solute (X_s) = \(\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}+\mathrm{n}_{\mathrm{s}}}\)
= \(\frac{1}{55.55+1}\)
= \(\frac{1}{56.55}\) = 0.0177
Mole fraction of water (X₀) = 1 – X_s
= 1 – 0.0177 = 0.9823
Vapour pressure of solution (P_s) = P₀X₀
= 12.3 × 0.9823 = 12.08 kP_a

Question 13.
Calculate the mass of a non-volatile solute (molar mass 40g mol^-1) which should be dissolved in 114g Octane to reduce its vapour pressure to 80%. [T.S. Mar. 16]
Answer:
Raoult’s law formula

Given vapour pressure reduced to 80%, when non-volatile solute is dissolved in octane.
P₀ = 1 atm
P_s = 0.8 atm
W = 114 gms
M = 114 gm/mole
w = ? ; m = 40
\(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{0}}=\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}}=\frac{\mathrm{w}}{\mathrm{m}} \times \frac{\mathrm{M}}{\mathrm{W}}\)
\(\frac{1-0.8}{1}=\frac{w}{40} \times \frac{114}{114}\)
w = 8 gms

Question 14.
A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water If freezing point of water is 273.15 K.
Answer:
Given 5% solution of cane sugar
w = 5gms; W = 95gms
m = 342; ∆T_f = 273.15 – 271 = 2.15
∆T_f = \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}}{\mathrm{m} \times \mathrm{W}}\)
2.15 =\(\frac{\mathrm{K}_{\mathrm{f}} \times 5}{342 \times 95}\) ……………… (1)
5% solution of glucose
w = 5gms; W = 95gms
m = 180, ∆T_f = ?
∆T_f = \(\frac{K_{f} \times 5}{180 \times 95}\) ………………… (2)
Dividing equation (2) by Equation (1)
\(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{2.15}=\frac{\mathrm{K}_{\mathrm{f}} \times 5}{180 \times 95} \times \frac{342 \times 95}{\mathrm{~K}_{\mathrm{f}} \times 5}\)
∆T_f = \(\frac{342 \times 2.15}{180}\) = 4.085 K
∴ The freezing point temperature for 5% glucose solution
= 273.15 – 4.085 = 269.07 K

Question 15.
If the osmotic pressure of glucose solution is 1.52 bar at 300 K. What would be its concentration if R = 0.083L bar mol^-1 K^-1 ?
Answer:
π = CRT
R = 0.0836.bar. mole^-1 K^-1
T = 300 K
π = 1.52 bar
C = \(\frac{\pi}{\mathrm{RT}}=\frac{1.52}{0.083 \times 300}=\frac{1.52}{24.9}\) = 0.061 M

Question 16.
Vapour pressure of of water at 293K is 17.535 mm Hg. Calculate the vapour pressure of the solution at 293K when 25g of glucose is dissolved in 450g of water? [A.P. Mar.19]
Answer:
Raoult1s law formula

Question 17.
How is molar mass related to the elevation in boiling point of a solution?
Answer:
The expression for elevation of boiling point is
∆T_b = \(\frac{\mathrm{K}_{\mathrm{b}} \times 1000 \times \mathrm{w}}{\mathrm{m} \times \mathrm{w}}\) = K_b = molal elevation constant
w = Weight of solute
W = Weight of solvent
m = molar mass of solute
Molar mass of solute m = \(\frac{K_{\mathrm{b}} \times 1000 \times \mathrm{w}}{\Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{w}}\)
∴ Molar mass of solute (m) and elevation of boiling point (∆T_b) are inversely related.

Question 18.
What is an ideal solution ?
Answer:
A solution of two or mote components which obeys Raoult’s law at all concentrations and at all temperatures is called ideal Solution. In ideal solution there should not be any association between solute and solvent, (i.e.) no chemical interaction between solute and solvent of solution.
Ex : The following mixtures form ideal solutions.

• Benzene + Toluene
• n – hexane + n – heptane
• ethyl bromide + ethyl iodide

Question 19
What is relative lowering of vapour pressure ? How is it useful to determine the molar mass of a solute ? [T.S. Mar. 15]
Answer:
Raoult’s law for volatile solute: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

Raoult’s law for non-volatile solute : The relative lowering of vapour pressure of dilute solution containing non-volatile solute is equal to the mole fraction of solute.

Relative lowering of vapour pressure \(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{0}}\) = X_s (mole fraction of solute)
\(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{0}}=\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}+\mathrm{n}_{\mathrm{s}}}\)
For very much dilute solutions n_s < < < …………… n₀
∴ \(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{0}}=\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}}=\frac{\mathrm{w}}{\mathrm{m}} \times \frac{\mathrm{M}}{\mathrm{W}}\)
W = Weight of solute
w = Weight of solvent
m = Molar mass of solute
M = Molar mass of solvent
Molar mass of solute m = \(\frac{W \times M}{W} \times \frac{P_{0}}{P_{0}-P_{s}}\)

Question 20.
How is molar mass related to the depression in freezing point of a solution ?
Answer:
The expression for depression in Freezing point is
∆T_f = \(\frac{\mathrm{K}_{\mathrm{f}} \times 1000 \times \mathrm{w}}{\mathrm{m} \times \mathrm{W}}\)
K_f = molal depression constant
W = Weight of solvent
w = Weight of solute
m = molar mass of solute
Molar mass of solute (m) = \(\frac{\mathrm{K}_{\mathrm{f}} \times 1000 \times \mathrm{w}}{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{W}}\)
∴ Molar mass ôf solute (m) and depression in freezing point (∆T_f) are inversely related.

Long Answer Questions

Question 1.
An aqueous solution of 2% non volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute ?
Answer:
Relative lowering of vapour pressure = \(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{0}}=\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}}\)
P₀ = 1.013 bar, P_s = 1.004 bar
w = 2gms
W = 98gms
M = 18
m = ?

