Yearly Archives: 2021

INTEXT Questions

Question 1.
Which of the ores mentioned in Table 6.1 (NCERT Textbook) can be concentrated by magnetic separation method?
Solution:
Ores in which one of the components (either the impurity or the actual ore) is magnetic can be concentrated by magnetic separation, e.g., ores containing iron (haematite, magnetite, siderite and iron pyrites).

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

Question 3.
The reaction, Cr₂O₃ + 2Al → Al₂O₃ + 2cr (∆G° = – 421 kJ) is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?
Solution:
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?
Solution:
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.

NCERT Exercises

Question 1.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Solution:
The E° of zinc (Zn²⁺/Zn = – 0.76 V) is lower luau that of copper (Cu²⁺/Cu = +0.34 V). Hence, zinc can displace Cu from solutions of Cu²⁺ ions.

But all these metals react with water forming their corresponding ions with the evolution of H₂ gas. Hence, Al, Mg, etc. cannot be used to displace zinc from the solution of Zn²⁺ ions. Thus, copper can be extracted by hydrometal-lurgy but not zinc.

Question 2.
What is the role of depressant in froth-floatation process?
Solution:
In froth -floatation process, the role of the depressant is to prevent one type of sulphide ore particles from forming the froth with air bubbles. For example, NaCN is used as a depressant to separate lead sulphide (PbS) ore from zinc sulphide (ZnS) ore. The reason is that NaCN forms a zinc complex, Na₂[Zn(CN)₄] on the surface of ZnS preventing it from the formation of the froth. Under these conditions, only PbS forms froth and therefore, it can be separated from ZnS ore.

Question 3.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Solution:
In the Ellingham diagram, the Cu₂O line is almost at the top. So it is quite easy to reduce oxide ores of copper directly to the metal by heating with coke (as can be seen in graph that the lines of C, CO and C, CO₂ are at much lower positions). But most of the ores are sulphide and some may also contain iron. So, the sulphide ores are roasted/smelted to give oxides :

Question 4.
Explain : (i) Zone refining (ii) Column chroma-tography.
Solution:
(i) This method 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 the impure metal. The molten zone moves along with the heater which is moved forward. As the heater moves forward, the pure metal crystallises out of the melt and the impurities pass on into the adjacent molten zone. The process is repeated several times and the heater is moved in the same direction. At one end, impurities get concentrated. This end is cut off. This method is very useful for producing semiconductor and other metals of very high purity, c.g., germanium, silicon, boron, gallium and indium.

(ii) Chromatographic method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The adsorbed components are removed (eluted) by using suitable (eluant). There are several chromatogrpahic techniques such as paper chromatography, column chromatography, gas chromatography etc.

Column chromatography : Column chroma-tography involves separation of a mixture over a column of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at its lower end (fig). The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube.

An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow the column slowly. Depending upon the degree to which the compounds are adsored, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distance in the column.

Question 5.
Out of C and CO, which is a better reducing agent at 673 K?
Solution:
When carbon acts as a reducing agent, it is either converted into CO or CO₂ or both.
2C + O₂ → 2CO
C + O₂ → CO₂
CO is oxidised to CO₂ when it is used as a reducing agent.
2CO + O₂ → 2CO₂
From the Ellingham diagram (refer answer number 3), it is clear that at the temperature 673 K, the AG° of the formation of CO₂ from CO is more negative than the formation of CO or CO₂ from carbon. Hence, at temperature 673 K, CO is a better reducing agent than C.

Question 6.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Solution:
Many of the metals such as copper, silver, gold, aluminium, lead, etc., are purified by this method. This is perhaps the most important method. The impure metal is made anode while a thin sheet of pure metal acts as a cathode. The electrolytic solution consists of generally an aqueous solution of a salt or a complex of the metal. On passing the current, the pure metal is deposited on the cathode and equivalent amount of the metal gets dissolved from the anode. Thus, the metal is transferred from anode to cathode through solution. The soluble impurities pass into the solution while the insoluble one, especially less electropositive impurities collect below the anode as anodic mud or anode sludge. Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum.

Question 7.
Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Solution:
Near the bottom of the furnace (zone of combustion, 2170 K), coke first combines with air to form CO₂ which then combines with more coke (zone of heat absorption, 1423 K) to form CO. The CO thus produced acts as the reducing agent and reduces iron oxide to spongy iron near the top of the furnace (zone of reduction, 823 K).

At the lower part of the furnace (zone of fusion, 1423-1673 K) the spongy iron melts and dissolves some carbon, S, P, SiO₂, Mn, etc.

The molten slag being less dense floats over the surface of the molten iron. The molten iron is then tapped off from the furnace and is then solidified to give blocks of iron called cast iron or pig iron.

Question 8.
Write chemical reactions taking place in the extraction of zinc from zinc blende.
Solution:
Concentration : When zinc blende is used, the powdered ore is concentrated by froth-floatation process.

Roasting : The concentrated ore is heated in excess of oxygen at about 900°C. Zinc sulphide is oxidised to zinc oxide. If some of the ore is oxidised to zinc sulphate, it also decomposes at 900°C into ZnO.

For roasting, a reverberatory furnace may be used.

Reduction : The principal reaction that takes place during reduction is the conversion of the oxide into the metal with the help of carbon.

Electrolytic refining : Purification of zinc is done by electrolytic refining using pure Zn as cathode and impure Zn as anode. The electrolyte is ZnSO₄.

Reaction at cathode : Zn2⁺_(aq)+ 2e^– → Zn_(s)
Reaction at anode : 2H₂O_(l) → O_2(g)+ 4H⁺_(aq)+ 4e^–
ZnSO₄ electrolyte is added from time to time.

Question 9.
State the role of silica in the metallurgy of copper.
Solution:
Iron present in pyrites has greater affinity for oxygen than copper. The copper oxide formed reacts with unchanged iron sulphide to form iron oxide so, most of the iron sulphide is oxidised to ferrous oxide.
2FeS + 3O₂ → 2FeO + 2sO₂
Ferrous oxide combines with silica which acts as flux and forms ferrous silicate. By this reaction most of the iron is removed as slag.

Question 10.
What is meant by the term “chromatography”?
Solution:
Chromatographic method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The adsorbed components are removed (eluted) by using suitable (eluant). There are several chromatogrpahic techniques such as paper chromatography, column chromatography, gas chromatography, etc.

Question 11.
What criterion is followed for the selection of the stationary phase in chromatography?
Solution:
The stationary phase is selected in such a way that the impurities are more strongly adsorbed or are more soluble in the stationary phase than element to be purified. Under these conditions, when the column is extracted, the impurities will be retained by the stationary phase whereas the pure component is easily removed.

Question 12.
Describe a method for refining nickel.
Solution:
Nickel is purified by Mond’s process. Impure nickel is treated with carbon monoxide at 60-80°C when volatile compound nickel carbonyl is formed. Nickel carbonyl decomposes at 180°C to form pure nickel and carbon monoxide.

