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Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements

Very Short Answer Questions

Question 1.
What inspired Bartlett for carrying out reaction between Xe and PtF₆ ?
Answer:

1. At first Bartlett prepared a red compound [latex]\mathrm{O}_2^{+}+\mathrm{PtF}_6^{-}[/latex]
2. As the first Ionisation Enthalpy of molecular oxygen is identical with that the Xe.
3. He made efforts to prepare same type of compound with Xe and was successful in preparing another red colour compound Xe⁺ [latex]\mathrm{PtF}_6^{-}[/latex] by mixing PtF₆ and Xe.

Question 2.
Which of the following does not exist ?
a) XeOF₄
b) NeF₂
c) XeF₂
d) XeF₆
Answer:
Given compounds are XeOF₄, NeF₂, XeF₂, XeF₆.
NeF₂ does not exist among the above compounds.

Reason : Due to small size and high I.E. of ‘Ne’. It does not form chemical compounds.

Question 3.
Why do noble gases have comparatively large atomic sizes ?
Answer:
Noble gases have comparatively large atomic sizes as they have vander waals radii which is larger than both the ionic and covalent radii.

Question 4.
List out the uses of Neon. [A.P. Mar. 18]
Answer:
Uses of Ne:

1. Ne is used in discharge tubes and fluor escent bulbs for advertisement display purpose.
2. ‘Ne’ – bulbs are used in botanical gardens and in green houses.

Question 5.
Write any two uses of argon.
Answer:
Uses of Ar :

1. ‘Ar’ is used to create inert atmosphere in high temperature net allurgical process.
2. ‘Ar’ is ued in filling electric bulbs.

Question 6.
In modern diving apparatus, a mixture of He and O₂ is used – Why ? [A.P. Mar. 16]
Answer:
In modem diving apparatus, a mixture of He and O₂ is used because He is very low soluble in blood.

Question 7.
Helium is heavier than hydrogen. Yet helium is used (instead of H₂) in filling baloons for meteorological observations – Why ?
Answer:
‘He’ is a non-inflammable and light gas. Hence it is used in filling baloons for meterological observations.

Question 8.
How is XeO₃ prepared ?
Answer:
XeF₆ on hydrolysis produce XeO₃
XeF₆ + 3H₂O → XeO₃ + 6HF

Question 9.
Give the preparation of
a) XeOF₄ and
b) XeOaF₂
Answer:
Partial hydrolysis of XeF₆ gives oxy fluorides XeOF₄ and XeO₂F₂
XeF₆ + H₂O → XeOF₄ + 2HF
XeF₆ + 2H₂O → XeO₂F₂ + 4HF

Question 10.
Explain the structure of XeO₃. [T.S. Mar. 16]
Answer:
Structure of XeO₃

1. Central atom is ‘Xe’
2. ‘Xe’ under goes sp³ hybridisation in 3^rd excited state.
   
3. ‘Xe’ forms 3σ-bonds and 3π-bonds with three oxygens.
4. Shape of molecule is pyramidal with bond angle 103°.

Question 11.
Noble gases are inert – explain.
Answer:
Noble gases are chemically inert:

1. Noble gases have stable electronic configuration coctet configuration except He)
2. Noble gas have high Ionisation energy values and have large positive values of electron gain Enthalpy.

Question 12.
Write the name and formula of the first noble gas compound prepared by Bertlett.
Answer:
The first noble gas compound prepared by Bertlett is XePtF6. Name of the compound is xenon hexa fluoro platinate.

Question 13.
ExplaIn the shape of XeF₄ on the basis of VSEPR theory
Answer:
Shape of XeF₄:

1. Central atom in XeF₄ is ‘Xe’.
2. Xe undergoes sp³d² hybridisation in it’s 2^nd excited. state.
   
3. Shape of the molecule is square planar with bond angle 90° and bond length 1 .95Å.
   
4. Xe – forms four o-bonds by the overlap of sp³d² – 2p_z(F) orbitals.

Question 14.
Give the outer electronic configuration of noble gases.
Answer:
The outer electronic configuration of noble gases is ns²np⁶ (except He (1s²)

Question 15.
Why do noble gases form compounds with fluorine and oxygen only ?
Answer:
Noble gases form compounds with flourine and oxygen only.
Reason : Oxygen and Fluorine are most electronegative elements.

Question 16.
How is XeOF₄ prepared ? Describe its molecular shape.
Answer:
Partial hydrolysis of XeF₆ gives XeOF₄

XeF₆ + H₂O → XeOF₄ + 2HF
XeOF₄ is a colourless volatile liquid it has a square pyramidal shape.

Question 17.
What is the major source of helium ?
Answer:
The major source of helium is natural gas.

Question 18.
Which noble gas is radioactive ? How is it formed ?
Answer:
Radon (Rn) is radio active noble gas. Radon is obtained as decay product of ₈₆Ra²²⁶
₈₆Ra²²⁶ → ₈₆Rn²²² + ₂He⁴

Question 19.
Name the following :
a) most abundant noble gas in atmosphere
b) radioactive noble gas
c) noble gas with least boiling point
d) noble gas forming large number of compounds
e) noble gas not present in atmosphere
Answer:
a) Argon is the most abundant noble gas in atmosphere.
b) Radon is the radio active noble gas.
c) Helium has lowest boiling point (4.2K)
d) Xenon forms large number of compounds.
e) Radon is not present in atmosphere.

Short Answer Questions

Question 1.
How are Xenon fluorides XeF₂, XeF₄ and XeF₆ obtained?
Answer:
Xenon forms the binary fluorides XeF₂, XeF₄, XeF₆ as follows. These are formed by direct combination of Xe and F₂.

Question 2.
How are XeO₃ and XeOF₄ prepared? [Mar. 14]
Answer:
XeF₆ on hydrolysis produce XeO₃
XeF₆ + 3H₂O → XeO₃ + 6HF
Partial hydrolysis of XeF₆ gives XeOF₄
XeF₆ + H₂O → XeOF₄ + 2HF

Question 3.
Give the formulae and describe the structures of a noble gas species, isoelectronic with
a) ICl₄
b) IBr₂
c) BrO^–₃
Answer:
a) ICl₄ is also electronic with XeF₄ and it has square planar shape.
b) IBr^–₂ is also electronic with XeF₂ and it has linear shape.
c) BrO^–₃ is also electronic with XeO₄ and it has tetrahedral shape.

Question 4.
Explain the reaction of the following with water.
a) XeF₂
b) XeF₄
c) XeF₆
Answer:
a) XeF₂ is hydrolysed to form Xe, HF and O₂
2XeF₂ + 2H₂O → 2Xe + 4HF + O₂

b) XeF₄ is hydrolysed to give XeO₃
6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24 HF + 3O₂

c) XeF₆ undergo hydrolysis to form XeO₃
XeF₆ + 3H₂O → XeO₃ + 6HF
XeF₆ undergo partial hydrolysis to form XeOF₄ + XeO₂F₂
XeF₆ + H₂O → XeOF₄ + 2HF
XeF₆ + 2H₂O → XeO₂F₂ + 4HF

Question 5.
Explain the structures of [A.P. Mar. 18] [Mar. 14]
a) XeF₂ and
b) XeF₄
Answer:
a) Structure of XeF₂ :

1. In XeF₂ central atom is ‘Xe’.
2. ‘Xe’ undergoes sp³d hybridisation in it’s 1^st excited state
   
3. Shape of molecule is linear
4. Xe form two σ – bonds with two fluorines (sp³ – 2p_z overlap)
   

b) Structure of XeF₄ :

1. Central atom in XeF₄ is ‘Xe’.
2. Xe undergoes sp³d² hybridisation in it’s 2^nd excited, state.
   
3. Shape of the molecule is square planar with bond angle 90° and bond length 1.95Å.
   
4. Xe – forms four a – bonds by the overlap of sp³d² – 2p_z(F) orbitals.

Question 6.
Explain the structures of
a) XeF₆ and
b) XeOF₄
Answer:
a) Structure of XeF₆ :

1. Central atom in XeF₆ is ‘Xe’
2. Xe under goes sp³d³ hybridisation in it’s 3^rd excited state.
   
3. Shape of molecule is distorted octahedral.
   

b) Structure of α XeOF₄

1. In XeOF₄ molecule ‘Xe’ under goes sp³d² hybridisation.
2. Shape of the molecule is square pyramid.
3. There is one Xe-O double bond containing
   pπ = dπ overlaping.
   Partial hydrolysis of XeF₆ gives XeOF₄
   XeF₆ + H₂O → XeOF₄ + 2HF
   XeOF₄ is a colourless volatile liquid it has a square pyramidal shape.
   

Question 7.
Complete the following.
a) XeF₂ + H₂O →
b) XeF₂ + PF₅ →
c) XeF₄ + SbF₅ →
d) XeF₆ + ASF₅ →
e) XeF₄ + O₂F₂ →
f) NaF + XeF₆ →
Answer:
a) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂
b) XeF₂ + PF₅ → [XeF]⁺ + A[F₆]^–
c) XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^–
d) XeF₆ + ASF₅ → [Xe₂F₁₁]⁺ [ASF₆]^–
e) XeF₄ + O₂F₂ → XeF₆ + O₂
f) NaF + XeF₆ → Na⁺ [XeF₇]^–

Question 8.
How are XeF₂ and XeF₄ prepared ? Give their structures. [T.S. Mar. 18]
Answer:
Xenon forms the binary fluorides XeF₂, XeF₄, XeF₆ as follows. These are formed by direct combination of Xe and F₂.

Structure of XeF₂

1. In XeF₂ central atom is Xe’.
2. ‘Xe’ undergoes sp³d hybridisation in it’s 1^st excited state
   
3. Shape of molecule is linear.
4. Xe form two σ bonds with two fluorines.

b) Structure of XeF₄ :

1. Central atom in XeF₄ is ‘Xe’.
2. Xe undergoes sp³d² hybridisation in it’s 2^nd excited state.
   
3. Shape of the molecule is square planar with bond angle 90° and bond length 1.95Å.
   
4. Xe – forms four σ-bonds by the overlap of sp³d² – 2p_z (F) orbitals.

Long Answer Question

Question 1.
How are XeF₂, XeF₄ and XeF₆ prepared ? Explain their reaction with water. Discuss their structures. [A.P. Mar. 15]
Answer:
Xenon forms the binary fluorides XeF₂, XeF₄, XeF₆ as follows. These are formed by direct combination of Xe and F₂.

Reaction with water:

1. XeF₂ is hydrolysed to form Xe, HF and O₂
   2XeF₂ + 2H₂O → 2Xe + 4HF + O₂
2. XeF₄ is hydrolysed to give XeO₃
   6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24 HF + 3O₂
3. XeFg undergo hydrolysis to form XeO₃
   XeF₆ + 3H₂O → XeOF₃ + 6HF
4. XeF₆ undergo partial hydrolysis to form XeOF₄ and XeO₂F₂
   XeF₆ + H₂O → XeOF₄ + 2HF
   XeF₆ + 2H₃O → XeO₂F₂ + 4HF

Structure of XeF₂:

1. In XeF₂ central atom is ‘Xe’.
2. ‘Xe’ undergoes sp³d hybridisation in it’s 1^st excited state
   
3. Shape of molecule is linear.
4. Xe form two σ-bonds with two fluorines.
   

b) Structure of XeF⁴ :

1. Central atom in XeF⁴ is ‘Xe’.
2. Xe undergoes sp³d² hybridisation in it’s 2^nd excited state.
   
3. Shape of the molecule is square planar with bond angle 90° and bond length 1.95Å.
   
4. Xe – forms four σ-bonds by the overlap of sp³d² – 2p_z (F) orbitals.

Structure of XeF₆ :

1. Central atom in XeF₆ is ‘Xe’
2. Xe under goes sp³d³ hybridisation in it’s 3^rd excited state.
   
3. Shape of molecule is distorted octahedral.
   

Textual Examples

Question 1.
Why are the elements of group 18 known as noble gases ?
Solution:
The elements present in Group 18 have their valence shell orbitals completely filled and, therefore, react with a few elements only under certain conditions. Therefore, they are now known as noble gases.

Question 2.
Noble gases have very low boiling points. Why ?
Solution:
Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.

Question 3.
Does the hydrolysis of XeF₆ lead to a redox reaction ?
Solution:
No, the products of hydrolysis are XeOF₄ and XeO₂F₂ where the oxidation states of all the elements remain the same as it was in the reacting state.

Intext Questions

Question 1.
Why is helium used in diving apparatus ?
Solution:
Helium is used in diving apparatus due to its very low solubility in blood.

Question 2.
Balance the following equation :
XeF₆ + H₂O → XeO₂F₂ + HF
Solution:
XeF₆ + 2H₂O → XeO₂F₂ + 4HF

Question 3.
Why has it been difficult to study the chemistry of radon ?
Solution:
Radon is a radioactive element with very short half-life of 3.82 days. That is why, the study of chemistry of radon is a difficult task.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(c) Group-17 Elements Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 6(c) Group-17 Elements

Very Short Answer Questions

Question 1.
Which halogen produces O₂ and O₃ on passing through water?
Answer:
Fluorine produces O₂ and O₃ on passing through water.
3F₂ + 3H₂O → 6HF + O₃
2F₂ + 2H₂O → 4HF + O₂

Question 2.
Interhalogen compounds are more reactive than the constituent halogens except fluorine – explain.
Answer:
Inter halogen compounds are more reactive than halogens. This is because X – X’ bond in interhalogens is weaker than X – X bond in halogens except F – F bond.