Question 2.
Heptane and Octane form an ideal solution. At 373 K the vapour pressure of the two liquid components are 105.2 kP_a and 46.8 kP_a respectively. What will be the vapour pressure of a mixture of 26.0 g heptane and 35g of octane?
Answer:
Number of moles of octane n₀ = \(\frac{\text { Weight }}{\text { GMwt }}=\frac{35}{114}\) = 0.307
Number of moles of heptane n_s = \(\frac{\text { Weight }}{\text { GMwt }}=\frac{26}{100}\) = 0.26
no 0.307
Mole fraction of octane X₀ = \(\frac{\mathrm{n}_{0}}{\mathrm{n}_{0}+\mathrm{n}_{\mathrm{s}}}=\frac{0.307}{0.307+0.26}\) = 0.541
Mole fraction of heptane X_s = \(\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}+\mathrm{n}_{\mathrm{s}}}=\frac{0.26}{0.307+0.26}\) = 0.459
Given vapour pressure of heptane P₁ = 105.2 kP_a
vapour pressure of Octane P₂ = 46.8 kP_a
26 gms of heptane and 35 gms of octane are mixed.
In that mixture
The vapour pressure of heptane (P₁₁) = P₁ × X_s
= 105.2 × 0.459 = 48.28 kP_a
The vapour pressure of Octane (P₂₂) = P₂ × X₀
= 46.8 × 0.541
= 25.32 kP_a
Total pressure of the mixture (P) = P₁₁ + P₂₂
= 25.32 + 48.28
= 73.6 kP_a

Question 3.
A solution containing 30g of non-volatile solute exactly in 90g of water has a vapour pressure of 2.8 kP_a at 298 K. Further 18g of water is then added to the solution and the new vapour pressure becomes 2.9 kP_a at 298 K. Calculate (i) The molar mass of the solute and (ii) Vapour pressure of water at 298 K.
Answer:
Calculation of molar mass of solute
Case – I:

Case – II :

Calculation of vapour pressure of water :
According to Raoult’s law
P_A = \(\mathrm{P}_{\mathrm{A}}^{0} \mathrm{X}_{\mathrm{A}}\)
2.8 = \(P_{A}^{0} \times \frac{M}{6+M}\)
M = 23
2.8 = \(\mathrm{P}_{\mathrm{A}}^{0} \times \frac{23}{6+23}\)
\(\mathrm{P}_{\mathrm{A}}^{0}=\frac{2.8 \times 29}{23}=\frac{81.2}{23}\) = 3.53 k.p_a

Question 4.
Two elements A and B from compounds having formula AB₂ and AB₄. When dissolved in 20g of Benzene (C₆H₆), 1g of AB₂ lowers the freezing point by 2.3 K whereas 1.0g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol_-1. Calculate atomic masses of A and B.
Answer:
Calculation of molecular masses of compounds AB₂ and AB₄
For AB₂ compounds
m = \(\frac{\mathrm{K}_{\mathrm{f}} \times 1000 \times \mathrm{W}}{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{W}}=\frac{5.1 \times 1000 \times 1}{2.3 \times 20}\) = 110.87 gms/mole
For AB₄ compound
m = \(\frac{5.1 \times 1000 \times 1}{1: 3 \times 20}\) = 196.15 gms/mole
Calculation of the atomic masses of elements
Atomic mass of element A = x
Atomic mass of element B = y
Molecular mass of AB₂ = x + 2y
Molecular mass of AB₄ = x + 4y
a + 2b = 110.87 ……………….. (1)
a + 4b = 196.15 ………………. (2)
Equation (2) – Eqiation (1)
x + 4y – x – 2y
196.15 – 110.87
2y = 85.28
y = 42.64
x + 2y = 110.87
x + 85.28 = 110.87
x = 25.59
Atomic mass of element A = 25.59 u
B = 42.64 u

Question 5.
Calculate the depression in the freezing point of water when 10g of CH₃CH₂CHClCOOH is added to 250g of water. K_a = 1.4 × 10^-3, K_f = 1.86 K kg mol^-1.
Answer:
Calculation of degree of dissociation :
Mass of solute = 10 gms
Molar mass of solute (CH₃ – CH₂ – CH – Cl – COOH) = 122.5 g/mole
Molality = \(\frac{10}{122.5} \times \frac{1000}{250}\) = 0.326 m

Calculation of depression in freezing point
∆T_f = i × K_f × m = 1.065 × 1.86 × 0.326 = 0.65 K

Question 6.
19.5g of CH₂FCOOH is dissolved in 500g of water. The depression In freezing point of water observed is 1.0°C. Calculate the Van’t Hoff factor and dissociation constant of fluoroacetic acid.
Answer:
Calculation of Vant Hoff factor of acid:
∆T_f = 1°C
K_f = 1.86 K kg/mole^-1
∆T_f = i K_f m
m = \(\frac{19.5}{78} \times \frac{1000}{500}\) = 0.5 m
i = \(\frac{1}{1.86 \times 0.5}\) = 1.0753
Calculation of degree of dissociation of the acid :

Question 7.
100 g of liquid A(molar mass 140g mol^-1) was dissolved in 1000 g of liquid B(molar mass 180g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.
Answer:
Calculation of vapour pressure of pure liquid A (\(\mathrm{P}_{\mathrm{A}}^{0}\))
Number of moles of liquid A
n_A = \(\frac{\mathrm{w}}{\mathrm{m}}=\frac{100}{140}\) = 0.7143
Number of moles of liquid B
n_B = \(\frac{\mathrm{w}}{\mathrm{m}}=\frac{1000}{180}\) = 5.5556
Mole fraction of A
X_A = \(\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{A}}+\mathrm{n}_{\mathrm{B}}}=\frac{0.7143}{0.7143+5.5556}=\frac{0.7143}{6.2699}\)
= 0.1139
Mole fraction of B (X_B) = 1 – 0.1139 = 0.8861
Vapour pressure of liquid (\(\mathrm{P}_{\mathrm{B}}^{0}\)) = 500 torr
Total vapour pressure of solution (P) = 475 torr
According to Raoult’s law
P = P\(\mathrm{P}_{\mathrm{A}}^{0}\) X_A + P\(\mathrm{P}_{\mathrm{B}}^{0}\) X_B
475 = P\(\mathrm{P}_{\mathrm{A}}^{0}\) (0.1139) + 500(0.8861)
475 = P\(\mathrm{P}_{\mathrm{A}}^{0}\) (0.1139) + 443.05
\(\mathrm{P}_{\mathrm{A}}^{0}\) = \(\frac{475-443.05}{0.1139}=\frac{31.95}{0.1139}\) = 280.5 torr
Calculation of vapour pressure of A in the solution (P_A)
P_A = P\(\mathrm{P}_{\mathrm{A}}^{0}\) X_A = 280.5 × 0.1139
P_A = 32 torr.