Question 13.
How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Solution:
Serpeck’s process is used when silica is present in considerable amounts in bauxite ore. The ore is mixed with coke and heated at 1800°C in presence of nitrogen, where AIN is formed.
Al₂O₃ + 3C + N₂ → 2AlN + 3CO
Silica is reduced to silicon which volatilises off at this temperature.
SiO₂ + 2C → Si + 2CO

Question 14.
Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Solution:
Calcination : It is the process of converting an ore into its oxide by heating it strongly below its melting point either in absence or limited supply of air.

Roasting : In roasting, the ore is heated in a regular supply of air in a furnace at a temperature below the melting point of the metal.
2 ZnS + 3O₂ → 2ZnO + 2SO₂
2 PbS + 3O₂ → 2PbO + 2SO₂

Question 15.
How is’cast iron’different from ‘pig iron’?
Solution:
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 and cast into variety of shapes.

Cast iron is different from pig iron 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.

Question 16.
Differentiate between “minerals”and “ores’.
Solution:
The natural substances in which the metal or their compounds occur in the earth are called minerals. The mineral has a definite composition. It may be a single compound or complex mixture. The minerals from which the metals can be conveniently and economically extracted are known as ores. All the ores arc minerals but all minerals cannot be ores, e.g., both bauxite (Al₂O₃.xH2O) and clay (Al₂O₃.2SiO₂.2H₂O) are minerals of aluminium. It is bauxite which is used for extraction of aluminium and not clay. Thus bauxite is an ore of aluminium.

Question 17.
Why copper matte is put in silica lined converter ?
Solution:
Copper matte consists of Cu2S and FeS. Wheia a blast of hot air is passed through molten matte taken in a silica lined converter, FeS present in matte is oxidised to FeO which combines with silica (SiO2) to form FeSiO3 (slag).

When complete iron has been removed as slag, some of the Cu2S undergoes oxidation to form Cu2O which then reacts with more Cu2S to form copper metal.
2CU₂S + 3O₂ → 2CU₂O + 2SO₂
2CU₂O + Cu₂S → 6Cu + SO₂
Thus, copper matte is heated in silica lined converter to remove FeS present in matte as FeSiO₃ (slag).

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Solution:
The role of cryolite is as follows :

1. It makes alumina a good conductor of electricity.
2. It lowers the fusion temperature of the bath from 2323 K to about 1140 K.

Question 19.
How is leaching carried out in case of low grade copper ores?
Solution:
The leaching of the low grade copper ores is carried out with acids in the presence of air when copper goes into solution as Cu²⁺ ions. Therefore,

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO?
Solution:
The standard free energy of formation (∆_fG°) of CO₂ from CO is higher than that of the formation of ZnO from Zn. Hence, CO cannot be used to reduce ZnO to Zn.

Question 21.
The value of ∆_fG° for formation of Cr₂O₃ is – 540 kJ mol^-1 and that of Al₂O₃ is – 827 kJ mol^-1. Is the reduction of Cr₂O₃ possible with Al?
Solution:

Question 22.
Out of C and CO, which is a better reducing agent for ZnO?
Solution:
The free energy of formation (∆_f G°) of CO from C becomes lower at temperatures above 1120 K whereas that of CO₂ from C becomes lower above 1323 K than ∆_fG° of ZnO. However, ∆_fG° of CO₂ from CO is always higher than that of ZnO. Therefore, C can reduce ZnO to Zn but not CO. Therefore, out of C and CO, C is a better reducing agent than CO for ZnO.
Reduction of ZnO is usually carried out around

Question 23.
The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Solution:
Some basic concepts of thermodynamics lielp us in understanding the theory of metallurgical transformations. Gibb’s energy is the most significant term.
The graphical representation of Gibb’s energy was first used by H.J.T. Ellingham. This
provides a sound basis for considering the k choice of reducing agent in the reduction of oxides. This is known as Ellingham diagram. (Tor diagram refer answer number 3) Such diagrams help us in predicting the feasibility of thermal reduction of an ore. The criterion of feasibility is that at a given temperature Gibb’s energy of the reaction must be negative.
Examples :
(i) Thermodynamics helps us to understand how coke reduces the oxide and why blast furnace is chosen. One of the main reduction steps in this process is :
FeO_(s) + C_(s) → Fe_(s/l) + CO_(g)… (1)
It can be seen as a couple of two simpler reactions. In one, the reduction of FeO is taking place and in the other, C is being oxidised to CO :

When both the reactions take place to yield the equation (1), the net Gibb’s energy change becomes :

Naturally, the resultant reaction will take place when the right hand side in equation (3) is negative. In ∆G° vs T plot representing reaction 2, the plot goes upward and that representing the change C → CO (C, CO) goes downward. At temperatures above 1073 K (approx), the C, CO line comes below the Fe, FeO line [∆G_(c,co) < ∆G_(Fe,FeO)] So in this range, coke will be reducing the FeO and will itself be oxidised to CO. In a similar way the reduction of Fe₃O₄ and Fe₂O₃ at relatively lower temperatures by CO can be explained on the basis of lower lying points of intersection of their curves with the CO, CO₂ curve in Ellinghan diagram.

(ii) In the graph of ∆r.G° vs T for formation of oxides, the Cu₂O line is almost at the top. So it is quite easy to reduce oxide ores of copper directly to the metal by heating with coke (both the lines of C, CO and C, CO₂ are at much lower position in the graph particularly after 500 – 600 K). However most of the ores are sulphide and some may also contain iron. The sulphide ores are roasted/smelted to give oxides:
2CU₂S + 3O₂ → 2Cu₂O + 2SO₂
The oxide can then be easily reduced to metallic copper using coke.
Cu₂O + C → 2Cu + CO

Question 24.
Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCI is subjected to electrolysis?
Solution:
Sodium metal is prepared by Down’s process. This process involves the electrolysis of a fused mixture of NaCI and CaCl₂ at 873 K. During electrolysis, sodium is discharged at the cathode and Cl₂ is obtained at the anode as a by-product.

Question 25.
What is the role of graphite rod in the electrometallurgy of aluminium?
Solution:
In this process, a fused mixture of alumina, cryolite and fluorspar (CaF₂) is electrolysed using graphite as anode and steel as cathode. During electrolysis, Al is liberated at the cathode whereas CO and CO₂ are liberated at the anode.

If some other metal is used as the anode other than graphite, then O₂ liberated will not only oxidise the metal of the electrode but would also convert some of the Al liberated at the cathode back to Al₂O₃. Since graphite is much cheaper than any metal, graphite is used as the anode. So, the role of graphite in electrometallurgy of Al is to prevent the liberation of O₂ at the anode which may otherwise oxidise some of the liberated Al back to Al₂O₃.

Question 26.
Outline the principles of refining of metals by the following methods :

1. Zone refining
2. Electrolytic refining
3. Vapour phase refining

Solution:
(i) Refer answer number 4(i).

(ii) In this method, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud.
Anode : M → Mⁿ⁺ ne^–
Cathode : Mⁿ⁺ + ne^– → M
Copper is refined using an electrolytic method. Anodes are of impure copper and pure copper strips are taken as cathode. The electrolyte is acidified solution of copper sulphate and the net result of electrolysis is the transfer of copper in pure form from the anode to the cathode :
Anode : Cu → Cu²⁺ + 2e^–
Cathode : Cu²⁺ + 2e^– → Cu
Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum; recovery of these elements may meet the cost of refining.
Zinc may also be refined this way.