Question 3.
What is the use of ClF₃?
Answer:
ClF₃ is very useful fluorinating agent and it is used for the production of VF₆ in the enrichment
of U²³⁵.
U_(s) + 3ClF_3(l) → UF_6(g) + 3ClF_(g)

Question 4.
Write two uses of ClO₂.
Answer:
Uses of ClO₂:

• ClO₂ is highly reactive oxidising agent.
• It is used as bleaching agent for paper pulp and textiles.
• It is used in water treatment.

Question 5.
Why are halogens coloured?
Answer:
Halogens are coloured due to the absorption of radiations in visible region which results in the excitation of outer electrons to higher energy level. Halogens absorb different quanta of radiation and display different colours.

Question 6.
Write the reactions of F₂ and Cl₂ with water. [Mar. 14]
Answer:
Fluorine produces O₂ and O₃ on passing through water.
3F₂ + 3 H₂O → 6HF + O₃
2F₂ +2 H₂O → 4HF + O₂

Chlorine dissolves in water giving a solution of chlorine water. A freshly prepared solution of chlorine water contains HCl and HOCl .
Cl₂ + H₂O → HCl + HOCl
HOCl is unstable and dissociates to give nascent oxygen
HOCl → HCl + (O)

Question 7.
With which neutral molecule, ClO^– is isoelectronic ? Is that molecule a Lewis base ? (Hint: ClF; Yes)
Answer:

• ClO^– is isoelectronic with the neutral molecule ClF.
• Yes. ClF is a Lewis base (Electron pair donar).

Question 8.
Arrange the following in the order of the property indicated for each set.
a) F₂, Cl₂, Br₂, I₂ – increasing bond dissociation enthalpy.
b) HF, HCl, HBr, HI – increasing acidic strength
c) HF, HCl, HBr, HI – increasing boiling points.
Answer:
a) Increasing bond’ dissociation enthalpy order
I₂ < F₂ < Br₂ < Cl₂

b) Increasing acidic strength order
HF < HCl < HBr < HI

c) Increasing boiling points order
HCl < HBr < HI < HF

Question 9.
Electron gain enthalpy of fluorine is less than that of chlorine – explain.
Answer:
The negative electron gain enthalpy of the fluorine is less them that of chlorine. This is due to small size of fluorine atom which results in strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and thus the incoming electron does not experience much attraction.

Question 10.
HF is a liquid while HCl is a gas – explain.
Answer:
HF is a liquid due to the presence of inter molecular hydrogen bonding where as HCl is a gas due to there is no such type of bonding in it.

Question 11.
Bond dissociation enthalpy of F₂ is less than that of Cl₂ – explain.
Ans. Bond dissociation enthalpy of F₂ is less than that of Cl₂
Explanation:
In F₂ molecule electron repulsions are greater among lone pairs because these lone pairs are much closer to each other than in case of Cl₂.

Question 12.
Write the formulae of the compounds, in which oxygen has positive oxidation states and mention the oxidation states of oxygen in them.
Answer:

• In OF₂ and O₂F₂ oxygen has positive oxidation states.
• In OF₂, oxygen oxidation state is + 2.
• In O₂F₂ oxygen oxidation state is + 1.

Question 13.
What is the use of O₂F₂ and I₂O₅?
Answer:
Use of O₂F₅:
O₂F₂ is a fluorinating agent. O₂F₂ oxidises plutonium to PUF₆ and the reaction is used in removing plutonium as PUF₆ from spent nuclear fuel.
Use of I₂O₅:
I₂O₅ is a good oxidising agent and is used in the estimation of carbon monoxide (CO).

Question 14.
Write two uses of hydrogen chloride.
Answer:
Uses of hydrogen chloride :

• It is used in medicines and as a laboratory reagent.
• It is used in the manufacturing of Cl₂, NH₄Cl and glucose (from com starch).
• It is used in extracting glue from bones-and purifying bone black.

Question 15.
Explain the reactions of Cl₂ with NaOH. [T.S. Mar. 16]
Answer:

1. Reaction with cold dilute NaOH: Chlorine reacts with cold dilute NaOH to give sodium hypochlorite and sodium chloride.
   2 NaOH (cold, dil.) + Cl₂ → NaCl + NaOCl + H₂O
2. Reaction with hot concentrated NaOH : Chlorine reacts with hot concentrated NaOH to give sodium chlorate and sodium chloride.
   3 Cl₂ + 6 NaOH (hot, Conc.) → 5 NaCl + NaClO₃ + 3 H₂O

Question 16.
What happens when Cl₂ reacts with dry slaked lime ? [A.P. Mar. 16, 15]
Answer:
Chlorine reacts with dry slaked lime and forms bleaching powder.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Question 17.
Chlorine acts as an oxidizing agent – explain with two examples.
Answer:
Chlorine acts as oxidising agent.
Example – 1: Cl₂ oxidises Iodine to Iodate .
I₂ + 6 H₂O + 5 Cl₂ → 2HIO₃ + 10 HCl
Example – 2 : Cl₂ oxidises Sodium Sulphite to Sodium Sulphate
Cl₂ + Na₂SO₃ + H₂O → Na₂SO₄ + 2 HCl

Question 18.
What is aqua regia ? Write its reaction with gold and platinum.
Answer:
A mixture of 3 parts of Cone. HCl and one part of Cone. HNO₃ constitutes aqua regia. It is used for dissolving noble metals.
It’s reaction with gold :
Au + 4H⁺ + [latex]\mathrm{NO}_3^{-}[/latex] + 4Cl^– → [latex]\mathrm{AuCl}_4^{-}[/latex] + NO + 2 H₂O
It’s reaction with Platinum
3Pt + 16H⁺ + 4[latex]\mathrm{NO}_3^{-}[/latex] + 18 Cl^– → [latex]3 \mathrm{PtCl}_6^{-2}[/latex] + 4 NO + 8H₂O

Question 19.
How is chlorine manufactured by Deacon’s method ? [T.S. & A.P. Mar. 19 & T.S. Mar. 16]
Answer:
Deacon’s process : In Deacon’s process chlorine is obtained by the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl₂ (catalyst) at 723 K.

Question 20.
Chlorine acts as a bleaching agent only in the presence of moisture – explain.
Answer:
Moist chlorine is a powerful bleaching agent. This bleaching property is due to oxidation.
Cl₂ + H₂O → 2HCl + (O)
Ex : Coloured substance + (O) → colourless substance.

Question 21.
The decreasing order of acidic character among hypohalogen acids is HClO > HBrO >HIO. Give reason.
Answer:
Given the decreasing order of acidic character among hypohalogen acids is

HaO > HBrO > HIO
From the above mentioned K_a values the given order of hypohalogen acids – acid strength order is
HClO > HBrO > HIO

Question 22.
The acidic nature of the oxoacids of chlorine is
HOCl < HClO₂ < HClO₃ < HClO₄ – explain.
(Hint: HA + H₂O ⇌ H₃O⁺ + A^– conjugate base, greater the stability of A^–, lesser will be its basic strength or greater will be the tendency of HA to release H⁺. In other words, stronger will be the acid HA. Among the conjugate bases of oxoacids of chlorine, the stability order is HOCl^– < [latex]\mathrm{ClO}_2^{-}[/latex] > [latex]\mathrm{ClO}_3^{-}[/latex] > [latex]\mathrm{ClO}_4^{-}[/latex])
Answer:
Given the acidic nature of oxoacids of chlorine is
HOCl^– < H[latex]\mathrm{ClO}_2^{-}[/latex] > H[latex]\mathrm{ClO}_3^{-}[/latex] > [latex]\mathrm{ClO}_4^{-}[/latex]
HA + H₂O ⇌ H₃O⁺ + A^– conjugate base, greater the stability of A^–, lesser will be its basic strength or greater will be the tendency of HA to release H⁺. In other words, stronger will be the acid HA. Among the conjugate bases of oxoacids of chlorine, the stability order is
HOCl^– < H[latex]\mathrm{ClO}_2^{-}[/latex] > H[latex]\mathrm{ClO}_3^{-}[/latex] > H[latex]\mathrm{ClO}_4^{-}[/latex]

Question 23.
What are interhalogen compounds ? Give two examples.
Answer:
The binary diamagnetic compounds of halogens which are formed by the reaction of halogens among themselves are called interhalogen compounds.
E.g.: IF₇, ClF₃, BrF₃, ClF, IF₃ etc.

Question 24.
Explain the structure of ClF₃.
Answer:
Structure of ClF₃:

• Central atom in ClF₃ is ‘Cl’
• Excited state electronic configuration of ‘Cl’ is
  
• Cl atom undergoes sp³d hybridisation
• It is a bent T-shaped molecule (or) trigonal bipyramidal with 2 – positions occupied by lone pairs.

Question 25.
OF₂ should be called oxygen difluolide and not fluorine oxide – Why ?
Answer:
OF₂ should be called oxygen difluoride and riot fluoride oxide.
The binary compotmds of oxygen with fluoride are not pronounced as oxides because the electronegativity of fluorine is greater than oxygen.

Question 26.
Iodine is more soluble in KI than in water – Explain.
Answer:
Iodine is more soluble in KI than in water.

Reason:
Iodine combine with KI forming soluble complex KI₃.
KI + I₂ → KI₃ (Soluble complex)
Iodine does not dissolve in water. This is due to positive free energy change (+ ∆G)

Question 27.
Among the hydrides of halogens
a) Which is most stable ?
b) Which is most acidic ?
c) Which has lowest boiling point ?
Answer:
a) Among hydrides of halogens HF is most stable.
b) Among hydrides of halogens HI is most acidic.
c) Among hydrides of halogens HCl (189K) has lowest boiling point.

Question 28.
Compare the bleaching action of Cl₂ and SO₂.
Answer:

• Cl₂ is a powerful bleaching agent. This bleaching action is due to oxidation.
  Cl₂ + H₂O → 2HCl+ (O)
  Coloured substance + (O) → colourless substance.
• It bleaches vegetable or organic matter in the presence of moisture. Bleaching effect of Cl₂ is permanent.
• SO₂ acts as bleaching agent in presence of moisture.
  SO₂ + 2 H₂O → H₂SO₄ + 2[H]
  
• SO₂ bleaches wool and silk.

Question 29.
Give the oxidation states of halogens in the following :
a) Cl₂O
b) [latex]\mathrm{ClO}_2^{-}[/latex]
c) KBrO₃
d) NaClO₄
Answer:
a) Cl₂O :
2x – 2 = 0
x = -+ 1
Oxidation state of chlorine in Cl₂O is + 1.

b) [latex]\mathrm{ClO}_2^{-}[/latex]
x + 2(-2) = -1
x = -1 + 4 = +3
In latex]\mathrm{ClO}_2^{-}[/latex] the oxidation state of ‘Cl’ is + 3

c) KBrO₃ :
1 + x + 3 (-2) = 0
x = + 5
In KBrO₃ the oxidation state of Br is + 5.

d) NaClO₄
1 + x + 4(-2) = 0
x = + 7
In NaClO₄ the oxidation state of Cl is + 7.

Question 30.
Describe the molecular shape of [latex]\mathrm{I}_3^{-}[/latex] .
(Hint: Central iodine is of sp³d. – linear)
Answer:

• In Tri iodide ion central iodine atom undergo sp³d hybridisation.
• It contains three lone pairs and two bond pairs.
• According to VSEPR theory it has linear shape.
  

Short Answer Questions

Question 1.
How can you prepare Cl₂ from HCl and HCl from Cl₂ ? Write the reactions.
Answer:
Preparation of Cl₂ from HCl:

• On heating MnO₂ with Conc. HCl, Cl₂ gas is liberated
  MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
• By the Õxidation of HCl gas by atmospheric oxygen in presence of CuCl₂ catalyst at 723 K.
  4 HCl + O₂ 2Cl₂ + 2 H₂O

Preparation of HCl from Cl₂:

• Cl₂ when reacted with H₂ to form HCl
  H_2(g) + Cl_2(g) → 2HCl_(g)

Question 2.
Write balanced equations for the following.
a) NaCl is heated with Conc.H₂SO₄ In the presence of MnO₂.
b) Chlorine is passed into a solution of Nal ¡n water.
Answer:
a) NaCl is heated with Conc.H₂SO₄ in presence of MnO₂ liberates Cl₂ gas.
4 NaCl + MnO₂ + 4 H₂SO₄ → MnCl₂ + 4 NaHSO₄ + 2 H₂O + Cl₂.

b) When chlorine water is added to a solution of sodium iodide a brown colour is formed.
Cl₂ + 2 NaI → 2 NaCl + I₂

Question 3.
Explain the structures of
a) BrF₅ and
b) IF₂.
Answer:
a) Structure of BrF₅.

• Central atom in BrF₅ is Br.
• ‘Br’ undergoes sp³d² hybridisation in 2^nd excited state.
  
• Shape of the molecule is octahedral with one position occupied by a lone pair (or) square pyramidal.

b) Structure of IF₇ :

• Central atom in IF₇ is T.
• T undergoes sp³d³ hybridisation in 3^rd excited state.
  