Question 8.
Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
Answer:
Varit Hoffs equation
Osmotic pressure (π) = i CRT
i = 2.47 .
V = 2.5 lit
R = 0.0821 lit. atm.k^-1. mol^-1
T = 27 + 273 = 300 K
π = 0.75 atm
π = i(\(\frac{n_{B}}{V}\))RT
n_B = \(\frac{\pi \mathrm{V}}{\mathrm{iRT}}=\frac{0.75 \times 2.5}{2.47 \times 0.0821 \times 300}\) = 0.0308
The amount of CaCl₂ dissolved = n_B × m_B
= 0.0308 × 111 = 3.42 gms

Question 9.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in two litre of water at 25°C assuming that it is completely dissociated.
Answer:
The amount of K₂SO₄ dissolved = 25 mg
Volume = 2 lit; T = 25°C = 298 K
Molecular weight of K₂SO₄ = 174 gms / mole

Total ions produced on dissociation = 3
i = 3
π = i CRT = \(\mathrm{i}\left(\frac{\mathrm{n}}{\mathrm{V}}\right) \mathrm{RT}=\mathrm{i}\left(\frac{\mathrm{w}}{\mathrm{m}}\right) \times \frac{\mathrm{RT}}{\mathrm{V}}\)
= \(\frac{3 \times 0.025 \times 0.0821 \times 298}{174 \times 2}\) = 5.27 × 10^-3 atm.

Question 10.
Benzene and Toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm of Hg and 32.06 mm of Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80g of benzene is mixed with 100g of toluene.
Answer:
Molecular weight of benzene (C₆H₆) = 78
Molecular weight of Toluene (C₇H₈) = 92
n_C6H6 = \(\frac{80}{78}\) = 1.026
n_C7H8= \(\frac{100}{92}\) = 1.087
X_C6H6 = \(\frac{1.026}{1.026+1.087}=\frac{1.026}{2.113}\) = 0.4855
X_C7H8 = 1 – 0.4855 = 0.5145
According to Raoult’s law
Partial pressure of benzene in solution
P_C6H6 = P_C6H6 × X_C6H6
= 50.71 × 0.4855 = 24.61 mm
Partial pressure of Toluene in solution
P_GH8 = × X_C7H8 = 32.06 × 0.5145 = 16.49 mm
Total vapour pressure of solution (P) = 24.61 + 16.49 = 41.1 mm
Mole fraction of benzene in vapour phase is
= \(\frac{\mathrm{X}_{\mathrm{C}_{6} \mathrm{H}_{6}} \times \mathrm{P}_{\mathrm{C}_{6} \mathrm{H}_{6}}^{0}}{\mathrm{P}_{\text {total }}}=\frac{0.4855 \times 50.71}{41.1}=\frac{24.61}{41.1}\) = 0.5987

Textual Examples

Question 1.
Calculate the mole fraction of ethylene glycol (C₂H₆O₂) In a solution containing 20% of C₂H₆O₂ by mass.
Solution:
Assume that we have 100g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20g of ethylene glycol and 80g of water.
Molar mass of C₂H₆O₂ = 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol^-1

Question 2.
Calculate the molarity of a solution containing 5g of NaOH in 500 mL solution. [T.S Mar.17]
Solution:
Moles of NaOH = \(\frac{5 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.125 mol
Volume of the solution in litres = 450 mL / 1000 mL L^-1
Using equation

= 0.278 mol L^-1 = 0.278 mol dm^-3

Question 3.
Calculate molality of 2.5 of ethanoic acid (CH₃COOH) In 75g of benzene. [T.S. Mar. 15]
Solution:
Molar mass of C₂H₄O₂: 12 × 2 + 1 × 4 + 16 × 2 = 60g mol^-1
Moles of C₂H₄O₂ = \(\frac{2.5 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.0417 mol
Mass of benzene in kg = 75g/1000 g kg^-1 = 75 × 10^-3 kg
Molality of C₂H₄O₂ = \(\frac{\text { Moles of } \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}}{\mathrm{~kg} \text { of benzene }}=\frac{0.0417 \times \mathrm{mol} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{75 \mathrm{~g}}\)
= 0.556 mol kg^-1

Question 4.
If N₂ gas is bubbled through water at 293 K, how many mililmoles of N₂ gas would dissolve in 1 litre of water ? Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henrys law constant for N₂ at 293 K is 76.48 k bar.
Solution:
The solubiity of gas is related to the mole fraction in aqueous solution.
The mole fraction of the gas in the solution is calculated by applying Henrys law. Thus:
x (nitrogen) = \(\frac{\mathrm{p} \text { (nitrogen) }}{\mathrm{K}_{\mathrm{H}}}=\frac{0.987 \mathrm{bar}}{76.480 \mathrm{bar}}\) = 1.29 × 10^-5
As 1 Litre of water contains 55.5 mol of it, therefore if n represents number of moles of N₂ in solution.
x(Nitrogen) = \(\frac{\mathrm{n} \mathrm{mol}}{\mathrm{n} \mathrm{mol}+55.5 \mathrm{~mol}}=\frac{\mathrm{n}}{55.5}\) = 1.29 × 10^-5
(n in denominator is neglected as it is << 55.5)
Thus n = 1.29 × 10^-5 55.5 mol 7.16 × 10^-4 mol
= \(\frac{7.16 \times 10^{-4} \mathrm{~mol} \times 1000 \mathrm{mmol}}{1 \mathrm{~mol}}\) = 0.716 m mol