(iii) In this method, the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal. So, the two requirements are :

(a) The metal should form a volatile compound with an available reagent.
(b) The volatile compound should be easily decomposable, so that the recovery is easy.

Following example will illustrate this technique.
Mond process for Refining Nickel : In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl :

The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal :

Question 27.
Predict conditions under which Al might be expected to reduce MgO.
Solution:
The two equations are :

INTEXT Questions

Question 1.
Which of the ores mentioned in Table 6.1 (NCERT Textbook) can be concentrated by magnetic separation method?
Solution:
Ores in which one of the components (either the impurity or the actual ore) is magnetic can be concentrated by magnetic separation, e.g., ores containing iron (haematite, magnetite, siderite and iron pyrites).

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

Question 3.
The reaction, Cr₂O₃ + 2Al → Al₂O₃ + 2cr (∆G° = – 421 kJ) is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?
Solution:
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?
Solution:
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.

NCERT Exercises

Question 1.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Solution:
The E° of zinc (Zn²⁺/Zn = – 0.76 V) is lower luau that of copper (Cu²⁺/Cu = +0.34 V). Hence, zinc can displace Cu from solutions of Cu²⁺ ions.

But all these metals react with water forming their corresponding ions with the evolution of H₂ gas. Hence, Al, Mg, etc. cannot be used to displace zinc from the solution of Zn²⁺ ions. Thus, copper can be extracted by hydrometal-lurgy but not zinc.

Question 2.
What is the role of depressant in froth-floatation process?
Solution:
In froth -floatation process, the role of the depressant is to prevent one type of sulphide ore particles from forming the froth with air bubbles. For example, NaCN is used as a depressant to separate lead sulphide (PbS) ore from zinc sulphide (ZnS) ore. The reason is that NaCN forms a zinc complex, Na₂[Zn(CN)₄] on the surface of ZnS preventing it from the formation of the froth. Under these conditions, only PbS forms froth and therefore, it can be separated from ZnS ore.

Question 3.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Solution:
In the Ellingham diagram, the Cu₂O line is almost at the top. So it is quite easy to reduce oxide ores of copper directly to the metal by heating with coke (as can be seen in graph that the lines of C, CO and C, CO₂ are at much lower positions). But most of the ores are sulphide and some may also contain iron. So, the sulphide ores are roasted/smelted to give oxides :

Question 4.
Explain : (i) Zone refining (ii) Column chroma-tography.
Solution:
(i) This method 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 the impure metal. The molten zone moves along with the heater which is moved forward. As the heater moves forward, the pure metal crystallises out of the melt and the impurities pass on into the adjacent molten zone. The process is repeated several times and the heater is moved in the same direction. At one end, impurities get concentrated. This end is cut off. This method is very useful for producing semiconductor and other metals of very high purity, c.g., germanium, silicon, boron, gallium and indium.

(ii) Chromatographic method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The adsorbed components are removed (eluted) by using suitable (eluant). There are several chromatogrpahic techniques such as paper chromatography, column chromatography, gas chromatography etc.

Column chromatography : Column chroma-tography involves separation of a mixture over a column of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at its lower end (fig). The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube.

An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow the column slowly. Depending upon the degree to which the compounds are adsored, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distance in the column.

Question 5.
Out of C and CO, which is a better reducing agent at 673 K?
Solution:
When carbon acts as a reducing agent, it is either converted into CO or CO₂ or both.
2C + O₂ → 2CO
C + O₂ → CO₂
CO is oxidised to CO₂ when it is used as a reducing agent.
2CO + O₂ → 2CO₂
From the Ellingham diagram (refer answer number 3), it is clear that at the temperature 673 K, the AG° of the formation of CO₂ from CO is more negative than the formation of CO or CO₂ from carbon. Hence, at temperature 673 K, CO is a better reducing agent than C.

Question 6.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Solution:
Many of the metals such as copper, silver, gold, aluminium, lead, etc., are purified by this method. This is perhaps the most important method. The impure metal is made anode while a thin sheet of pure metal acts as a cathode. The electrolytic solution consists of generally an aqueous solution of a salt or a complex of the metal. On passing the current, the pure metal is deposited on the cathode and equivalent amount of the metal gets dissolved from the anode. Thus, the metal is transferred from anode to cathode through solution. The soluble impurities pass into the solution while the insoluble one, especially less electropositive impurities collect below the anode as anodic mud or anode sludge. Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum.

Question 7.
Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Solution:
Near the bottom of the furnace (zone of combustion, 2170 K), coke first combines with air to form CO₂ which then combines with more coke (zone of heat absorption, 1423 K) to form CO. The CO thus produced acts as the reducing agent and reduces iron oxide to spongy iron near the top of the furnace (zone of reduction, 823 K).

At the lower part of the furnace (zone of fusion, 1423-1673 K) the spongy iron melts and dissolves some carbon, S, P, SiO₂, Mn, etc.

The molten slag being less dense floats over the surface of the molten iron. The molten iron is then tapped off from the furnace and is then solidified to give blocks of iron called cast iron or pig iron.

Question 8.
Write chemical reactions taking place in the extraction of zinc from zinc blende.
Solution:
Concentration : When zinc blende is used, the powdered ore is concentrated by froth-floatation process.

Roasting : The concentrated ore is heated in excess of oxygen at about 900°C. Zinc sulphide is oxidised to zinc oxide. If some of the ore is oxidised to zinc sulphate, it also decomposes at 900°C into ZnO.

For roasting, a reverberatory furnace may be used.

Reduction : The principal reaction that takes place during reduction is the conversion of the oxide into the metal with the help of carbon.

Electrolytic refining : Purification of zinc is done by electrolytic refining using pure Zn as cathode and impure Zn as anode. The electrolyte is ZnSO₄.

Reaction at cathode : Zn2⁺_(aq)+ 2e^– → Zn_(s)
Reaction at anode : 2H₂O_(l) → O_2(g)+ 4H⁺_(aq)+ 4e^–
ZnSO₄ electrolyte is added from time to time.

Question 9.
State the role of silica in the metallurgy of copper.
Solution:
Iron present in pyrites has greater affinity for oxygen than copper. The copper oxide formed reacts with unchanged iron sulphide to form iron oxide so, most of the iron sulphide is oxidised to ferrous oxide.
2FeS + 3O₂ → 2FeO + 2sO₂
Ferrous oxide combines with silica which acts as flux and forms ferrous silicate. By this reaction most of the iron is removed as slag.

Question 10.
What is meant by the term “chromatography”?
Solution:
Chromatographic method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The adsorbed components are removed (eluted) by using suitable (eluant). There are several chromatogrpahic techniques such as paper chromatography, column chromatography, gas chromatography, etc.

Question 11.
What criterion is followed for the selection of the stationary phase in chromatography?
Solution:
The stationary phase is selected in such a way that the impurities are more strongly adsorbed or are more soluble in the stationary phase than element to be purified. Under these conditions, when the column is extracted, the impurities will be retained by the stationary phase whereas the pure component is easily removed.

Question 12.
Describe a method for refining nickel.
Solution:
Nickel is purified by Mond’s process. Impure nickel is treated with carbon monoxide at 60-80°C when volatile compound nickel carbonyl is formed. Nickel carbonyl decomposes at 180°C to form pure nickel and carbon monoxide.