• Shape of the molecule is Penta gonal bipyramid structure
  

Question 4.
Write a short note on the hydrides of halogens.
Answer:
Hydrides of halogens :
Formation :

These compounds dissolve in water to form hydrohalic acids.
Boiling points HF – 293 K
HCl – 189K
HBr – 206 K
HI – 238 K
The decreasing order of stability of these compounds is
HF > HCl > HBr > HI
The increasing order of acidic strength of these compounds
HF < HCl < HBr < HI

Question 5.
How is chlorine obtained in the laboratory ? How does it react with the following ? [T.S. Mar. 15]
a) cold dil. NaOH
b) excess NH₃
c) KI
Answer:
In the laboratory chlorine is prepared by the oxidation of HCl with MnO₂.
4 HCl + MnO₂ → MnCl₂ + 2 H₂O + Cl₂ ↑

a) Chlorine reacts with cold dil. NaOH to form sodium hypochlorite.
Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O

b) Cl₂ reacts with excess of NH3, Nitrogen and ammonium chloride are formed.
8 NH₃ + 3 Cl₂ → 6 NH₄Cl + N₂ ↑

c) Cl₂ reacts with KI to liberate iodine
Cl₂ + 2KI → 2 KCl + I₂ ↑

Question 6.
What are interhalogen compounds ? Give some examples to illustrate the definition. How are they classified ?
Answer:
The binary diamagnetic compounds of halogens which are formed by the reaction of halogens among themselves are called interhalogen compounds.
E.g.: IF₇, ClF₃, BrF₃, ClF, IF₃ etc.
The above examples are binary diamagnetic compounds and formed by* combination of halogens only. .
Inter halogen compounds are classified into four types.

1. AX – Type : Eg : ClF, BrF
2. AX₃ – Type : Eg: ClF₃, IF₃
3. AX₅ – Type : Eg : ClF₅, BrF₅
4. AX₇ – Type : Eg : IF₇

‘A’ is less electronegative halogen.
X is more electronegative halogens.

Long Answer Questions

Question 1.
How is ClF₃ prepared ? How does it react with water ? Explain its structure.
Answer:
Preparation of ClF₃: Chlorine reacts with excess of fluorine to form ClF₃.

Reaction with H₂O: ClF₃ reacts with water explosively and oxidises water to give oxygen or in controlled quantities oxygen diflouride (OF₂) as well as HF and HCl.
ClF₃ + 2H₂O → 3 HF + HCl + O₂
ClF₃ + H₂O → 4 HF + HCl + OF₂
Structure of ClF₃:

• Central atom in ClF₃ is ‘Cl’
• Excited state electronic configuration of ‘Cl’ is
  
• Cl atom undergoes sp³d hybridisation
• It is a bent T-shaped molecule (or) trigonal bipyramidal with 2 – positions occupied by lone pairs.

Question 2.
How is chlorine prepared in the laboratory ? How does it react with the following ? [T.S. Mar. 19, 15]
a) Iron
b) hot, cone. NaOH
c) acidified FeSO₄
d) Iodine
e) H₂S
f) Na₂S₂O₃
Answer:
In the laboratory chlorine is prepared by the oxidation of HCl with MnO₂.
4 HCl + MnO₂ → MnCl₂ + 2 H₂O + Cl₂ ↑

a) Cl₂ reacts with Iron to form FeCl₃
2 Fe + 3Cl₂ → 2 FeCl₃

b) Cl₂ reacts with hot Cone. NaOH to form sodium chloride and sodium chlorate
3Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O

c) Cl₂ reacts with acidified FeSO₄ and ferric ions are formed.
2 FeSO₄ + H₂SO₄ + Cl₂ → Fe₂(SO₄)₃ + 2 HCl

d) Cl₂ reacts with Iodine to form ICl

e) Cl₂ reacts with H₂S and forms HCl and ‘S’
Cl₂ + H₂S → 2 HCl + S

f) ‘S’ is precipitated by the reaction of Cl₂ with Na₂S₂O₃.
‘Na₂S₂O₃ + Cl₂ + H₂O → Na₂SO₄ + 2 HCl + S

Question 3.
Discuss the anomalous behaviour of fluorine.
Answer:
Abnormal behaviour of fluorine : Fluorine being the first element in the group differs considerably from other halogens. The reasons may be traced out to one or more of the following, characteristics.
F₂ has

1. a small size
2. The highest electronegativity amongst the known elements.
3. No d-orbital are available in its valence shell.
4. Low enthalpy of F-F bond dissociation and
5. Only two electrons are present in the penultimate shell while other halogens 8 electrons.

The abnormal characteristics of F₂ can be summarised as follows.

1. F₂ exhibits -1 oxidation state because it is the most electronegative element known. Therefore, no higher oxidation state for it in its compounds.
2. In it’s hydride HF, it has hydrogen bonding, but other halides of hydrogen do not show this property. HF can form [latex]\mathrm{HF}_2^{-}[/latex] ion. No such ions are known with other halogens.
3. It combines with carbon while others do not combine even under drastic conditions.
4. F₂ has a lower EA compared to that of Cl₂ even though F₂ is the most electronegative element.
5. Fluorides have the highest ionic character among the halides.
   Example : AlF₂ is ionic while AlCl₃ is covalent compound.

Question 4.
How is chlorine prepared by electrolytic method ? Explain its reaction with [A.P. Mar. 16, 15]
a) NaOH and
b) NH₃ under different conditions.
Answer:
Preparation of chlorine by electrolytic method : Chlorine is obtained by the electrolysis of brine solutions (Cone. NaCl). Cl₂ gas is liberated at anode.
2 NaCl → 2 Na⁺ + 2Cl^–
2 H₂O + 2e^– → 2 OH^– + H₂ (cathode)
2 Cl^– → Cl₂ + 2e^– (anode)
a)

1. NaOH:
   Chlorine reacts with cold dil. NaOH to form sodium hypochlorite.
   Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O
2. Cl₂ reacts with hot cone. NaOH to form sodium chloride and sodium chlorate.
   3 Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O

b)

1. Cl₂ reacts with excess of NH₃, Nitrogen and ammonium chlorate are formed.
   8 NH₃ + 3 Cl₂ → 6 NH₄Cl + N₂ ↑
2. NH₃ reacts with excess of Cl₂ to form NCl₃ and HCl.
   

Question 5.
Write the names and formulae of the oxoacids of chlorine. Explain their structures and relative acidic nature.
Answer:
Four oxyacids of chlorine are known. They are
Hypochlorous acid – HOCl
Chlorous acid – HClO₂
Chloric acid – HClO₃
Perchloric acid – HClO₄
Structure of HClO : In this chlorine atom is sp³ hybridised. Outer electronic configuration of Cl in ClO^– after sp³ hybridisation.

Shape is tetrahedral with 3 lone pairs (or) linear.
No π – bonds.

Chlorous acid : (HClO₂) : Chlorine is in sp³ hybrid state, in first excited state. Shape is tetrahedral with 2 lone pairs (or) angular one π_d-p bond is present.
First excited state

Chloric acid (HClO₃) : The central chlorine atom undergoes sp³ hybridisation in second excited state.

Shape is tetrahedral with one lone pair (or) pyramidal.
Two π_d-p bonds present.

Perchloric acid (HClO₄) : The central chlorine atom undergoes sp³ hybridisation in third excited state.

Shape is perfect tetrahedral. No lone pairs.
Three π_d-p bonds present.

Textual Examples

Question 1.
Halogens have maximum negative electron gain enthalpy in the respective periods of the periodic table. Why ?
Solution:
Halogens have the smallest size in their respective periods and therefore high effective nuclear charge. As a consequence, they readily accept one electron to acquire noble gas electronic configuration.

Question 2.
Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent than chlorine. Why ?
Solution:
It is due to

1. Low enthalpy of dissociation of F-F bond
2. High hydration enthalpy of F^–

Question 3.
Fluorine exhibits only – 1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Explain.
Solution:
Fluorine is the most electronegative element and cannot exhibit any positive oxidation .state. Other halogens have d orbitals and therefore, can expand their octets and show +1, +3, +5 and +7 oxidation states also.

Question 4.
Write the balanced chemical equation for the reaction of Cl₂ with hot and concentrated NaOH. Is this reaction a disproportionation reaction ? Justify.
Solution:
3 Cl₂ + 6 NaOH → 5NaCl + NaClO₃ + 3H₂O
Yes. Chlorine from zero oxidation state is changed to -1 and +5 oxidation states.

Question 5.
When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why ?
Solution:
Its reaction with iron produces H₂.
Fe + 2HCl → FeC₂ + H₂
Liberation of hydrogen prevents the formation of ferric chloride.

Question 6.
Discuss the molecular shape of BrF₃ on the basis of VSEPR theory.
Solution:
The central atom Br has seven electrons in the valence shell. Three of these will form electron pair bonds with three fluorine atoms leaving behind four electrons. Thus, there are three bond pairs and two lone pairs.

According to VSEPR theory, these will occupy the comers of a trigonal bipyramid. The two lone pairs will occupy the equatorial positions to minimise lone pair- lone pair and the bond pair-lone pair repulsions which are greater than the bond pair-bond pair repulsions. In addition, the axial flourine atoms will be bent towards the equatorial fluorine in order to minimise the lone pair-lone pair repulsions. The shape would be that of a slightly bent T.

Intext Questions

Question 1.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.
Oxidizing power is a combined effect of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
Solution:
Comparing F₂ and Cl₂ with the given parameters.

From the data given above, it is clear that the bond dissociation enthalpy and electron gain enthalpy are higher for chlorine but hydration energy is much higher for fluorine. It compensates the effect of other two and thus, makes flourine more oxidizing than chlorine.

The relative oxidizing power of the halogens can be further illustrated by their reactions with water.
2 F₂(g) + 2 H₂O(l) → 4H⁺ (aq) + 4F^– (aq) + O₂(g) ……………. (i)
Cl₂(g) + H₂O(l) → HCl (aq) + HOCl (aq) ……………… (ii)

Question 2.
Give two examples to show the anomalous behaviour of fluorine.
Solution:
The two anomalous behaviour of flourine are as follows :
a) It forms only one oxo-acid while other halogens form a number of oxo-acids.
b) Hydrogen flouride (HF) is a liquid (b.p. 293 K) due to strong hydrogen bonding whereas other hydrogen halides are gases.

Question 3.
Sea is the greatest source of some halogens. Comment.
Solution:
Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium. NaCl is the most abundant (2.5% by mass) among all of these. Certain forms of marine life contain iodine in their systems; Various sea weeds, for example, contain upto . 0.5% of iodine and chile saltpetre contains upto 0.2% of sodium iodate. NaCl and camalite, KCl, MgCl₂. 6H₂O etc are present in the deposits of dried up seas.

Question 4.
Give the reason for bleaching action of Cl₂ ?
Solution:
Bleaching action of chlorine is due to its oxidizing property. When chlorine reacts with water, it gives nascent oxygen which decoloures is the coloured substance.
Cl₂ + H₂O → 2HCl + [O]
Coloured substance + [O] → Colourless substance
Bleaching action of chlorine creates permanent effect. It bleaches the vegetable or organic matter in the presence of moisture.

Question 5.
Name some poisonous gases which can be prepared from chlorine gas.
Solution:

1. Phosgene (COCl₂)
2. Tear gas (CCl₃NO₂)
3. Mustard gas (ClCH₂CH₂SCH₂CH₂Cl)

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements

Very Short Answer Questions

Question 1.
Why is dioxygen a gas but sulphur a solid?
Answer:
Dioxygen is a gas but sulphur a solid.
Explanation:

• Due to small atomic size and high electronegativity oxygen forms Pπ – Pπ multiple bond and exists as O₂ molecules held together by weak vander waal’s forces. Thus oxygen exists as a gas at room temperature.
• Due to large atomic size and less electronegativity sulphur forms strong S—S single bonds and exists as S₈ molecules with puckered ring structure. Hence sulphur is a solid at room temperature.

Question 2.
What happens when
a) KClO₃ is heated with MnO₂
b) O₃ is passed through KI soultion
Answer:
a) When KClO₃ is heated with Mn02 and liberates oxygen gas.

b) O₃ is passed through KI solution I₂ is liberated
2KI + O₃ + H₂O → 2KOH + I₂ + 2O₂

Question 3.
Give two examples each for amphoteric oxides and neutral oxides.
Answer:

• Examples of amphoterc oxides – Al₂O₃, SiO₂, PbO.
• Examples of neutral oxides – CO, NO and N₂O.

Question 4.
Oxygen generally exhibits an oxidation state of -2 only while the other members of the group show oxidation states of +2, +4 and +6 also – explain.
Answer:

• Oxygen exhibits -2 oxidation state due to its high electronegativity. The tendency of exhibiting -2 oxidation state decrease down the group.
• Due to decrease of electronegativity down the group the other elements exhibit +2, +4 and +6 oxidation states also.

Question 5.
Write any two compounds, in which oxygen shows an oxidation state different from -2. Give the oxidation states of oxygen in them.
Answer:
OF₂ and O₂ F₂ are two compounds in which oxygen shows an oxidation state different from-2.

• In OF₂ the oxidation state of oxygen is +2
• In O₂F₂ the oxidation state of oxygen is +1.

Question 6.
Oxygen molecule has the formula O₂ while sulphur has S₈ – explain.
Answer:
Due to small atomic size and high electronegativity oxygen forms Pπ – Pπ multiple bond and exists as O₂ molecule.
Due to large atomic size and less electronegativity sulphur forms strong S – S single bonds and exists S₈ molecule.

Question 7.
Why is H₂O a liquid while H₂S is a gas ?
Answer:
H₂O is liquid due to the presence of intermolecular hydrogen bonding. While H₂S is gas because it is not having such type of bonding.