Question 5.
Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K and (ii) mole fractions of each component in vapour phase.
Solution:
i) Molar mass of CH₂Cl₂ = 12 × 1 + 1 × 2 + 35.5 × 2 = 85 g mol^-1
Molar mass of CHCl₃ = 12 × 1 + 1 × 1 +35.5 × 3 = 119.5 g mol^-1
Moles of CH₂Cl₂ = \(\frac{40 \mathrm{~g}}{85 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.47 mol
Moles of CHCl₃ = \(\frac{25.5 \mathrm{~g}}{119.5 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.2 13 mol
Total number of moles = 0.47 + 0.2 13 = 0.683 mol
X_CH2Cl2 = \(\frac{0.47 \mathrm{~mol}}{0.683 \mathrm{~mol}}\) = 0.688
X_CHCl3 = 1.00 – 0.688 = 0.3 12
Using equation :
p_total = \(\mathrm{p}_{1}^{0}+\left(\mathrm{p}_{2}^{0}-\mathrm{p}_{1}^{0}\right)\) x₂ = 200 + (415 – 200) × 0.688
= 200 + 147.9 = 347.9mm Hg

ii) Using the relation, y₁ = p₁/p_total we can calculate the mole fraction of the components in gas phase (y₁)
P_CH2Cl2 = 0.688 × 415 mm Hg = 285.5 mm Hg
P_CHCl3 = 0.3 12 × 200 mm Hg = 62.4 mm Hg
y_CH2Cl2 = 285.5 mm Hg/347.9 mm Hg = 0.82
y_CHCl3 = 62.4 mm Hg/347.9 mm Hg = 0.18

Note : Since CH₂Cl₂ is a more volatile component than CHCl₃, [\(\mathrm{p}_{\mathrm{CH}_{2} \mathrm{Cl}_{2}}^{0}\) = 415 mm Hg and
\(\mathrm{p}_{\mathrm{CHCl}_{3}}^{0}\) = 200 mm Hg] and the vapour phase is also richer in CH₂Cl₂ [y_CH2Cl2 = 0.82 and y_CHCl3 = 0.18], it may thus be concluded that at equilibrium, vapour phase will be always rich in the component which is more volatile.

Question 6.
The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5g when added to 39.0 g of benzene (molar mass 78 g mol^-1), vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance ? [A.P. Mar. 18; 16]
Solution:
The various quantities known to us are as follows.
\(\mathrm{p}_{1}^{0}\) = 0.850 bar
P = 0.845bar
M₁ = 78 g mol^-1
w₂ = 0.5 g
w₁ = 39 g
Substituting these values in equation \(\frac{\mathrm{P}^{0}-\mathrm{P}}{\mathrm{P}_{1}^{0}}=\frac{\mathrm{w}_{2} \times \mathrm{M}_{1}}{\mathrm{M}_{2} \times \mathrm{w}_{1}}\) we get
\(\frac{0.850 \mathrm{bar}-0.845 \mathrm{bar}}{0.850 \mathrm{bar}}=\frac{0.5 \mathrm{~g} \times 78 \mathrm{~g} \mathrm{~mol}^{-1}}{\mathrm{M}_{2} \times 39 \mathrm{~g}}\)
Therefore, M₂ = 170 g mol^-1.

Question 7.
18g of glucose, C₆H₁₂O₆, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar ? K_b for water is 0.52 K kg mol^-1.
Solution:
Moles of glucose = 18g/180 g mol^-1 = 0.1 mol
Number of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0.1 mol kg^-1
For water, change in boiling point
∆T_b = K_b × m = 0.52 K kg mol^-1 × 0.1 mol kg^-1 = 0.052 K .
Since water boils at 373.15 K at 1.013 bar pressure, therefore, the boiling point of solution will be 373.15 + 0.052 = 373.202 K.

Question 8.
The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_b for benzene is 2.53 K kg mol^-1.
Solution:
The elevation of (∆T_b) in.the boiling point = 354.11 K – 353.23 K. = 0.88 K
Substituting these values in expression we get, M₂ = \(\frac{1000 \times \mathrm{w}_{2} \times \mathrm{K}_{\mathrm{b}}}{\Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{w}_{1}}\)
M₂ = \(\frac{2.53 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 1.8 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{0.88 \mathrm{~K} \times 90 \mathrm{~g}}\)
Therefore, molar mass of the solute, M₂ = 58 g mol^-1.

Question 9.
45 g of ethylene glycol (C₂H₆O₂) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution.
Solution:
Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol = \(\frac{\text { moles of ethylene glycol }}{\text { mass of water in kilogram }}\)
Moles of ethylene glycol = \(\frac{45 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.73 mol
Mass of water in kg = \(\frac{600 \mathrm{~g}}{1000 \mathrm{~g} \mathrm{~kg}^{-1}}\) = 0.6 kg
Hence molality of ethylene glycol = \(\frac{0.73 \mathrm{~mol}}{0.60 \mathrm{~kg}}\) = 1.2 mol kg^-1
Therefore freezing point depression,
∆T_f = 1.86 K kg mol^-1 × 1.2 mol kg^-1 = 2.2 K
Freezing point of the aqueous solution 273.15 K – 2.2 K = 270.95 K

Question 10.
1.00 g of a non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol^-1. Find the molar mass of the solute.
Solution:
Substituting the values of various terms involved in equation
M₂ = \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}_{2} \times 1000}{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{w}_{1}}\) we get
M₂ = \(\frac{5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 1.00 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{0.40 \times 50 \mathrm{~g}}\) = 256 g mol^-1
Thus, molar mass of the solute = 256 g mol^-1

Question 11.
200 cm³ of an aqueous solution of a protein contains 1.26 g of the protein. The osmotiç pressure of such a solution at 300 K is found to be 2.57 × 10^-3 bar. Calculate the molar mass of the protein.
Solution:
The various quantities known to us are as follows : π = 2.57 × 10^-3 bar.
V = 200 cm³ = 0.200 litre
T = 300K
R = 0.083 L bar mol^-1 K^-1
Substituting these values in equation,
we get M₂ = \(\frac{w_{2} R T}{\pi V}\)
M₂ = \(\frac{1.26 \mathrm{~g} \times 0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{2.57 \times 10^{-3} \mathrm{bar} \times 0.200 \mathrm{~L}}\)
= 61.022 g mol^-1