Question 13.
How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Solution:
Serpeck’s process is used when silica is present in considerable amounts in bauxite ore. The ore is mixed with coke and heated at 1800°C in presence of nitrogen, where AIN is formed.
Al₂O₃ + 3C + N₂ → 2AlN + 3CO
Silica is reduced to silicon which volatilises off at this temperature.
SiO₂ + 2C → Si + 2CO

Question 14.
Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Solution:
Calcination : It is the process of converting an ore into its oxide by heating it strongly below its melting point either in absence or limited supply of air.

Roasting : In roasting, the ore is heated in a regular supply of air in a furnace at a temperature below the melting point of the metal.
2 ZnS + 3O₂ → 2ZnO + 2SO₂
2 PbS + 3O₂ → 2PbO + 2SO₂

Question 15.
How is’cast iron’different from ‘pig iron’?
Solution:
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 and cast into variety of shapes.

Cast iron is different from pig iron 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.

Question 16.
Differentiate between “minerals”and “ores’.
Solution:
The natural substances in which the metal or their compounds occur in the earth are called minerals. The mineral has a definite composition. It may be a single compound or complex mixture. The minerals from which the metals can be conveniently and economically extracted are known as ores. All the ores arc minerals but all minerals cannot be ores, e.g., both bauxite (Al₂O₃.xH2O) and clay (Al₂O₃.2SiO₂.2H₂O) are minerals of aluminium. It is bauxite which is used for extraction of aluminium and not clay. Thus bauxite is an ore of aluminium.

Question 17.
Why copper matte is put in silica lined converter ?
Solution:
Copper matte consists of Cu2S and FeS. Wheia a blast of hot air is passed through molten matte taken in a silica lined converter, FeS present in matte is oxidised to FeO which combines with silica (SiO2) to form FeSiO3 (slag).

When complete iron has been removed as slag, some of the Cu2S undergoes oxidation to form Cu2O which then reacts with more Cu2S to form copper metal.
2CU₂S + 3O₂ → 2CU₂O + 2SO₂
2CU₂O + Cu₂S → 6Cu + SO₂
Thus, copper matte is heated in silica lined converter to remove FeS present in matte as FeSiO₃ (slag).

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Solution:
The role of cryolite is as follows :

1. It makes alumina a good conductor of electricity.
2. It lowers the fusion temperature of the bath from 2323 K to about 1140 K.

Question 19.
How is leaching carried out in case of low grade copper ores?
Solution:
The leaching of the low grade copper ores is carried out with acids in the presence of air when copper goes into solution as Cu²⁺ ions. Therefore,

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO?
Solution:
The standard free energy of formation (∆_fG°) of CO₂ from CO is higher than that of the formation of ZnO from Zn. Hence, CO cannot be used to reduce ZnO to Zn.

Question 21.
The value of ∆_fG° for formation of Cr₂O₃ is – 540 kJ mol^-1 and that of Al₂O₃ is – 827 kJ mol^-1. Is the reduction of Cr₂O₃ possible with Al?
Solution:

Question 22.
Out of C and CO, which is a better reducing agent for ZnO?
Solution:
The free energy of formation (∆_f G°) of CO from C becomes lower at temperatures above 1120 K whereas that of CO₂ from C becomes lower above 1323 K than ∆_fG° of ZnO. However, ∆_fG° of CO₂ from CO is always higher than that of ZnO. Therefore, C can reduce ZnO to Zn but not CO. Therefore, out of C and CO, C is a better reducing agent than CO for ZnO.
Reduction of ZnO is usually carried out around

Question 23.
The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Solution:
Some basic concepts of thermodynamics lielp us in understanding the theory of metallurgical transformations. Gibb’s energy is the most significant term.
The graphical representation of Gibb’s energy was first used by H.J.T. Ellingham. This
provides a sound basis for considering the k choice of reducing agent in the reduction of oxides. This is known as Ellingham diagram. (Tor diagram refer answer number 3) Such diagrams help us in predicting the feasibility of thermal reduction of an ore. The criterion of feasibility is that at a given temperature Gibb’s energy of the reaction must be negative.
Examples :
(i) Thermodynamics helps us to understand how coke reduces the oxide and why blast furnace is chosen. One of the main reduction steps in this process is :
FeO_(s) + C_(s) → Fe_(s/l) + CO_(g)… (1)
It can be seen as a couple of two simpler reactions. In one, the reduction of FeO is taking place and in the other, C is being oxidised to CO :

When both the reactions take place to yield the equation (1), the net Gibb’s energy change becomes :

Naturally, the resultant reaction will take place when the right hand side in equation (3) is negative. In ∆G° vs T plot representing reaction 2, the plot goes upward and that representing the change C → CO (C, CO) goes downward. At temperatures above 1073 K (approx), the C, CO line comes below the Fe, FeO line [∆G_(c,co) < ∆G_(Fe,FeO)] So in this range, coke will be reducing the FeO and will itself be oxidised to CO. In a similar way the reduction of Fe₃O₄ and Fe₂O₃ at relatively lower temperatures by CO can be explained on the basis of lower lying points of intersection of their curves with the CO, CO₂ curve in Ellinghan diagram.

(ii) In the graph of ∆r.G° vs T for formation of oxides, the Cu₂O line is almost at the top. So it is quite easy to reduce oxide ores of copper directly to the metal by heating with coke (both the lines of C, CO and C, CO₂ are at much lower position in the graph particularly after 500 – 600 K). However most of the ores are sulphide and some may also contain iron. The sulphide ores are roasted/smelted to give oxides:
2CU₂S + 3O₂ → 2Cu₂O + 2SO₂
The oxide can then be easily reduced to metallic copper using coke.
Cu₂O + C → 2Cu + CO

Question 24.
Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCI is subjected to electrolysis?
Solution:
Sodium metal is prepared by Down’s process. This process involves the electrolysis of a fused mixture of NaCI and CaCl₂ at 873 K. During electrolysis, sodium is discharged at the cathode and Cl₂ is obtained at the anode as a by-product.

Question 25.
What is the role of graphite rod in the electrometallurgy of aluminium?
Solution:
In this process, a fused mixture of alumina, cryolite and fluorspar (CaF₂) is electrolysed using graphite as anode and steel as cathode. During electrolysis, Al is liberated at the cathode whereas CO and CO₂ are liberated at the anode.

If some other metal is used as the anode other than graphite, then O₂ liberated will not only oxidise the metal of the electrode but would also convert some of the Al liberated at the cathode back to Al₂O₃. Since graphite is much cheaper than any metal, graphite is used as the anode. So, the role of graphite in electrometallurgy of Al is to prevent the liberation of O₂ at the anode which may otherwise oxidise some of the liberated Al back to Al₂O₃.

Question 26.
Outline the principles of refining of metals by the following methods :

1. Zone refining
2. Electrolytic refining
3. Vapour phase refining

Solution:
(i) Refer answer number 4(i).