Question 8.
H₂O is neutral while H₂S is acidic – explain.
Answer:
H₂O is neutral while H₂S is acidic.
Reason: The O-H bond dissociation Enthalpy is greater than the S – H bond dissociation Enthalpy.

Question 9.
Name the most abundant element present in earth’s crust.
Answer:
The most abundant element present in earth’s crust is oxygen (about 46.6%).

Question 10.
Which element of group-16 shows highest catenation ?
Answer:
Sulphus shows highest catenation among group – 16 elements and exists as S₉ molecule with puckered ring structure.

Question 11.
Among the hydrides of chalcogens, which is most acidic and which is most stable ?
Answer:

• Among hydrides of chalcogens, H₂Te is most acidic.
• Among hydrides of chalcogens, H₂O is most stable.

Question 12.
Give the hybridization of sulphur in the following.
a) SO₂
b) SO₃
c) SF₄
d)SF₆
Answer:
a) Hybridisation of’S’ in SO₂ is Sp²
b) Hybridisation of ‘S’ in SO₃ is Sp²
c) Hybridisation of ‘S’ in SF₄ is Sp³d
d) Hybridisation of ‘S’ in SF₆ is Sp³d²

Question 13.
Write the names and formulae of any two oxyacids of sulphur. Indicate the oxidation state of sulphur in them.
Answer:

• Peroxy mono sulphuric acid – H₂SO₅ ‘S’ oxidation state +6 .
• Peroxy di sulphuric acid – H₂S₂O₈ ‘S’ oxidation state +6

Question 14.
Explain the structures of SF₄ and SF₆.
Answer:
Structure of SF₄:

• It SF₄ ‘S’ undergoes sp³d hybndisation.
• It has trigonal bipyramidal structure in which one of the equitorial positions is occupied by a lone pair of electrons. This geometry is also known as see – saw geometry.

Structure of SF₆:

• In SF₆ ‘S’ undergoes sp³d² hybridisation.
• It has octahedral structure.

Question 15.
Give one example each for
a) a neutral oxide
b) a peroxide
c) a super oxide
Answer:
a) CO, N₂O are neutral oxides.
b) Na₂O₂, BaO₂ are pernxides.
c) KO₂, RbO₂ are super oxides.

Question 16.
What is tailing of mercury? How is it removed? [A.P. & T.S. (Mar. 15)]
Answer:
Mercury loses it’s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.
2Hg + O₃ → Hg₂O + O₂
It is removed by shaking it with water which dissolves Hg₂O.

Question 17.
Write the principle involved in the quantitative estimation of ozone gas.
Answer:
When ozone reacts with an excess of Kl solution buffered with a borate buffer (pH 9.2) iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. In this way O₃ is estimated quantitatively.

Question 18.
Write the structure of ozone.
Answer:
Structure of Ozone:

• O₃ is angular molecule with bond angle 117°
• O – O bond length is 128 pm.

Question 19.
SO₂ can be used as an anti-chior. Explain.
Answer:
SO₂ gas is used as anti-chlor. Anti-chior means the substance which removes the excess of on clothes. SO₂ reacts with chlorine in presence of charcoal to give suiphuryl chloride
SO_2(g) + Cl_2(g) → SO₂Cl_2l

Question 20.
How is ozone detected ?
Answer:
Ozone is a pale blue gas, dark blue liquid and violet black solid and it has characterstic smell.

• It is detected by the reaction with ‘Hg’ which is called as tailing of mercury.
  2Hg + O₃ → Hg₂O + O₂
• Ozone turns benzidene paper to brown colour.

Question 21.
How does ozone react with Ethylene ?
Answer:
Ethylene reacts with ozone to form Ethylene ozonoid followed by the hydrolysis to form formaldehyde.

Question 22.
Out of O₂ and O₃, which is paramagnetic ?
Answer:

• O₂ is paramagnetic due to presence of unpaired electrons
• O₃ (gaseous) is diamagnetic due to absence of unpaired electrons.

Question 23.
Between O₃ and O₂, ozone is a better oxidizing agent – why ?
Answer:
Ozone is next to fluorine in the oxidising capacity. It is best oxidising agent than O₂. (Fluorine is the powerful oxidising agent). Ozone liberates nascent oxygen easily.

Question 24.
Write any two uses each for O₃ and H₂SO₄.
Answer:
Uses of O₃:

• Ozone is used in sterilisation of water.
• Ozone is used in manufacture of artificial silk and camphor etc.
• Ozone is used to identify unsaturation in carbon compounds.

Uses of H₂SO₄:

• H₂SO₄ is used in the manufacture of fertilisers.
• H₂SO₄ is used in petrol refining.
• H₂SO₄ is used in detergent industry.

Question 25.
Which form of sulphur shows paramagnetism ?
Answer:
In vapour state sulphur partly exists as S₂ molecule which has two unpaired electrons in the antibonding π (π*)orbitals like O₂. Hence exhibits paramagnetism.

Question 26.
How is the presence of SO₂ detected ?
Answer:
SO₂ has a pungent odour SO₂ presence can be detected by the following tests.

1. SO₂ changes the colour of acidified potassium dichromate solution from orange to green.
   
2. SO₂ decolourises acidified KMnO₄ solution.
   

Question 27.
Why are group – 16 elements called chalcogens ?
Answer:
Chalcogens means mineral forming (or) ore forming elements. Most of elements exist in earth crust as oxides, sulphides, selinides, telurids etc. So Group – 16 elements are called as chalcogens.

Question 28.
Among chalcogens, which has highest eletronegativity and which has highest electron gain enthalpy?
Answer:

• Among chalcogens Oxygen has high electronegativity.
• Among chalcogens Sulphur has high electron gain Enthalpy.

Question 29.
Which hydride of group – 16 has highest boiling point and weakest acidic character ?
Answ:

• Among group – 16 hydrides water (H₂O) has high bioling point.
• Among group – 16 hydrides water (H₂O) has weakest acidic, character.

Short Answer Questions

Question 1.
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation states and hydride formation.
Answer:
1) Electronic configurations :
Oxygen (O) – [He] 2s² 2p²
Sulphur (S) – [Ne] 3s² 3p²
Selenium (Se) – [Ar] 3d¹⁰ 4s²4p⁴
Tellurium (Te) – [Kr] 4d¹⁰ 5s² sp⁴
Polonium (Po) – [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁴
All the above elements has general outer electronic configuration ns²np⁴.

2) Oxidation states :
All the gives elements (chalcogens) exhibits common oxidation state of-2
O^-2, S^-2, Se^-2 etc.

3) Hydride formation:
All these elements (chalcogens) forms hydrides of type EH₂(E = chalcogen)
Eg : H₂O, H₂S, H₂Se, H₂Te, H₂Po.
The above mentioned concepts evident that the elements O, S, Se, Te and Po are present in the same group of periodic table.

Question 2.
Describe the manufacture of H₂SO₄ by contact process.
Answer:
Manufacture of H₂SO₄ by contact process:
Manufacturing of H₂SO₄ involves three main steps.
Step – 1.
SO₂ production : The required SO₂ for this process is obtained by burning S(or) Iron pyrites in oxygen.
S + O₂ → SO₂
4FeS₂ + 15O₂ → 2Fe₂O₃ + 8SO₃
Step – 2
SO₃ formation: SO₂ is oxidised in presence of catalyst with atmosphric air to form SO₃

Step – 3
Formation of H₂SO₄: SO₃ formed in the above step absorbed in 98% H₂SO₄ to get oleum (H₂S₂O₇). This oleum is diluted to get desired concentration of H₂SO₄.
SO₃ + H₂SO₄ → H₂S₂O₇
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 3.
How is ozone prepared ? How does it react with the following? [Mar. 14]
a) PbS
b) KI
c) Hg
d) Ag
Answer:
Preparation of Ozone:
A slow dry stream of oxygen under silent electric discharge to form ozone (about 10%). The product obtained is known as ozonised oxygen.
3O₂ → 2O₃ ∆ H° = 142kJ/mole

• The formation of ozone is an endothermic reaction.
• It is necessary to use silent electric discharge in the preparation of O₃ to prevent its decomposition

a) Reaction with PbS : Black lead suiphide oxidised to white lead sulphate in presence of ozone.
PbS + 4O₃ → PbSO₄ + 4O₂

b) Reaction with KI: Moist Kl is oxidised to Iodine in presence of ozone.
2KI + H₂O + O₃ -→ 2KOH + I₂ + O₂

c) Reaction with Hg : Mercury loses it’s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.
2Hg + O₃ → Hg₂ O + O₂
It is removed by shaking it with water which dissolves Hg₂O.

d) Reaction with Ag : Ag metal oxidised to Ag₂O (Ag metal is tarnished):
2Ag + O₃ → Ag₂O + O₂

Question 4.
Write a short note on the allotropy of sulphur.
Answer:
The important allotropes of sulphur are
a) yellow rhombic (α. sulphur)
b) Monoclinic (β – sulphur)

• The stable from is α-sulphur (at room temperature)

Rhombic sulphur (α – Sulphur):

• Colour : Yellow. ,
• Melting point: 385.8K.
• Specific gravity : 2.06.
• It is insouble in water and partially soluble in alcohol, benzene etc. and readily soluble in CS₂.

Monoclinic sulphur (β – Sulphur):

• Melting point: 392K
• Specific gravity : 1.98.
• It is soluble in CS₂.
• Rhombic sulphur transforms to monoclinic sulphur by heating above 369K. This temperature is called transition temperature.

Question 5.
How does SO₂ react with the following ?
a) Na₂SO₃(aq)
b) Cl₂
c) Fe⁺³ ions
d) KMnO₄
Answer:
a) Sodium sulphite (aq) reacts with So₂ to form sodium hydrogen sulphite.
Na₂SO₃ + H₂O + SO₂ → 2NaHSO₃

b) SO₂ as reacts with chlorine gas in the presence of charcoal to form sulphuryl chloride.
SO_2(g) + Cl_2(g) → SO₂Cl_2(l)

c) Fe⁺³ ions are reduced to Fe⁺² ions by SO₂.
2Fe⁺³ + SO₂ + 2H₂O → 2Fe⁺² + SO^-2₄ + 4H⁺

d) SO₂ gas decolourises acidified potassium permanganate (VII) solution.
5SO₂ + 2MnO^–₄ + 2H₂O → 5SO^-2₄ + 4H⁺+ 2Mn⁺²

Question 6.
Starting from elemental sulphur, how is H₂SO₄ prepared ?
Answer:
Manufacture of H₂SO₄ by contact process :
Manufacturing of H₂SO₄ involves three main steps.
Step – 1.
SO₂ production : The required SO₂ for this process is obtained by burning S(or) Iron pyrites in oxygen.
S + O₂ → SO₂
4FeS₂ + 15O₂ → 2Fe₂O₃ + 8SO₃
Step – 2
SO₃ formation: SO₂ is oxidised in presence of catalyst with atmosphric air to form SO₃

Step – 3
Formation of H₂SO₄: SO₃ formed in the above step absorbed in 98% H₂SO₄ to get oleum (H₂S₂O₇). This oleum is diluted to get desired concentration of H₂SO₄.
SO₃ + H₂SO₄ → H₂S₂O₇
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 7.
Describe the structures (shapes) of SO^-2₄ and SO₃.
Answer:
Structure of SO₃:

• In SO₃ sulphur undergoes sp² hybridisation.
• Shape : Trigonal planait s
• Bondangle: 120°.
• S -0 bond length: 143 pm.

Structure of SO^-2₄:

• In SO^-2₄ sulphur undergo sp³ hybridisation.
• Shape : Tetrahedral.
• It has several resonance structures.
• In this two Pπ – dπ bonds are present.

Question 8.
Which oxide of sulphur can act as both oxidizing and reducing agent? Give one example each.
Answer:
Sulphur dioxide (SO2) acts as both oxidising as well as reducing agent.
SO₂ as Oxidising agent:
Sodium suiphide oxidises to hypo with SO₂.
2Na₂S + 3SO₂ → 2Na₂S₂O₂ + S
SO₂ as Reducing agent:
SO₂ reduces Fe⁺³ ions to Fe⁺² ions.
2Fe⁺³ + SO₂ + 2H₂O → 2Fe⁺² + SO^-2₄ + 4H⁺

Question 9.
Explain the conditions favourable for the formation of SO₃ from SO₂ in the contact process of H₂SO₄.
Answer:
Le Chatlier’s principle – Application to produce SO₃:
The oxidation of SO₂ to SO₃ in the presence of a catalyst is a reversible reaction. The thermochemical equation for the conversion is written as
2SO_2(g) + O_2(g) ⇌ 2SO_3(g); ∆H = -196 kJ
The equation reveals the following points :

1. 3 volumes of the reactants convert into 2 volumes of SO₃. i.e., a decrease of volume accompanies the reaction.
2. the reaction is an exothermic change.
3. the catalyst may be present to increase the SO₃ yields.

According to Le Chatlier’s principle,
i) a decrease in volume of the system is favoured at high pressures. But in practice only about 2 bar pressure is used. The reason for not using high pressures is acid resisting towers which can withstand high pressures cannot be built.

ii) exothermic changes are favoured at low temperatures. It is not always convenient in the industry to work at low temperatures. In such situations an optimum temperature is maintained. At this temperature considerable amounts of the product are obtained. In the manufactrue of H₂SO₄, the optimum temperature suitable for the conversion of SO₂ into SO₃ is experimentally found to be 720K.

iii) The rate of formation of SO₃ is enhanced by the use of a catalyst. (V₂ O₅ (or) Pt – asbestos).
Favourable Conditions :
Temperature : 720K
Pressure : 2 bar
Catalyst: V₂O₅ (or) platinized asbestos.

Question 10.
Complete the following
a) KCl + H₂SO₄ (Conc) →
b) Sucrose
c) Cu + H₂SO₄ (Conc) →
d) C + H₂SO₄ (Cone) →
Answer:
a) 2KCl + Cone. H₂SO₄ → 2HCl + K₂SO₄
b) C₁₂H₂₂O₁₁ 12C + nH₂O
c) Cu + 2H₂SO₄(Conc) → CuS0₄ + SO₂ + 2H₂O
d) C + 2H₂SO₄(Conc) → CO₂ + 2SO₂ + 2H₂O

Question 11.
Which is used for drying ammonia ?
Answer:
For drying ammonia quick line (CaO) is used.
For drying ammonia cone. H₂SO₄, P₄O₁₀ and anhydrous CaCl₂ cannot be used because they react with ammonia and forms (NH₄)₂SO₄, (NH₄)₃PO₄ and CaCl₂. 8NH₃
H₂SO₄ + 2NH₃ → (NH₄)₂ SO₄

CaCl₂ + 8NH₃ → CaCl₃8NH₃.

Question 12.
Why cone H₂SO₄, P₄O₁₀ and anhydrous CaCl₂ cannot be used to dry ammonia ?
(Hint: ammonia reacts with them forming (NH₄)₂ SO₄; (NH₄)₃ PO₄ and CaCl₂, 8NH₂)
Answer:
For drying ammonia quick line (CaO) is used.
For drying ammonia cone. H₂SO₄, P₄O₁₀ and anhydrous CaCl₂ cannot be used because they react with ammonia and forms (NH₄)₂SO₄, (NH₄)₃PO₄ and CaCl₂. 8NH₃
H₂SO₄ + 2NH₃ → (NH₄)₂ SO₄

CaCl₂ + 8NH₃ → CaCl₃8NH₃.

Long Answer Questions

Question 1.
Explain in detail the manufacture of sulphuric acid by contact process. [T.S. Mar. 18, 16]
Answer:
Manufacture of H₂SO₄ by contact process :
Manufacturing of H₂SO₄ involves three main steps.
Step – 1.
SO₂ production : The required SO₂ for this process is obtained by burning S(or) Iron pyrites in oxygen.
S + O₂ → SO₂
4FeS₂ + 15O₂ → 2Fe₂O₃ + 8SO₃
Step – 2
SO₃ formation: SO₂ is oxidised in presence of catalyst with atmosphric air to form SO₃

Le Chatlier’s principle – Application to produce SO₃ :
The oxidation of SO₂ to SO₃ in the presence of a catalyst is a reversible reaction. The thermochemical equation for the conversion is written as
2SO_2(g) + O_2(g) ⇌ 2SO_3(g); ∆H = -196 kJ
The equation reveals the following points : .

1. 3 volumes of the reactants convert into 2 volumes of SO₃. i.e., a decrease of volume accompanies the reaction.
2. the reaction is an exothermic change.
3. the catalyst may be present to increase the SO₃ yields.

According to Le Chatlier’s principle, .
i) a decrease in volume of the system is favoured at high pressures. But in practice only about 2 bar pressure is used. The reason for not using high pressures is acid resisting towers which can withstand high pressures cannot be built.

ii) exothermic changes are favoured at low temperatures. It is not always convenient in the industry to work at low temperatures. In such situations an optimum temperature is maintained. At this temperature considerable amounts of the product are obtained. In the manufactrue of H₂SO₄, the optimum temperature suitable for the conversion of SO₂ into SO., is experimentally found to be 720K.

iii) The rate of formation of SO₃ is enhanced by the use of a catalyst. (V₂ O₅ (or) Pt – asbestos).
Favourable Conditions :
Temperature : 720K .
Pressure : 2 bar
Catalyst: V₂O₅ (or) platinized asbestos.

Step – 3 :
Formation of H₂SO₄: SO₃ formed in the above step absorbed in 98% H₂SO₄ to get oleum (H₂S₂O₇). This oleum is diluted to get desired concentration of H₂SO₄.
SO₃ + H₂SO₄ → H₂S₂O₇
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 2.
How is ozone prepared from oxygen ? Explain its reaction with [A.P. Mar. 19, 18, 17, 16] [Mar. 14]
a) C₂H₄
b) KI
c) Hg
d) PbS.
Answer:
Preparation of Ozone :
A slow dry stream of oxygen under silent electric discharge to form ozone (about 10%). The product obtained is known as ozonised oxygen.
3O₂ → 2O₃; ∆H° = 142kJ/mole

• The formation of ozone is an endothermic reaction.
• It is necessary to use silent electric discharge in the preparation of O₃ to prevent its decomposition

a) Reaction with C₂H₃ : Ethylene reacts with ozone to form Ethylene ozonoid followed by the hydrolysis to form formaldehyde.

) Reaction with KI: moist KI is oxidised to Iodine in presence of ozone.
2KI + H₂O + O₃ → 2KOH + I₂ + O₂

c) Reaction with Hg : Mercury loses it’s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.
2Hg + O₃ → Hg₂O + O₂
It is removed by shaking it with water which dissolves Hg₂O.

d) Reaction with PbS : Black lead sulphide oxidised to white lead sulphate in presence of ozone.
PbS + 4O₃ → PbSO₄ + 4O₂.

Intext Questions

Question 1.
List the important sources of sulphur.
Answer:
Occurrence or sources of sulphur in the earth’s crust, percentage of sulphur is only 0.03 to 0. 1%. In combined state, it occurs

1. In the form of sulphates, eg., gypsum (CaSO₄ . 2H₂O). epsom salt (MgSO₄ . 7H₂O), baryte
   (BaSO₄).
2. In the form of sulphides, eg., galena (PbS), zinc blende (ZnS), Copper pyrites (CuFeS₂). In volcanoes, traces of sulphur occur as H2S. In organic materials such as eggs, proteins, garlic, onion,mustard, hair and wool, sulphur is present in trace amounts.

Question 2.
Write the order of thermal stability of the hydrides of group 16 elements.
Answer:
The thermal stability of hydrides of group 16 elements is directly proportional to the bond dissociation enthalpy of H – E bond. On moving down the group, bond dissociation energy decreases because bond length increases.
Thus, the order of bond dissociation energy is
H₂O > H₂S > H₂Se > H₂Te > H₂Po
This is also the order of thermal stability.

Question 3.
Why is H₂O a liquid and H₂S a gas ?
Answer:
The difference in electronegativity values of 0(3.5) and H(2.1) is more than the difference between the electronegativity values of H(2.1) and S(2.5) i.e., O – H bond is more polar than S – H bond. That is why H-bonding is present among water molecules but absent in H₂S. Thus,
strong intermolecular interactions causes water to exist as a liquid but due to weak van der waals’ force H₂S exists as a gas.

Question 4.
Which of the following does not react with oxygen directly ? Zn, Ti, Pt, Fe.
Answer:
Platinum (Pt).

Question 5.
Complete the following reactions.
i) C₂H₄ + O₂ →
ii) 4Al + 3O₂ →
Answer:

Question 6.
Why does O₃ act as a powerful oxidising agent ?
Answer:
Because ozone liberates nascent oxygen very easily.

Question 7.
How is Oa estimated quantitatively ?
Answer:
When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. In this way O₃ can be estimated quantitatively.

Question 8.
What happens when sulphur dioxide is passed through an aqueous solution of Fe (III) salt ?
Answer:
When SO₂ is passed through an aqueous solution of Fe (III), i.e., ferric salt, it is reduced to Fe (II) i,e., ferrous salt. Here, SO₂ acts as a reducing agent.

Question 9.
Comment on the nature of two S – O bonds formed in SO₂ molecule. Are the two S – O bonds in this molecule equal ?
Answer:
The two S – O bonds in SO₂ molecule are covalent in nature. These are equal with bond length = 143 pm. The resonating structures of SO₂ are as follows.

SO₂ is a resonance hybrid of these two canonical forms.

Question 10.
How is the presence of SO₂ detected ?
Answer:
SO₂ has a pungent odour. Two tests to detect the presence of SO₂ are as follows :

1. SO₂ decolourises acidified KMnO₄ solution.
   
2. SO₂ Changes the colour of acidified potassium dichromate solution from orange to green.
   

Question 11.
Mention three areas in which H2S04 plays an important role.
Answer:
Uses of sulphuric acid.

1. It is used in the manufacture of pigments, paints and dyestuff intermediate.
2. It is used in petroleum refining.
3. It is also used in fertilizer industry.

Question 12.
Write the conditions to maximize the yield of H₂SO₄ by contact process.
Answer:
The key step in the manufacture of H₂SO₄ is catalytic oxidation of SO₂ to produce SO₃ in presence of V₂O₅.

The reaction is exothermic, reversible and the forward reaction results in the decrease in volume. Thus according to Le-Chatelier’s principle, the forward reaction should be favoured by low temperature and high pressure. But the temperature should not be very low otherwise the. rate of reaction will become very slow.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements

Very Short Answer Questions

Question 1.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:
Nitrogen gas exists as a diatomic molecule. Due to the presence of a triple bond between Base N – Atoms bond dissociation energy is high (941.4 KJ /mol). Hence nitrogen is inert and unreactive.
Phosphorus is a tetra atomic molecule and the P-P single bond is weaker than N ≡ N. P – P bond dissociation energy is 213 KJ/mole. Hence phosphorus is more reactive than Nitrogen.

Question 2.
How is nitrogen prepared in the laboratory?
Write the chemical equations of the reactions involved.
Answer:
Preparation of di Nitrogen :

• Very pure nitrogen is obtained by the thermal decomposition of sodium (or) barium a-zide.
  Ba(N₃)₂ → Ba + 3N₂
• In the laboratory dinitrogen is prepared by treating an aqueous solution of NH₄Cl with NaNO₂
  NH₄Cl_(aq) + NaNO_2(aq) → N_2(g) + 2H₂O_(I) + NaCl_(aq)
• Nitrogen can also be obtained by the thermal decompostion of ammonium dichromate.
  (NH₄)₃ Cr₂O₇ N₂ + 4H₂O + Cr₂O₃

Question 3.
Nitrogen exists as diatomic molecule and phosphorus as P₄ – Why ?
Answer:
Nitrogen exists as diatomic molecule :

• Nitrogen has small size and high electronegativity and nitrogen atom forms Pπ – Pπ multiple bonds with it self (triple bond). So it exists as a discrete diatomic molecule in elementary state.

Phosphorus exists as tetra atomic molecule :

• Phosphorus has large size and less electronegative and it forms P-P single bonds. So it exists as tetra atomic i.e., P₄.

Question 4.
Why does nitrogen show catenation properties less than phosphorus ?
Answer:
Explanation :
The single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non-boriding electrons in N₂ because of small bond length. Therefore the catenation property is weaker in nitrogen as compared to phosphorus.

Question 5.
Nitrogen molecule is highly stable – Why ?
Answer:
Nitrogen Molecule is more stable because in between two nitrogen atoms of N₂, a triple bond is present. To break this triple bond high energy is required (941.4KJ/mole).

Question 6.
Why are the compounds of bismuth more stable in +3 oxidation state ?
Answer:
Bismuth compounds are more stable in +3 oxidation state because ‘Bi’ exhibits +3 stable oxidation state instead of +5 due to inert pair effect.

Question 7.
What is allotropy ? Explain the different allotropic forms of phosphorus.
Answer:
Allotropy: The existance of an element in different physical forms having similar chemical properties is called allotropy.
Allotropes of ‘P’: → White ‘P’ (or) Yellow ‘P’.

• Red ‘P’
• Scarlet ‘P’
• Violet ‘P’
• α – black ‘P’
• β – black ‘P’.

White phosphorus :

• It is poisonous and insoluble in water and soluble in carbon disulphide and glows in dark. It is a translucent white waxy solid. ‘
•  It dissolves in boiling NaOH solution and gives PH₃.
  P₄ + 3NaOH + 3H₂ → PH₃ + 3NaH₂PO₂. (sodium hypo phosphite)
•  It is more reactive than other solid phases.
• Bond angle is 60° and it readily catches fire.

Red phosphorus :

• Red ‘F possesses iron grey lustre.
• In is odour less, non poisonous and insoluble in water as well as CS₂.
• Red F’ is much less reactive than white ‘P’.
  Black ‘P’:
• α – Black ‘P’: It is formed when red ‘P’ is heated in a sealed tube 803K.
• β – Black F : It is prepared by heating white P’ at 473 K under high pressure.

Question 8.
How do you account for the inert character of dinitrogen ?
Answer:
Di Nitrogen is chemically inert. .
Explanation : In nitrogen molecule there exists a triple bond between two nitrogen atoms. To break this triple bond high energy (bond dissociation energy) is required (941.4 KJ/mole).

Question 9.
Explain the difference in the structures of white and red phosphorus.
Answer:
White ‘P’ molecule has tetrahedral structure (discrete molecule). Discrete ‘P’ molecules are held by vander waal’s forces

White phosphorus
Red ‘P’ is polymeric consisting of chains of P₄ tetrahedron linked together through covalent bonds.

Question 10.
How is α – black phosphorus prepared from red phosphorus?
Answer:
α – Black ‘P: It is formed when red? is heated in a sealed tube 803K.

Question 11.
Write the difference between the properties of white phosphorus and red phosphorus.
Answer:
White ‘P’

• It is white waxy solid (translucent).
• It is insolable in water and solable in CS₂.
• It is more reactive.
• It is poisonous.

Red ‘P’

• It possesses iron grey lustre.
• It is in soluble in CS₂ as cool water.
• It is more reactive than white ‘P’.
•  It is non – poisonous.

Question 12.
What is inert pair effect ?
Answer:
Inert pair effect: The reluctance of ns pair of electrons to take part in bond formation is called inert pair effect.
Bi exhibits +3 oxidation state instead of +5 due to inert pair effect.

Question 13.
Explain why is NH₃ basic while BiH₃ is only feebly basic.
Answer:
NH₃ is basic while BiH₃ is only feebly basic.

Explanation :
Due to small atomic size of nitrogen, the electron density on nitrogen atom is greater than that on Bi atom. So electron releasing tendency is greater in NH₃.
Hence NH₃ is basic and BiH₃ is feebly basic.

Question 14.
Arrange the hydrides of group – 15 elements in the increasing order of basic strength and decreasing order of reducing character.
Answer:

1. Increasing order of basic strength of Group – 15 elements hydrides is
   BiH₃ < SbH₃ < ASH₃ < PH₃ < NH₃.
2. Decreasing order of reducing character of Group – 15 elements hydrides is
   BiH₃ > SbH₃ > ASH₃ > PH₃ > NH₃.

Question 15.
PH₃ is a weaker base than NH₃ – Explain.
Answer:
PH₃ is a weaker base than NH₃.

• In NH₃ nitrogen atom undergoes SP³ hybridisation and due to small size it has high electron density than in ‘P’ of PH₃.
• Due to large size of ‘P’ atom and availability of large surface area, lonepair of electron spread in PH₃. Hence PH₃ is weaker base then NH₃.

Question 16.
A hydride of group -15 elements dissolves in water to form a basic solution. This solution dissolves the AgCl precipitate. Name the hydride. Write the chemical equations involved.
Answer:
Given a hydride of group – 15 elements dissolves in water to form a basic solution;This solution dissolves the Agcl precipitate.

The given hydride is ammonia. It forms basic solution when dissolved in water due to formation of OH^– ions.
NH_3(g) + H_xO_l ⇌ [latex]\mathrm{NH}_4^{+}(\mathrm{aq})[/latex] + OH^–_(aq)
This solution dissolves AgCl. Ppt due to formation of complex compound.
AgCl_(s) (white Ppt) + 2NH_3(aq) → [Ag(NH³)²]Cl_(aq) (colourless)

Question 17.
What happens when white phosphorus is heated with cone. NaOH solution in an inert atmosphere of CO₂? [A.P. Mar. 19, 15]
Answer:
When white phosphorus heated with con. NaOH solution in an inert atomosphere of CO₂ forms PH₃.
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂.

Question 18.
NH₃ forms hydrogen bonds but PH₃ does not – Why ?
Answer:
NH₃ forms hydrogen bonds but PH₃ does not.
Reason : Ammonia forms hydrogen bonds because it it a polar molecule and N-H bond is highly polar. Nitrogen has more electronegativity than phosphrous. In case of PH₃ P-H bond polarity decreases.

Question 19.
The HNH angle is higher than HPH, HAsH and HSbH angles – Why?
Answer:
The central atom ε (where, ε = N, P, As, Sb, Bi) in all given hydrides is Sp³ hybridized. However, its electronegativity decreases and atomic size increase on moving down the group. Therefore is a gradual decrease in the force of repulsion between the shared electron pairs around the central atom. Thus, bond angle decreases as we move down the group.

Question 20.
How do calcium phosphide and heavy water react?
Answer:
Calcium phosphide reacts with heavy water to form Deutero phosphine.
Ca₃P₂ + 6D₂O → 3 Ca (OD)₂ + 2PD₃

Question 21.
Ammonia is a good complexing agent – Explain with an example. [Mar. 14]
Answer:
NH₃ is a lewis base and it donates electron pair to form dative bond with metal ions. This results in the formation of complex compound.

Question 22.
A mixture of Ca₃P₂ and CaC₂ is used in making Holme’s signal – Explain. [A.P. Mar. 16]
Answer:
A Mixture of Ca₃P₂ and CaC₂ is used in Holme’s signal.This Mixture containing containers are pierced and thrown in the sea, when the gas is evolved bum and serve as a signal.
The spontaneous combustion of PH₃ is the technical use of Holme’s signal.

Question 23.
Which chemical compound is formed in the brown ring test of nitrate ions ?
Answer:
In the brown ring test of nitrate salts a brown ring is formed. It’s formula is [Fe(H₂O)₅N0]⁺².

Question 24.
Give the resonating structures of NO₂ and N₂O₅.
Answer:
Resonance structure of NO₂:

Resonance structure of N₂O₅

Question 25.
Why does R₃P = O exist but R₃N = O does not (R = alkyl group) ?
Answer:
R₃P = O exist but R₃N = O does not.
Explanation :
Nitrogen does not form dπ – Pπ multiple bond with oxygen because of lack of d – orbitals in Nitrogen atom. But in case of R₃N = O the value of nitrogen should be 5. So these compounds do not exist where as in case of ‘P’ atom d-orbitals are available. So P-atom can able to form dπ – Pπ multiple bonds hence R₃P = O exist.

Question 26.
How is nitric oxide (NO) prepared ?
Answer:
The catalytic oxidation of NH₃ by atmospheric oxygen gives nitric oxide.

Question 27.
Give one example each of normal oxide and mixed oxide of nitrogen.
Answer:

• Nitric oxide (NO) is an example of normal oxide of Nitrogen.
• Dinitrogen trioxide (N₂O₃) is an example of mixed oxide of nitrogen.

Question 28.
NO is paramagnetic in gaseous state but diamagnetic in liquid and solid states – Why ?
Answer:
In gaseous state NO₂ exists as a Monomer and contains one unpaired electron but in solid state it dimerises to N₂O₄ so it doesnot contain unpaired electron.
Hence NO₂ is parce magnetic is geseous state but diamagnetic in solid state.

Question 29.
Give an example of
a) acidic oxide of phosphorus
b) neutral oxide of nitrogen.
Answer:
a) P₂O₅ (or) P₄O₁₀, phosphorus pentoxide is an example of acidic oxide of phosphorus
b) Nitrous oxide (N₂O) and Nitric oxide (NO) are neutral oxides of nitrogen.

Question 30.
Explain the following
a) reaction of alkali with red phosphorus.
b) reaction between PCl₃ and H₃PO₃.
Answer:
a) Red phosphorus reacts with an alkali to form hypophosphoric acid (H₄P₂O₆)
b) PCl₃ undergoes hydrolysis to form H₃PO₃
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Question 31.
How does PCl₃ react with
a) CH₃COOH
b) C₂H₅OH and
c) water.
Answer:
a) PCl₃ reacts with CH₃COOH and form phosphorous acid, acetyl chloride.
3CH₃COOH + PCl₃ → 3CH₃COCl + H₂PO₃

b) PCl₃ reacts with phosphorous acid and forms phosphorus acid, Ethyl chloride
3C₂H₅OH + PCl₃ → 2C₂H₅Cl₃ + H₃PO₃

c) PCl₃ reacts with water (hydrolysis) to form phosphorus acid. It undergo hydrolysis in presence of moisture.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Question 32.
PCl₃ can act as an oxidizing as well as a reducing agent – Justify.
Answer:
PCl₃ acts as reducing agent. It is evidented by the following reaction.

PCl₃ acts as oxidising agent. It is evidented by the following reaction.

Question 33.
Which of the following are not known ?
PCl₃, AsCl₃, SbCl₃, NCl₅, BiCl₅, PH₅
Answer:
NCl₅, BiCl₅, PH₅ are not known in the given compounds.

Question 34.
Which of the following is more covalent – SbCl₅ or SbCl₃ ?
Answer:
SbCl₅ (Penta halide) is more covalent than SbCl₃ (Tri halide). Because Sb in the higher oxidation state exert more polarising power.

Question 35.
Write the oxidation states of phosphorus in solid PCl₅.
Answer:

• In solid state PCl₅ exists as an ionic solid [PCl₄)⁺ [PCl₆]^–
• ‘P’ exhibits +5 oxidation state in [PCl₄)⁺ [PCl₆]^–

Question 36.
Illustrate how copper metal can give different products on reaction with HNO₃.
Answer:
Reaction of Cu metal with dilute HNO₃.

Reaction of Cu Metal with cone. HNO₃

Question 37.
Which oxide of nitrogen has oxidation number of N same as that in nitric acid ?
Answer:

• In HNO₃, ‘N’ has oxidation state +5.
• Among oxides of nitrogen N₂O₅ exhibits ‘+5’ oxidation state.

Question 38.
Write the chemical reactions that occur in the manufacture of nitric acid.
Answer:
Chemical reactions involved in the manufacturing of HNO₃.

• Oxidation of NH₃
  
• NO₂ formation :
  2NO + O₂ ⇌ 2NO₂
• Formation of HNO₃:
  3NO₂ + H₂O → 2HNO₃ + NO

Question 39.
Iron becomes passive in cone. HNO₃ – Why ?
Answer:
Iron becomes passive in cone. HNO₃ due to formation of a passive film of oxide on the surface of iron.

Question 40.
Give the uses of
a) nitric acid and
b) ammonia.
Answer:
Use of HNO₃:

• HNO₃ is used in the manufacture of ammonium nitrate for fertilisers and other nitrates for use in explosives and pyrotechnics.
• HNO₃ is used in the pickling of stainless steel.
• HNO₃ is used as oxidiser in rocket fuels.

Question 41.
What are the oxidation states of phosphorus in the following ?
i) H₃PO₃
ii) PCl₃
iii) Ca₃P₂
iv) Na₃PO₄
v) POF₃
Answer:
i) H₃PO₃
3(1) + x + 3(-3) = 0
x = +3

ii) PCl₃
x + 3(-1) = 0
x = 3

iii) Ca₃P₂
3(+2) + 2x = 0
x = -3

iv) Na₃PO₄
3(1) + x + 4(-2) = 0
x = +5

v) POF₃
x + (-2) + 3(-1) = 0
x = +5

Question 42.
H₃PO₃ is diprotic while H₃PO₂ is monoprotic – Why ?
Answer:
H₃PO₃ is diprotic :
Structure of H₃PO₃:

H₃PO₂ is monoprotic :
Structure of H₃PO₂ :

Question 43.
Give the disproportionation reaction of H₃PO₃.
Answer:
Orthophosphoric acid (H₃PO₃) on heating disproportionates to give orthophosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃.

Question 44.
H₃PO₂ is a good reducing agent – Explain with an example.
Answer:
In H₃PO₂, two H-atoms are bonded directly to P-atom which imparts reducing character to the acid.

Question 45.
Draw the structures of
a) Hypo phosphoric acid
b) Cyclic meta phosphoric acid.
Answer:
a) Structure of hypo phosphoric acid (H₄P₂O₆) :

b) Structure of cyclic meta phosphoric acid (HPO₃)₃

Short Answer Questions

Question 1.
Discuss the general characteristics of Group – 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization enthalpy and electronegativity.
Answer:
1) Electronic Configuration : All these elements of this group have s²p³ configuration in their respective outmost orbits.
Nitrogen (N) : [He] 2s²2p³; phosphorus (P) : [Ne] 3s²3p³, Arsenic (As) : [Ar] 3d¹⁰4s²4p³
Atimony (Sb) : [Kr] 4d¹⁰ 5s² 5p³ and Bismuth (Bi) = [Xe] 4f¹⁴ 5d¹⁰ 6s²6p³

2) Oxidation Spates : As per their configurations, these elements may utilize either the three electrons iti the p-orbitals or the five electrons present in both s -and p -orbitals in the exhibition of oxidation states. Consequently, they exhibit common oxidation states +III and +V.

3) Atomic size: In Group-15 elements atomic size increase down the group. A considerable increase in covalent radius observed from N to P and from As to Bi only a small increase in atomic size observed.

4) Ionisation enthalpy : In group – 15 elements ionisation enthalpy decreases down the group due to gradual increase in atomic size.

5) Electronegativity : In group-15 elements electronegativity values decrease down the group with increase in atomic size.

Question 2.
Discuss the trends in chemical reactivity of group – 15 elements.
Answer:
Nitrogen gas exists as diatomic molecule. Dut to the presence of triple bond between Base N – Atoms bond dissociation energy is high (941.4 KJ /mol). Hence nitrogen is inert and unreactive.

Phosphorus is a tetra atomic molecule and P-P single bond is weaker than N ≡ N. P – P bond dissociation energy U 213 KJ/mole. Hence phosphorus is more reactive than Nitrogen.

Explanation :
The single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non-bonding electrons in N₂ because of small bond length. There fore the catenation property is weaker in nitrogen as compoud to phosphorares.
R₃ P = O exist but R₃N = O does not.

Explanation :
Nitrogen does not form dπ – Pπ multiple bond with oxygen because of lack of d – orbitals in Nitrogen atom. But in case of R₃N = 0 the value of nitrogen should be 5. So these compounds do not exist where as in case of P’ atom d-orbitals are available. So P-atom can able to form dπ – Pπ multiple bonds hence R₃P = 0 exist.

1. Reactivity towards hydrogen: Group -15 elements forms EH₃ type hydrides (E = Group – 15 elements) .
   Eg : PH₃, NH₃, AsH₃, BiH₃, SbH₃.
   • Among above hydrides NH₃ is mild reducing agent while BiH₃ is strong reducing agent.
   • Stability of hydrides decreases from NH₃ to BiH₃.
   • Basicity of hydrides decreases as follows.
     NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃.
2. Reactivity towards Oxygen : These forms two types of oxides E₂O₃ and E<sub2O₅.
   Eg : P₂O₃ > N₂O₅, N₂O₃, P₂O₅.
   • Acidic character of oxides decrease down the group.
   • E₃O₃ of N and P are acidic, As and Sb are amphoteric while ‘Bi’ is basic.
3. Reactivity towards halogens : These elements forms two types of halides EX₃ and EX₅.
   ‘N’ – does not form penta halides due to lack of the d-orbitals.
   Penta halides are more covalent than tri halides because the elements in higher oxidation state have more polarising power.
4. Reactivity towards metals : All these elements react with metals to form their binary compounds containing – 3 oxidation state.
   Eg. : Ca₃N₂, Ca₃P₂ etc.

Question 3.
How does P₄ react with the following ?
a) SOCl₂
b) SO₂Cl₂
Answer:
a) When P₄ reacts with SOCl₂ to form phosphorus trichloride.
P₄ + 8 SOCl₂ → 4 PCl₃ + 4 SO₂ + 2 S₂Cl₂
b) When P₄ reacts with SOCl₂ to form phosphorus pentachloride.
P₄ + 10 SOCl₂ → 4 PCl₅ + 10 SO₂

Question 4.
Explain the anomalous nature of nitrogen in group -15.
Answer:
Anomalous properties of Nitrogen : Nitrogen differs from the remaining elements of this group due to its small size, high electronegativity, high ionisation enthalpy and non-availability of d-orbitals.

Nitrogen gas exists as diatomic molecule. Due to the presence of triple bond between base N – Atoms bond dissociation energy is high (941.4 KJ /mol). Hence nitrogen is inert and un-reactive.

Explanation :

• The single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non-bonding electrons in N₂ because of small bond length. Therefore the catenation property is weaker in nitrogen as compound to phosphorus.
• Nitrogen does not form penta halides due to lack of d-orbitals.
  R₃ P = O exist but R₃N = O does not.

Explanation :
Nitrogen does not form dπ – Pπ multiple bond with oxygen because of lack of d – orbitals in Nitrogen atom. But in case of R₃N = O the value of nitrogen should be 5. So these compounds do not exist where as in case of ‘P atom d-orbitals are available. So P-atom can able to form dπ – Pπ multiple bonds hence R₃P = O exist.

Question 5.
Complete the following reactions.
a) Ca₃P₂ + HaO →
b) P₄ + KOH →
c) CuSO₄ + NH₃ →
d) Mg + N₂ →
d) (NH₄)₂ + Cr₂O₇
f) Decomposition of nitrous acid →
Answer:
a) Ca₃P₂ + 6 H₂O → 3 Ca(OH)₂ + 2PH₃
b) P₄ + 3 KOH + 3 H₂O → PH₃ + 3KH₂ PO₂
c) CuSO_4(aq) (blue) + 4NH₃ → [Cu (NH₃)₄] SO₄ (deep blue)
d) 3 Mg + N₂ → Mg₃N₂
e) (NH₄)₂ + Cr₂O₇ N₂ + 4 H₂O + Cr₂O₃
f) 3HNO₂ → HNO₃ + 2 NO + H₂O

Question 6.
How does PCl₅ react with the following?
a) Water
b) C₂H₅OH
c) CH₃COOH
d) Ag
Answer:
a) PCl₅ undergo hydrolysis to form phosphoric acid.
PCl₅ + H₂O → POCl₃ + 2HCl
POCl₃ + 3H₂O → H₃PO₄ + 3 HCl

b) PCl₅ reacts with C₂H₅OH to form Ethyl chloride.
C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl

c) PCl₅ reacts with CH₃COOH to form acetyl chloride.
CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl

d) PCl₅ reacts with Ag to form PCl₃ and AgCl
PCl₅ + 2 Ag → 2 AgCl + PCl₃

Question 7.
Complete the following.
a) NH₄NO₃
b) HNO₃ + P₄O₁₀ →
c) Pb(NO₃)₂
d) Zn + dil.HNO₃ →
e) P₄ + conc.HNO₃ →
f) HgCl₂ + PH₃ →
Answer:
a) NH₄NO₃ N₂O (Nitrous oxide) + 2 H₂O
b) 12 HNO₃ + P₄O₁₀ → 6 N₂O₅ + 4 H₃PO₄
c) 2 Pb(NO₃)₂ 2 PbO + 4 NO₂ + O₂
d) 4 Zn + 10 HNO₃ (dil) → 4 Zn (NO₃)₂ + 5 H₂O + N₂O
e) P₄ + 20 HNO₃ → 4H₃PO₄ + 20 NO₂ + 4 H₂O
f) 3 HgCl₂ + 2 PH₃ → Hg₃P₂ + 6 HCl

Long Answer Questions

Question 1.
How is ammonia manufactured by Habes process? Explain the reactions of ammonia with [A.P. Mar. 18]
a) ZnSO_4(aq)
b) CuSO_4(aq)
c) AgCl_(s)
Answer:
In Haber process ammonia is directly synthesised from elements (nitrogen and hydrogen). The principle involved in this is
N₂ (g) + 3H₂O (g) ⇌ 2NH₃ + 92.4kJ
This is a reversible exothermic reaction.
According to Le Chatelier’s principle favourable conditions for the better yield of ammonia are low temperature and high pressure. But the optimum conditions are
Temperature : 720k
Pressure : 200 atmospheres
Catalyst : Finely divided iron in the presence of molybdenum (Promoter).
Procedure : A mixture of nitrogen and hydrogen in the volume ratio 1 : 3 is heated to 725 – 775K at a pressure of 200 atmospheres is passed over hot finely divided iron mixed with small amount of molybdenum as promotor. The gases coming out of the catalyst chamber consists of 10 – 20% ammonia gas are cooled and compressed, so that ammonia gas is liquified, and the uncondensed gases are sent for recirculation.

b) Aq.CuSO₄ reacts with ammonia to form a deep blue complex,

c) Solid AgCl reacts with ammonia to form a colourless complex.

Question 2.
How is nitric acid manufactured by Ostwald’s process ? How does it react with the following ? [T.S. Mar. 19]
a) Copper
b) Zn
c) S8
d) P4
Answer:
Ostwald’s process : Ammonia, mixed with air in 1 : 7 or 1 : 8, when passed over a hot platinum gauze catalyst, is oxidised to NO mostly (about 95%).
The reaction is

The liberated heat keeps the catalyst hot. The ‘NO’ is cooled and is mixed with oxygen to give the dioxide, in large empty towers (oxidation chamber). The product is then passed in to warm water, under pressure in the presence of excess of air, to give HNO₃.
4NO₂ + O₂ + 2H₂O → 4HNO₃.
The acid formed is about 61% concentrated.

a) Copper reacts with dil. HNO₃ and cone. HNO₃ and liberates Nitric Oxide and Nitrogen dioxide respectively
3 Cu + 5 HNO₃ (dil) → 3 Cu (NO₃)₂ + 2 NO + 4 H₂O
Cu + 4 HNO₃ (cone.) → Cu (NO₃)₂ + 2 NO₂ + 2 H₂O

b) Zn reacts with dil. HNO₃ and Cone. HNO₃ and liberates Nitrous oxide and Nitrogen dioxide respectively.
4 Zn + 10 HNO₃ (dil) → 4 Zn (NO₃)₂ + 5 H₂O + N₂O
Zn + 4 HNO₃ (cone.) → Zn (NO₃)₂ + 2 H₂ + 2 NO₂

c) S₈ reacts with cone, nitric acid to form Sulphuric acid, NO₂ gas.
S₈ + 48 HNO₃ → 8 H₂SO₄ + 48 NO₂ + 16 H₂O
d) P₄ reacts with cone, nitric acid to form phosphoric acid and NO₂ gas.
P₄ + 20 HNO₃ → 4 H₃PO₄ + 20 NO₂ + 4 H₂O

Textual Examples

Question 1.
Though nitrogen exhibits +5 oxidation state, it does not form peniahalide. Give reason.
Answer:
Nitrogen with n = 2, has s and p orbitals only. It does not have d orbitals to expand its covalence beyond four. That is why it does not form pentahalide.

Question 2.
PH₃ has lower boiling point than NH₃. Why ? [T.S. Mar. 16]
Answer:
Unlike NH₃. PH₃ molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of PH₃ is lower than NH₃.

Question 3.
Write the reaction of thermal decomposition of sodium azide.
Answer:
Thermal decomposition of sodium azide gives dinitrogen gas.
2NaN₃ → 2Na + 3N₂

Question 4.
Why does NH₃ act as a Lewis base ?
Answer:
Nitrogen atom in NH₃ has one lone pair of electrons with is available for donation. Therefore, it acts as a Lewis base.

Question 5.
Why does NO₂ dimerise ?
Answer:
NO₂ contains odd number of valence electrons. It be haves as a typical odd molecule. On dimerisation. It is converted to stable N₂O₄ molecule with even number of electrons.

Question 6.
In what way can it be proved that PH₃ is basic in nature ?
Answer:
PH₃ reacts with acids like HI to form PH₄I which shows that it is basic in nature.
PH₃ + HI →PH₄I
Due to lone pair on phosphorus atom, PH₃ is acting as a Lewis base in the above reaction.

Question 7.
Why does PCl₃ fume in moisture ?
Answer:
PCl₃ hydrolyses in the presence of moisture giving fumes of HCl.
PCl₃+ 3H₂O → H₃PO₃ + 3HCl

Question 8.
Are all the five bonds in PCl₅ molecule equivalent ? Justify your answer.
Answer:
PCl₅ has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are equivalent. While the two axial bonds are different and longer than equatorial bonds.

Intext Questions

Question 1.
Why are pentahalides more covalent than trihalides ?
Answer:
Higher the positive oxidation state of central atom, more will be its polarising power, which in turn increases the covalent character of the bond formed between the central atom and the halogen atom. In pentahalides, the central atom is in +5 oxidation state while in trihalides, it is in +3 oxidation state. Therefore, pentahalides are more covalent than trihalides.

Question 2.
Why is BiH₃ the strongest reducing agent amongst all the hydrides of Group 15 elements ?
Answer:
Among the 15 group elements, the size of Bi atom is largest and hence, the Bi-H bond length is largest or Bi-H bond dissociation energy is lowest. That’s why Bi-H bond dissociates more readily than the other hydrides of the group and hence, BiH₃ is the strongest reducing agent.

Question 3.
Why is N₂ less reactive at room temperature ?
Answer:
Dinitrogen is inert of less reactive because the bond enthalpy of N ≡ N bond is very high.

Question 4.
Mention the conditions required to maximise the yield of ammonia.
Answer:
Ammonia is produced by Haber’s process as.

Yield of ammonia is favoured by high pressure according to Le-Chatelier’s principle, Other conditions, that favour the production of ammonia are as follows :

1. Temperature – approximately 700 K
2. Pressure – 200 atm or 200 × 10⁵ Pa
3. Catalyst – Iron oxide
4. Promotor – Molybdenum of K₂O and Al₂O₃

Question 5.
How does ammonia react with a solution of Cu²⁺?
Answer:
When ammonia (aqueous solution is ammonium hydroxide) reacts with a solution of Cu²⁺, a deep blue solution is obtained due to the formation of tetraamine copper (II) ion.

Question 6.
What is the covalence of nitrogen in N₂O₅?
Answer:
Structural formula of N₂O₅

Since, N atom has 4 shared electron pairs; the valence of N is 4.

Question 7.
Bond angle in PH₄⁺ is higher than that in PH₃. Why ?
Answer:
In both PH₄⁺ and PH₃, phosphorus atom is sp³ hybridized. In PH₄⁺, all the four orbitals are bonded whereas in PH₃, there is a lone pair of electrons too. Due to lone pair-bonded pair repulsion in PH₃, the bond angle is less than 109.5°.

Question 8.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂? .
Answer:
White phosphorus dissolves in boiling NaOH in an inert atmosphere of CO₂ producing phosphine (PH₃) gas and sodium hypophosphite (NaH₂PO₂).

Question 9.
What happens when PCl₅ is heated ?
Answer:
In PCl₃, there are 5P – Cl bonds, out of which three are equatorial (longer) and two are axial (shorter). When PCl₅ is heated strongly, two less stable axial bonds break and phosphorus trichloride (PCl₃) is formed.

Structure of PCl₅ showing axial and equatorial bonds.

Question 10.
Write a balanced equation for the hydrolytic reaction of PCl₅ in heavy water.
Answer:
Reaction of PCl₅ in heavy water (D₂O)

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
[latex]\frac{4}{3}[/latex] Cr_(s) + O_2(g) → [latex]\frac{2}{3}[/latex] Cr₂O_2(s), ∆G° – 540 KJ ……………….. (1)
[latex]\frac{4}{3}[/latex] Al_(s) + O_2(g) → [latex]\frac{2}{3}[/latex] Al₂O_3(s), ∆G° – 827 KJ ……………….. (2)
Equation (1) – Equation (2)
[latex]\frac{2}{3}[/latex] Cr₂O_3(s) + [latex]\frac{4}{3}[/latex] Al_(s) → [latex]\frac{2}{3}[/latex] Al₂O_3(s) + [latex]\frac{4}{3}[/latex] 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
[latex]\frac{4}{3}[/latex]Al_(S) + O_2(g) → 2Mg_(S) + [latex]\frac{2}{3}[/latex] O_2(g) [latex]\frac{4}{3}[/latex] 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) + [latex]\frac{4}{3}[/latex] Al_(S) ⇌ 2Mg_(S) + [latex]\frac{2}{3}[/latex] 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
[latex]\frac{4}{3}[/latex]Al_(S) + O_2(g) → [latex]\frac{2}{3}[/latex] 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) + [latex]\frac{4}{3}[/latex]Al_(S) → 2Mg_(S) + [latex]\frac{2}{3}[/latex] 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) [latex]\frac{4}{3}[/latex] Al + O₂ → [latex]\frac{2}{3}[/latex] 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 :
[latex]\frac{2}{3}[/latex] Al₂O₃ + 2Mg → 2MgO + [latex]\frac{4}{3}[/latex] 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 [latex]\frac{x}{m}[/latex] = 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 [latex]\frac{1}{n}[/latex] = 0, [latex]\frac{x}{m}[/latex] = 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
[latex]\frac{x}{m}[/latex] = 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 [latex]\frac{x}{m}[/latex] = log k + [latex]\frac{1}{n}[/latex] log P.

• A graph is plotted taking log [latex]\frac{x}{m}[/latex] 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 [latex]\frac{1}{n}[/latex] value.
• The intercept on the y-axis gives value of log k.
• [latex]\frac{1}{n}[/latex] has values between 0 and 1
• When [latex]\frac{1}{n}[/latex], = 0, [latex]\frac{x}{m}[/latex] = constant, the adsorption is independent of pressure
  [latex]\frac{1}{n}[/latex] = 1, [latex]\frac{x}{m}[/latex] = kP i.e., [latex]\frac{x}{m}[/latex] ∝ 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
  [latex]\frac{\mathrm{x}}{\mathrm{m}}[/latex] = 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.
  
  [latex]\frac{\mathrm{x}}{\mathrm{m}}[/latex] = k.P^1/4
  Applying logarithm
  Log [latex]\frac{\mathrm{x}}{\mathrm{m}}[/latex] = log k + [latex]\frac{\mathrm{1}}{\mathrm{n}}[/latex] log p.
• The validity of Freundlich isotherm can be verified by plotting log [latex]\frac{\mathrm{x}}{\mathrm{m}}[/latex] 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 [latex]\frac{\mathrm{1}}{\mathrm{n}}[/latex]
• 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.
[latex]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 {) }}[/latex]
Answer:
Given reaction is
[latex]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 {) }}[/latex]
Rate of reaction = [latex]=-\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}}[/latex]

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 + [latex]\frac{1}{2}[/latex] 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 = [latex]\frac{[R]_0-[R]}{t}[/latex]
Eg : Decomposition of NH₃

Rate = k [NH₃]⁰ = k
∴ [latex]\frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}[/latex] = 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

1. 
2. 

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 = [latex]\frac{2.303}{t} \log \frac{[R]_0}{[R]}[/latex]
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 + [latex]\frac{1}{2}[/latex] 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 = ([latex]\frac{2.303}{t}[/latex]) log ([latex]\frac{\mathrm{p}_0}{\mathrm{p}_{\mathrm{A}}}[/latex])
k = ([latex]\frac{2.303}{t}[/latex]) log ([latex]\frac{\mathrm{p}_0}{2 \mathrm{p}_{\mathrm{i}}-\mathrm{p}_{\mathrm{t}}}[/latex])
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) = [latex]\frac{0.693}{k}[/latex]
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] = [latex]\frac{[\mathrm{R}]_0}{2}[/latex]
∴ t_1/2 = [latex]\frac{[\mathrm{R}]_0 / 2}{\mathrm{k}}[/latex]
∴ t_1/2 = [latex]\frac{\left[\mathrm{R}_0\right]}{2 \mathrm{k}}[/latex]
Half life of a first order reaction
t_1/2 = [latex]\frac{0.693}{\mathrm{k}}[/latex]
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:
[latex]\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][/latex]
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) + [latex]\mathrm{BrO}_{3(\mathrm{aq})}^{-}[/latex] + [latex]6 \mathrm{H}_{(\mathrm{aq})}^{+}[/latex] → 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)
[latex]\frac{\Delta[\mathrm{R}]}{\Delta t}[/latex] = k [NO]² [O₂]

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

iii) CH₃COOC₂H_5(l) + H₂O_(l) → CH₃COOH_(aq) + C₂H₅OH_(aq)
[latex]\frac{\Delta[\mathrm{R}]}{\Delta t}[/latex] = 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 + [latex]\frac{1}{2}[/latex] 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 ∝ [latex]\frac{1}{a^{n-1}}[/latex]
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.
[latex]\left(\frac{t_{1 / 2}^{\prime}}{t_{1 / 2}^n}\right)=\left(\frac{a^n}{a^{\prime}}\right)^{n-1}[/latex]
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 = [latex]\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}[/latex] = k [R]°
Rate = [latex]\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}[/latex]; 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 = [latex]\frac{\left[R_0\right]-[R]}{t}[/latex]
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 [latex]\frac{[\mathrm{R}]}{\left[\mathrm{R}_0\right]}[/latex] = -kt ………………. (2)
k = [latex]\frac{1}{\mathrm{t}} \ln \frac{\left[\mathrm{R}_0\right]}{[\mathrm{R}]}[/latex]
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) = [latex]\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_2}-\frac{1}{\mathrm{~T}_1}\right)[/latex]
[latex]\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][/latex]
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 = [latex]\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}[/latex] = k [R]°
Rate = [latex]\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}[/latex]; 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 = [latex]\frac{\left[R_0\right]-[R]}{t}[/latex]
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 :

1. 
2. 

Examples of first order reaction :
N₂O_5(g) → N₂O_4(g) + [latex]\frac{1}{2}[/latex] 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) = [latex]\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_2}-\frac{1}{\mathrm{~T}_1}\right)[/latex]
[latex]\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][/latex]
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 = [latex]\frac{0.693}{k}[/latex]
∴ k = [latex]\frac{0.693}{t_{1 / 2}}=\frac{0.693}{10}[/latex] = 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 = [latex]\frac{2.303}{t} \log _{10} \frac{a}{(a-x)}[/latex]
k = [latex]\frac{2.303}{t}[/latex] log [latex]\frac{0.6}{0.2}[/latex] = 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 = [latex]\frac{1}{t}[/latex] [[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 = [latex]\frac{1}{t}[/latex] [0.10 – 0.075]
t = [latex]\frac{0.025}{0.0030}[/latex] = 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 = [latex]\frac{2.303}{t} \log \frac{a}{(a-x)}[/latex]
k = [latex]\frac{2.303}{40} \log \frac{100}{70}[/latex] = 0.0576 [2.0000 – 1.8451]
= 0.0576 (0.1549) = 8.922 × 10^-3
[latex]\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\mathrm{k}}=\frac{0.693}{8.922 \times 10^{-3}}[/latex] = 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
[latex]\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\mathrm{k}}[/latex]
[latex]\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{200 \mathrm{~s}^{-1}}[/latex] = 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.
[latex]\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\mathrm{k}}=\frac{0.693}{2.4 \times 10^{-3}}[/latex]
= 288.75 sec.
To convert into [latex]\frac{3}{4}[/latex] 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 = [latex]\frac{2.303}{t} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}[/latex]
k = [latex]\frac{2.303}{15} \log \frac{100}{80}[/latex] = 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 = [latex]\frac{0.693}{1_{1 / 2}}=\frac{0.693}{5 \times 10^{-6}}[/latex] = 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 = [latex]\frac{2.303}{t} \log \frac{a}{a-x}[/latex]
1.06 x 10 = [latex]\frac{2.303}{t} \log \frac{100}{85}[/latex]
t = [latex]\frac{2.303}{1.06 \times 10^{-3}} \log \frac{100}{85}[/latex] = 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 = [latex]\frac{0.693}{k}[/latex]
For 99.9% → a = 100
a – x = 100 – 99.9 = 0.1
k = [latex]\frac{2.303}{\mathrm{k}} \log \frac{100}{0.1}[/latex]
But k = [latex]\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{2.303}{0.693}[/latex] × 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 = [latex]\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}[/latex] or log x = [latex]\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}[/latex]
or log x = -[latex]\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}}[/latex] = -18.8323
x = Antilog [-18.8324] = Antilog [latex]\overline{19} .1677[/latex] = 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 [latex]\frac{0.03-0.02}{25 \times 60}=\frac{0.01}{25 \times 60}[/latex]
= [latex]\frac{0.01}{1500}[/latex] = 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 = [latex]\frac{-1}{2} \frac{\mathrm{d}}{\mathrm{dt}}[/latex] [A] = [latex]\frac{1}{2}\left[\frac{(0.5-0.4)}{10}\right][/latex] = 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 = [latex]\frac{1}{2}[/latex] + 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) = [latex]\frac{0.693}{60 \mathrm{~min}}=\frac{0.693}{(60 \times 60) \mathrm{s}}[/latex] = 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] = [latex]\frac{\text { Rate }}{[\mathrm{NO}]^2}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^2}[/latex] = mol^-1 L s^-1

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

iii) Rate = k[CH₃CHO]^3/2
order of reaction = [latex]\frac{3}{2}[/latex]
Dimensions of k = [latex]\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}}[/latex] = mol^-1/2 L^1/2 s^-1

iv) Rate = k[C₂H₅Cl]
order of reaction = 1
Dimensions of k = [latex]\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}}[/latex] = 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 = [latex]\frac{1}{2}[/latex] × 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 = [latex]\frac{\mathrm{dx}}{\mathrm{dt}}[/latex] = 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 = [latex]\frac{\text { Rate }}{\left[\left(\mathrm{PCH}_3 \mathrm{OCH}_3\right]^{3 / 2}\right.}[/latex]
= [latex]\frac{\text { bar } \min ^{-1}}{[\mathrm{bar}]^{3 / 2}}[/latex] = 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 [latex]\frac{1}{2}[/latex] i.e., [A] = [latex]\frac{1}{2}[/latex] a
Rate = k [[latex]\frac{a}{2}[/latex]]² = [latex]\frac{1}{4}[/latex] ka²
Rate of reaction becomes [latex]\frac{1}{2}[/latex] 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 = [latex]\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}[/latex] = 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 = [latex]\frac{2.303}{k} \log \frac{a}{(a-x)}[/latex] …………….. (i)
Given (a – x) [latex]\frac{a}{16}[/latex], 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 = [latex]\frac{(2.17+2.24) \times 10^{-3} \mathrm{~s}^{-1}}{2}[/latex]
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 – [latex]\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}[/latex]
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 + [latex]\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}[/latex]
= log (2.418 × 10^-5 s^-1) + [latex]\frac{179900 \mathrm{Jmol}^{-1}}{2.303 \times\left(8.314 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right) \times 546 \mathrm{k}}[/latex]
= -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_{[latex]\frac{1}{2}[/latex]} = 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
[latex]-\frac{E_{\mathrm{a}}}{\mathrm{RT}}=\frac{-28000 \mathrm{k}}{\mathrm{T}}[/latex]
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 – [latex]\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}[/latex] …………………. (i)
log K = 14.34 – [latex]\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}[/latex] …………………. (ii)
on comparing both equations.
[latex]\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}=\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}[/latex]
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
[latex]\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}[/latex]
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 [latex]\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}[/latex]
log k = log (4 × 10¹⁰) – [latex]\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})}[/latex]
log k = 10.6021 – 12.5870 = – 1.9849
k = Antilog (-1.9849) = Antilog [latex](\overline{2} .0151)[/latex] = 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 = [latex]\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][/latex]
= 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 = [latex]\frac{1}{4}\left\{\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}\right\}[/latex]
[latex]\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}[/latex] = 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) + [latex]\frac{1}{2 \mathrm{O}_2}[/latex] (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
[latex]\log \frac{[\mathrm{R}]_1}{[\mathrm{R}]_2}=\frac{\mathrm{k}\left(\mathrm{t}_2-\mathrm{t}_1\right)}{2.303}[/latex]

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 = [latex]\frac{0.693}{k}[/latex]
t_1/2 = [latex]\frac{0.693}{5.5 \times 10^{-14} \mathrm{~S}^{-1}}[/latex] = 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 = [latex]\frac{2.303}{t} \log \frac{C_0}{C}[/latex] 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
= [latex]-\frac{1}{2} \frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}[/latex]
= [latex]-\frac{1}{2} \frac{(0.4-0.5) \mathrm{mol} \mathrm{L}^{-1}}{10 \mathrm{~min}}[/latex]
= 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 = [latex]\frac{1}{2}[/latex] + 2 = [latex]\frac{1}{2}[/latex]
Order = [latex]\frac{5}{2}[/latex]

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) .
[latex]\frac{\mathrm{r}^{\prime}}{\mathrm{r}}=\frac{\mathrm{k} \times[9 \mathrm{x}]^2}{\mathrm{k} \times[\mathrm{x}]^2}=9[/latex]
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.