Question 12.
2 g of benzoic acid (C₆H₅COOH) dissolved in 25g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol^-1. What is the percentage association of acid if it forms dimer in solution ?
Solution:
The given quantities are :
w₂ = 2g; K_f = 4.9 K kg mol^-1; w₁ = 25 g
∆T_f = 1.62 K
Substituting these values, in equation, we get:
M₂ = \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}_{2} \times 1000}{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{w}_{1}}\)
M₂ = \(\frac{4.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 2 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{25 \mathrm{~g} \times 1.62 \mathrm{~K}}\) = 241.98 g mol^-1
Thus, experimental molar mass of benzoic acid in benzene is = 241.98 g mol^-1
Now consider the following equilibrium for the acid :
2 C₆H₅COOH ⇌ (C₆H₅COOH)₂
If x represents the degree of association of the solute then we would have (1 – x) mol of benzoic acid left in unassociated form and correspondingly \(\frac{x}{2}\) as associated moles of benzoic acid at equilibrium.
Therefore, total number of particles at equilibrium is :
1 – x + \(\frac{x}{2}\) = 1 – \(\frac{x}{2}\).
Thus, total number of moles of particles at equilibrium equals van’t Hoff factor i.
But i = \(\frac{\text { Normal molar mass }}{\text { Abnormal molar mass }}=\frac{122 \mathrm{~g} \mathrm{~mol}^{-1}}{241.98 \mathrm{~g} \mathrm{~mol}^{-1}}\)
or = \(\frac{x}{2}\) = 1 – \(\frac{122}{241.98}\) = 1 – 0.504 = 0.496
or x = 2 × 0.496 = 0.992
Therefore, degree of association of benzoic acid in benzene is 99.2%.

Question 13.
0.6 mL of acetic acid (CH₃COOH), having density 1.06 g mL^-1, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van’t Hoff factor and the dissociation constant of acid.
Answer:
Number of moles of acetic acid = \(\frac{0.6 \mathrm{~mL} \times 1.06 \mathrm{~g} \mathrm{~mL}^{-1}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.0106 mol = n
Molality = \(\frac{0.0106 \times \mathrm{mol}}{1000 \mathrm{~mL} \times 1 \mathrm{~g} \mathrm{~mL}^{-1}}\) = 0.0106 mol kg^-1
Using equation ∆T_f = \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{w}_{1}}\)
∆T_f = 1.86 K kg mol^-1 × 0.0106 mol kg^-1 = 0.0197 K
van’t Hoff factor (i) = \(\frac{\text { Observed freezing point }}{\text { Calculated freezing point }}=\frac{0.0205 \mathrm{~K}}{0.0197 \mathrm{~K}}\) = 1.041
Acetic acid is a weak electrolyte and will dissociate into two ions: acetate and hydrogen ions per molecule of acetic acid. If x is the degree of dissociation of acetic acid, then we would have n (1 — X) moles of undissociated acetic acid, mc moles of CH₃COO^– and nx moles of H⁺ ions.

Intext Questions

Question 1.
Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachlodie (CCl₄) if 22g of benzene is dissolved in 122g of carbon tetrachioride.
Then, calculate the mass percentage from the formula
Mass % = × 100
Solution:
Mass of benzene = 22g; Mass of CCl₄ = 122g
Mass of solution = 22 + 122 = 144g
Mass % of benzene = \(\frac{22}{144}\) × 100 = 15.28%
Mass of CCl₄ = 100 – 15.28 = 84.72%
Note: Mass percent of CCl₄ can also be calculated by using the formula as:
Mass % of CCCl₄ = \(\frac{122}{144}\) × 100 = 84.72%

Question 2.
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
• Then calculate the mole fraction by using the formula
Mole fraction of a component = \(\frac{\text { Number of moles of the component }}{\text { Total number of moles of all components }}\)
x_A = \(\frac{\mathbf{n}_{\mathbf{A}}}{\mathbf{n}_{\mathbf{A}}+\mathbf{n}_{\mathbf{B}}}\)
Solution:
For 100 g of the solution
Mass of benzene = 30 g
Mass of carbon tetrachioride = 100 – 30 = 70g
Molar mass of benzene (C₆H₆) (12 × 6) + (6 × 1) = 72 + 6 = 78 g mol^-1
Moles of benzene, n_C6H6 = \(\frac{\text { Mass }}{\text { Molar mass }}=\frac{30}{78}\) = 0385 mol
Molar mass of carbon tetrachioride (CCl₄) = 12 + (35.5 × 4)
= 12 + 142.0 = 154 g mol^-1
Moles of CCl₄,n_CCl4 = \(\frac{70 \mathrm{~g}}{\left(154 \mathrm{~mol}^{-1}\right)}\) = (154 mol^-1)
Mole fraction of benzene, X_C6H6 = \(\frac{\mathrm{n}_{\mathrm{C}_{6} \mathrm{H}_{6}}}{\mathrm{n}_{\mathrm{C}_{6} \mathrm{H}_{6}}+\mathrm{n}_{\mathrm{CC}_{4}}}=\frac{0.385 \mathrm{~mol}}{(0.385+0.454) \mathrm{mol}}\) = 0.459

Question 3.
Calculate the molarity of each of the following solution:
a) 30g of CO(NH₃)₂. 6H₂O in 4.3 L of solution.
b) 30 mL of 0.5 MH₂SO₄ diluted to 500 mL.

So, first find molar mass by adding atomic masses of different elements, then find moles of solute and then molarity.
b)Use molarity equation for dilution.
M₁V₁
(Before dilution)
M₂V₂
(After dilution)
Solution:
a) Molar mass of CO(NO₃)₂. 6H₂O = (58.7) + 2(14 + 48) + (6 × 18) g mol^-1
= 58.7 + 124 + 108 = 290.7 = 291g mol^-1
Mole of CO(NO₃)₂ 6 H₂O = \(\frac{30 \mathrm{~g}}{291 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.103
Volume of solution = 4.3 L
Molarity (M) = \(\frac{0.103 \mathrm{~mol}}{4.3 \mathrm{~L}}\) = 0.024 mol L^-1 = 0.024 M

b) Volume of undiluted H₂SO₄ solution (V₁) = 30 mL
Molarity of undiluted H₂SO₄ solution (M₁) = 0.5 M
Volume of diluted H₂SO₄ solution (V₂) = 500 mL
We know that M₁V₁ = M₂V₂
∴ M₂ = \(\frac{\mathrm{M}_{1} \mathrm{~V}_{1}}{\mathrm{~V}_{2}}=\frac{(0.5 \mathrm{M})(30 \mathrm{~mL})}{500 \mathrm{~mL}}\) = 0.03 M

Question 4.
Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molar aqueous solution.

So, find the molar mass of solute by adding atomic masses of different element present in it and mass by using the formula,
Molality =
Mass of solvent in kg
Solution:
Molality of the solution = 0.25 m = 0.25 mol kg^-1
Molar mass of urea (NH₂CONH₂) = (14 × 2) + (1 × 4) + 12 + 16
= 60 g mol^-1
Mass of solvent (water) = 2.5 kg
Molality =
(0.25 mol kg^-1) = \(\frac{\text { Mass of urea }}{\left(60 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(2.5 \mathrm{~kg})}\)
Mass of urea = (0.25 mol kg^-1) × (60 g mol^-1) × (2.5 kg) = 37.5 g

Question 5.
Calculate a) molality b) molarity and c) mole fraction of KI if the density of 20% (mass / mass) aqueous Kl is 1.202 g mL^-1.
• As density and % by mass is given, so find the mass of solute and solvent (as x % solution contains x g solute in (100 – x) g solvent).
Find volume of the solution, by using,
Volume = \(\frac{\text { Mass }}{\text { Density }}\)
Recall the formulae of molality, molarity and mole fraction, to calculate them.
Molality =
Solution:
a) Molality
Weight of KI in 100 g of water = 20 g
Weight of water in the solution = 100 – 20 = 80 g = 0.08 kg
Molar mass of KI = 39 + 127 = 166 g mol^-1
Molarity of the solution (m) = \(\frac{\text { Number of moles of KI }}{\text { Mass of water in } \mathrm{kg}}\)
= \(\frac{(20 \mathrm{~g}) /\left(166 \mathrm{~g} \mathrm{~mol}^{-1}\right)}{(0.08 \mathrm{~kg})}\) = 1.506 mol kg^-1 = 1.506 m

b) Molarity
Weight of the solution = 100 g
Density of the solution = 1.202g m^-1

c) Mole fraction of KI

Question 6.
H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195m, calculate Henry’s law constant.
Solution:
a) Calculation of mole fraction of H₂S 0.195 m means that 0.195 mole of H₂S are dissolves in 1000 g of water.
Number of moles of water in 1000g, (n_H2O) = \(\frac{(1000 \mathrm{~g})}{\left(18 \mathrm{~g} \mathrm{~mol}^{-1}\right)}\) = 55.55 mol
Mole fraction H₂S (x_H2S) = \(\frac{\mathrm{n}_{\mathrm{H}_{2} \mathrm{~s}}}{\mathrm{n}_{\mathrm{H}_{2} \mathrm{~s}}+\mathrm{n}_{\mathrm{H}_{2} \mathrm{O}}}\)
= \(\frac{(0.195 \mathrm{~mol})}{(0.195+55.55) \mathrm{mol}}=\frac{(0.195 \mathrm{~mol})}{(55.745 \mathrm{~mol})}\) = 0.0035

b) Calculation of Henry’s law constant:
According to Henry’s law

Question 7.
Henrys law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at’298 K.
Solution:
Step I : Calculation of number of moles of CO₂
According to Henrys law,

n_CO2 = x_CO2 × (27.78 mol) (1.52 10^-3) × (27.78 mol) = 0.0422 mol
Note: n_CO2 is considered negligible due to its little solubility in water.
Step II : Calculation of mass of dissolved CO₂ in water
Mass of CO₂ = No. of moles CO₂ × Molar mass of CO₂
= (0.0422 mol) × (44 g mol^-1) = 1.857 g

Question 8.
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
. Apply Raoults law p_T = \(\mathbf{p}_{\mathrm{A}}^{0} \mathbf{x}_{\mathrm{A}}+\mathbf{p}_{\mathrm{B}}^{0} \mathbf{x}_{\mathrm{B}}=\mathbf{p}_{\mathrm{B}}^{0} \mathbf{x}_{\mathrm{A}}+\mathbf{p}_{\mathrm{B}}^{0}\left(1-\mathbf{x}_{\mathrm{A}}\right)\) to calculate mole fraction of A(x_A) and B(x_B).
In vapour phase, partial pressure are used instead of number of moles.
Solution:
Step I: ComposItion In liquid phase
Vapour pressure of pure liquid A (\(p_{A}^{0}\)) = 450 mm
Vapour pressure of pure liquid B (\(p_{B}^{0}\)) = 700 mm
Total vapour pressure of the solution (p) = 600 mm
According to Raoult’s law,
p = \(\mathbf{p}_{\mathrm{A}}^{0} \mathbf{x}_{\mathrm{A}}+\mathbf{p}_{\mathrm{B}}^{0} \mathbf{x}_{\mathrm{B}}=\mathbf{p}_{\mathrm{B}}^{0} \mathbf{x}_{\mathrm{A}}+\mathbf{p}_{\mathrm{B}}^{0}\left(1-\mathbf{x}_{\mathrm{A}}\right)\)
(600 mm) = 450 mm × x_A + 700 mm (1 – x_A)
=700 mm + x_A + (450 – 700)mm
= 700 – x_A (250 mm)
x_A = \(\frac{(600-700)}{-(250 \mathrm{~mm})}\) = 0.40
Mole fraction of A(x_A) = 0.40
Molefraction of B(x_B) = 1 – 0.40 = 0.60

Step II: Composition in vapour phase
P_A = \(\mathrm{p}_{\mathrm{A}}^{0} \mathrm{x}_{\mathrm{A}}\) = (450 mm) × 0.40 = 180 mm
P_B = \(\mathrm{p}_{\mathrm{B}}^{0} \mathrm{x}_{\mathrm{B}}\) = (700 mm) × 0.60 = 420 mm
Mole fraction of A in vapour phase = \(\frac{\mathrm{p}_{\mathrm{A}}}{\mathrm{p}_{\mathrm{A}}+\mathrm{p}_{\mathrm{B}}}\)
= \(\frac{(180) \mathrm{mm}}{(180+420) \mathrm{mm}}\) = 0.30
Mole fraction of B in vapour phase = \(\frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{p}_{\mathrm{A}}+\mathrm{p}_{\mathrm{B}}}=\frac{(420) \mathrm{mm}}{(180+420) \mathrm{mm}}\) = 0.70

Question 9.
Vapour pressure of pure water at 298 K Is 23.8 mm Hg. 50g urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and Its relative lowering.
Consider Raoult’s law and formula for relative lowering in vapour pressure,
\(\frac{\mathbf{p}_{\mathrm{A}}^{0}-\mathbf{p}_{\mathrm{s}}}{\mathbf{p}_{\mathrm{A}}^{0}}=\frac{\mathbf{n}_{\mathrm{B}}}{\mathbf{n}_{\mathrm{A}}}=\frac{\mathbf{W}_{\mathrm{B}}}{\mathbf{M}_{\mathrm{B}}} \times \frac{\mathbf{M}_{\mathrm{A}}}{\mathbf{W}_{\mathrm{A}}}\)
Where, \(\frac{\mathbf{p}_{\mathrm{A}}^{0}-\mathbf{p}_{\mathrm{s}}}{\mathbf{p}_{\mathrm{A}}^{0}}\) is called relative lowering in vapour pressure.
Solution:
Step I : Calculation of vapour pressure of water for this solution.
According to Raoult’s law,
\(\frac{\mathrm{p}_{\mathrm{A}}^{0}-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}_{\mathrm{A}}^{0}}=\frac{\mathrm{n}_{\mathrm{B}}}{\mathrm{n}_{\mathrm{A}}}=\frac{\mathrm{W}_{\mathrm{B}} / \mathrm{M}_{\mathrm{B}}}{\mathrm{W}_{\mathrm{A}} / \mathrm{M}_{\mathrm{A}}}=\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}} \times \frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{A}}}\) ………………. (1)
(Pure water) \(\mathrm{p}_{\mathrm{A}}^{0}\) = 23.8 mm;
W_B (urea) = 50 g; W_A (water) = 850 g
M_B (urea) = 60 g mol^-1; M_A (water) = 180 g mol^-1
Placing the values in eq. (i)
\(\frac{\mathrm{p}_{\mathrm{A}}^{0}-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}_{\mathrm{A}}^{0}}=\frac{(50 \mathrm{~g}) \times\left(18 \mathrm{~g} \mathrm{~mol}^{-1}\right)}{\left(60 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(850 \mathrm{~g})}=0.01762\)
\(\frac{23.8-\mathrm{p}_{\mathrm{s}}}{238}\) = 0.01762; 23.8 – p_s = 0.4194
p_s = 23.3806 ≈ 23.38 mm Hg

Step II : Calculation of relative lowering of vapour pressure
Relative lowering in vapour pressure = \(\frac{\mathrm{p}_{\mathrm{A}}^{0}-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}_{\mathrm{A}}^{0}}=\frac{(23.8-23.38) \mathrm{mm}}{(23.8 \mathrm{~mm})}\) = 0.0176

Question 10.
Boiling point of water 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
[K_b for water is 0.52 K kg mol^-1]
i) Since boiling point is changing, apply the formula for elevation In boiling point,
∆T_b = K_bm

iii) Find ∆T_b as ∆T_b = T_b = T_b – \(T_{b}^{0}\)
T_b = Boiling point of solution
T_b = Boiling point of pure solvent
Solution:
Mass of water (W_A) = 500 g = 0.5 kg
Elevation in boiling point (∆T_b) = 100°C – 99.63°C = 037°C = 0.37 K
Molal elevation constant (K_b) = 0.52 K kg mol^-1
Molar mass sucrose C₁₂H₂₂O₁₁
(M_B) = (12 × 12) + (22 × 1) + (16 × 11)
= 342 g mol^-1
W_B = \(\frac{\mathrm{M}_{\mathrm{B}} \times \Delta \mathrm{T}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{b}}}\)
= \(\frac{\left(342 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0.37 \mathrm{~K}) \times(0.5 \mathrm{~kg})}{\left(0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)}\) = 121.7 g

Question 11.
Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75g of acetic acid to lower its melting point by 1.5°C. K_f = 3.9 K kg mol^-1.
. Since, lowering of melting point is given apply the formula for lowering of melting point, i.e.,
∆T_f = K_f.m
∆T_f = \(\frac{\mathbf{K}_{f} \cdot \mathbf{W}_{\mathbf{B}}}{\mathbf{M}_{\mathbf{B}} \times \mathbf{W}_{\mathbf{A}}}\) or W_B = \(\frac{\Delta \mathbf{T}_{\mathrm{f}^{*}} \mathbf{M}_{\mathrm{B}} \cdot \mathbf{W}_{\mathrm{A}}}{\mathbf{K}_{\mathrm{f}}}\)
Solution:
Mass of ascorbic acid (W_A) = 75 g = 0.075 kg
Depression in melting point (∆T_f) = 1.5° C = 1.5 K
Molar mass of ascorbic acid (M_B) = (12 × 6) + (8 × 1) + (16 × 6) = 176 g mol^-1
Molal depression constant (K_f) = 3.9 K kg mol^-1
W_B = \(\frac{\left(176 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(1.5 \mathrm{~K}) \times(0.075 \mathrm{~kg})}{\left(3.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)}\) = 5.08 g

Question 12.
Calculate the osmotic pressure in Pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 1,85,000 in 450 mL of water at 37°C.
Use the formula for osmotic pressure (π) = CRT and C = \(\frac{\mathbf{n}}{\mathbf{V}}\) and n = \(\frac{\mathbf{W}_{\mathbf{B}}}{\mathbf{M}_{\mathbf{B}}}\)
Solution:
Mass of polymer (W_B) = 1.0 g
Molar mass of polymer (M_B) = 185000 g mol^-1
Volume of solution (V) = 450 mL = 0.450 L
Temperature (T) = 37 + 273 = 310 K
Solution constant (R) = 8.314 × 10³ Pa L K^-1 mol^-1
Osmotic pressure (π) = CRT
= \(\frac{\mathrm{W}_{\mathrm{B}} \times \mathrm{R} \times \mathrm{T}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{V}}\)
π = \(\frac{(1.0 \mathrm{~g}) \times\left(8.314 \times 10^{3} \mathrm{~Pa} \mathrm{~L} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(310 \mathrm{~K})}{\left(185000 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0.450 \mathrm{~L})}\)
= 30.96 Pa

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 1st Lesson Solid State Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 1st Lesson Solid State

Very Short Answer Questions

Question 1.
Define the term amorphous.
Answer:
An amorphous (no form) solid is a compound which does not have an orderly arrangement of particles. In amorphous solids, the constituent particles, atoms, and molecules have short-range order only.
E.g.: Glass, rubber, plastics, etc.

Question 2.
What makes a glass different from quartz?
Answer:

• Glass is an amorphous solid in which the constituent particles have only a short range order.
• Quartz is a crystalline form of silica in which constituent particles have long range order.

Question 3.
Classify the following solids as ionic, metallic, molecular, covalent network or amorphous.
i) Si
ii) I₂
iii) P₄
iv) Rb
v) SiC
vi) LiBr
vii) Ammonium .Phosphate (NH₄)₃ PO₄
viii) Plastic
ix) graphite
x) Tetra phosphorous decoxide
xi) brass
Answer:
i) Si – Covalent network solid
ii) I₂ – Molecular solid with covalent bonds
iii) P₄ – Molecular solid with covalent bonds ,
iv) Rb – Metallic solid
v) SiC – Giant molecular Network Solid with covalent bonds
vi) LiBr – Ionic solids
vii) Ammonium Phsophate (NH₄)₃ PO₄ – Ionic Solids
viii) Plastic – Amorphous solid
ix) Graphite – Hexagonal network solid with covalent bonds (giant molecule)
x) Tetra phosphorous decoxide (P₄O₆) – Molecular solid with covalent bonds.
xi) Brass – Metallic solid

Question 4.
What is meant by the term coordination number ?
Answer:
The number of nearest neighbouring particles of a particle is defined as the co-ordination number.
(Or)
The number of nearest oppositely charged ions surrounding a particular ion is also called as co-ordination number.
E.g.: Co-ordination no. of Na⁺ in NaCl lattice is ‘6’.

Question 5.
What is the co-ordination number of atoms in a cubic close – pack structure ?
Answer:
The co-ordination number of atoms in a cubic close pack structure is ’12’.

Question 6.
What is the co-ordination number of atoms in a body – centered cubic structure ?
Answer:
The co-ordination number of atoms in a body – centered cubic structure is ‘8’.

Question 7.
Stability of a crystal is reflected in the magnitude of its melting point. Comment.
Answer:
Stability of a crystal is reflected in the magnitude of its melting point.
Explanation :

• The stability of a crystal mainly depends upon the magnitude of forces of attraction between the constituent particles.
• As the attractive forces between the constituent particles increases stability of the crystal aslo increases.
• As the stability of crystal increases melting point of solid will be higher.

Question 8.
How are the intermolecular forces amonj the molecules affect the melting point ?
Answer:

• As the intermolecular forces between constituent particles of solid increases, stability of the compound increases.
• As the stability of crystal increases belting point of solid also increases (high).

Question 9.
How do you distinguish between hexagonal close – packing and cubic close – packing structures ?
Answer:
Hexagonal close packing : The spheres of the 3^rd layer are exactly aligned with those of first layer. This pattern of spheres is repeated in alternate layers. Tetrahedral voids of the 2^nd layer may be covered by the spheres of 3^rd layer. This structure is called hexagonal close packed (hep) structure.
Cubic close packing : The spheres of 3^rd layer cover the octahedral voids of 2^nd layer. But the spheres of 4th layers are aligned with those of first layer. This structure is called cubic close packing.

Question 10.
How do you distinguish between crystal lattice and unit cell ?
Answer:
Crystal lattice : A regular arrangement of the constituent particles of a crystal in the three dimensional space is called crystal lattice.
Unit cell: The simple unit of crystal lattice which when repeated again and again gives the entire crystal of a given substance is called unit cell.

Question 11.
How many lattice points are there in one unit cell of face centered cubic lattice ?
Answer:
In face centered cubic unit cell.
The number of comer atoms per unit cell.
= 8 comer atoms × \(\frac{1}{8}\) atom per unit cell = 8 × \(\frac{1}{8}\) =1 atom
Number of face centered atoms per unit cell
= 6 face centered atoms × \(\frac{1}{2}\) atom per unit cell
= 6 × \(\frac{1}{8}\) = 3 atoms
∴ Total no. of lattice points = 1 + 3 = 4.

Question 12.
How many lattice points are there in one unit cell of face – centered tetragonal lattice ?
Answer:
In face centered tetragonal unit cell
Number of face centered atoms per unit cell
= 6 face centered atoms × \(\frac{1}{2}\) atom per unit cell
= 6 × \(\frac{1}{2}\) = 3 atoms
∴ Total no. of lattice points = 1 + 3 = 4.

Question 13.
How many lattice points are there in one unit cell of body centered cubic lattice ?
Answer:
In body – centered cubic unit cell
The number of comer atoms per unit cell
= 8 comers × \(\frac{1}{8}\) per comer atom
= 8 × \(\frac{1}{8}\) = 1 atom
Number of atoms at body center = 1 × 1 = 1 atom
∴ Total no. of lattice points = 1 + 1 = 2.

Question 14.
What is a semi conductor ?
Answer:
The solids which are having moderate conductivity between insulators and conductors are called semi conductors.
These have the conductivity range from 10^-6 to 10⁴ Ohm^-1 m^-1.
By doping process the conductivity of semi conductors increases.
E.g. : Si, Ge crystal.

Question 15.
What is Schottky defect ?
Answer:
Schottky defect:

1. It is a point defect in which an atom or ion is missing from its normal site in the lattice”.
2.  In order to maintain electrical neutrality, the number of missing cations and anions are equal.
3. This sort of defect occurs mainly in highly ionic compounds, where cationic and anionic sizes are similar.
   In such compounds the co-ordination number in high.
   Ex.: NaCl, CsCl etc.
4. Illustration:
   
5. This defect decreases the density of the substance.