(ii) In this method, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud.
Anode : M → Mⁿ⁺ ne^–
Cathode : Mⁿ⁺ + ne^– → M
Copper is refined using an electrolytic method. Anodes are of impure copper and pure copper strips are taken as cathode. The electrolyte is acidified solution of copper sulphate and the net result of electrolysis is the transfer of copper in pure form from the anode to the cathode :
Anode : Cu → Cu²⁺ + 2e^–
Cathode : Cu²⁺ + 2e^– → Cu
Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum; recovery of these elements may meet the cost of refining.
Zinc may also be refined this way.

(iii) In this method, the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal. So, the two requirements are :

(a) The metal should form a volatile compound with an available reagent.
(b) The volatile compound should be easily decomposable, so that the recovery is easy.

Following example will illustrate this technique.
Mond process for Refining Nickel : In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl :

The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal :

Question 27.
Predict conditions under which Al might be expected to reduce MgO.
Solution:
The two equations are :

 INTEXT Questions

Question 1.
Why are substances like platinum and palladium often used for carrying out electrolysis of aqueous solutions?
Solution:
Due to their inert nature, these metals do not affect the products of electrolysis. They have good adsorbing capacity for hydrogen.

Question 2.
Why does physisorption decrease with the increase of temperature?
Solution:
Physical adsorption of a gas by a solid is generally reversible. Thus,
Solid + Gas  Gas / Solid + Heat
Since the adsorption process is exothermic, the physical adsorption occurs readily at low temperature and decreases with increasing temperature (Le Chatelier’s principle).

Question 3.
Why are powdered substances more effective adsorbents than their crystalline forms?
Solution:
The extent of adsorption increases with increase in surface area of the adsorbent. Finely powdered substances have large, porous areas and act as good adsorbents.

Question 4.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process?
Solution:
CO is a catalytic poison. It reacts with iron to form iron carbonyl thus inhibiting the activity of catalyst.

Question 5.
Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?
Solution:
The acid formed during the reaction provides hydrogen ions which act as catalyst for the reaction and it becomes faster.

Question 6.
What is the role of desorption in the process of catalysis?
Solution:
In catalysis the products formed are desorbed and detached from the surface so that, more reactants can get adsorbed on the surface of catalyst.

Question 7.
What modification can you suggest in the Hardy Schulze law?
Solution:
The Hardy Schulze law considers the coagulation of sols because of neutralisation of their charges. Since coagulation can also occur by mixing two oppositely charged sols, it should also include “when oppositely charged sols are mixed in proper proportions to neutralize the charges of each other, the coagulation of both the sols occurs.”

Question 8.
Why is it essential to wash the precipitate with water before estimating it quantitatively?
Solution:
Few impurities which are soluble in water and are adsorbed on the surface of the precipitate are removed by washing them with water.

NCERT Exercises

Question 1.
Distinguish between the meaning of the terms adsorption and absorption. Give one example of each.
Solution:
The phenomenon of accumulation of the molecules of a substance on a solid or liquid surface resulting in the increased concentration of the molecules on the surface is called adsorption. In absorption, the substance is uniformly distributed throughout the bulk of the solution. A distinction can be made by taking an example of water vapours. Water vapours are absorbed by anhydrous calcium chloride but adsorbed by silica gel.

Question 2.
What is the difference between physisorption and chemisorption?
Solution:

Question 3.
Give reason why a finely divided substance is more effective as an adsorbent.
Solution:
The extent of adsorption increases with increase in surface area of the adsorbent. Thus, finely divided metals and porous substances having large surface areas are good adsorbents.

Question 4.
What are the factors which influence the adsorption of a gas on a solid?
Solution:
Factors affecting adsorption of a gas on solids are :

Nature of the adsorbent : The same gas is adsorbed to different extents by different solids at the same temperature. Also, greater the surface area of the adsorbent, more is the gas adsorbed.

Nature of the adsorbate : Different gases are adsorbed to different extents by different solids at the same temperature. Higher the critical temperature of the gas, greater is its amount adsorbed.

Temperature : Since adsorption is an exothermic process, applying Le Chatelier’s principle, we can find out that adsorption decreases with an increase in temperature.

Specific area of the adsorbent : Surface area available for adsorption per gram of the adsorbent increases the extent of adsorption. Greater the surface area, higher would be the adsorption therefore, porous or powdered adsorbents are used.

Pressure : At constant temperature, the adsorption of gas increases with pressure.

Activation of adsorbent : It means increasing the adsorbing power of an adsorbent by increasing its surface area. It is done by :

1. making the adsorbent’s surface rough
2. removing gases already adsorbed
3. subdividing the adsorbent into smaller pieces.

Question 5.
What is an adsorption isotherm? Describe Freundlich adsorption isotherm.
Solution:
Adsorption isotherm is a graph between the amount of the gas adsorbed by an adsorbent and equilibrium pressure of the adsorbate at constant temperature. Freundlich obtained an empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. It is mathematically represented in the following way :

Question 6.
What do you understand by activation of adsorbent? How is it achieved?
Solution:
Activation of an adsorbent means increasing its adsorbing power by increasing the surface area of the adsorbent by making its surface rough, and removing adsorbed gases from it. With an increase in surface area the adsorption increases.

Question 7.
What role does adsorption play in heterogenous catalysis?
Solution:
Adsorption of reactants on solid surface of the catalysts increases the rate of reaction. There are many gaseous reactions of industrial importance involving solid catalysts. Manufacture of ammonia using iron as a catalyst, manufacture of H₂SO₄ by Contact process and use of finely divided nickel in the hydrogenation of oils are excellent examples of heterogeneous catalysis.

Question 8.
Why is adsorption always exothermic?
Solution:
During adsorption, there is always a decrease in residual forces of the surface, i.e., there is decrease in surface energy which appears as heat. Adsorption therefore, is invariably an exothermic process. In other words, ∆H of adsorption is always negative to keep the value of ∆G negative for the reaction to be spontaneous as ∆S decreases during adsorption.

Question 9.
How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?
Solution:
Colloids can be classified into eight types depending upon the physical state of the dispersed phase and the dispersion medium.

Question 10.
Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Solution:

1. Adsorption decreases with an increase in, temperature because it is an exothermic process and according to Le Chatelier’s principle the reaction will proceed in backward direction with increase in temperature.
2. At a constant temperature, adsorption increases with pressure.

Question 11.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Solution:
There are two types of colloidal sols :

(i) Lyophilic sols : The word lyophilic means solvent loving. They are obtained by directly mixing the dispersed phase and the dispersion medium, e.g., sols of gum, gelatin, starch, etc. They are solvent attracting hence quite stable and cannot be coagulated easily.

(ii) Lyophobic sols : They cannot be prepared by directly mixing the dispersed phase and dispersion medium but are prepared by special methods, e.g., sols of metals. They are solvent repelling. Hydrophobic sols are easily coagulated due to repulsion between water and dispersed phase.

Question 12.
What is the difference between multimolecular and macromoiecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Solution:
Depending upon the type of particles of the dispersed phase, colloids are classified as : multimolecular, macromoiecular and associated colloids.

(i) Multimolecular colloids : On dissolution, a large number of atoms or smaller molecules of a substance aggregate together to form species having size in the colloidal range (diameter < 1 nm). The species thus formed are called multimolecular colloids. For example, a gold sol may contain particles of various sizes having many atoms. Sulphur sol consists of particles containing a thousand or more of S8 sulphur molecules.

(ii) Macromoiecular colloids : Macromolecules in suitable solvents form solutions in which the size of the macromolecules may be in the colloidal range. Such systems are called macromoiecular colloids. These colloids are quite stable and resemble true solutions in many respects. Examples of naturally occurring macromolecules are starch, cellulose, proteins and enzymes; and those of man-made macromolecules are polythene, nylon, polystyrene, synthetic rubber, etc.

(iii) Associated colloids (Micelles) : There are some substances which at low concentrations behave as normal strong electrolytes but, at higher concentrations exhibit colloidal behaviour due to the formation of aggregates. The aggregated particles thus formed are called micelles. These are also known as associated colloids. The formation of micelles takes place only above a particular temperature called Kraft temperature (Tk) and above a particular concentration called critical micelle concentration (CMC). On dilution, these colloids revert back to individual ions. Surface active agents such as soaps and synthetic detergents belong to this class. For soaps, the CMC is 10^-4 to 10^-3 mol L^-1. These colloids have both lyophobic and lyophilic parts. Micelles may contain as many as 100 molecules or more.

Question 13.
What are enzymes? Write in brief the mechanism of enzyme catalysis.
Solution:
Enzymes are complex nitrogenous organic compounds which act as biological catalysts and increase the rate of cellular processes. According to the lock and key model, like every lock has a specific key, similarly every enzyme acts at a specific substrate.

When the substrate fits the active site (lock) of the enzyme, the chemical change begins. But it has also been noticed that enzyme changes shape, when substrate lands at the active site. This induced-fit model of enzyme action pictures the substrate inducing the 1 active site to adopt a perfect fit, rather than a rigid shaped lock and key. Therefore, the new model for enzyme action is called induced fit model.

Question 14.
How are colloids classified on the basis of

1. physical states of components
2. nature of dispersion medium and
3. interaction between dispersed phase and dispersion medium?

Solution:

1. Refer answer number 9.
2. Depending upon the nature of dispersion medium colloids can be classified as sol if the dispersion medium is liquid, gel if the dispersion medium is solid. If the dispersion medium is water, the sol is called hydrosol and if the dispersion medium is alcohol, it is called alcosol. A colloid in which the dispersion medium as well as dispersed phase are liquids, is called emulsion.
3. Refer answer number 11.

Question 15.
Explain what is observed

1. when a beam of light is passed through a colloidal sol.
2. an electrolyte, NaCI is added to hydrated ferric oxide sol.
3. electric current is passed through a colloidal sol?

Solution:
(i) When a beam of light is passed through colloidal particles, its path becomes clearly visible and is known as Tyndall effect. It is due to scattering of light by colloidal particles. The bright cone of the light is called Tyndall cone.

(ii) When NaCI is added to hydrated ferric oxide sol coagulation takes place. Since ferric oxide is a positive sol, it is coagulated by the negative chloride ions.

(iii) When electric potential is applied across two platinum electrodes dipped in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied electric potential is called electrophoresis.

Positively charged particles move towards the cathode while negatively charged particles move towards the anode. Since all the colloidal particles in a given colloidal solution carry the same charge, the particles move to one or the other electrode depending on the charge.

Question 16.
What are emulsions? What are their different types? Give example of each type.
Solution:
These are liquid-liquid colloidal systems, i.e., the dispersion of finely divided droplets in another liquid. If a mixture of two immiscible or partially miscible liquids is shaken, a coarse dispersion of one liquid in the other is obtained which is called emulsion. Generally, one of the two liquids is water. There are two types of emulsions :

1. Oil dispersed in water (O/W type) and
2. Water dispersed in oil (W/O type).

In the first system, water acts as dispersion medium. Examples of this type of emulsion are milk and vanishing cream. In milk, liquid fat is dispersed in water. In the second system, oil acts as dispersion medium. Common examples of this type are butter and cream.

Question 17.
What is demulsification? Name two demulsifiers.
Solution:
The process of converting the emulsion back into two distinct components, oil and water is called demulsification. This can be done by

1. boiling
2. freezing
3. changing pH
4. electrostatic precipitation.

Question 18.
Action of soap is due to emulsification and micelle formation. Comment.
Solution:
Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOO^– Na⁺ (e.g., sodium stearate 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 anions are pulled into the bulk of the solution and aggregate 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. An aggregate thus formed is known as ‘ionic micelle’.

The cleansing action of soap is due to the fact that soap molecules form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles

(a) Grease on cloth
(b) Stearate ions (from soap) arranging around the grease droplets
(c) Micelle formed

Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats. Tne negatively charged sheath around the globules prevents them from coming together and forming aggregates.

Question 19.
Give four examples of heterogeneous catalysis.
Solution:

Question 20.
What do you mean by activity and selectivity of catalysts?
Solution:
(a) Activity : 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 active. But adsorption must not be so strong that they are immobilised. It is observed that maximum activity is shown by elements of groups 7 – 9 of the periodic table
2H₂ + O₂  2H₂O

(b) Selectivity : The selectivity of a calatyst is its ability to yield a particular product in the reaction e.g.,

Thus, a selective catalyst can act as a catalyst in one reaction and may fail to catalyse another reaction.

Question 21.
Describe some features of catalysis by zeolites.
Solution:
(a) Zeolites are hydrated aluminosilicates which have a three dimensional network structure containing water molecules in their pores.
(b) The pores are made vacant by heating before catalysis.
(c) The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules and also on the pores and cavities in them, e.g., ZSM-5 converts alcohols to hydrocarbons by dehydrating them.

Alcohols  Hydrocarbons

Question 22.
What is shape selective catalysis?
Solution:
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 catalysis. Zeolites are good shape-selective catalysts because of their honeycomb-like structures. They are microporous aluminosilicates with three dimensional network of silicates in which some silicon atoms are replaced by aluminium atoms giving Al-O-Si framework. The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as upon the pores and cavities of the zeolites. They are found in nature as well as synthesised for catalytic selectivity.

Question 23.
Explain the following terms :

1. Electrophoresis
2. Coagulation
3. Dialysis
4. Tyndall effect

Solution:
(i) Electrophoresis : Refer answer number 15 (iii)

(ii) Coagulation or precipitation : The stability of the lyophobic sols is due to the presence of charge on colloidal particles. If somehow, the charge is removed, the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. The process of settling down of colloidal particles is called coagulation.

(iii) Dialysis : It is the process of removing dissolved substances from a colloidal solution by means of diffusion through a suitable membrane. Since particles (ions or smaller molecules) in a true solution can pass through animal membrane (bladder) or parchment paper or cellophane sheet but not the colloidal particles, the membrane can be used for dialysis. The apparatus used for this purpose is called dialyser. A bag of suitable membrane containing the colloidal solution is suspended in a vessel through which fresh water is continuously flowing. The molecules and ions diffuse through membrane into the outer water and pure colloidal solution is left behind.

(iv) Tyndall effect : Refer answer number 15 (i)

Question 24.
Give four uses of emulsions.
Solution:

1. Some of the medicines are effective as emulsions.
2. Paints are emulsions which are used in our daily life.
3. Soaps and detergents act as cleansing agents, action of which is based on emulsification.
4. Photographic films are coated with emulsion of AgBr on its surface.

Question 25.
What are micelles? Give an example of a micelle system.
Solution:
Micelles are substances that behave as normal strong electrolytes at low concentration but at high concentrations behave as colloids due to formation of aggregates. They are also called associated colloids, e.g., soaps and detergents. They can form ions and may contain 100 or more molecules to form a micelle.

Question 26.
Explain the terms with suitable examples :

1. Alcosol
2. Aerosol
3. Hydrosol

Solution:
(i) Alcosol : The sol in which alcohol is used as dispersion medium is called alcosol e.g., sol of cellulose nitrate in ethyl alcohol.

(ii) Aerosol : The sol in which dispersion medium is gas and dispersed phase is either solid or liquid, the colloidal system is called aerosol e.g., fog, insecticides, sprays, etc.

(iii) Hydrosol : The sol in which dispersion medium is water is called hydrosol e.g., starch sol.

Question 27.
Comment on the statement that’colloid is not a substance but a state of substance’.
Solution:
Colloid is not a substance, but a state of substance because the same substance may exist as a colloid or crystalloid under different conditions e.g., sulphur. Colloidal solution of sulphur consists of sulphur molecules dispersed in water. In this state, sulphur atoms combine to form multimolecules whose size lies between 1 nm to 1000 nm and form colloidal state. Sulphur forms true solution in carbon disulphide. Similarly soap is a solution at low concentration but a colloid at higher concentration.

 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 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Solution:

Question 2.
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:

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:

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:
The reaction, X → Y, follows second order kinetics hence the rate law equation will be
Rate = kC², where C = [X]
If concentration of X increases three times, now, [X] = 3C mol L^-1
∴ Rate = k(3C)² = 9kC²
Thus the rate of reaction will become 9 times. Hence, the rate of formation of Y will increase 9 times.

Question 5.
A first order reaction has a rate constant 1.15 × 10^-3 s^-1. How long will 5 g of this reactant take to reduce to 3 g?
Solution:

Question 6.
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:

Question 7.
What will be the effect of temperature on rate constant?
Solution:
It has been found that for a chemical reaction, with rise in temperature by 10°, the rate constant is nearly doubled.

The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation.
K = Ae^-Ea/RT
Where A is the Arrhenius factor or the frequency factor. It is also called pre exponential factor. It is a constant specific to a particular reaction. R is gas constant and E_a is activation energy measured in joules/mole (J mol^-1).

Question 8.
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_a.
Solution:

Question 9.
The activation energy for the reaction 2Hl_(g) → H_2(g) + l_2(g) is 209.5 kJ mol^-1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
Solution:
Fraction of molecules having energy equal to or greater than activation energy is given as

 NCERT Exercises

Question 1.
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

Solution:

Question 2.
For the reaction : 2A + B → A2B the rate = k[A][B]2 with k = 2.0 × 10^-6 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^-1.
Solution:

Question 3.
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 L^-1 s^-1 ?
Solution:

Question 4.
The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by
Rate = k[CH₃OCH₃]^3/2
The rate of reaction is followed by increase in pressure in closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
Rate = K(pCH₃OCH₃)^3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants ?
Solution:
In terms of pressure,
Units of rate = bar min^-1

Question 5.
Mention the factors that affect the rate of a chemical reaction.
Solution:
Following are the factors on which rate of reaction depends.
(i) Nature of the reactant : Rate of reaction depends on nature of reactant.
Example : Reactions of ionic compounds are faster than that of covalent compounds.

(ii) State of reactants : Solid reactions are slow, reactions of liquids are fast whereas that of gases are very fast.

(iii) Temperature : Rate of reaction largely depends on temperature. It has been observed that every 10°C rise in temperature increases rate of reaction by 2-3 times.
 = 2 – 3 . This ratio is called temperature coefficient.
There are two reasons for increasing rate of reaction with increasing temperature.

(a) Increase in temperature increases average kinetic energy of reactant molecules. Hence, rate of collision increases.
(b) With increase in temperature number of molecules having threshold energy also increases i.e. number of active molecules increases. As a result, number of effective collisions increases. Hence, rate of reaction increases.

(iv) Concentration : Rate of reaction also depends on concentration of reactants.
Rate = k × C^”, where n = order of reaction, C = concentration of reactant.

(v) Presence of catalyst : Rate of reaction also depends on presence of catalyst. Catalyst increases rate of reaction by any of the following ways:
(a) Increasing surface area of reaction.
(b) Adsorbing the reactants on its surface and thus increasing chance of collision.
(c) By forming unstable intermediate with the substrate.
(d) By providing alternate path of lower activation energy.

Question 6.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

1. doubled
2. reduced to half ?

Solution:

Question 7.
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Solution:
The rate constant increases with increase in temperature and becomes almost double for every 10° increase in temperature. Swedish chemist, Arrhenius derived a quantitative relation between rate of reaction and temperature. According to Arrhenius,

Question 8.
In a pseudo first order hydrolysis of ester in water, the following results were obtained :

1. Calculate the average rate of reaction between the time interval 30 to 60 seconds.
2. Calculate the pseudo first order rate constant for the hydrolysis of ester.

Solution:

Question 9.
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 concentration of 6 three times?
3. How is the rate affected when the concentrations of both A and B are doubled?

Solution:
(i) Reaction is first order in A and second order in B, hence differential rate equation is

Question 10.
In a reaction between A and B, the initial rate of reaction (r₀) 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:

Question 11.
The following results have been obtained during the kinetic studies of the reaction :
2 A + B → C + D

Determine the rate law and the rate constant for the reaction.
Solution:

Question 12.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table :

Solution:

Question 13.
Calculate the half-life of the first order reaction from their rate constants given below:

1. 200 s^-1
2. 2 min^-1
3. 4 year^-1

Solution:

Question 14.
The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.
Solution:
Radioactive decay follows first order kinetics. Therefore,

Question 15.
The experimental data for the decomposition of N₂O₅
[2N₂O₅ → 4NO₂ + O₂]
in gas phase at 318K are given below :

1. Plot [N₂O₅] against t.
2. Find the half-life period for the reaction.
3. Draw a graph between log [N₂O₅] and t.
4. What is the rate law?
5. Calculate the rate constant.
6. Calculate the half-life period from k and compare it with (ii).

Solution:

Question 16.
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:

Question 17.
During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If 1 µg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically ?
Solution:

Question 18.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution:

Question 19.
A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.
Solution:

Question 20.
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Calculate the rate constant
Solution:

 Question 21.
The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.

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

Question 22.
The rate constant for the decomposition of N₂O₅ at various temperatures is given below :

Draw a graph between In k and 1/7 and calculate the value of A and Ea. Predict the rate constant at 30°C and 50°C.
Solution:
The values of rate constants for the decomposition of N₂O₅ at various temperatures are given below :

Question 23.
The rate constant for the decomposition of a hydrocarbon 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:

Question 24.
Consider a certain reaction A → Products with k = 2.0 × 10^-2 s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.
Solution:

Question 25.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?
Solution:
Sucrose decomposes according to first order rate law, hence

Question 26.
The decomposition of a hydrocarbon follows the equation

Solution:

Question 27.
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 Ea for this reaction and at what temperature will its half-period be 256 minutes?
Solution:

Question 28.
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:

Question 29.
The time required for 10% completion of the 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, calculated at 318 K and E_a.
Solution:

Question 30.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution:

INTEXT Questions

Question 1.
How would you determine the standard electrode potential of the system Mg²⁺ | Mg?
Solution:
Set up an electrochemical cell consisting of MglMgSO₄(1 M) as one electrode by dipping a magnesium wire in 1 M MgSO₄ solution and standard hydrogen electrode Pt, H₂ (1 atm) | H⁺(1 M) as the second electrode. Measure the EMF of the cell and also note the direction of deflection in the voltmeter. The direction of deflection shows that the electrons flow from magnesium electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode.

Question 2.
Can you store copper sulphate solution in a zinc pot ?
Solution:

Question 3.
Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Solution:
Oxidation of ferrous ions means :

Only those substances can oxidise Fe²⁺ to Fe³⁺ which are stronger oxidising agents and have positive reduction potentials greater than 0.77 V so that EMF of the cell reaction is positive. This is for elements lying below Fe³⁺/Fe²⁺ in the electrochemical series, for example, Br₂, Cl₂ and F₂.

Question 4.
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Solution:

Question 5.
Calculate the emf of the cell in which the following reaction takes place :
Ni_(S) + 2Ag⁺ (0.002 M) → Ni²⁺(0.160 M) + 2Ag_(S)
Given that E°cell = 1.05 V
Solution:

Question 6.
The cell in which the following reaction occurs: 2Fe³⁺_(aq) + 2l^–_(aq) → 2Fe²⁺_(aq) + l_2(s) has E°cell = 0.236 V at 298 K. Calculate the standard Gibb’s energy and the equilibrium constant of the cell reaction.
Solution:

Question 7.
Why does the conductivity of a solution decrease with dilution?
Solution:
Conductivity of solution decreases with dilution because number of ions per unit volume decreases.

Question 8.
Suggest a way to determine the A°m value of water.
Solution:
Water is a weak electrolyte. Its A°m value can be determined with the help of Kohlrausch’s law.

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 λ (H⁺) = 349.6 S cm² mol^-1 and λ (HCOO^–) = 54.6 S cm² mol^-1
Solution:

Question 10.
lf a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow though the wire?
Solution:

Question 11.
Suggest; a list of metals that are extracted electrolytically.
Solution:
Ca, Na, K, Al are extracted electrolytically.

Question 12.
Consider the reaction:
Cr₂O^2-₇ + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O,
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂ O^2-₇?
Solution:

Question 13.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Solution:
Chemical reactions while recharging :
2PbSO₄ + 2H₂O → PbO₂ + Pb + 2H₂SO₄
Electricity is passed through the electrolyte PbSO₄ which is converted into PbO₂ and Pb.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Solution:
CH₄ and CO can be used in fuel cell instead of hydrogen.

Question 15.
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Solution:
The following reactions take place at the surface of iron metal which acts as an electrochemical cell.

The water layer present on the surface of iron dissolves acidic oxides of air like C0₂ to form acids which dissociate to give H⁺ ions. Fe starts losing electrons in presence of H⁺ ions.
H₂O + CO₂ → H₂CO₃  2H⁺ + C0₃^2-

NCERT Exercises

Question 1.
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Solution:
Mg, Al, Zn, Fe, Cu

Question 2.
Given the standard electrode potentials,
K⁺/ K = – 2.93 V, Ag⁺/ Ag = 0.80 V, Hg²⁺/ Hg = 0.79 V, Mg²⁺/ Mg = -2.37 V, Cr³⁺/ Cr = – 0.74 V
Arrange these metals in their increasing order of reducing power.
Solution:
Higher the oxidation potential, more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be
Ag < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction Zn_(S) + 2Ag⁺_(aq) Zn²⁺_(aq), + 2Ag_(s) takes place. Further show :

1. Which of the electrode is negatively charged?
2. The carriers of the current i n the cell.
3. Individual reaction at each electrode.

Solution:

Question 4.
Calculate the standard cell potentials of galvanic cell in which the following reactions take place :

Solution:

Question 5.
Write the Nernst equation and emf of the following cells at 298 K :

Solution:

Question 6.
In the button cells widely used in watches and other devices the following reaction takes place :

Solution:

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Solution:
The reciprocal of resistivity is known as specific conductance or simply conductivity. It is denoted by K (kappa). Thus, if K is the specific conductance and G is the conductance of the solution, then

Now, if I = 1 cm and A = lsq.cm, then K = G.

Hence, conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross¬section. Alternatively, it may be defined as conductance of one centimetre cube of the solution of the electrolyte.

Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from 1 mole of the electrolyte dissolved in V cm³ of the solution when the electrodes are one cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by ∆_m.

Variation of conductivity and molar conductivity with concentration: Conductivity always decreases with decrease in concentration, for both weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity increases with decrease in concentration. This is because that total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume.

Question 8.
The conductivity of 0.20 M solution of KCI at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity.
Solution:

Question 9.
The resistance of a conductivity cell containing 0.001 M KCI solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCI solution at 298 K is 0.146 × 10^-3 S cm^-1.
Solution:
Cell constant = K × R = 0.146 × 10^-3 × 1500 = 0.219 cm^-1

Question 10.
The conductivity of sodium chloride at 298K has been determined at different concentrations and the results are given below:
Concentration/M 0.001 0.010 0.020 0.050 0.100
10² × K/S m^-1 1.237 11.85 23.15 55.53 106.74
Calculate ∆_m for all concentrations and draw a plot between ∆_m and c^1/2. Find the value of ∆°_m
Solution:

Question 11.
Conductivity of 0.00241 M acetic acid is 7.896 × 10^-5 S cm^-1. Calculate its molar conductivity. If ∆°_m for acetic acid is 390.5 S cm² mol^-1, what is its dissociation constant?
Solution:

Question 12.
How much charge is required for the following reductions:

1. 1 mol of Al³⁺to Al ?
2. 1 mol of Cu²⁺ to Cu ?
3. 1 mol of MnO₄^– to Mn²⁺ ?

Solution:

Question 13.
How much electricity in terms of Faraday is required to produce

1. 20.0 g of Ca from molten CaCl₂?
2. 40.0 g of Al from molten Al₂O₃?

Solution:

Question 14.
How much electricity is required in coulomb for the oxidation of

1. 1 mol of H₂O to O₂?
2. 1 mol of FeO to Fe₂O₃?

Solution:

Question 15.
A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
(At. mass of Ni = 58.7)
Solution:

Question 16.
Three electrolytic cells A, B, C containing solutions of ZnSO₄, AgNO₃ and CuSO₄ respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
(At. wt. of Ag = 108, Cu = 63.5, Zn = 65.3)
Solution:

Question 17.
Using the standard electrode potentials predict if the reaction between the following is feasible :

Solution:

Question 18.
Predict the product of electrolysis in each of the following:

1. An aqueous solution of AgNO₃ with silver electrodes.
2. An aqueous solution of AgNO₃ with platinum electrodes.
3. A dilute solution of H₂SO₄ with platinum electrodes.
4. An aqueous solution of CuCI₂ with platinum electrodes.

Solution: