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Students get through the TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Answer the following questions.

Question 1.
Classify the following aliphatic or aromatic nitro compounds.
(i) 1, 2, dimethyl-1-nitropropane
(ii) 2-nitro-1 -methyl benzene
(iii) 1, 3, 5 trinitro benzene
(iv) 2-phenyl-1-nitro ethane
(v) 2-methyl-2-nitro propane
Answer:

Question 2.
What are nitro compounds? How are they classified? Give one example for each type.
Answer:
(i) Nitro compounds are considered as the derivaties of hydrocarbons. If one of the hydrogen atom of hydrocarbon is replaced by the —NO₂ group, the resultant organic compound is called a nitrocompound.
(ii) Examples for primary nitro compound.

(iii) Example for secondary nitro compound.

(iv) Example for a tertiary nitro compound.

Question 3.
Write the structural formula of the isomers of the following compound and indicate the type of isomerism involved, (i) 1-nitrobutane,
(ii) nitroethane.
Answer:
(i) (a) 1-nitrobutane and 2-methyl-1-nitropropane are- chain isomers. These two isomers differ in the length of carbon chain and hence exhibit chain isomerism.

(b) 1-nitrobutane, 2-nitrobutane and 2-methyl 2-nitropropane are position isomers. They differ in the position of the functional group and hence exhibit positfon isomerism.

(c) 1-nitrobutane and butyl nitrite are functional isomers. They differ in the nature of the functional group and hence functional isomerism.

(ii) Nitromethane and methyl nitrite exhibit tautomerism.

Question 4.
Methyl nitrite and nitro methane exhibits tautomerism. How will you distinguish between these form?
Answer:

Question 5.
Between 2-nitrobenzene, and 2-methyl – 2- nitro propane which does not exhibit tautomerism? Why?
Answer:
Tautomerism is exhibited by nitro alkanes which contain one or more ‘α’ hydrogen atoms. Then 1° and 2° nitro alkanes exhibit tautomerism while 3° nitro alkanes which do not have a ‘α’ hydrogen atom does not exhibit tautomerism.
2-nitrobutane has a hydrogen atom and hence exhibit tautomerism. While 2-methyl -2- nitropropane does not contain an a hydrogen atom and hence does not exhibit tautomerism.

Question 6.
Explain the acidic nature of nitroalkanes.
Answer:
Primary and secondary nitro alkanes show acidic nature because of the presence of a hydrogen atom.
Because of the presence of electron withdrawing nitro group, the ‘α’ hydrogen atom is abstracted by a base.

Hence, they are acidic. When the number of alkyl groups attached to α- carbon acidity decreases because of +1 effect of alkyl groups.

Question 7.
How will you prepare nitro benzene from (i) ethyl bromide (ii) methane, (iii) α – chloroacetic acid (iv) tert-butyl amine, (v) acetaldoxine.
Answer:
(i) Alkyl bromides (or) iodides on heating with ethanolic solution of potassium nitrite gives nitroethane.

(ii) Gaseous mixture of methane and nitric acid passed through a red hot metal tube to give nitromethane.

(iii) α – chloroacetic acid when boiled with aqueous solution of sodium nitrite gives nitromethane.

(iv) tert-butyl amine is oxidised with aqueous KMnO₄ to give tert – nitro alkanes.

(v) Oxidation of acetaldoxime and acetoneoxime with trifluoroperoxy acetic acid gives nitroethane (1°) and 2- nitropropane (2°) respectively.

Question 8.
Give the equation for the reduction of nitromethane in (a) acid medium and (b) neutral medium.
Answer:
(a) Nitromethane on reduction in acid medium gives methylamine.

(b) Reduction under neutral condition using Zn / NH₄OH or Zn /CaCl₂, hydroxylamine is formed.

Question 9.
Complete the following equation. Identify A, B, and C.


Answer:

Question 10.
How does ethyl nitrite react with (i) Sn / HCl and (ii) HCl / H₂O.
Answer:

Question 11.
What is Nef carbonyl synthesis? Give equation.
Answer:
The formation of an aldehyde by the alkaline hydrolysis of a nitro alkane is known as Nef carbonyl synthesis.

Question 12.
How are the following conversion made?
(i) Benzene to mcta dinitro benzene, (ii) Nitro benzene to nitroso benzene, (iii) Nitro benzene to azo benzene, (iv) Nitro benzene to hydrazo benzene.
Answer:

Question 13.
Identify the reagents used in the following conversions. Write complete equations.
(i) nitrobenzene to aniline
(ii) metadinitrobenzene to metanitroaniline
(iii) nitrobenzene to 3-nitrobenzene sulphonic acid.
Answer:
(i) Any reducing agent: Ni (or) Pt or LiAlH₄

(ii) Ammonium poly sulphide

(iii) Conc. H₂SO₄

Question 14.
Explain why nitro group in nitro benzene is meta directing.
Answer:
The structure of nitro benzene is a resonance hybrid of the following cannonical structures.

As a result of electron withdrawing nature of the NO₂ group, the electron density at ortho and para position deefeases compafed to meta position i.e., the meta position is relatively electron rich compound to ortho and para positions. Hence the electrophile attacks the meta position.

Question 15.
How will you effect the following conversions?
(i) Benzene to m-chloro nitro benzene,
(ii) Benzene to 1, 2, 3 tri nitro benzene
(iii) m-chloro nitro benzenetto m-chloro aniline,
(iv) 1, 3, di nitrobenzene to 3- nitro aniline.
Answer:

Question 16.
Give the IUPAC names of the following:

Answer:
(i) N -phenyl benzenamine
(ii) N, N -diphenyl benzenamine
(iii) N -methyl-N-phenylethanamine
(iv) prop -2- en-1-amine
(v) hexane-1, 6-diamine
(vi) N -methyl-propan-1-amine
(vii) N, N -diethyl-butan-1-amine
(viii) N -ethyl- N-methyl propan-2-amine
(ix) N, N -dimethyl-benzenamine
(x) phenyl-methanamine

Question 17.
Write the structures of the chain isomers of Butan-1-amine.
Answer:

Question 18.
Explain ‘metamerism’ with a suitable example of amines.
Answer:
Aliphatic amines having the same molecular formula but different alkyl groups on either side of the nitrogen atoms show metamerism,
eg: CH₃CH₂NHCH₂CH₃ (diethyl amine) and CH₃NHCH₂CH₂CH₃ (methyl-n-propyl amine)
CH₃NHCH(CH₃₂)₂ (Isopropyl methylamine) are metamers.

Question 19.
What are all the possible isomers of an amine having the molecular formula C₃H₉N and C₄H₁₁N.
Answer:

These amines exhibit functional isomerism.
Amine (C₄H₁₁N): CH₃CH₂CH₂CH₂NH₂ (Butan-1-amine) exhibit chain as well as position isomerism.

Question 20.
Write the structure of the following:
(i) 4N-dimethyl pentan-2-amine
(ii) 2 (N, N-dimethyl) butanamine
(iii) 2-aminoethanol
(iv) 4-aminobutanoicacid
(v) N-methyl-2-nitro pentanamine
(vi) prop-2-en-l -amine.
(vii) N-ethyl-N-methylbenzenamine
(viii) 3-methylbenzenamine
(ix) 2-methoxybenzenamine
(x) N-phenylbenzenamine
Answer:

Question 21.
How is ethanamine prepared from (i) nitroethane, (ii) ethanenitrile, (iii) acetamide, (iv) ethyl bromide?
Answer:
(i) Reduction of nitroethane using H₂ in the presence of Ni or Sn/HCl or Pd/H₂.

(ii) Reduction of ethanenitrile using sodium amalgam in ethanol.

(iii) Reduction of acetamide using LiAlH₄.

(iv) By Gabriel’s phthalimide synthesis.

Question 22.
Identify the products:

Answer:
(i) C₆H₅NH₂ (aniline)
(ii) CH₃NHCH₃ (N-methyl methanamine)
(iii) C₆H₅NH₂ (aniline)
(iv) C₆H₅CH₂NH₂ (benzyl amine)

Question 23.
Using sodium azide (NaN₃) convert methyl bromide to methylamine.
Answer:

Question 24.
How will you prepare aniline from (i) chlorobenzene and (ii) phenol?
Answer:

Question 25.
What happens when
(i) Vapours of ethanol and ammonia are passed over alumina.
(ii) Ethanamide is treated with LiAlH₄. Give equation.
Answer:

Question 26.
Explain Why?
(i) Amines have higher boiling points than hydrocarbons of comparable molecular mass.
(ii) Among isomeric amines 3° amines have the lowest melting point.
(iii) The boiling point of amines are lower than those of alcohols and acids of comparable molecular mass.
(iv) Aliphatic amines with maximum six carbon atoms are soluble in water to some extent, while aromatic amines are insoluble in water.
Answer:
(i) This is due to the reason that amines being polar form inter molecular hydrogen bonding (except tertiary amines which do not have a H atom linked to N atom) and exist as associated molecules.

(ii) The degree of association depends on the extent of hydrogen bonding. Since 1° amines have two, 2° amines have one while 3° amines have no hydrogen linked to nitrogen. Therefore, among isomeric amine 1° amines have highest while 3° amine have the lowest boiling points.
(iii) The electronegativity of nitrogen is lower than oxygen, amines form weaker H—bonding compared to alcohol and carboxylic acids.i.e., The extent of association is less compared to that of alcohols or carboxylic acids. Hence amines will have lower boiling point compared to alcohol or carboxylic acid of comparable molecular mass.
(iv) All the three classes of amines (1°, 2° and 3°) form H bonds with water as a result lower aliphatic amines are soluble in water. As the size of the alkyl group increases (with increase in molecular mass), the solubility decreases due to a corresponding increase in hydrocarbon part of the molecule. The borderline solubility is reached with amines of about six carbon atoms in the molecule. Aromatic amines, on the other hand are insoluble in water. This is due to larger hydrocarbon part which tend to retard the formation of H-bonds.

Question 27.
Account for the fact that among ethyl amines, the decreasing order of basicity in aqueous solution is (CH₃CH₂)₂NH > (CH₃CH₂)₃N > CH₃CH₂NH₂.
Answer:
The basicity of an amine in aqueous solution primarily depends upon the stability of the ammonium cation or the conjugate acid formed by accepting a proton from water. The stability of the ammonium cation in turn depends on the combination of the following factors.
(i) +1 effect of alkyl groups.
(ii) Extent of hydrogen bonding with water molecules.
(iii) Steric effects of the alkyl groups.
In the case of alkyl groups bigger than ‘CH₃’ group i.e., ethyl, propyl, etc there will be steric hindrance to hydrogen bonding. As a result, stability due to +1 effect predominates over the stability due to H—bonding and hence 3° amines because more basic than 1° amines. In all overall decreasing basic strength of ethyl amine follows the sequence (CH₃CH₂)₂NH > (CH₃CH₂)₃N > CH₃CH₂NH₂ > NH₃.

Question 28.
How does nitrous acid (a mixture of sodium nitrite and dil. HCl) react with (i) CH₃NH₂ (ii) (CH₃)₂NH and (iii) (CH₃)₃N? Give equations.
Answer:
Aliphatic primary amines react with nitrous acid give an alcohol and nitrogen gas is produced. Methyl amine is an exception.

Secondary amines will form nitroso alkyl amine.

Tertiary amines react with nitrous acid and form nitrites.

Question 29.
How does nitrous acid react with the following? Give equations.
(i) ethylamine, (ii) di-ethylamine, (iii) triethyl amine, (iv) aniline, (v) N-methyl aniline, (vi) N, N- dimethylaniline.
Answer:
(i) Ethylamine reacts with nitrous acid to give ethyl diazonium chloride, which is unstable and it is converted to ethanol by liberating N₂.

(ii) A yellow dye of N-nitrosodiethylamine is formed.

(iii) Triethylammoniumnitrate which is soluble in water formed.

(iv) Aniline reacts with nitrous acid at low temperature (273 – 278 K) to give benzene diazonium chloride which is stable for a short time and slowly decomposes even at low temperatures.

This reaction is known as diazotisation.
(v)

This reaction is known as Libermann’s nitroso test.
(vi) Aromatic tertiary amine reacts with nitrous acid at 273K to give p-nitroso compound.

Question 30.
Explain why amino group is ortho para directing in electrophilic substitution reactions.
Answer:
The NH₂ group is a strong activating group. In aniline the NH₂ is directly attached to the benzene ring, the lone pair of electrons on the nitrogen is in conjugation with benzene ring which increases the electron density at ortho and para position, thereby facilitating the electrophilic attack at ortho and para.

Question 31.
Aniline gives 2, 4, 6 tribromo aniline when treated with bromine water, but not a mono bromo aniline. Explain why?
Answer:
Aniline is a more powerful activating group. The lone pair of electrons enters into conjugation with the benzene ring, as a result both ortho and para position become rich in electron density. At the same time Br⁺ is also a strong electrophile. Hence, all the three positions are attacked by the electrophile.

Question 32.
Why is it necessary to acetylate aniline to get a mono bromo aniline?
Answer:
Acetylation of aniline reduces the activity of the amino group. The activity of the acetylated amine is reduced because the lone pair of electron on nitrogen is delocalised by the neighbouring carbonyl group by resonance. Hence, it is not easily available for conjugation.

The acetyl amino group is less activating than amino group.

Question 33.
Explain why direct nitration of aniline gives a mixture of ortho, meta and para isomers.
Answer:

Nitration of aniline involves acidic medium. So protonation of aniline takes place forming mainly m-nitro aniline. Hence, direct nitration of aniline forms the following products.

Question 34.
How will you convert aniline?
(i) to p-bromo aniline
(ii) to 2,4,6 tribromo aniline
(iii) to para nitro aniline
(iv) to benzene diazonium chloride
(v) to sulphanilic acid
Answer:

Question 35.
What is zwitter ion? Explain with an example.
Answer:
Zwitter ion is a compound in which both acidic and basic groups are present in the same molecule, eg: sulphanilic acid. It has an acidic (SO₂OH) and a basic group (NH₂). Thus a proton from the acid leaves and protonates the basic group. As a result it exists as a dipolar ion.

Question 36.
How will you distinguish between primary, secondary and tertiary amine?
Answer:

Heisenberg reagent: Benzene sulphoanyl chloride in the presence of excess aqueous KOH solution.

Question 37.
Complete the following reactions:

Answer:

Question 38.
Give examples for (i) Sandmeyer’s reaction (ii) Gattermann reaction.
Answer:
(i) Sandmeyer’s reaction: Benzene diazonium chloride solution on reaction with cuprous chloride in hydrochloric acid (CuCl / HCl) or cuprous bromide in hydrochloric acid (CuBr / HCl) gives corresponding chloro benzene or bromo benzene. This reaction is known as Sandmeyer’s reaction. Aryl cyanides can also be prepared by this method.

(ii) Gattermann reaction: This reaction involves the treatment of benzene diazonium chloride with Cu / HCl or Cu / HBr respectively instead of cuprous halide and hydrogen halide.

Question 39.
How does the following reagents react with benzene diazonium chloride? Give equations.
(i) HBF₄ and the product formed is heated.
(ii) HBF₄ and die product formed is heated with an aqueous solution of sodium nitrite in the presence of copper.
(iii) HBF₄ and the product is heated with CH₃COOH.
(iv) Cuprous cyanide in the presence of KCN.
Answer:

Question 40.
Accomplish die following conversions.
(i) Nitrobenzene to benzene,
(ii) 4-Nitro aniline to 1, 2, 3 -tribromo benzene,
(iii) p-toludine to 2-bromo-4-methyl aniline,
(iv) m-nitro aniline to m-chloroaniline,
(v) p-nitroaniline to p-iodomtro benzene,
(vi) benzyl chloride to 2-phenyl ethanamine.
Answer:
(i) Nitrobenzene to benzene:

(ii) 4 -nitro aniline to 1, 2, 3 – tribromo benzene.

(iii) p-toludine to 2-bromo 4-methyl aniline.


(vi) Benzyl chloride to 2- phenyl ethylamine.

Question 41.
Give equations for the following reactions.
(i) Ethyl bromide is treated with KCN
(ii) Heating acetamide with P₂O₅
(iii) Heating acetaldoxime with P₂O₅
(iv) Treating methyl magnesium bromide with cyanogen chloride.
Answer:

Question 42.
Write the IUPAC names of the following:


Answer:

Question 43.
What is the reducing agent used in the following reduction reactions? Give equations.
(i) Ethanenitrile to ethanamine
(ii) Benzonitrile to benzylamine
(iii) Ethane nitrile to Acetaldimine hydrochloride which on hydrolysis gives acetaldehyde.
Answer:
(a) Complete reduction or strong reduction when reduced with H₂ in the presence of Pt or Ni as catalyst or by using LiAlH₄, or sodium and alcohol alkyl or aryl cyanides yield primary amines.

This specific type of reduction of alkyl or aryl cyanides using LiAlH₄ or Na/ alcohol is named as mendices reduction.
(b) Partial reduction or mild reduction: When SnCl₂ / HCl is used as a reducing agent at room temperature alkyl or aryl cyanides are reduced to amine hydrochloride which on subsequent hydrolysis gives aldehyde. This type of reduction is named as Stephen’s reduction.
SnCl₂ + 2HCl → SnCl₄ + 2 [H]

Question 44.
Give examples for (i) Thrope nitrile condensation and (ii) Levine and Hauser acetylation.
Answer:
(i) Thrope nitrile condensation: Self condensation of two molecules of alkyl nitrile (containing α—H atom) in the presence of sodium to form iminonitrile.

(ii) Levine and Hauser acetylation: The nitriles containing α-hydrogen also undergo condensation with esters in the presence of sodamide in ether to form ketonitriles. This reaction is known as “Levine and Hauser” acetylation. This reaction involves replacement of ethoxy (OC₂H₅) group by methylnitrile (-CH₂CN) group and is called as cyanomethylation reaction.

Question 45.
Write the electronic structure of alkyl cyanides and isocyanides.
Answer:
The electronic structure of alkyl cyanides is

The carbon atom of CN group is sp hybridised. CN group contain one σ and two pi bonds.
The electronic structure of isocyanides is

Structurally isocyanides are represented as a resonance hybrid of the following two forms.

The dipolar form (I) makes higher contribution.

Question 46.
Name the reagents used in the following reactions.

Answer:

Question 47.
What happens when methyl isocyanide is:
(a) treated with dilute HCl.
(b) treated with sodium and alcohol
(c) heated at 250°C
(d) treated with sulphur in the presence of ozone.
Answer:

Question 48.
Mention the uses of (a) Nitro alkanes,
(b) Nitrobenzene,
(c) Cyanides and isocyanides
Answer:
(a) Nitroalkanes:

1. Nitromethane is used as a fuel for cars.
2. Chloropicrin (CCl₃NO₂) is used as an insecticide.
3. Nitroethane is used as a fuel additive and precursor to explosive and they are good solvents for polymers, cellulose ester, synthetic rubber and dyes etc.,
4. 4% solution of ethylnitrite in alcohol is known as sweet spirit of nitre and in used as diuretic.

(b) Nitrobenzene:

1. Nitrobenzene is used to produce lubricating oils in motors and machinery.
2. It is used in the manufacture of dyes, drugs, pesticides, synthetic rubber, aniline and explosives like TNT, TNB.

(c) Cyanides and isocyanides;

1. Alkyl cyanides are important intermediates in the organic synthesis of larger number of compounds like acids, amides, esters, amines etc.
2. Nitriles are used in textile industry in the manufacture of nitrile rubber and also as a solvent particularly in perfume industry.

Question 49.
An organic compound (A) having the molecular formula C₂H₇N is treated with nitrous acid to give (B) of molecular formula C₂H₆O which answers the iodoform test. Identify A and B and explain the reactions.
Answer:

Since, ‘B’ is an alcohol formed from A by the action of nitrous acid, A should be a primary amine which contains ‘NH₂’ group.
C₂H₅N—NH₂=C₂H₅ i.e. The alkyl group is C₂H₅ and hence A is C₂H₅NH₂ ethyl amine. B is ethyl alcohol and it is confirmed by the fact that it answer iodo form test.

Question 50.
An organic compound (A) with molecular formula C₆H₇N gives (B) with HNO₂ / HCl at 273K. The aqueous solution of (B) on heating gives compound (C) which gives violet colour with neutral FeCl₃. Identify the compounds A, B and C. Write the equations.
Answer:
A = C₆H₅NH₂ (aniline);
B = C₆H₂N₂Cl (benzene diazonium chloride)
C = C₆H₅OH (phenol)

(i) Since ‘C’ gives a violet colouration with neutral FeCl₃ it should be phenol.
(ii) Since phenol is obtained by boiling aqueous solution of (B), it should be benzene diazonium chloride.
(iii) Hence, A should be aniline.
Equation:

Choose the correct answer:

1.1, 2 dimethyl- 1-nitropropane. Choose the
incorrect statement about this compound.
(a) It is a primary nitro compound.
(b) It is also called nitroneopentane.
(c) The ‘NO₂’ group is attached to a secondary carbon atom.
(d) It is an aliphatic compound.
Answer:
(c)

2. 1-nitrobutane and 2-methyl-1-nitro propane:
(a) chain isomers
(b) position isomers
(c) functional isomers
(d) metamers
Answer:
(a)
Hint:

3. Which of the following nitro compounds does not exhibit tautomerism?

Answer:
(d)
Hint: Tertiary nitro alkanes do not exhibit tautomerism due to absence of a-H atoms.

4. The incorrect statement between nitroform and acidform of nitromethane is:
(a) nitroform is less acidic whereas aciform is more acidic.
(b) nitroform dissolves in NaOH slowly while aciform dissolves in NaOH instantly.
(c) Both decolorise FeCl₃ solution
(d) They are tautomers
Answer:
(c)
Hint: Nitroform decolorises FeCl₃ solution while aciform gives a reddish brown colour with FeCl₃ solution.

5.

Identify A and B.
(a) A = CH₃CH₂NO₂ ; B = CH₃CH₂NH₂
(b) A- CH₃NO₂ ; B = CH₃NH₂
(c) A = CH₃CHO ; B = CH₃CH₂OH
(d) A = CH₃CH₂NH₂ ; B = CH₃CH₂OH
Answer:
(a)

6.

Identify A and B.


Answer:
(b)

7. The products obtained by the reduction of nitromethane in acid medium and neutral medium are:
(a) methylamine and N-methylhydroxylamine
(b) methylamine and ethanol
(c) N-methylhydroxylamine and methylamine
(d) N-methylhydroxylamine in both cases.
Answer:
(a)
Hint:

8. An amine is boiled with HCl and H₂O. Which of the following will give acetone:
(a) CH₃NH₂
(b) (CH₃)₂CHNO₂
(c) (CH₃)₃CNO₂
(d) both (b) and (c)
Answer:
(b)
Hint: CH₃NH₂ gives CH₃COOH (CH₃)₂
CHNO₂ gives (CH₃)₂CO
(CH₃)₃ CNO₂ no reaction

9. The correct IUPAC name for CH₂=CH—CH₂NH—CH₃ is:
(a) Allylmethyl amine
(b) 2-amino-4-pentane
(c) 4 amino pent-l-ene
(d) N-methyl prop-2 en-1 amine
Answer:
(d)

10. Amongst the following the strongest base in aqueous medium is:
(a) CH₃NH₂
(b) NC—CH₂NH₂
(c) (CH₃)₂NH
(d) C₆H₅NH—CH₃
Answer:
(c)
Hint: 2° amines are more basic than 1° amines, i. e., (CH₃)₂NH is more basic than CH₃NH₂. Due to -I effect of CN group NC—CH₂NH₂ is less basic than even CH₃NH₂. C₆H₅NHCH₃ is less basic than both C₆H₅NH₂ and (CH₃)₂NH due to delocalisation of lone pair of electrons on the N atom into the benzene ring.

11. In order to prepare a 1 ° amine from an alkyl halide with simultaneous addition of one CH₂ group in the carbon chain, the reagent used a source of nitrogen is:
(a) Sodium amide, NaNH₂
(b) Sodium azide, NaN₃
(c) Potassium cyanide, KCN
(d) Potassium phthalimide C₆H₄(CO₂) N^–K^–
Answer:
(c)
Hint: Cyanides on reduction gives 1: amines with a CH₂ group

12. The correct increasing order of basic strength for the following compounds is:

(a) II < III < I
(b) III < I < II
(c) III < II < I
(d) II < I < III
Answer:
(b)

13. Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is:
(a) Hoffmann bromamide reaction
(b) Gabriel’s phthalimide synthesis
(c) Sandmeyer’s reaction
(d) Reaction with NH₃
Answer:
(b)

14. Which of the following compounds is the weakest Bronsted base?

Answer:
(c)
Hint: Amines (a) and (b) have a stronger tendency to accept a proton and hence stronger Bronsted bases than phenol (c) and alcohol (d) since phenol is more acidic than alcohol, phenol (c) has the least tendency to accept a proton and hence the weakest Bronsted base.

15. Which of the following compounds cannot be prepared by Sandmeyer’s reaction?
(I) Chlorobenzene
(II) Bromobenzene
(III) Iodo benzene
(IV) Fluoro benzene
(a) (IV)
(b) (III)
(c) (I) and (II)
(d) (III) and (IV)
Answer:
(a)

16. The products of the following reaction are:


Answer:
(c)
Hint: —NHCOCH₃ is an o/p directing group.

17. An organic compound ‘A’ on treatment with NH₃ gives ‘B’ which on heating gives ‘C’. ‘C’ when treated with Br₂ in the presence of KOH produces ethylamine. The compound ‘A’ is:

Answer:
(d)
Hint:

18. In a set of reactions, meta mono benzoic acid gave a product ‘D’. Identify D.



Answer:
(d)
Hint:

19.

Answer:
(b)
Hint: Diazonium salts of benzylamine is not stable. It decomposes insitu, to form benzyl alcohol.

20. Predict the product:


Answer:
(d)

21. Aniline in a set of following reactions yielded a coloured product Y. The structure of Y would be:

Answer:
(b)
Hint:

22. Butylamine (I), diethyl amine (II) and N, N diethyl amine, (III) have the same molar mass. The increasing order of their boiling points is:
(a) III < II < I
(b) I < II < III
(c) II < III < I
(d) III < I < II
Answer:
(a)
Hint: Extent of H-bonding increases as the number of H-atoms increases and hence boiling point increases in the order 3°<2°< 1°.

23. Assertion: Ammonolysis of alkyl halides is not a suitable method for the preparation of primary amines.
Reason: Ammonolysis of alkyl halides mainly produces 2° amines.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true and reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(c)

24. Assertion: Gabriel phthalimide reaction can be used to prepare aryl and arylalkyl amines.
Reason: Aryl halides are as reactive as alkyl halides towards nucleophilic substitution reactions.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true and reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(d)
Hint: Correct assertion: Gabriel phthalimide reaction can be used to prepare alkyl and aryl alkyl primary amines.
Correct reason: Alkyl and aralkyl halides undergo nucleophilic substitution reaction but aryl halides do not.

25. Assertion: Aniline does not undergo Friedel Craft’s reaction.
Reason: Friedel crafts reaction is an electrophilic substitution reaction.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true and reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(b)
Hint: Correct explanation: AlCl₃ forms a salt with aniline (C₆H₅NH₂⁺ AlCl₃^–) which deactivates the benzene ring thereby preventing Friedel craft’s reaction.

Students get through the TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Answer the following questions.

Question 1.
Write the IUPAC names of the following:

Answer:
(i) 3-phenyl prop-2-en-l-al
(ii) cyclohexanecarbaldehyde
(iii) 3-oxopentan-l-al
(iv) But-2-en-l-al

Question 2.
Give the structure of the following compounds.
(i) 4 – Nitropropiophenone
(ii) 2 – Hydroxycyclopentane carbaldehyde
(iii) phenyl acetaldehyde
Answer:

Question 3.
Write the IUPAC names of
(i) Diacetone alcohol
(ii) Crotonaldehyde
Answer:
(i) 4-Hydroxy-4-methylpentan-2-ol
(ii) But-2-en-l-al

Question 4.
Write the structure of
(i) 3 – oxopentanal
(ii) 1 – phenylpentan – 1 – one
Answer:

Question 5.
Name the following compounds according to IUPAC system of nomenclature.
(i) CH₃CH(CH₃)CH₂CH₂CHO
(ii) CH₃CH₂CO(C₂H₅)CH₂CH₂Cl
(iii) CH₃CH=CHCHO
(iv) CH₃COCH₂COCH₃
(v) CH₃CH(CH₃)CH₂(CH₃)₂COCH₃
(vi) (CH₃)₃CCH₂COOH
(vii) OHCC₆H₄CHO(P)
Answer:
(i) 4-methyl pentanal
(ii) 6-chloro-4-ethylhexan-3 -one
(iii) But-2-en-l-al
(iv) Pentane 2,4, dione
(v) 3, 3, 5 Trimethylhexan – 2 – one
(vi) 3, 3, Dimethyl butanoic acid
(vii) Benzene 1,4, dicarbaldehyde

Question 6.
Draw the structures of the following compounds.
(i) 3-methyl butanal
(ii) 4-nitro propiophenone
(iii) p-methyl benzaldehyde
(iv) 4-methylpent-3-en-2-one
(v) 4-chloropentan-2-one
(vi) 3-Bromo-4-phenyl pentanoic acid
(vii) p, p’ dihydroxy benzo phenone
(viii) Hex-2-en-4 yonic acid
Answer:

Question 7.
Write IUPAC names of the following aldehydes and ketones. Wherever possible give also common names.
(i) CH₃CO(CH₂)₄CH₃
(ii) CH₃CH₂CH(Br)CH₂CH(CH₃)CHO
(iii) CH₃(CH₂)₄CHO
(iv) Ph—CH=CH.CHO
(v)

(vi) PhCOPh
Answer:

Question 8.
Give the product of ozonolysis of the following alkenes.

Answer:
To identify the product of ozonolysis, draw a line in between C=C bond and ‘O’ on either side of the double bond

Question 9.
Identify the products of hydration of the following:
(i) ethyne
(ii) prop-1-yne
(iii) Hex-1-yne
(iv) Diphenyl acetylene
Answer:
Hydration means addition of water. The reagent used is mercuric sulphate and dilute sulphuric acid at 333K. Ketones are formed on hydration of alkynes. MarkovnikofFs mle is followed in water addition.

Question 10.
How are the following compounds formed by distillation of calcium salts of their carboxylic acids?
(i) Formaldehyde,
(ii) Acetaldehyde,
(iii) Acetone,
(iv) Butan-2-one,
(v) Cyclopentanone,
(vi) Benzaldehyde
Answer:
(i) Calcium formate alone is heated.

(ii) By dry distillation of calcium acetate and calcium formate.

(iii) Dry distillation of calcium acetate alone.

(iv) Calcium acetate and calcium propionate are distilled together.

(v) Cyclic ketones are formed when calcium salts of dibasic acids are heated.

(vi) By dry distillation of a mixture of calcium benzoate and calcium formate.

Question 11.
Complete the following equations:

Answer:

Question 12.

(i) Identify the product.
(ii) Name the reaction.
(iii) What is the intermediate formed in the reaction.
Answer:
(i) CH₃CHO (acetaldehyde)
(ii) Stephen’s reaction
(iii) CH₃—CH=NH (an imine)
Complete reaction:

Question 13.

(i) Name the reagent i.e., X used for the above reduction.
(ii) Why other reducing agents like H₂, in the presence of a catalyst are not used.
(iii) Write the IUPAC name of the product formed.
Answer:
(i) Diisobutyl aluminium hydride (DIBAL – H) is used as a reducing agent.
X = H₂O
(ii) It selectively reduces alkyl cyanides to imines, which on hydrolysis gives aldehydes. The double bond in the nitrile is not reduced. The other reducing agents reduce the double bond also.
(iii) hex-4-enal is the IUPAC name of the product.

Question 14.
How is benzaldehyde prepared from (i) methyl benzene, (ii) benzene, (iii) benzyl alcohol. Give equations.
Answer:
(i) Oxidation of methyl benzene, using chromyl chloride as oxidising agent gives benzaldehyde. Acetic anhydride and CrO₃ can also be used for reaction.

(ii) Benzene is treated with CO and HCl Benzaldehyde is formed.

This reaction is known as Gattermann- koch reaction.
(iii) Oxidation of benzyl alcohol using PCC (pyridinium chromo chromate) gives benzaldehyde.

Question 15.
Identify the product of the following reactions. Write the complete equation.

Answer:

Question 16.
Give examples for Friedel-crafts acetylation reaction.
Answer:
Refer Question 15 (ii) and (iii). These reaction are called Friedel-crafts reaction. When acetyl chloride is used it is known as acetylation reaction. When benzoyl chloride is used, it is called benzoylation reaction.

Question 17.
The boiling points of aldehydes and ketones are high compared to hydrocarbons and ethers of comparable molecular mass. Explain why?
Answer:
It is due to the weak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions.

Question 18.
Explain nucleophilic addition reaction with an example.
Answer:
The carbonyl carbon carries a small degree of positive charge. Nucleophile such as CN^– can attack the carbonyl carbon and uses its lone pair to form a new carbon – nucleophile ‘σ‘ bond, at the same time two electrons from the carbon – oxygen double bond move to the most electronegative oxygen atom. This results in the formation of an alkoxide ion. In this process, the hybridisation of carbon changes from sp² to sp³.

The tetrahedral intermediate can be protonated by water or an acid to form an alcohol.

eg: Addition of HCN
Attack of CN^– on carbonyl carbon followed by protonation gives cyanohydrins.

Question 19.
Explain why the carbonyl group in aldehydes and ketones is polar.
Answer:
The carbonyl group of aldehydes and ketones contains a double bond between carbon and oxygen. Oxygen is more electronegative than carbon and it attracts the shared pair of electron which makes the carbonyl group as polar and hence aldehydes and ketones have high dipole moments.

Question 20.
What is meant by the following terms? Give an example in each case.
(i) Cyanohydrin,
(ii) Acetal,
(iii) Semicarbazone,
(iv) Aldol
(v) Hemiacetal,
(vi) Oxime
(vii) Ketal,
(viii) Imine,
(ix) 2,4, DNP derivative
(x) Schiff’s base
Answer:
(i) gem – Hydroxy nitriles i.e., compounds containing hydroxyl and cyano groups on the same carbon atom are called cyano hydrins. These are produced by the addition of HCN to aldehydes and ketones in weakly basic medium.

(ii) gem – dialkoxy alkanes in which two alkoxy groups attached to the terminal carbon atom are called acetals. They are produced by the action of an aldehyde with two equivalent of monohydric alcohol in the presence of dry HCl gas.

(iii) Semi carbazones are derivatives of aldehydes and ketones and are produced by the action of semicarbazides on them in a weakly acidic medium.

(iv) Aldols are β – hydroxy aldehydes or ketones and are produced by the condensation of two molecules of the same or one molecule each of two different aldehydes or ketones (containing an alpha hydrogen atom) in the presence of dilute, aqueous NaOH.

(v) α – alkoxy alcohols are called hemi acetals. They are produced by the addition of one molecule of a mono hydric alcohol to an aldehyde in the presence of dry HCl gas.

(vi) Oximes are produced when aldehydes or ketones react with hydroxyl amine in weakly acidic medium.

(vii) gem-dialkoxy alkanes in which two alkoxy groups present on the same carbon atom are called ketals.
They are produced when a ketone is heated with ethylene glycol in the presence of dry HCl gas.


(viii) Compounds containing >C=N group are called imines. These are produced when aldehydes and ketones react with ammonia and its derivatives.

When ‘Z’ = H, alkyl, aryl,—NH, —OH, C₆H₅ etc..,
(ix) 2, 4, dinitro phenyl hydrazones (i.e., 2, 4, DNP derivative) are produced when aldehydes or ketones react with 2,4 dinitro phenyl hydrazine in weakly acidic medium.

(x) Aldehydes and ketones react with primary aliphatic or aromatic amines in the presence of acid form azomethenes or schifTs bases. Schiffs bases may also be regarded as imines.

Question 21.
How does acetaldehyde react with (i) hydroxyl amine, (ii) hydrazine, (iii) phenyl hydrazine, (iv) semicarbazide? Give equations.
Answer:

Question 22.
Give equations for the reaction between
(i) acetaldehyde and sodium bisulphite
(ii) acetone and sodium bisulphite
Answer:

Question 23.
Write the structure of the prodcuts formed when acetone reacts with (i) hydrazine, (ii) phenyl hydrazine, (iii) semicarbazide.
Answer:

Question 24.
What is urotropine? How it is formed? Write its structure.
Answer:
Hexamethelenetetramine is called urotropine. It is formed when formaldehyde reacts with ammonia.
6HCHO + 4NH₃ → (CH₂)₆N₄ + 6H₂O
Structure:

Question 25.
Mention the uses of urotropine.
Answer:

1. Urotropine is used as a medicine to treat urinary infection.
2. Nitration of Urotropine under controlled condition gives an explosive RDX (Research and development explosive). It is also called cyclonite or cyclotri methylenetrinitramine.

Question 26.
What is popoffs rule? How it is used to predict the oxidation products of unsymmetrical ketones.
Answer:
It states that during the oxidation of an unsymmetrical ketone, a (C—CO) bond is cleaved in such a way that the keto group stays with the smaller alkyl group.
Eg:

Question 27.
How will you prepare
(i) diacetone amine from acetone.
(ii) aldimine from acetaldehyde.
(iii) hydrobenzamide from benzaldehyde.
Answer:
(i)

Two molecules of acetone, combine with ammonia to form diacetoneamine.
(ii)

(iii) With ammonia, benzaldehyde forms a complex condensation product called hydrobenzamide.

Question 28.
What is a haloform reaction? Explain with suitable example.
Answer:
Aldehydes and ketones containing CH₃CO group (methyl carbonyl group), on treatment with excess halogen in the presence of an alkali produce a haloform (chloro form, bromo form, iodoform). This reaction is known as haloform reaction.

Question 29.
What is iodoform test? Explain.
Answer:
Compounds containing CH₃CO (methyl carbonyl) group or compounds on oxidation give compounds containing CH₃CO group, when treated with iodine and NaOH gives a yellow precipitate is called iodoform. This is known as iodoform test.

Question 30.
How will you distinguish between by means of a chemical test.
(i) Propanal and propanone
(ii) acetaldehyde and benzaldehyde
(iii) Ethanal and propanal
Answer:
All these pairs of compounds can be distinguished by iodoform test.
(i) Propanone (CH₃ CO CH₃) will answer iodoform test. Propanal will not undergo iodo form test.

(ii) Acetaldehyde (CH₃CHO) will answer iodoform test, whereas benzaldehyde does not answer iodoform test

(iii) Ethanal (acetaldehyde) will answer iodoform test, but not propanal acetaldehyde

Question 31.
Write a short note an aldol condensation.
Answer:
The carbon attached to carbonyl carbon is called α – carbon and the hydrogen atom attached to α – carbon is called α – hydrogen. In presence of dilute base NaOH, or KOH, two molecules of an aldehyde or ketone having α – hydrogen add together to give β – hydroxyl aldehyde (aldol) or β – hydroxyl ketone (ketol). The reaction is called aldol condensation reaction. The aldol or ketol readily loses water to give α, β – unsaturated compounds which are aldol condensation products.
Acetaldehyde when warmed with dil.NaOH gives β – hydroxyl butraldehyde (acetaldol)

Question 32.
Give the mechanism of aldol condensation.
Answer:
Mechanism: The mechanism of aldol condensation of acetaldehyde takes place in three steps.
Step 1: The carbanion is formed as the α – hydrogen atom is removed as a proton by the base.

Step 2: The carbanion attacks the carbonyl carbon of another unionized aldehyde to form an alkoxide ion.

Step 3: The alkoxide ion formed is protonated by water to form aldol.

The aldol rapidly undergoes dehydration on heating with acid to form α – β unsaturated aldehyde.

Question 33.
Give the oxidation products obtained when pentan-2-one is oxidised by (conc.HNO₃).
Answer:
During these oxidations, rupture of the carbon-carbon bonds occur on either side of the keto group giving a mixture of carboxylic acid, each containing less a number of carbon atom. Than the original ketone.

Question 34.
What is crossed aldol condensation? Give an example.
Answer:
Aldol condensation can also takes place between two different aldehydes or ketones or between one aldehyde and one ketone such an aldol condensation is called crossed or mixed aldol condensation. This reaction is not very useful as the product is usually a mixture of all possible condensation products and cannot be separated easily.

Question 35.
What happens when benzaldehyde reacts with (i) acetaldehyde in the presence of dilute NaOH (ii) acetone in the presence of dilute NaOH.
[OR] Explain with examples Claisen – Schmidt condensation.
Answer:
Benzaldehyde condenses with aliphatic aldehyde or methyl ketone in the presence of dil. alkali at room temperature to form unsaturated aldehyde or ketone. This type of reaction is called Claisen – Schmidt condensation.

Question 36.
Write a short note on Cannizaro reaction.
Answer:
In the presence of concentrated aqueous or alcoholic alkali, aldehydes which do not have α – hydrogen atom under go self oxidation and reduction (disproportionation) to give a mixture of alcohol and a salt of carboxylic acid. This reaction is called Cannizaro reaction.
Benzaldehyde on treatment with concentrated NaOH (50%) gives benzyl alcohol and sodium benzoate.

This reaction is an example of disproportion reaction.

Question 37.
Write the mechanism of Cannizaro reaction.
Answer:
Cannizaro reaction involves three steps.
Step 1: Attack of OH^– on the carbonyl carbons.

Step 2: Hydride ion transfer

Step 3: Acid – base reaction

Cannizaro reaction is a characteristic of aldehyde having no α – hydrogen.

Question 38.
Give an example for crossed Cannizaro reaction.
Answer:
When Cannizaro reaction takes place between two different aldehyde (neither containing an α hydrogen atom), the reaction is called as cross cannizaro reaction.

In crossed cannizaro reaction more reactive aldehyde is oxidized and less reactive aldehyde is reduced.

Question 39.
Give an example for benzoin condensation.
Answer:
Benzaldehyde reacts with alcoholic KCN to form benzoin

Question 40.
Complete the following equation.


Answer:

Question 41.
How will you convert ethanal into following compounds?
(i) Butane 1, 3 diol
(ii) But -2-enal
(iii) But-2-enoic acid.
Answer:

Question 42.
Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Acetophenone and benzophenone
(ii) Phenol and benzoic acid
(iii) Pentan -2- one and pentan – 3-one
(iv) Benzaldehyde and acetophenone
Answer:
(i) Acetophenone and benzophenone: Acetophenone being a methyl ketone, when treated with I₂ / NaOH (NaOI) gives a yellow precipitate of iodoform but benzophenone does not give a yellow precipitate.

(ii) Phenol and benzoic acid: NaHCO₃ test: Benzoic acid being a stronger acid than carbonic acid (H₂CO₃) decomposes NaHCO₃ to evolve CO₂, but phenol being weaker than carbonic acid does not.

FeCl₃ Test: Phenol gives a violet colouration with neutral ferric chloride solution, while benzoic acid gives a buff coloured precipitate of ferric benzoate.

(iii) Pentan-2-one and pentan-3-one:
Sodium bisulphite test: 2-Pentanone being a methyl ketone when treated with a saturated solution of sodium bisulphite gives a white precipitate of 2 – pentanone sodium bisulphite addition compound where as 3 – pentanone does not.


(Note: Sterically hindered ketones, do not undergo this reaction, eg: acetophenone)
Iodoform Test: 2 – pentanone, being a methyl ketone, when treated NaOI (I₂ / NaOH) gives a yellow precipitate of iodoform but 3-pentanone does not.

(iv) Benzaldehyde and acetophenone:
AgNO₃ test: Benzaldehye reduces Tollen’s reagent to give a silver mirror (Ag) but acetophenone does not reduce Tollen’s reagent.

Iodoform test: Acetophenone gives a yellow precipitate when treated with I₂ / NaOH but benzaldehyde does not.

Question 43.
What happens when benzaldehyde is treated with
(i) Br₂ in the presence of FeBr₃
(ii) a mixture of conc.H₂SO₄ and conc.HNO₃
(iii) conc.H₂SO₄
(iv) chlorine
(v) chlorine in the presence of FeCl₃
(vi) methyl bromide in the presence of anhydrous AlCl₃.
Answer:
(i) m-bromobenzaldehyde is formed.

(ii) m-nitrobenzaldehyde is formed.

(iii) m- benzaldehyde sulphonic acid is formed:

(iv) Benzoyl chloride is formed:

(v) m-chloro benzaldehyde is formed:

Question 44.
Mention the uses of (i) formaldehyde (ii) urotropine (iii) acetaldehyde (iv) acetone (v) benzaldehyde (vi) acetophenone and benzophenone.
Answer:
Formaldehyde:

1. 40% aqueous solution of formaldehyde is called formalin. It is used for preserving biological specimens.
2. Formalin has hardening effect, hence it is used for tanning.
3. Formalin is used in the production of thermosetting plastic known as bakelite, which is obtained by heating phenol with formalin.

Urotropine:

1. Urotropine is used as a medicine to treat urinary infection.
2. Nitration of Urotropine under controlled condition gives an explosive RDX (Research and development explosive). It is also called cyclonite or cyclotri methylene trinitramine.

Acetaldehye:

1. Acetaldehyde is used for silvering of mirrors.
2. Paraldehyde is used in medicine as a hypnotic.
3. Acetaldehyde is used in the commercial preparation of number of organic compounds like acetic acid, ethyl acetate etc.,

Acetone:

1. Acetone is used as a solvent, in the manufacture of smokeless powder (cordite).
2. It is used as a nail polish remover.
3. It is used in the preparation of sulphonal, a hypnotic.
4. It is used in the manufacture of thermosoftening plastic Perspex.

Benzaldehyde is used:

1. as a flavoring agent
2. in perfumes
3. in dye intermediates
4. as starting material for the synthesis of several other organic compounds like cinnamaldehyde, cinnamic acid, benzyl chloride etc.

Aromatic Ketones:

1. Acetophenone has been used in perfumery and as a hypnotic under the name hyphone.
2. Benzophenone is used in perfumery and in the preparation of benzhydrol drop.

Question 45.
Give equation for the following reactions.
(i) Nitration of acetophenone
(ii) Bromination of benzophenone
(iii) Friedel – crafts alkylation of benzo phenone
Answer:

Question 46.
Explain the structure of carboxylic acid group.
Answer:
A carboxyl group is made up of a carboxyl group joined to a hydroxyl group.

In the carboxyl group, the carbon atom is attached to two carbon atoms: one by a double bond and the other by a single bond which in turn linked to a hydrogen atom by a single bond. The remaining free valency of the carbon atom of the carboxyl is satisfied by a H atom or an alkyl group. Thus, the structure of carboxylic acids.

where ‘R’ is ‘H’ or alkyl group.
The carbon atom and the two oxygen atoms are sp2 hybridised. The two sp² hybridesed orbitals of the carboxyl carbon overlap with one sp² hybridised orbital of each oxygen atom while the third sp² hybridised orbital of carbon overlaps with either a ‘s’ orbital of H – atom or a sp³ hybridised orbital of ‘C’ atom of the alkyl group to form three a bonds. Each of the two oxygen atoms are left with one unhybridised orbital which is perpendicular to the a bonding skeleton.

All the three ‘p’ orbitals being parallel, overlap to form n bond which is partly delocalised between carbon and oxygen atoms on one side and carbon and oxygen of the ‘OH’group on the other side.

In other words, RCOOH may be represented as a resonance hybrid of the following two canonical structures.

As a result of resonance (i) the C—O single bond length in carboxylic acids is shorter than the normal C—O single bond in alcohols and ethers and (ii) C=O bond length in carboxylic acids is slightly larger than the normal C=O bond length in aldehydes and ketones.

Question 47.
How is acetic acid prepared from (i) ethanol (ii) methyl cyanide (iii) ethyl acetate (iv) methyl magnesium bromide (v) acetyl chloride (vi) acetic anhydride? Give equation.
Answer:
(i) Oxidation of ethanol by acidified K₂Cr₂O₇.

(ii) Acid hydrolysis of methyl cyanide.

(iii) Acid hydrolysis of ethylacetate

(iv) Methyl magnesium bromide reacts with carbondioxide and the product formed on hydrolysis gives acetic acid.

(v) Hydrolysis of acetyl chloride

(vi) Hydrolysis of acetic anhydride

Question 48.
Identify X and Y


Answer:
(i) X = C₆H₅COOMg Br; Y = C₆H₅COOH
(ii) X = C₆H₅COOH; Y = C₆H₅COCl
(iii) X = C₆H₅COOH; Y = C₆H₅COONa

Question 49.
Explain why?
(i) Carboxylic acids have higher boiling point than aldehydes, ketones, or even alcohols of comparable molecular mass.
(ii) Lower aliphatic / carboxylic acid are miscible with water while higher carboxylic acids are immiscible with water.
Answer:
(i) Carboxylic acids have higher boiling point ’ than aldehydes, ketones and even alcohols of comparable molecular masses. This is due to more association of carboxylic acid molecules through intermolecular hydrogen bonding.

(ii) Lower aliphatic carboxylic acids (up to four carbon) are miscible with water due to the formation of hydrogen bonds with water. Higher carboxylic acid are insoluble in water due to increased hydrophobic interaction of hydrocarbon part. The simplest aromatic carboxylic acid, benzoic acid is insoluble in water.

Question 50.
How will you convert
(i) Ethyl benzene to benzoic acid
(ii) Isopropyl benzene to benzoic acid
(iii) p-nitro toluene to p-nitrpbenzoic acid
(iv) o-xylene to phthalic acid
(v) But-2-ene to ethanoic acid
(vi) Benzonitrile to benzoic acid
(vii) Ethyl magnesium iodide to propanoic acid
(viii) Ethyl benzoate to benzoic acid
(ix) Benzamide to benzoic acid
(x) Propan – 2 – one to butanoic acid.
Answer:

Question 51.
Which of the following compounds will undergo aldol condensation? Which the Cannizaro reaction, and which neither? Write the structure of the expected products of aldol condensation and Cannizaro reaction.
(i) Methanal
(ii) 2 – methylpentanal
(iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vii) Phenyl acetaldehyde
(viii) Butan-1-ol
(ix) 2, 2-dimethyl butanal
Answer:
I. 2 – methyl pentanal
Cyclohexanone
1 – phenyl propanone and phenyl acetaldehyde
contain one or more a hydrogen and hence undergo aldol condensation.



II. Methanal, benzaldehyde, and 2,2 dimethyl butanal do not contain a hydrogen atoms and hence undergo cannizaro reaction. The reaction and structures are:

III. Benzophenone is a ketone with no a hydrogens while butan-1-ol is an alcohol. Both of these neither undergo aldol condensation nor cannizaro reaction.

Question 52.
Give equations for reactions of acetic acid with the following reagents.
(i) Na (ii) NaOH (iii) Na₂CO₃ (iv) PCl₅ (v) SOCl₂ (vi) C₂H₅OH (vii) LiAlH₄ (viii) Red P and HI (ix) Sodalime (x) NH₃ followed by heating (xi) Cl₂ / Red P.
Answer:

Question 53.
Show how each of the following could be converted to benzoic acid:
(i) Ethyl benzene,
(ii) acetophenone,
(iii) benzophenone,
(iv) phenyl ethene.
Answer:

Question 54.
What is esterification? Give an example.
Answer:
The formation of an ester by the action of carboxylic acid and alcohol in the presence of an acid is called esterification.

In these reaction the C—OH bond in acids is cleaved.

Question 55.
Give a brief accounts of decarboxylation reaction.
Answer:
Removal of CO₂ from carboxyl group in called as decarboxylation. Carboxylic acids lose carbon di oxide to form hydrocarbon when their sodium salts are heated with soda lime (NaOH and CaO in the ratio 3:1)

Question 56.
Suggest a suitable reagent to bring about the following conversions. Give equations.
(i) Acetic acid to acetyl chloride
(ii) Benzoic acid to ethyl benzoate
(iii) m – nitro benzoic acid to m – nitro methyl benzoate
(iv) Ethanoic acid to ethanol
(v) Acetic acid to ethane
Answer:
(i) PCl₅ or SOCl₂

(ii) C₂H₅OH in the presence of dil H₂SO₄

(iii) Methyl alcohol in the presence of dil. H₂SO₄

(iv) Lithium aluminium hydride

(v) HI in the presence of Red phosphorus

Question 57.
Give the products of the following:


Answer:
All these reactions are decarboxylation reactions. In these reactions the COONa group is replaced by ‘H’.

Question 58.
What is HVZ reaction? Give an example.
Answer:
Carboxylic acids having an α – hydrogen are halogenated at the α – position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to form α halo carboxylic acids. This reaction is known as Hell-Volhard-Zelinsky reaction (HVZ reaction) The α-Halogenated acids are convenient starting materials for preparing α – substituted acids, eg:
Substitution reaction in the hydrocarbon part α – Halogenetion:

Question 59.
Explain why the ‘COOH’ group in benzoic acid in meta directing.
Answer:
The structure of benzoic acid is a resonance hybrid of the following canonical structure.

Because of -M effect of carboxylic acid, the ortho and para positions becomes less in electron density relative to the meta position, i.e., meta-position becomes rich in electron density compared to ortho and para positions. Hence, the electrophile attacks the meta position, i.e., the COOH group deactivates the benzene ring and is the meta directing group.

Question 60.
How are the following compounds prepared from benzoic acid?
(i) m- Bromo benzoic acid
(ii) m- nitro benzoic acid
(iii) m-sulpho benzoic acid.
Answer:
(i) Bromination of benzene gives m – bromo benzoic acid.

(ii) Nitration of benzene using nitrating mixture gives m – nitro benzoic acid.

(iii) Sulphonation of benzene, using fuming sulphuric acid gives m – sulpho benzoic acid.

Question 61.
Formic acid is a reducing agent. Substantiate this statement with examples.
Answer:
Formic acid contains both an aldehyde as well as an acid group. Hence, like other aldehydes, formic acid can easily be oxidised and therefore acts as a strong reducing agent.

(i) Formic acid reduces Tollens reagent (ammonical silver nitrate solution) to metallic silver.

(ii) Formic acid reduces Fehlings solution. It reduces blue coloured cupric ions to red coloured cuprous ions.

Question 62.
Give the tests for carboxylic acid group in an organic compound.
Answer:

1. In an aqueous solution, carboxylic acid turns blue litmus red.
2. Carboxylic acids give brisk effervescence with sodium bicarbonate due to the evolution of carbon dioxide.
3. When a carboxylic acid is warmed with alcohol and conc.H₂SO₄ it forms an ester, which is detected by its fruity odour.

Question 63.
Give a brief account of the acidity of carboxylic acids.
Answer:
Carboxylic acids undergo ionization to produce H⁺ and carboxylate ions in an aqueous solution. The carboxylate anion is stabilized by resonance which makes the carboxylic acid to donate the proton easily.

The resonance structure of carboxylate ion are given below.

The strength of carboxylic acid can be expressed in terms of the dissociation constant(K_a):

The dissociation constant is generally called acidity constant because it measures the relative strength of an acid. The stronger the acid, the large will be its K_a value.
The dissociation constant of an acid can also be expressed in terms of p^K_a value.
pK_^a = log K_a
A stronger acid will have higher K_a value but smaller pK_^a value, the reverse is true for weaker acids.

Question 64.
Briefly discuss the effect of substituents on the acidity of carboxylic acids.
Answer:
Effect of substituents on the acidity of carboxylic acid.
(i) Electron releasing alkyl group decreases the acidity: The electron-releasing groups (+1 groups) increase the negative charge on the carboxylate ion and destabilize it and hence the loss of proton becomes difficult. For example, formic acid is more stronger than acetic acid,
eg:

(ii) Electron withdrawing substituents increases the acidity: The electron-withdrawing substituents decrease the negative charge on the carboxylate ion and stabilize it. In such cases, the loss of proton becomes relatively easy.
Acidity increases with increasing electronegativity of the substituents. For example, the acidity of various halo acetic acids follows the order F—CH₂—COOH > Cl—CH₂—COOH > Br—CH₂—COOH > I —CH₃—COOH
Acidity increases with increasing number of electron – withdrawing substituents on the a – carbon.
For example
Cl₃C—COOH > Cl₂CH—COOH > ClCH₂COOH > CH₃COOH
The effect of various, electron withdrawing groups on the acidity of a carboxylic acid follows the order,
—NO₂ > —CN >—F >—Cl >—Br >—I >Ph
The relative acidities of various organic compounds are
RCOOH > ArOH > H₂O > ROH >RC ≡ CH

Question 65.
Fluorine is more electronegative than chlorine even their p- fluoro benzoic acid in weaker than p- chloro benzoic aicd. Explain.
Answer:
Since halogens are far more electronegative than carbon and also possess lone pairs of electrons, they exert both -1 and + M effects. Now, the fluorine, the lone pairs of electrons are present in ‘2p’ orbitals and in chlorine, they are present in ‘3p’ orbitals. Since ‘2p’ orbitals of fluorine and chlorine are of almost equal size, the + M effect is more pronounced in p fluoro benzoic acid than in p-chloro benzoic effect.

Question 66.
Explain why p – nitrobenzoic acid has a higher K_a value than benzoic acid.
Answer:
Higher the K_a value, the stronger is the acid. Thus, p-nitro benzoic acid is a stronger acid than benzoic acid. This is due to

1. Because of — I and -M effect of NO₂ group the electron density in O—H bond decreases. As a result, the —O—H bond becomes weak and hence, p – nitrobenzoic acid easily loses a proton than benzoic acid.
   
2. Due to -1 and -M effect of nitrogroup dispersal of negative charge occurs and hencep -nitro benzoate ion becomes more stable than benzoate ion.
   

Question 67.
Give the IUPAC names of the following compounds,
(i) PhCH₂CH₂COOH
(ii) (CH₃)₂C=CHCOOH CH

Answer:
(i) 3 -phenyl propanoic acid
(ii) 3 -methyl but – 2-enoic acid
(iii) 2 -methylcyclopentane carboxylic acid
(iv) 2, 4, 6 -trinitro benzoic acid or 2, 4, 6-trinitro benzene carboxylic acid.

Question 68.
Explain the mechanism for the reaction.

Answer:
Esterification of carboxylic acids with alcohols is a nucleophilic acyl substitution.
In such reactions the nucleophile first adds to a carbonyl group to give a tetrahedral intermediate which then readily loses the leaving group to give the substitution products. The mechanism involves the following steps.
Step 1: Protonation of the carbonyl group:
In the presence of mineral acids (H₂SO₄ or HCl), the carbonyl group of the carboxylic acid accepts a proton to form protonated carboxylic acid.

Step 2: Nucleophilic attack of the alcohol molecule. As a result of protonation, the carbonyl carbon becomes electrophilic (i.e. more electropositive) and hence readily undergoes nucleophilic attack by lone pairs of electrons on the oxygen atom of the alcohol molecule to form a tetrahedral intermediate (II).

Step 3: Loss of a molecule of water and a proton. The tetrahedral intermediate (II) undergoes a proton transfer to give another tetrahedral intermediate (III). During this proton transfer, the OH group gets converted to – OH₂ group which being a good leaving group is lost as a neutral water molecule. The protonation ester (IV), thus formed, finally loses a proton to give the ester (V).

Question 69.
Give the uses of (i) formic acid (ii) acetic acid (iii) benzoic acid (iv) acetyl chloride (v) acetic anhydride (vi) ethyl acetate.
Answer:
Formic acid: It is used

1. for the dehydration of hides
2. as a coagulating agent for rubber latex
3. in medicine for the treatment of gout
4. as an antiseptic in the preservation of fruit juice.

Acetic acid: It is used

1. as table vinegar
2. for coagulating rubber latex
3. for manufacture of cellulose acetate and poly vinylacetate.

Benzoic acid: It is used

1. as food preservation either in the pure form or in the form of sodium benzoate
2. in medicine as an urinary antiseptic
3. for manufacture of dyes.

Acetyl chloride: It is used

1. as acetylating agent in organic synthesis
2. in detection and estimation of —OH, — NH₂ groups in organic compounds.

Acetic anhydride: It is used

1. acetylating agent
2. in the preparation of medicine like asprin and phenacetin
3. for the manufacture plastics like cellulose acetate and poly vinyl acetate.

Ethyl acetate is used

1. in the preparation of artificial fruit essences
2. as a solvent for lacquers
3. in the preparation of organic synthetic reagent like ethyl acetoacetate.

Uses
Acetamide is used in the preparation of primary amines.

Question 70.
A compound with molecular formula, C₄H₁₀O₃ on acetylation with acetic anhydride gives a compound with molecular weight 190. Find out the number of hydroxyl groups present in the compound.
Answer:
During acetylation one ‘H’ atom (at mass = 1 amu) of the OH group is replaced by an acetyl group (CH₃CO – molar mass = 43 amu)
—OH + (CH₃CO)₂O → —O—COCH₃ + CH₃COOH
In other words acetylation of each OH group increases the molecular mass by 43 – 1 = 42 amu. Now that the molecular mass of the compound C₄H₁₀O₃ = 106 amu, while that of the acetylated product is 190 amu. Therefore the number of ‘OH’ groups present in the compound is \(\frac{190-106}{42}\) =2.

Question 71.
Explain (i) Perkins’ reaction (ii) Knoevenagal reaction.
Answer:
(i) Perkins’ reaction:
When an aromatic aldehyde is heated with an aliphatic acid anhydride in the presence of the sodium salt of the acid corresponding to the anhydride, condensation takes place and an α, β unsaturated acid is obtained. This reaction is known as Perkin’s reaction.

(ii) Knoevenagal reaction:

Benzaldehyde condenses with malonic acid in presence of pyridine forming cinnamic acid, Pyridine act as the basic catalyst.

Question 72.
Give names of reagents which bring about the following conversions.
(i) Hexan-1-ol to hexanal
(ii) Cyclohexanol to cyclohexanone
(iii) p-fluorotoluene to p-fluorobenzaldehyde,
(iv) Ethanenitrile to ethanal
(v) allyl alcohol to propenal
(vi) But-2-ene to ethanal
Answer:

Question 73.
Bring out Che following conversions.
(i) Benzyl alcohol to phenyl ethanoic acids.
(ii) 3 – nitrobromo benzene to – 3 – nitrobromo benzoic acid
(iii) Methyl acetophenone to benzene 1, 4, dicarboxylic acid
(iv) cyclohexene to hexane 1, 6 dicarboxylic acid.
Answer:

Question 74.
Identify A – E in the following reactions:
Answer:

Question 75.
Write a short note on the electrophilic substitution reaction of benzaldehyde.
Answer:
The ‘CHO’ group in benzaldehyde deactivates the benzene ring due to its —M effect. As a result the electron densities in ortho and para position are decreased compared to the meta position, i.e, The meta position is electron rich compared to ortho and para position. Hence electrophilic substitution reaction occur at meta position.

Question 76.
Which acid of each pair shown here would you expect to be stronger?


Answer:
(i)

(ii) Due to much stronger -1 effect of F over Cl FCH₂ COO^– ion is much more stable than ClCH₂COO ion and hence FCH₂COOH is a stronger acid than ClCFI₂COOH.
(iii)

Inductive effect decreases with increasing distance, therefore, -I effect of is somewhat stronger in 3-fluorobutanoic acid than in 4-fluorobutanoic acid. In other words, CH₃CHCH₂COCT ion is in more stable than
FCH₂CH₂CH₂COO^– ion and hence  is a stronger acid than FCH₂CH₂CH₂COOH.
Therefore due to greater stability of CF₃—C₆H₄—COO^– (p) ion over CH₃—C₆H₄COOH^– (p) is a stronger acid than H₃CC₆H₄—COOH (p).
(iv)

Question 77.
Identify A to E in the following reactions:
Answer:

(Note: NaBH₄ reduces COCl to CH₂OH, but does not reduce NO₂ to NH₂ group.)

Question 78.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene
(ii) Benzoic acid to benzaldehyde
(iii) Ethanol to 3 – hydroxy butanal
(iv) Benzene to m – nitrobenzene
(v) Benzaldehyde to benzophenone
(vi) bromo benzene to 1-phenyl ethanol
(vii) Benzaldehyde to 3 phenylpropan-1-ol
(viii) Benzaldehyde to a – hydroxy phenyl acetic acid
(ix) Benzoic acid to p-nitrobenzyl alcohol.
Answer:
(i) Propanone to propene:

(ii) Benzoic acid to benzaldehyde

(iii) Ethanol to 3 – hydroxy butanal:

(iv) Benzene to m – dinitrobenzene:

(v) Benzaldehyde to benzophenone:

(vi) Bromobenzene to 1 – phenyl ethanol:

(vii) Benzaldehyde to 3 – phenylpropan – 1 – ol:

(viii) Benzaldehyde to a – hydroxyphenyl acetic acid:

(ix) Benzoic acid to m-nitrobenzyl alcohol:

Question 79.
What is transesterification? Give an example.
Answer:
Esters of an alcohol can react with another alcohol in the presence of a mineral acid to give the ester of second alcohol. The interchange of alcohol portions of the esters is termed transesterification.

Question 80.
Give an example for Claisen condensation.
Answer:
Esters containing at least one α – hydrogen atom undergo self-condensation in the presence of a strong base such as sodium ethoxide to form β – keto ester. This reaction is known as Claisen condensation.

Question 81.
How are the following compounds prepared?
(i) acetyl chloride and ethyl chloride from ethyl acetate
(ii) acetamide from acetyl chloride
(iii) Acetamide from acetic anhydride.
Answer:

Question 82.
Complete the following equation:
(i) CH₃COCl + CH₃NH₂ →
(ii) CH₃COCl + (CH₃)₂ NH →
(iii) (CH₃CO)₂O + PCl₅ →
Answer:

Question 83.
How are the following compounds prepared from acetamide?
(i) acetic acid (ii) sodium acetate (iii) methyl cyanide (iv) methyl amine (v) ethyl amine
Answer:

Question 84.
Write the structure and IUPAC names of the following acid derivatives.
(i) acetyl chloride (ii) propionyl chloride (iii) Benzoyl chloride (iv) acetic anhydride (v) propionic anhydride (vi) benzoic anhydride (vii) methyl acetate (viii) Ethyl acetate (ix) phenyl acetate (x) acetamide (xi) propionamide (xii) Benzamide.
Answer:

Question 85.
Briefly explain amphoteric nature of acid amide.
Answer:
Amides behave both as weak acid as well as weak base and thus show amphoteric character. This can be proved by the following reactions.
Acetamide (as base) reacts with hydrochloric acid to form salt.

Acetamide (as acid) reacts with sodium to form sodium salt and hydrogen gas is liberated.

Question 86.
An unknown aldehyde (A) on reaction with alkali gives a β – hydroxy aldehyde which loses water to form an unsaturated aldehyde, 2-butanal. Another aldehyde (B) undergoes a disproportionation reaction in the presence of conc.alkali to form products (C) and (D). (C) is an aryl alcohol with formula C₇H₈O. Identify (A) and (B).
(ii) Write the sequence of chemical reaction involved.
(iii) Name the product, when (B) reacts with zinc amalgam and hydrochloric acid.
Answer:
Since the aldehyde (A) oft reaction With alkali gives a β – hydroxy aldehyde (i.e, an aldol), which loses water to form 2 – butanal (an unsaturated aldehyde) ‘A’ must be acetaldehyde.

Since aldehyde (B) on treatment with conc.alkali undergo disproportionation (i.e., Cannizaro reaction) to give two products (C) and (D). One of these products must be an alcohol and the other must be the corresponding acid. Further, since, the product (C) is an aryl alcohol with molecular C₇H₅O, (C) must be benzyl alcohol, (i.e, C₆H₅ aryl group) OH is alcoholic group C₆H₅O. C₇H₈O—C₆H₆O = CH₂. i.e, C₆H₅CH₂OH. (D) must be sodium benzoate and aldehyde (B) must be benzaldehyde.

Thus aldehyde (A) is acetaldehyde, (B) is benzaldehyde.
(ii) The sequence of reaction is already explained above.
(iii) When (B) reacts with Zn—Hg / HCl, toluene obtained.

Question 87.
A compound ‘X’ (C₂H₄O) on oxidation gives ‘Y’ (C₂H₄O₂). X undergoes a haloform reaction. On treatment with HCN, (X) forms (Z) which on hydrolysis gives 2 – hydroxy propanoic acid.
(i) Write down the structures of ‘X’ and ‘Z’.
(ii) Name the product when ‘X’ is treated with dilute NaOH.
(iii) Write down the equation for the reaction involved.
Answer:
(a) Since the MF C₂H₄O of X corresponds to the general formula C_nH_2nO, characteristic of aldehydes or ketones. (X) may either be an aldehyde or ketone. Since ketone must have at least ‘3’ carbon atoms but X (C₂H₄O) has only ‘2’ two. Therefore (X) must be an aldehyde. The only possible aldehyde is CH₃CHO (MF C₂H₄O) (acetaldehyde).
If (X) is acetaldehyde, than the compound (Z) is acetic acid (CH₃COOH).
(b) (X) is acetaldehyde is confirmed by the following reaction.
(i) It undergoes haloform reaction
(i.e.,)

(ii) On treatment with KCN, acetaldehyde forms a cyanohydrin (Z) which on hydrolysis gives 2 – hydroxy propanoic acid.

(iii) In the presence of dilute NaOH, acetaldehyde undergoes aldol condensation to form 3- hydroxy butanal commonly called aldol.

Choose the correct answer:

1. The IUPAC name of the compound is

(a) Mesityl oxide
(b) 4- methyl pent – 3 – en – 2 – one
(c) 4 – methyl pent – 2 – en – 4 – one
(d) 2 – methyl pent – 4 – one
Answer:
(b)

2. Ozonolysis of 2-methyl but-2-ene followed by treatment with Zn/H₂O gives:
(a) ethanal
(b) propanone
(c) propanal and prop – 2 – one
(d) ethanal and propan – 2 – one
Answer:
(d)
Hint:

3. Identify the product ‘Y’ in the following reaction sequence.

(a) pentane
(b) cyclobutane
(c) cyclopentane
(d) cyclopentanone
Answer:
(c)
Hint:

4. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having molecular mass 44 u. The alkene is:
(a) 2 – butene
(b) ethene
(c) propene
(d) 1 – butene
Answer:
(a)
Hint: The aldehyde with molecular mass 44 u is CH₃CHO (acetaldehyde). Therefore the symmetrical alkene is 2 – butene.

5. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:
(a) a vinyl group
(b) an isopropyl group
(c) an acetylene triple bond
(d) two ethylenic double bonds
Answer:
(a)
Hint:

6. Identify the compound ‘X’ in the following reaction:

Answer: (a)
Hint:

7. The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds.
I. CH₃CHO,
II. (CH₃)₂CO and
III. PhCOPh is:
(a) III > II > I (b) II > I > III
(c) I > III > II (d) I > II > III
Answer:
(d)
Hint: Reaction between RMgX with carbonyl compound is nucleophilic addition. The greater the positive charge on the carbonyl carbon, the faster is the rate of nucleophilic addition.

8. A carbonyl compound reacts with hydrogen cyanide to form a cyanohydrin which on hydrolysis forms a racemic mixture of a – hydroxy acids. The carbonyl compound is:
(a) formaldehyde
(b) acetaldehyde
(c) acetone
(d) diethyl ketone
Answer:
(b)
Hint: Acetaldehyde cyanohydrin on hydrolysis gives lactic acid
This contains a chiral carbon and exists as a racemic mixture. All other compounds are symmetrical and hence, their cyanohydrins and their corresponding α – hydroxy acids are not optically active because they do not contain chiral carbon.

9. In a set of reactions, acetic acid yielded the product D

The structure of ‘D’ would be:

Answer:
(a)
Hint:

10. (CH₃)₂C=CHCOCH₃ can be oxidised to (CH₃)₂C=CHCOOH by:
(a) Chormic acid
(b) NaOI
(c) Cu at 300°C
(d) KMnO₄
Answer:
(b)
Hint:

11. A compound ‘A’ (C₅H₁₀C₃₁₂) on hydrolysis gives C₅H₁₀O which reacts with NH₂OH, forms iodoform but does not give Fehling’s test A is:

Answer:
(a)
Hint: Since the compound ‘A’ on hydrolysis gives C₅H₁₀O does i ot reduce Fehlings solution but reacts with NH₂OH, forms iodoform, it must be methyl ketone.

12. Acetophenone, when reacted with a base, C₂H₅ONa, yields a stable compound that has the structure:

Answer:
(c)
Hint:

13.

The structure of ‘B’ is:

Answer: (a)
Hint:

14. Self-condensation of two moles of ethyl acetate in presence of sodium ethoxide yields:
(a) ethyl propionate
(b) ethyl butyrate
(c) acetoacetic ester
(d) methyl acetoacetate
Answer:
(c)
Hint:

15. m- chlorobenzaldehyde on reaction with conc. KOH at room temperature gives:
(a) potassium-m-chloro benzoic acid and m-hydroxy benzaldehyde.
(b) m-hydroxy benzaldehyde and m-chloro benzyl alcohol.
(c) m-chloro benzyl alcohol and m-hydroxy benzyl alcohol.
(d) potassium-m-chloro benzoate and m-chloro benzyl alcohol.
Answer: (d)
Hint:

16. Consider the following compounds:

Which will give the iodoform test?
(a) only I
(b) both I and II
(c) only II
(d) all
Answer:
(c)
Hint: Although contains a CH₃CO group linked to carbon, it does not undergo iodoform test. This is because iodination occurs at the more reactive CH₂ group rather than terminal CH₃ which is essential for iodoform test to occur

17. Fehling solution will oxidise:

(a) All
(b) only I and IV
(c) only II and IV
(d) only III and IV
Answer:
(d)
Hint: Fehling solution oxidises only aliphatic aldehydes

18.

The product ‘Z’ is:
(a) benzaldehyde
(b) benzoic acid
(c) benzene
(d) toluene
Answer:
(b)
Hint:
X = Benzene
Y = Toluene
Z = Benzoic acid

19. Match entries of column I with appropriate entries of column II.

(a) (A) – (p)- (B) – (s)- (C) – (q); (D) – (r)
(b) (A) – (q); (B) – (s); (C) – (p); (D) – (r)
(c) (A) – (r); (B) – (s); (C) – (p); (D) – (q)
(d) (A) – (r); (B) – (p); (C) – (q); (D) – (s)
Answer:
(c)

20. Assertion: Fehling solution oxidizes acetaldehyde to acetic acid but not benzaldehyde to benzoic acid.
Reason: The C—H bond in benzaldehyde is stronger than acetaldehyde.
(a) Both assertion and reason are correct and the reason is the correct explanation of assertion.
(b) Both assertion and reason are correct but the reason is not the correct explanation of assertion.
(c) Assertion is true but reason is wrong.
(d) Both assertion and reason are wrong.
Answer:
(a)

21. Assertion: Carboxylic acids contain a carbonyl group but do not give characteristic reactions of the carbonyl group.
Reason: Due to resonance, the electrophilic nature of the carboxyl cation is greatly reduced as compared to carbonyl carbon in aldehydes and ketones.
(a) Both assertion and reason are correct and the reason is the correct explanation of assertion.
(b) Both assertion and reason are correct but the reason is not the correct explanation of assertion.
(c) Assertion is true but the reason is wrong.
(d) Both assertion and reason are wrong.
Answer:
(a)

22. Among the following acids which has the lowest pK_a value?
(a) CH₃COOH
(b) HCOOH
(c) (CH₃)₂CHCOOH
(d) CH₃CH₂COOH
Answer:
(b)
Hint: HCOOH is the strongest acid and hence the lowest pK_a value.

23. Which of the following presents the correct order of the acidity in the given compounds?
(a) FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH
(b) CH₃COOH > BrCH₂COOH > ClCH₂COOH > FCH₂COOH
(c) FCH₂COOH > CH₃COOH > BrCH₂COOH > ClCH₂COOH
(d) BrCH₂COOH > ClCH₂COOH > FCH₂COOH > CH₃COOH
Answer:
(a)

24. The correct acidity order of the following is

(a) III > IV > II > I
(b) IV > III > I > II
(c) III > II > I > IV
(d) II > III > IV > I
Answer: (a)
Hint: Carboxylic acids are stronger than phenols. Further, electrons donating groups decrease the acidity of phenol/carboxylic acid while electron-withdrawing groups increase the acidity of phenol/carboxylic acid. Thus, the over all order.
III > IV > II > I

25. Which one of the following pairs gives effervescence with aq.NaHCO₃
I. CH₃COCI
II. CH₃COCH₃
III. CH₃COOCH₃
IV. CH₃COOCOCH₃
(a) I and II
(b) I and IV
(c) II and III
(d) I and III
Answer:
(b)
Hint: Both CH₃COCl and CH₃COOCOCH₃ react with water to produce acids and hence give effervescence with NaHCO₃.

26. Propionic acid with Br₂/P yields a dibromo product. Its structure would be:
(a) CH(Br)₂—CH₂COOH
(b) CH₂(Br)CH₂COBr
(c) CH₃C(Br)₂COOH
(d) CH₂(Br)CH(Br)COOH
Answer:
(c)
Hint:Bromination occurs at a position.

27. When benzoic acid is treated with LiAlH₄, it forms:
(a) benzaldehyde
(b) benzyl alcohol
(c) benzene
(d) toluene
Answer:
(b)

28. Sodium ethoxide reacts with ethanoyl chloride. The compound that is produced in the above reaction is:
(a) 2-butanone
(b) ethyl chloride
(c) ethyl ethanoate
(d) diethyl ether
Answer:
(c)
Hint: CH₃COCl + C₂H₅ONa → CH₃COOC₂H₅ + NaCl

29. Consider the following:

The correct decreasing order of their reactivity towards hydrolysis is:
(a) II > IV > I > III
(b) II > IV > III > I
(c) I > II > III > IV
(d) IV > II > I > III
Answer:
(a)
Hint: Electron donating groups decrease while electron-withdrawing groups increase the reactivity of acid chlorides towards hydrolysis.

The reactivity depends on the positive charge on the cabonyl carbon.

Students get through the TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Answer the following questions.

Question 1.
Classify the following as primary, secondary and tertiary alcohols.


Answer:
Primary alcohols: (i), (ii), (iii)
Secondary alcohols: (iv), (v)
Tertiary alcohol: (vi)

Question 2.
Identify allyl alcohols in the above (Q.No. 1) examples.
Answer:
(ii) and (vi).

Question 3.
Name the following compounds according to IUPAC system:
Answer:

Question 4.
Write IUPAC names of the following:
Answer:

Question 5.
Write the structures of the compounds whose IUPAC names are as follows:
Answer:

Question 6.
Complete the following reactions:


Answer:

Question 7.
Using suitable Grignard reagent and a carbonyl compound, how will you prepare
(i) phenyl methanol,
(ii) butan-2-ol,
(iii) 2-methylhexan-2-ol,
(iv) propan-2-ol.
Answer:
(i) phenyl methanol:

(ii) butan-2-ol:

(iii) 2-methylhexan-2-ol:

(iv) propan-2-ol:

Question 8.
Give equation for the preparation of 2-methyl-2-propanol using a suitbale Grignard reagent and a carbonyl compound.
Answer:
Tertiary containing at least two identical groups can be prepared by the addition of a Grignard reagent to an ester other the formic ester followed by acid hydrolysis.

Question 9.
Name the reducing agents that will convert a carbonyl compound to an alcohol. Give an example for each.
Answer:
Raney Ni, / Sodium analgen in H₂O (Na—Hg/H₂O), H₂ in the presence of Pt.

where ‘R’ is an alkyl group.

Question 10.
Why is LiAlH₄ is used to reduce crotonaldehyde?
Answer:
Li crotonaldehyde is an unsaturated aldehyde (CH₃CH=CHCHO). Lithium aluminium hydride selectively reduces ‘CHO’ group to ‘CH₂OH’ group without affecting the C=C.

Question 11.
Name the product formed when sodium borohydride is used as a reducing agent for the reduction of a compound containing both carbonyl and carboxyl group, (or) Identify the product:
Answer:

Question 12.
Complete the following reactions:


Answer:

Question 13.
How is glycerol prepared from triglycerides or fats?
Answer:
The alkaline hydrolysis of these fats gives glycerol and the reaction is known as saponification.

Question 14.
Complete the following equations:

Answer:

Question 15.
Give mechanism of reaction between alcohols and halogen acids.
Answer:
Primary alcohols react by S_N2 mechanism while secondary and tertiary alcohol react by S_N1 mechanism.

Question 16.
How will you distinguish between

Answer:

(ii) Secondary alcohol will give a blue colour as shown above whereas a tertiary alcohol will not produce any colour.

Question 17.
Give the structures of the products you would expect when each of the following alcohol reacts with (a) HCl-ZnCl₂, (b) HBr, (c) SOCl₂ (i) Butan-l-ol and (ii) 2-methylbutan-2-ol.
Answer:
(a) with HCl-ZnCl₂ (Lucas reagent):
[2-methylbutan-2-ol being a tertiary alcohol reacts with Lucas reagent to produce turbidity immediately due to formation of insoluble tert-alkyl-halide, while butan-l-ol being a primary alcohol does not react with Lucas reagent at room temperature.]

(b) with HBr: Both alcohols produce the corresponding alkyl bromides.

(c) with SOCl₂: Both alcohols react to form their corresponding alkyl chlorides.

Question 18.
Arrange the following compounds in increasing order of boiling points, pentan-1 -ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol. Give reason for your answer.
Methanol < ethanol < propan-1-ol < butan-2- ol < butan-1-ol < pentan-1-ol.
Answer:
Boiling point increases regularly as the molecular mass increases due to the corresponding increase in their vander waals forces of attraction. The boiling point increases in the order.
methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol.
Among isomeric alcohol, 2° alcohols have lower boiling point than 1° alcohols due to corresponding decrease in the extent of hydrogen bonding because of steric hindrance. Thus the boiling point of butan-2-ol is lower than butan-1-ol.

Question 19.
Explain why propanol has higher boiling point than that of hydrocarbon butane?
Answer:
The molecuels of butane are held together by weak vander waals forces of attraction while these of propanol are held together by stronger hydrogen bonding.

Therefore boiling point of propanol is higher than that of butane.

Question 20.
Write the mechanism for the reaction

Answer:
The reaction follows S_N1 mechanism.
Step 1: Formation of protonated alcohol.

Step 2: Protonated alcohols forms carbocation.

Step 3: Elimination of a proton to form alkane.

Question 21.
Give mechanism for the formation of prop-1 – ene from propan-1-ol.
Answer:
The reaction is

Mechanism:
Step 1: Formation of protonated alcohol.

Step 2: Formation of carbocation.

Step 3: Elimination of a proton.

Question 22.
State Saytzeff rule.
Answer:
During intramolecular dehydration, if there is a possibility to form a carbon-carbon double bond at different locations, the preferred location is the one that gives the more (highly) substituted alkene i.e., the stable alkene.

Question 23.
Name the reagents in the following reactions:
(i) oxidation of a primary alcohol to a carboxylic acid.
(ii) oxidation of a primary alcohol to an aldehyde.
(iii) benzyl alcohol to benzoic acid.
(iv) butan-2-one to butan-2-ol.
Answer:
(i) Acidified potassium dichromate or neutral or acidic or alkaline potassium permanganate (followed by hydrolysis with dilute H₂SO₄).
(ii) Pyridinium chloro chromate(PCC) in CH₂Cl₂ or pyridinium dichromate (PDC) in CH₂Cl₂.
(iii) Acidified or alkaline KMnO₄ (followed by hydrolysis with dil H₂SO₄).
(iv) Ni/H₂ or NaBH₄ or LiAlH₄.

Question 24.
What happen when,
(i) vapours of ethanol is passed over heated copper at 573 K.
(ii) vapours of propan-2-ol is passed over heated copper at 573 K.
(iii) vapours of 2-methyl propan-2-ol is passed over heated copper at 573 K. Give equation.
Answer:
(i) Ethanal is formed.

(ii) Propanone is formed.

(iii) 2-methyl prop-1-ene is formed.

Question 25.
What is nitroglycerine? How it is formed? Give equation.
Answer:
Glycerotrinitrate (1,2,3, trinitroxy propane) is nitroglycerine. It is formed by nitration of glycerine in the presence of conc. H₂SO₄.

Question 26.
How is acrolein formed from glycerol?
Answer:
When glycerol is heated with dehydrating agents like conc. H₂SO₄, KHSO_4, or P₂O₅ acrolein is formed.

Question 27.
Mention the cause for the acidic nature of alcohols.
Answer:
The acidic character of alcohols is due to the electronegative oxygen atom which withdraws electrons of the O—H bond towards itself. As a result, the O—H bond becomes weak and hence a proton can be easily abstracted by a base, i.e., ionization of alcohol to yield alkoxide ion and H+ ions are facilitated.
ROH ⇌ RO^– + H⁺

Question 28.
Mention the uses of (i) methanol, (ii) ethanol, (iii) ethylene glycol, (iv) glycerol.
Answer:
(i) Uses of methanol:
(a) Methanol is used as a solvent for paints, varnishes, shellac, gums, cement, etc.
(b) In the manufacture of dyes, drugs, perfumes, and formaldehyde.

(ii) Uses of ethanol:
(a) Ethanol is used as an important beverage.
(b) It is also used in the preparation of

• Paints and varnishes.
• Organic compounds like ether, chloroform, iodoform, etc.,
• Dyes, transparent soaps.

(c) As a substitute for petrol under the name power alcohol used as fuel for airplane.
(d) It is used as a preservative for biological specimens.

(iii) Uses of ethylene glycol:
(a) Ethylene glycol is used as an antifreeze in an automobile radiators.
(b) Its dinitrate is used as an explosive with DNG.

(iv) Uses of glycerol:
(a) Glycerol is used as a sweetening agent in confectionery and beverages.
(b) It is used in the manufacture of cosmetics and transparent soaps.
(c) It is used in making printing inks and stamp pad ink and lubricant for watches and clocks.
(d) It is used in the manufacture of explosives like dynamite and cordite by mixing it with china clay.

Question 29.
Alcohols are easily protonated in comparison to phenols. Explain why.
Answer:
In phenols, the lone pair of electrons on the oxygen atom is delocalized over the benzene ring due to resonance and hence, not easily available for protonation. In contrast, in alcohols, the lone pair of electrons on the oxygen atom is localized due to the absence of resonance and hence are easily available for protonation.

Question 30.
Identify the major product obtained in the following reaction:

Answer:

Question 31.
Identify the product of the reaction:

Answer:
In the presence of a weak nucleophile such as ethanol, neopentyl bromide ionizes to form first a 1° carbocation which rearranges to form a more stable 3° carbocation. This is then attacked by the weak nucleophile ethanol followed by a less proton to yield ethyl neopentyl ether.

Question 32.
Show how 2-methyl propan-1-ol is prepared by the reaction of a suitable Grignard reagent.
Answer:

This part that has been enclosed in the box comes from methanal while the remaining part comes from the Grignard reagent. Therefore the Grignard reagent should be isopropyl magnesium bromide. Thus,

Question 33.
Write the structure of the products of the following reactions.

Answer:

(ii) NaBH₄ (sodium bromohydride) is a reducing agent. It reduces aldehydes and ketones but not esters.

(ii) NaBH₄ reduces, -CHO group to CH₂OH group. Thus,

Question 34.
What is esterification? Give an example.
Answer:
Formation of an ester from an alcohol and an organic acid in the presence of a catalyst is known as esterification. In the reaction, the ‘OH’ of the carboxylic acid and ‘H’ of the alcoholic group are removed as water.
Eg:

Question 35.
Why are alcohols are weaker acids than water?
Answer:
In water, ‘O’ atom is attached to two ‘H’ atom, but in alcohols, ‘O’ atom is attached to one ‘H’ atom and one ‘R’ group (alkyl group). Since ‘R’ group has an electron-donating inductive effect (+I), it increases the electron density in the O—H bond compared to that O—H bond in water. In other words, —O—H bond in alcohols is stronger than the —O—H bond in water. As a result, proton removal from alcohol is more difficult than in water. Hence, alcohols are weaker acids than water.

Question 36.
The acidic nature of alcohols is in the order 1° alcohol > 2° alcohol > 3° alcohol. Explain.
Answer:

Alkyl (R) groups are electron releasing (i.e., +I effect) the electron density in the O—H bond in tertiary alcohols is maximum followed by secondary alcohol, while in primary alcohols, it is minimum. This means that O—H bond in tertiary alcohol is the strongest and hence most difficult to break followed by O—H bond in the secondary alcohols, while the O—H bond in primary alcohols is the weakest and hence most easy to break. Thus the acidic strength is in the order.
Primary > Secondary > Tertiary

Question 37.
Arrange the following compounds in increasing order against each.
(i) CH₃CH₂OH, CF₃CH₂OH. CCl₃CH₂OH (acid strength)
(ii) 2-methyl-2-propanol, 1-butanol, 2-butanol (reactivity towards sodium)
Answer:
(i) Due to -I effect of halogens, the electron density in the O—H bond decreases. As a result, the O—H bond strength becomes weak and proton removal is easier compared to CH₃CH₂OH. Further, since fluorine has a stronger -I effect, than chlorine, CF₃CH₂COOH is a stronger acid than CCl₃CH₂COOH. Hence, CF₃CH₂OH is the strongest acid while CH2CH2OH is the weakest acid. Hence increasing order of acid strength is
CH₃CH₂OH < CCl₃CH₂OH < CF₃CH₂OH.
(ii) It is an acid-base reaction since alcohols are acidic and sodium is a strong base. As such, the reactivity of these alcohols towards sodium increases as the acidic character of alcohols increases. Since the acidic character increases in the order 3° < 2° < 1°. Therefore the reactivity towards alcohol increases in the same order.

Question 38.
How will you convert phenol to (i) benzene, (ii) aniline, (iii) phenylacetate, (iv) phenyl benzoate, (v) anisole?
Answer:
(i) Benzene: Phenol is converted to benzene on heating with zinc dust. In this reaction, the hydroxyl group which is attached to the aromatic ring is eliminated.

(ii) Aniline: Phenol on heating with ammonia in presence of anhydrous ZnCl₂ gives aniline.

(iii) Phenyl acetate:

(iv) Phenyl benzoate:

(v) Anisole:

Question 39.
Identify A and B in the following:

Answer:

(ii) A= HNO₂ (273 -278) K
B = H₂O

Question 40.
What is picric acid? How it is prepared? 2,4,6 trinitrophenol is called picric acid.
Answer:
Nitration with cone. HNO₃ and conc.H₂SO₄ gives picric acid.

Question 41.
Give equation for the following:
(i) nitration of phenol,
(ii) sulphonation of phenol,
(iii) nitrosation of phenol.
Answer:
(i) Nitration of phenol: Phenol can be nitrated using 20% nitric acid even at room temperature, a mixture of ortho and para nitro phenols are formed.

(ii) Sulphonation of phenol: Phenol reacts with cone. H₂SO₄ at 280 K to form o-phenol sulphonic acid as the major product. When the reaction is carried out at 373 K the major product is p-phenol sulphonic acid.

(iii) Nitrosation of phenol: phenol can be readily nitrosated at low temperature with HNO₂.

Question 42.
What happens when phenol is treated with
(i) acidified potassium dichromate,
(ii) hydrogen in the presence of nickel as a catalyst,
(iii) bromine water,
(iv) bromine in the presence of CCl₄.
Answer:
Phenol treated with
(i) acidified potassium dichromate (oxidation):
Phenol undergoes oxidation with air or acidified K₂CrO₇ with conc. H₂SO₄ to form 1, 4-benzoquinone.

(ii) hydrogen in the presence of nickel as catalyst (reduction): Phenol on catalytic hydrogenation gives cyclohexanol.

(iii) bromine water (halogenation): phenol reacts with bromine water to give a white precipitate of 2,4,6-tribromo phenol.

(iv) bromine in the presence of CCl₄: If the reaction is carried out in CS₂ or CCl₄ at 278 K, a mixture of ortho and para bromo phenols are formed.

Question 43.
Explain why the phenolic group is ortho para directing.
Answer:
The lone pair of electrons on the oxygen atom of the phenolic group enter into conjugation with the benzene ring due +M effect of the phenolic group. As a result, the ortho and para position become electron-rich compounds to the meta position. Hence the electrophilic attack the ortho and para position.

Question 44.
Write a short note on Reimer-Tiemann reaction.
Answer:
On treating phenol with CHCl₃ / NaOH, a -CHO group is introduced at the ortho position.
This reaction proceeds through the formation of substituted benzal chloride intermediate.

Question 45.
How will you distinguish between alcohol and phenol?
Answer:

1. Phenol reacts with benzene diazonium chloride to form a red-orange dye, but ethanol has no reaction with it.
2. Phenol gives purple coloration with neutral ferric chloride solution, alcohols do not give such coloration with FeCl₃.
3. Phenol reacts with NaOH to give sodium phenoxide. Ethyl alcohol does not react with NaOH.

Question 46.
Mention the uses of phenol.
Answer:

1. About half of the world production of phenol is used for making phenol-formaldehyde resin. (Bakelite).
2. Phenol is a starting material for the preparation of
   (a) drugs such as phenacetin, salol, aspirin, etc.
   (b) phenolphthalein indicator.
   (c) explosive like picric acid.
3. It is used as an antiseptic-carbolic lotion and carbolic soaps.

Question 47.
Give the IUPAC names of the following ethers.


Answer:
(i) 1-ethoxy-2-methyl propane
(ii) 2-chloro-l-methoxy ethane
(iii) 4-nitroanisole
(iv) 1-methoxy propane
(v) 4-ethoxy-1, 1, dimethyl cyclohexane
(vi) ethoxy benzene

Question 48.
Ethers are soluble in water. Give reason.
Answer:
The oxygen of ether can also form hydrogen bonds with water and hence they are miscible with water. Ethers dissolve a wide range of polar and non-polar substances.

Question 49.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis.
(i) 1 -propoxypropane
(ii) Ethoxy benzene
(iii) 2-methyl-2-methoxy propane
(iv) 1-methoxyethane
Answer:
(i)

Question 50.
Give the mechanism for the conversion of (i) ethanol to ethoxyethane, (ii) dehydration of 1-propanol to 1-proproxy propane.
Answer:
(i)

[Acid catalyzed 1° alcohol to ethers occurs via S_N2 reaction, involving a nucleophilic attack by the alcohol molecule on the protonated alcohol molecule.]

Question 51.
Explain why the boiling point of ethers is slightly higher than that of alkanes and lower than that of alcohols of comparable molecular mass.
Answer:
Due to weak dipole-dipole interactions, the boiling point of ethers is only slightly higher than those of u-alkanes having comparable molecular masses.
The boiling points of ethers are less than that of alcohols is that ethers do not form hydrogen bonds.

Question 52.
Mention the uses of (i) Diethyl ether and (ii) Anisole.
Answer:
Uses of Diethyl ether:

1. Diethyl ether is used as a surgical anesthetic agent in surgery.
2. It is a good solvent for organic reactions and extraction.
3. It is used as a volatile starting fluid for diesel and gasoline engines.
4. It is used as a refrigerant.

Uses of anisole:

1. Anisole is a precursor to the synthesis of perfumes and insecticide pheromones
2. It is used as a pharmaceutical agent.

Question 53.
Discuss the mechanism of the reaction
CH₃OCH₂CH₃ + HI → CH₃I + CH₃CH₂OH.
Answer:
This reaction is an S_N2 reaction. The cleavage of ethers by halogen acids occurs by the following mechanism.
Step 1: Ethers being Lewis acids, undergo protonation to form oxonium ions.

This reaction takes place with HBr or HI because they are sufficiently acidic.
Step 2: Iodide ion is a good nucleophile. Due to steric hindrance, it attacks the smaller alkyl group of oxonium ion formed in step 1 and displaces the alcohol molecule by S_N2 mechanism as shown below:

Step 3: when excess HI is used, ethanol formed reacts with another molecule of HI to form ethyl iodide.

Question 54.
Explain why ‘OCH₃’ in the anisole group is ortho-para directing.
Answer:
The lone pair of electrons on the oxygen atom of the alkoxy group enters into the conjugation with the benzene ring due to its +M effect.

As a result, the ortho and para position become rich in electron density compared to the meta position. Hence, the electrophile attacks ortho and para position.

Question 55.
Briefly account of dipolar nature of C—O bonds in ethers.
Answer:
Because of the greater electronegativity of oxygen and carbon, the C—O bond is slightly polar thus have a dipole moment. Since the two C—O bonds in ethers are inclined to each other at an angle of 110°, the two dipoles do not cancel each other. As a result, ethers have a dipole moment of 1.15 to 1.3D

Question 56.
Give the equation for the reaction between diethyl ether with an excess of oxygen.
Answer:

Question 57.
Give examples for (i) phthalein fusion and (ii) coupling reaction.
Answer:
(i) Phthalein fusion: On heating phenol with phthalic anhydride in presence of conc. H₂SO₄, phenolphthalein is obtained.

(ii) Coupling reaction: Phenol couples with benzene diazonium chloride in an alkaline solution to form p-hydroxy azobenzene (a red-orange dye).

Choose the correct answer:

1. The IUPAC name of the compound is:

(a) 2-chloro-5- hydroxyhexane
(b) 2-hydroxy-5-chlorohexane
(c) 5-chlorohexan-2-ol
(d) 2-chlorohexan-5-ol
Answer:
(c)

2. The IUPAC name of the compound is:

(a) 1-methoxy-1-methyl ethane
(b) 2-methoxy-2-methyl ethane
(c) 2-methoxypropane
(d) isopropyl-methyl ether
Answer:
(c)

3. Arrange the following compounds in increasing order of boiling point:
I – propan-1-ol
II – butan-1-ol
III – butan-2-ol
IV – pentan-1-ol
(a) I, III, II, IV
(b) I, II, III, IV
(c) IV, III, II, I
(d) IV, II, III, I
Answer:
(a)
Hint: The boiling point increases as the molecular mass increases. Further among isomeric alcohols 1° alcohols have higher boiling points than 2° alcohols.

4. Acid catalysed hydration of alkenes except ethene, leads to the formation of:
(a) primary alcohol.
(b) secondary or tertiary alcohol.
(c) a mixture of primary and secondary alcohol.
(d) a mixture of secondary and tertiary alcohol.
Answer:
(b)
Hint: Hydration of ethene gives a 1° alcohol while all others give a either a 2° alcohol or 3° alcohol.

5. Which of the following Grignard reagent is suitable for the preparation of 3-methyl-2- butanol?
(a) 2-butanone + methyl magnesium bromide
(b) Acetone + ethyl magnesium bromide
(c) Acetaldehyde + isopropyl magnesium bromide
(d) Ethyl propionate + methyl magnesium bromide
Answer:
(c)
Hint:

6. Which of the following esters shown, after reduction with LiAlH₄ and aqueous workup, will yield two molecules of only a single alcohol?
(a) C₆H₅COOC₆H₅
(b) CH₃CH₂COOCH₂CH₃
(c) CH₃COOCH₃
(d) C₆H₅COOCH₂C₆H₅
Answer:
(d)
Hint: Acyl and arylalkyl groups contain the same number of carbon atoms.

7. Which of the following will exhibit highest boiling point?

Answer:
(b)
Hint: Among isomeric alcohols, the alcohol with no branching has the highest boiling point.

8. Consider the reactions:

The mechanisms of reactions (i) and (ii) are respectively:
(a) S_N1 and S_N2
(b) S_N1 and S_N1
(c) S_N2 and S_N2
(d) S_N2 and S_N1
Answer:
(c)

9.

which of the following compounds will be formed?

Answer:
(a)
Hint: Nucleophilic attack occurs on the smaller alkyl group.

10. The reaction of with RMgX leads to the formation of:

Answer:
(d)
Hint:

11.

X is:
(a) bromine in benzene
(b) bromine in water
(c) potassium bromide solution
(d) bromine in CCl₄ at 0° C
Answer:
(b)
Hint:

12. Phenol can be distinguished from ethanol by the following reagent except.
(a) sodium
(b) I₂ / NaOH
(c) neutral FeCl₃
(d) Br₂ / H₂O
Answer:
(a)
Hint: Na reacts with both ethapol and phenol.

13. An ether is more volatile than an alcohol having the same molecular formula. This is due to:
(a) intermolecular hydrogen bonding in alcohols
(b) dipolar character of ethers
(c) alcohols having resonance structures
(d) intermolecular hydrogen bonding in ethers
Answer:
(a)

14. Mark the correct increasing order of reactivity of the following compounds with HBr / HI.

(a) I < II < III
(b) II < I < III
(c) II < III < I
(d) III < II < I
Answer:
(c)
Hint: All the three benzyl alcohols react with HBr or HI through the formation of intermediate carbocation. Obviously more stable the carbocation, more reactive is the alcohol. Now, electron withdrawing group i.e, NO₂, Cl, etc., decrease the stability of carbocations. Since, NO₂ is a stronger, electron withdrawing group, then Cl, stability of carbocation increases in the order.

Therefore, the reactivity of the benzyl alcohols increases in the same orders, i.e., II < III < I.

15. Which of the following reagents can be used to oxidize primary alcohols to aldehydes?
1. CrO₃ in acidic medium,
2. KMnO₄ in acidic medium,
3. Pyridinium chlorochromate,
4. Heat in the presence of Cu at 573 K.
(a) 1, 3, 4
(b) 1, 3
(c) 1, 4
(d) none of the above
Answer:
(a)
Hint: KMnO₄ will oxidise the aldehyde formed to carboxylic acid.

16. Assertion: Bond angles in ethers is slightly more than the tetrahedral angle.
Reason: There are repulsions between two bulky ‘R’ groups.
(a) Assertion and reason both are correct, and reason is the correct explanation of assertion.
(b) Assertion and reason both are correct, but reason is not the correct explanation of assertion.
(c) Assertion is wrong, but reason is correct.
(d) Both assertion and reason are wrong.
Answer:
(c)
Hint: Correct assertion: Bond angles in ethers are slightly more than the tetrahedral bond angle.

17. Assertion: Bromination of benzene does not require Lewis acid is catalyst.
Reason: Lewis acid polarises the bromine molecule in Friedel craft’s reaction.
(a) Assertion and reason both are correct, and reason is the correct explanation of assertion.
(b) Assertion and reason both are correct, but reason is not the correct explanation of assertion.
(c) Assertion is true, but reason is wrong.
(d) Both assertion and reason are wrong.
Answer:
(b)

18. Match the entries in column I with appropriate entries in column II and choose the correct option using the codes given:



Answer:
(b)

19. Match the entries in column I with appropriate entries in column II and choose the correct option using the codes given:


Answer:
(d)

20. In the following sequence of reactions,

(a) CH₃CH₂CH₂OH
(b) CH₃CH(OH)CH₃
(c) CH₃CH₂CH₂CH₂OH
(d) (CH₃)₃CCH₂OH
Answer:
(b)
Hint: Since alkenes give Markovnikoff addition products, therefore 1° alcohols (a), (c) and (d) cannot be obtained by hydration. Thus, ‘Z’ is 2-propanol.
i.e.

21. Consider the following reaction:

The compound ‘Z’ is:
(a) CH₃CH₂OCH₂CH₃
(b) CH₃CH₂OSO₃H
(c) CH₃CH₂OH
(d) CH₂=CH₂
Answer:
(c)
Hint:

22. Formation of tert-butyl ether by reaction of sodium tert-butoxide and methyl bromide involves:
(a) elimination reaction
(b) electrophilic addition reaction
(c) nucleophilic addition reaction
(d) nucleophilic substitution reaction
Answer:
(d)

23.

Answer:
(a)

24. The red coloured compound formed during Victor-Meyer’s test is:

Answer:
(b)
Hint:

25. Following compounds are given:
(i) CH₃CH₂OH
(ii) CH₃COCH₃
(iii) CH₃CH(OH)CH₃
(iv) CH₃OH
Which of the above compounds on being warmed with iodine solution andNaOH, will give iodoform?
(a) (i), (iii) and (iv)
(b) only (i)
(c) (i), (ii) and (iii)
(d) (i) and (ii)
Answer:
(c)
Hint: Compounds containing CH₃CO group or CH₃CH(OH) group on warming with I₂ and NaOH will form iodoform. Hence, (i), (ii) and (iii) will give iodoform.

Students get through the TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Answer the following questions.

Question 1.
Give three examples of adsorption.
Answer:

1. Charcoal adsorbs ammonia.
2. Silica gel adsorbs water.
3. Charcoal adsorbs colorants from sugar.

Question 2.
Distinguish between adsorption and absorption.
Answer:
Adsorption is a surface phenomenon while absorption is a bulk phenomenon, i.e., the adsorbate molecules are distributed throughout the adsorbent.

Question 3.
What are adsorbent and adsorbates? Give examples.
Answer:
Adsorbent is the material on which adsorption take place. Adsorbed substance is called adsorbate.
Examples for adsorbates: The gaseous molecules like He, Ne, O₂, N₂, SO₂ and NH₃ and solutions of NaCl or KCl.
Examples for adsorbents: Silica gel, metals like Ni, Cu, Pt, Ag and Pd and certain colloids.

Question 4.
What is an interface?
Answer:
The surface of separation of two phases where the concentration of adsorbed molecules is high is known as interface.

Question 5.
Mention the characteristics of adsorption.
Answer:

1. Adsorption can occur in all interfacial surfaces i.e., the adsorption can occur in between gas-solid, liquid solid, liquid- liquid, solid- solid and gas-liquid.
2. Adsorption is always accompanied by decrease in free energy. When ΔG reaches zero, the equilibrium is attained.
3. Adsorption is a spontaneous process.
4. When molecules are adsorbed, there is always a decrease in randomness of the molecules.
   We know, ΔG = ΔH – T ΔS where ΔG is change in free energy.
   ΔH is change in enthalpy and ΔS is change in entropy.
   Hence, ΔH = ΔG + TΔS Adsorption is exothermic as there is an interaction between adsorbate and adsorbent.

Question 6.
What do you understand by the term ‘sorption’?
Answer:
The term represents simultaneous adsorption and absorption.

Question 7.
What is the term used for sorption of gases on metal surfaces?
Answer:
Occlusion.

Question 8.
Give three examples for chemical adsorption or chemisorption.
Answer:

1. Adsorption of O₂ on tungsten.
2. Adsorption of H₂ on nickel.
3. Adsorption of ethyl alcohol vapours on nickel.

Question 9.
Give two examples for physical adsorption.
Answer:

1. Adsorption of N₂ on mica.
2. Adsorption of gases on charcol.

Question 10.
Explain the nature of the force of attraction that exists between an adsorbent and an adsorbate in (i) physical adsorption and (ii) chemical adsorption.
Answer:
The forces of attraction that exists between adsorbate and adsorbent in
(i) Physical adsorption are (a) vander waals force of attraction, (b) dipole-dipole interaction and (c) dispersion forces.
(ii) In chemisorption gas molecules are held to the surface by formation of chemical bonds. Since strong bonds are formed, nearly 400 kJ mole is given out as heat of adsorption.

Question 11.
Why is adsorption exothermic?
Answer:
When adsorption occurs, entropy decreases, i.e., ΔS = negative. As ΔG = ΔH – TΔS, for a spontaneous process, AG must be negative. This can only be so, if ΔH is negative, i.e., the process is exothermic.

Question 12.
How do size of particles of adsorbent, pressure of gas and prevailing temperature influence the extent of adsorption of a gas on a solid?
Answer:

1. The smaller the size of the particles of the adsorbent, greater is the surface area, greater is the adsorption.
2. At constant temperature, adsorption first increases with increase in pressure and then attains equilibrium at high pressures.
3. In physical adsorption, it decreases with increase in temperatures, but in chemisorption first, it increases and then decreases.

Question 13.
What are adsorption isotherms and isobars? Explain.
Answer:
A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called adsoiption isotherm.
The plot of the amount of adsorption verses temperature at constant pressure is called adsorption isobar. The adsorption isobars of physisorption and chemisorption as shown in figure.

In physical adsorption, \(\frac{x}{m}\) decreases with increases in T, But in chemical adsorption, \(\frac{x}{m}\) increases with rise in temperature and then decreases. The increase illustrates the requirement of activation of the surface for adsorption is due to fact that the formation of activated complex requires certain energy.
The decrease at high temperature is due to desorption, as the kinetic energy of the adsorbate increases.

Question 14.
Derive Freundlich adsorption isotherm.
Answer:
A plot between the amount of adsorbate adsorbed and the pressure or concentration of adsorbate at constant temperature is called adsorption isotherms.
In order to explain these isotherms, various equations were suggested as follows: Freundlich adsorption isotherm.
According to Freundlinch,
\(\frac{x}{m}\) = Kp^{\(\frac{1}{n}\)}
where V is the amount of adsorbate or adsorbed on ‘m’ gm of adsorbent at a pressure of p. K and n are constants.
Value n is always less than unity.
This equation is applicable for adsoiption of gases on solid surfaces. The same equation becomes \(\frac{x}{m}\) = Kc^{\(\frac{1}{n}\)} when used for adsorption in solutions with c as concentration.
This equation quantitively predict the effect of pressure (or concentration) on the adsorption of gases (or adsorbates) at a constant temperature.
Taking log on both sides of the equation

Hence the intercept represents the value of log K and the slope \(\frac{b}{q}\) gives \(\frac{1}{n}\).
This equation explains the increase of \(\frac{x}{m}\) with an increase in pressure. But experimental values show the deviation at low pressure.

Question 15.
Explain the application of adsorption (i) in the use of gas masks, (ii) in the use of softening hard water, (iii) in decolourisation of sugar.
Answer:
(i) Gas masks are devices containing suitable adsorbents so that poisonous gases present in the atmosphere are preferentially adsorbed and the air for breathing is purified.
Activated charcoal is one of the best adsorbents.
(ii) Permutit is employed for this process which adsorbs Ca²⁺ and Mg²⁺ ions in its surface, there is an ion exchange as shown below it occurs on the surface.

Exhausted permutit is regenerated by adding a solution of common salt.

(iii) Sugar prepared from molasses is decolourised to remove coloured impurities by adding animal charcoal which acts as decolourising material.

Question 16.
What are adsorption indicators?
Answer:
In the precipitation titrations, the endpoint is indicated by an external indicator that changes its colour after getting adsorbed on precipitate. It is used to indicate the endpoint of the titration.

Question 17.
Distinguish between homogeneous catalysis and heterogeneous catalysis with suitable examples.
Answer:

Question 18.
Name the catalysts used in the following:
(i) Haber’s process in the manufacture of ammonia.
(ii) Oxidation of ammonia.
(iii) Hydrogenation of alkenes.
(iv) Decomposition of H₂O₂.
(v) Formation of acetophenone from benzene and ethanoyl chloride. Also, give equations for the reactions.
Answer:
(i) The catalyst used is Iron,

(ii) The catalyst is platinum gauze.

(iii) The catalyst is finely divided nickel.

(iv) The catalyst is platinum

(v) The catalyst is anhydrous aluminium chloride

Question 19.
Mention the characteristics of catalysts.
Answer:

1. For a chemical reaction, the catalyst is needed in very small quantities. Generally, a pinch of catalyst is enough for a reaction in bulk.
2. There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.
3. A catalyst itself can not initiate a reaction. It means it can not start a reaction that is not taking place. But, if the reaction is taking place at a slow rate it can increase its rate.
4. A solid catalyst will be more effective if it is taken in a finely divided form.
5. A catalyst can catalyse a particular type of reaction, hence they are said to be specific in nature.
6. In an equilibrium reaction, the presence of catalyst reduces the time for attainment of equilibrium and here it does not affect the position of equilibrium and the value of the equilibrium constant.
7. A catalyst is highly effective at a particular temperature called as optimum temperature.
8. The presence of a catalyst generally does not change the nature of products
   eg: 2SO₂ + O₂ → 2SO₃
   This reaction is slow in the absence of a catalyst, but fast in the presence if Pt catalyst.

Question 20.
What are promoters? Explain with an example.
Answer:
In a catalysed reaction the presence of a certain substance increases the activity of a catalyst. Such a substance is called a promoter. For example in the Haber’s process of manufacture ammonia, the activity of the iron catalyst is increased by the presence of molybdenum. Hence molybdenum is called a promoter. In the same way, Al₂O₃ can also be used as a promoter to increase the activity of the iron catalyst.

Question 21.
Explain the action of anhydrous aluminium chloride in Friedel-Crafts reaction.
Answer:
The mechanism of Friedel crafts reaction is given below

The action of catalyst is explained as follows:

It is an intermediate.

Question 22.
Give the mechanism of the following reactions based on intermediate formation theory.
Answer:

1. Thermal decomposition of KClO₃ in the presence of MnO₂.
   Steps in the reaction
   2KClO₃ → 2KCl + 3O₂
   can be given as
   2KClO₃ + 6MnO₂ → 6MnO₃ + 2KCl
   It is an intermediate,
   6MnO₃ → 6MnO₂ + 3O₂
2. The reaction between H2 and O2 in the presence of copper.
   2Cu + \(\frac{1}{2}\) O₂ → Cu₂O It is an intermediate.
   Cu₂O + H₂ → H₂O + 2Cu
3. Oxidation of HCl by air in the presence of CuCl₂.
   2CuCl₂ → Cl₂ + Cu₂Cl₂
   2Cu₂Cl₂ + O₂ → 2Cu₂OCl₂
   It is an intermediate.
   2Cu₂OCl₂ + 4HCl → 2H₂O + 4CuCl₂
   This theory describes
   (a) the specificity of a catalyst and
   (b) the increase in the rate of the reaction with an increase in the concentration of a catalyst.

Question 23.
Mention the limitations of the intermediate compound formation theory of catalysis.
Answer:

1. The intermediate compound theory fails to explain the action of catalytic poison and activators (promotors).
2. This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 24.
Explain the salient features of the adsorption theory of catalysis.
Answer:
According to this theory, the reactants are adsorbed on the catalyst surface to form an activated complex which subsequently decomposes and gives the product.
The various steps involved in a heterogeneous catalysed reaction are given as follows:

1. Reactant molecules diffuse from bulk to the catalyst surface.
2. The reactant molecules are adsorbed on the surface of the catalyst.
3. The adsorbed reactant molecules are activated and form an activated complex which is decomposed to form the products.
4. The product molecules are desorbed.
5. The product diffuses away from the surface of the catalyst.

Question 25.
What are active centres? How does the presence of active centres influence the rate of catalysed reaction?
Answer:
The surface of a catalyst is not smooth. It bears steps, cracks and comers. Hence the atoms on such locations of the surface are coordinatively unsaturated. So, they have much residual force of attraction. Such sites are called active centres. So, the surface carries high surface free energy.
The presence of such active centres increases the rate of reaction by adsorbing and activating the reactants.
The adsorption theory explains the following

1. Increase in the activity of a catalyst by increasing the surface area. An increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.
2. The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.
3. A promoter or activator increases the number of active centres on the surfaces.

Question 26.
What is the role of a promoter in heterogeneous catalysis?
Answer:
A promoter increases the number of active centres on the surfaces.

Question 27.
What are catalytic poisons? How do they affect a catalysed reaction?
Answer:
Substances that destroy the activity of catalysts are known as catalytic poisons. They affect the catalytic reaction by occupying the active centres of the catalyst as a result active centres will not available for adsorption by reactants.

Question 28.
Give some examples of enzyme catalyses. Some common examples for enzyme catalysis.
Answer:

1. The peptide glycyl L-glutamyl L-lyrosin is hydrolysed by an enzyme called pepsin.
2. The enzyme diastase hydrolyses starch into maltose.
   2(C₆H₁₀O₅)_n + nH₂O → nC₂H₂₂O₁₁
3. The yeast contains the enzyme zymase which converts glucose into ethanol.
   C₆H₁₂O₆ + H₂O → 2C₂H₅OH + 2CO₂
4. The enzyme micoderma aceti oxidises alcohol into acetic acid.
   C₂H₅OH + O₂ → CH₃COOH + H₂O
5. The enzyme urease present in soya beens hydrolyses the urea.
   

Question 29.
Briefly outline the characteristics of enzyme catalysed reaction.
Answer:

1. An enzyme catalyst is specific in nature, i.e., one catalyst cannot catalyse more than one reaction.
2. The enzyme catalysed reactions are very fast compared to other catalysed reactions.
3. The rate of an enzyme-catalysed reaction depends on temperature. At an optimum temperature, the rate of the reaction becomes maximum. Beyond the optimum temperate, the rate decrease.
4. The enzyme activity is maximum at a particular pH called optimum pH. The favourable pH range for enzymatic reactions is 5-7.
5. The enzymatic activity is increased by the presence of containing substances, known as coenzymes.
6. The activity of enzymes can also be inhibited or poisoned due to the presence of certain substances.

Question 30.
Explain phase transfer catalysis with an example.
Answer:
Substitution of Cl^– and CN^– in the following reaction.

R-Cl = 1-chlorooctane
R-CN = 1-cyanooctane
By direct heating of two-phase mixture of organic 1 -chlorooctane with aqueous sodium cyanide for several days, 1-cyanooctane is not obtained. However, if a small amount of quaternary ammonium salt like tetraalkyl ammonium chloride is added, a rapid transition of 1 -cyanooctane occurs in about 100% yield after 1 or 2 hours. In this reaction, the tetraalkyl ammonium cation, which has hydrophobic and hydrophilic ends, transports CN from the aqueous phase to the organic phase using its hydrophilic end and facilitates the reaction with 1-chloroocatne as shown below:

So phase transfer catalyst speeds up the reaction by transporting one reactant from one phase to another.

Question 31.
Name the dispersed phase and dispersed medium in the following: fog, dust, sharing cream, milk, paints.
Answer:

Question 32.
Give the name of the colloid and one example for (i) a gas dispersed in solid, (ii) a liquid dispersed in a solid, (iii) a solid dispersed in a solid.
Answer:
(i) A gas dispersed in a solid is known as a solid foam. Pumice stone, rubber, bread are examples.
(ii) A liquid dispersed in a solid is known as a gel. Butter and cheese are examples.
(iii) A solid dispersed in another solid is known as a solid sol. Pearls, coloured glass, alloys are examples of solid sols.

Question 33.
Briefly outline the principle involved in the preparation of colloidal solution by dispersion methods.
Answer:
In the dispersion method, the particles whose size is higher than that of colloidal particles are disintegrated to that of the colloidal particles subjecting it to a mechanical disintegration in a colloid mill or by electro-dispersion.

Question 34.
Explain how colloidal particles of metals are prepared by electro dispersion method.
Answer:
An electrical arc is struck between electrodes dispersed in the water surrounded by ice. When a current of 1 amp /100V is passed an arc produced forms vapours of metal which immediately condense to form a colloidal solution. By this method colloidal solution of many metals like copper, silver, gold, platinum, etc. can be prepared Alkali hydroxide is added as a stabilising agent for the colloidal solution.

Question 35.
Explain peptisation with an example.
Answer:
The conversion of a freshly precipitated substance into a colloid by the addition of an electrolyte (peptising agent) is known as peptisation. For example, freshly precipitated silver chloride is converted into a colloidal solution by the addition of HCl.

Question 36.
Explain the principle involved in the preparation of colloids by the condensation method.
Answer:
Particles whose size are lesser than the colloidal particles are brought together to the size of colloidal particles.

Question 37.
What is the principle involved in the purification of a colloidal solution by dialysis?
Answer:
The separation of crystalloids (electrolytes) from colloids is based upon the principle that particles of crystalloid pass through the animal membrane (bladder) or parchment paper, or cellophane sheet whereas those of colloid do not.

Question 38.
Explain why? (i) colloidal particles are not visible to the naked eye, (ii) colloidal particles pass through ordinary filter paper, (iii) the sky appears blue, (iv) finest gold sol is red in colour.
Answer:
(i) No particle is visible to the naked eye if its diameter is less than half the wavelength of light used. The shortest wavelength of light is about 4000Å or 400 mμ. Hence no particle of diameter less than 200 mμ can be seen. The size of the colloidal particles is less than 200 mμ.
(ii) The size of the colloidal particles are smaller than the pores in the filter paper and hence, they pass through the filter paper.
(iii) This is due to the Tyndall effect. The colloidal particles absorb light energy and then scatter in all directions.
(iv) The colour of the colloidal solution depends as the wavelength of scattered radiation by dispersed particles. The wavelength of light further depends on the size and nature of particles. The finest gold sol is red in colour. As the size of particles increase, it becomes purple, then blue and finally golden yellow.

Question 39.
Explain the Tyndall effect.
Answer:
Colloids have optical properties. When a homogeneous solution is seen in the direction of light, it appears clear but it appears dark, in a perpendicular direction.
But when light passes through a colloidal solution, it is scattered in all directions. The colloidal particles absorb a portion of light and the remaining portion is scattered from the surface of the colloid. Hence the path of light is made clear. This phenomenon is known as the Tyndall effect.

Question 40.
Give an account of the Brownian movement.
Answer:
The zig-zag random motion of colloidal particles, when viewed through ultramicroscope, is known as Brownian movement. The reason for Brownian movement is that the colloidal sol particles are continuously bombarded with the molecules of the dispersion medium and hence they follow a zigzag, random, continuous movement.
Brownian movement enables us,
I. to calculate Avogadro number.
II. to confirm kinetic theory which considers the ceaseless rapid movement of molecules that increases with an increase in temperature.
III. to understand the stability of colloids: As the particles are in continuous rapid movement they do not come close and hence not get condensed. That is Brownian movement does not allow the particles to be acted on by force of gravity.

Question 41.
Explain the term electrophorosis.
Answer:
The movement of colloidal particles towards cathode or anode depending on the charge of the particles under the influence of electric current is known as electrophorosis.
This experiment is used to detect the charge on the colloidal particles. If the colloidal particles move towards the cathode, it is negatively charged. If they are positively charged, they migrate towards anode.

Question 42.
Give examples for positively charged and negatively charged colloids.
Answer:

Question 43.
Account for the charge on colloidal particles.
Answer:
The colloidal particles, in whichever method is prepared, contain traces of electrolytes. It preferentially adsorbs either positive or negative ions from the solution. If the particles adsorb positive ions from the solution, it becomes a positively charged colloid. If it adsorbs a negative ion from the solution, it becomes a negatively charged colloid.

Question 44.
Account for the stability of a colloid.
Answer:
Since the colloidal particles are either positive or negatively charged, they do not come close together. They repel each other. Thus, the charge on the colloidal particles is responsible for its stability.

Question 45.
Explain the term coagulation or flocculation.
Answer:
The conversion of a colloid to a precipitate is known as coagulation or precipitation. Coagulation results in the removal of charges on the colloid by any one of the following:

1. Addition of electrolytes
2. electrophoresis
3. mixing oppositely charged sols
4. boiling.

These methods help remove charges from the particles. Thus, the particles come closer, form an aggregate of particles and finally settles down as a precipitate due to gravity.

Question 46.
The peptising agent is added to convert a precipitate into a colloidal solution. Give reason.
Answer:
Ions (either positive or negative) of the peptising agent (electrolyte) are adsorbed on the particles of the precipitate. They repel and hit each other breaking the particles of the precipitate into colloidal size.

Question 47.
Explain what is observed (i) when a beam of light is passed through a colloidal solution.
(ii) an electrolyte, NaCl is added to ferric hydroxide solution,
(iii) electric current is passed through a colloidal solution.
Answer:
(i) Scattering of light by colloidal particles take place and the path of light becomes visible.
(ii) The positively charged colloidal particles of Fe(OH)₃ gets coagulated by the oppositely charged Cl^– ions provided by NaCl.
(iii) On passing electric current, colloidal particles move towards the oppositely charged electrode where they lose their charge Cl^– get coagulated.

Question 48.
What happens when gelatin is mixed with gold sol?
Answer:
Gold sol is lyophobic. Gelatin forms lyophilic sol and acts as a protective colloid. On adding gelatin to gold sol, the latter becomes more stable.

Question 49.
A colloid is formed by adding FeCl₃ in excess of hot water. What will happen if an excess ferric chloride is added to this colloid?
Answer:
On adding FeCl₃ to water, a colloidal sol of hydrated ferric oxide is formed. This is a positively charged colloid as it adsorbs Fe⁺³ ions on its surface. On adding NaCl, Cl^– ions bring about coagulation.

Question 50.
How does the boiling of a colloid, help the coagulation of a colloidal solution?
Answer:
Boiling vessels in increased collisions between the colloidal particles. The sol particles combine and settle down as a precipitate.

Question 51.
Explain the protective action of lyophilic colloids.
Answer:
The addition of a lyophobic colloid into lyophilic colloid protects the latter from coagulation. Hence, lyophilic colloids act as protective colloids.

Question 52.
Define gold number. Mention its use.
Answer:
‘Gold number’ as a measure of protecting the power of a colloid. The gold number is defined as the number of milligrams of hydrophilic colloid that will just prevent the precipitation of 10ml of gold sol on the addition of 1ml of 10% NaCl solution. Smaller the gold number greater the protective power.

Question 53.
What are emulsions? Mention various types of emulsions.
Answer:
Emulsions are a colloidal solutions in which a liquid is dispersed in another liquid.
Generally, there are two types of emulsions.

1. Oil in Water (O/W)
2. Water in Oil (W/O)

Question 54.
Give one example each for (i) oil in water and (ii) water in oil emulsions.
Answer:
(i) In Oil in Water (O/W) emulsions, the dispersed phase is oil and the dispersion medium is water.
eg: Milk is an emulsion of liquid fat dispersed in water and vanishing cream.
(ii) Emulsion of Water in Oil (W/O) in which water is the dispersed phase and oil is the dispersion medium.
eg: Cod liver oil, butter and cold cream.

Question 55.
What is emulsification?
Answer:
The process of preparation of emulsion by the dispersal of one liquid in another liquid is called Emulsification.

Question 56.
What is an emulsifier or emulsification agent?
Answer:
It is a substance added to have a stable emulsion. The type of emulsion formed depends upon the nature of the emulsifying agent. For example, the presence of soluble soaps as an emulsifying agent favours the formation of oil in water type of emulsions, whereas insoluble soaps (containing non alkali metal atoms) favours the formation of water in oil emulsions.

Question 57.
Mention the properties of the emulsion.
Answer:

1. Emulsions exhibit all the properties like the Tyndall effect, Brownian movement, Electrophoresis, Coagulation on the addition of electrolytes.
2. Emulsions can be separated into their constituent liquids by boiling, freezing, centrifuging, electrostatic precipitation by adding large amounts of the electrolytes to precipitate out the dispersed phase or by chemical destruction of emulsifying agent.
3. Emulsions can be diluted by adding any amount of dispersion medium.

Choose the correct answer:

1. Which of the following is true with respect of adsorption?
(a) ΔG < 0, ΔS > 0, ΔH < 0
(b) ΔG < 0, ΔS < 0, ΔH < 0
(c) ΔG > 0, ΔS > 0, ΔH < 0
(d) ΔG < 0, ΔS < 0, ΔH > 0
Answer:
(b)

2. Which of the following statements is incorrect regarding physisorption?
(a) It occurs because of vander Waals forces.
(b) More easily liquefiable gases are adsorbed readily.
(c) Under high pressure, it results in multimolecular layer on the adsorbent surface.
(d) Enthalpy of adsorption in low and positive.
Answer:
(d)
Hint: Enthalpy of adsorption is always negative.

3. Physical adsorption of a gaseous species may change to chemical adsorption with
(a) decrease in temperature
(b) increase in temperature
(c) increase in surface area of the adsorbent
(d) decrease in surface area of the adsorbent
Answer:
(b)
Hint: At high temperatures, gaseous species may dissociate into atoms that are chemisorbed, eg: N₂ an iron surface at ≥ 773 K.

4. In physisorption, adsorbent does not show specificity for any particular gas because:
(a) involved van der Waals forces are universal
(b) gases involved behave like ideal gases
(c) enthalpy of adsorption is low
(d) it is a reversible gas
Answer:
(a)

5. Which of the following is an example of adsorption?
(a) water on silica gel
(b) water on calcium carbide
(c) hydrogen on finely divided nickel
(d) oxygen on the metal surface
Answer:
(c)

6. On the basis of data given below predict which of the following gases show least adsorption on a definite amount of charcoal

(a) CO₂
(b) SO₂
(c) CH₄
(d) H₂
Answer:
(d)
Hint: Higher the critical temperature, the greater is the adsorption. H₂ has the least critical temperature and hence least adsorbed.

7. Which of the following are the characteristics of chemisorption?
(i) High heat of adsorption
(ii) Irreversibility
(iii) Low activation energy
(a) (i) and (ii) only
(b) (i) and (iii) only
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)
Answer:
(a)

8. The plot of \(\frac{x}{m}\) (along Y-axis) versus log C (along X-axis) in the Freundlich adsorption isotherm is a horizontal line parallel to X-axis when,
(a) n = 0
(b) n = 1
(c) n = ∞
(d) such a plot is impossible
Answer:
(c)
Hint:

i.e., constant parallel to log C axis.
[Note: This is the equation for adsorption of solids in solution.]

9. Match the following:

Which of the following is the correct option?

Answer:
(b)

10. Which of the following will show the Tyndall effect?
(a) Aqueous solution of soap below critical miscelle concentration.
(b) Aqueous solution of soap above critical miscelle concentration.
(c) Aqueous solution of sodium chloride
(d) Aqueous solution of sugar
Answer:
(b)

11. Method by which lyophobic sol can be protected:
(a) addition of oppositely charged sol
(b) addition of an electrolyte
(c) addition of lyophilic sol
(d) by boiling
Answer:
(c)

12. Freshly prepared precipitate sometimes gets converted to colloidal solution by:
(a) coagulation
(b) electrolysis
(c) diffusion
(d) peptisation
Answer:
(d)

13. Which of the following electrolytes will have maximum coagulating value for AgI / Ag⁺ Sol?
(a) Na₂S
(b) Na₃PO₄
(c) Na₂SO₄
(d) NaCl
Answer:
(b)
Hint: AgI / Ag⁺ is a positively charged sol. Electrolytes whose negative ion (anion) has the least negative charge will require the maximum amount and will have the maximum coagulating value. Hence, Cl^– in NaCl.

14. A colloidal system having a solid substance as a dispersed phase and liquid as dispersion medium is classified as:
(a) solid sol
(b) gel
(c) emulsion
(d) sol
Answer:
(d)

15. Which of the following process is responsible for the formation of the delta at places where the river meets the sea?
(a) Emulsification
(b) Colloid formation
(c) Coagulation
(d) Peptisation
Answer:
(c)
Hint: Formation of the delta-shaped heap of sand clay etc where the river falls into the sea due to coagulation of sand/clay by NaCl present in seawater.

16. When an excess of a very dilute aqueous solution of KI is added to a very dilute aqueous solution of silver nitrate, the colloidal particles of silver iodide are associated with which of the following Helmholtz double layer?

Answer:
(d)
Hint:
As excess KI has been added, I^– ions are adsorbed on AgI forming a fixed layer (giving negative charge). It then attracts the counter ion (K⁺) from the medium forming a second layer, (diffused layer).

17. The ratio of a number of moles of AgNO₃, Pb(NO₃)₂ and Fe(NO₃)₃ required for coagulation of a definite amount of colloidal sol of silver iodide prepared by mixing AgNO₃ with an excess of KI will be:
(a) 1:2:3
(b) 3:2:1
(c) 6:3:2
(d) 2:3:6
Answer:
(c)
Hint:
With excess KI, colloidal particles will be [AgI]I^–.

Molar ratio required for coagulation of same amount of [AgI]I^– is 1 : \(\frac{1}{2}\) : \(\frac{1}{3}\) = 6 : 3 : 2.

18. Gelatin is mostly used in making ice creams in order to:
(a) prevent the formation of colloidal sol.
(b) enrich fragrance.
(c) prevent crystallisation and stabilise the mix.
(d) modify the taste.
Answer:
(c)
Hint: Gelatin is a protective colloid. It stabilises the mix and prevents crystallisation.

19. Which of the following has a minimum gold number?
(a) Starch
(b) Sodium oleate
(c) Gum arabic
(d) Gelatin
Answer:
(d)

20. Match the following:

Which of the following is the correct option?

Answer:
(a)

Students get through the TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Answer the following questions.

Question 1.
Define the following terms:
(i) Resistivity,
(ii) conductivity,
(iii) cell constant,
(iv) specific conductance,
(iv) molar conductivity,
(vi) equivalent conductance.
Answer:
(i) Resistivity is defined as the resistance of an electrolyte confined between two electrodes having unit cross-sectional area and are separated by unit distance.
(ii) The reciprocal of specific resistance \(\left(\frac{1}{\rho}\right)\) is called specific conductance or conductivity.
(iii) The ratio between the distance between the two electrodes in a conductivity cell to their area of cross-section is called cell constant.
(iv) The specific conductance is defined as the conductance of a cube of an electrolytic solution of unit dimension.
(v) Molar conductance is defined by the conducting power of all the ions produced by 1 mole of the electrolyte.
(vi) Equivalent conductance is defined as the conductance of ‘V’ m³ of the electrolyte solution containing one gram equivalent of electrolyte in a conductivity cell in which the electrodes are 1 meter apart.

Question 2.
Give the relationship between resistance and specific resistance of an electrolytic conductor.
Answer:
R = ρ × \(\frac{l}{a}\)
where ‘R’ is the resistance, ρ is the specific resistance or resistivity, ‘l’ is the distance between the electrodes in a conductivity cell, and ‘a’ is the area of cross-section of electrodes.

Question 3.
Mention the units of (i) resistance, (ii) specific resistance, (iii) conductance, (iv) specific conductance, (v) molar conductance and (vi) equivalent conductance.
Answer:

Question 4.
Give the relationship between molar conductance and specific conductance of an electrolyte.
Answer:

Question 5.
Give the relationship between equivalent conductance and specific conductance of an electrolyte.
Answer:

Question 6.
Mention the factors which influence electrolytic conductance.
Answer:

1. A solvent of a higher dielectric constant shows high conductance in solution.
2. Conductance is inversely proportional to the viscosity of the medium, i.e., conductivity increases with the decrease in viscosity.
3. If the temperature of the electrolytic solution increases, conductance also increases. An increase in temperature increases the kinetic energy of the ions and decreases the attractive force between the oppositely charged ions and hence conductivity increases.
4. Molar conductance of a solution increases with an increase in dilution. This is because, for a strong electrolyte, interionic forces of attraction decrease with dilution. For a weak electrolyte, the degree of dissociation increases with dilution.

Question 7.
Explain how does the molar conductance of the electrolytes varies with the concentration of the electrolyte.
Answer:
The molar conductance of a strong electrolyte increases on dilution. The variation of molar conductance with dilution is given by an empirical formula

where Λ_m⁰ is called limiting molar conductivity and C is the concentration of the electrolyte.

For strong electrolytes such as KCl, NaCl, etc., the plot, Λ_m vs √C, gives a straight line as shown in the graph. It is also observed that the plot is not a linear one for weak electrolytes.
For strong electrolyte, at a high concentration, the number of constituent ions of the electrolyte in a given volume is high and hence the attractive force between the oppositely charged ions is also high. Moreover, the ions also experienced a viscous drag due to greater solvation. These factors attribute to the low molar conductivity at high concentrations. When the dilution increases, the ions are far apart and the attractive forces decrease. At infinite dilution the ions are so far apart, the interaction between them becomes insignificant, and hence, the molar conductivity increases and reaches a maximum value at infinite dilution.
For a weak electrolyte, at high concentration, the plot is almost parallel to the concentration axis with a slight increase in conductivity as the dilution increases. When the concentration approaches zero, there is a sudden increase in the molar conductance and the curve is almost parallel to Λ_m⁰ axis. This is due to the fact that the dissociation of the weak electrolyte increases with the increase in dilution (Ostwald dilution law). Λ_m⁰ values for strong electrolytes can be obtained by extrapolating the straight line. But the same procedure is not applicable for weak electrolytes, as the plot is not a linear one, A°m values of the weak electrolytes can be determined using Kohlraush’s law.

Question 8.
Flow much charge is required for the following reactions?
(i) 1 mole of Al³⁺ to Al
(ii) 1 mole of Cu²⁺ to Cu
(iii) 1 mole of MnO₄^– to Mn²⁺
Answer:
(i) Al³⁺ +3e → Al
Reduction of 1 mole of Al³⁺ requires 3 moles of electrons.
∴ Charge on 3 moles of electrons
= 3 × 96500 C
= 289500 coulomb
(ii) Cu²⁺ + 2e → Cu
Reduction of 1 mole of Cu²⁺ requires 2 mole of electrons
= 2 × 96500 C
= 193000 coulomb
(iii) MnO₄^– + 8H⁺ + 5e → Mn²⁺ + 4H₂O
Reduction of MnO₄^– ions require 5 mole of electrons
= 5 × 96500 C
= 482500 Coulomb

Question 9.
A solution of CuSO₄ is electrolyzed for 10 minutes with a current of 1.5 amperes. What mass of copper deposited at the cathode?
Answer:

∴ Mass of copper deposited at the cathode = 0.296 g

Question 10.
How many hours does it take to reduce 3 mole of Fe³⁺ to Fe²⁺. With 2 ampere of current? (F = 96500 coulomb)
Answer:
The required equation is Fe³⁺ + e → Fe²⁺
3 mole of Fe³⁺ requires 3 mole of electrons.
∴ Required charge = 3 × 96500 coloumbs
= 289500 coloumbs
charge = current × time
289500 = 2 × t
t = \(\frac{289500}{2}\)
= 1.475 × 10⁵ sec
= 40 hour.

Question 11.
In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold salt and the second cell contains copper sulfate solution.
If the oxidation number of gold is +3, find the amount of copper deposited on the cathode in the second cell. Also, calculate the magnitude of the current in amperes.
Answer:

Let Z be the electrochemical equivalent of copper

Question 12.
The specific conductance of a decinormal solution of KCl is 0.00112 ohm^-1 cm^-1. The resistance of a cell containing the solution was found to be 56 ohms. What is the value of the cell constant?
Answer:
Specific conductance = \(\frac{Cell constant}{Resistance}\)
Cell constant = Specific conductance × Resistance
= 0.0112 × 56
= 0.6272 cm^-1

Question 13.
1.0 N solution of salt surrounding two platinum electrodes 2.1 cm apart and 4.2 sq.cm in area was found to offer a resistance of 50 ohms. Calculate the equivalent conductivity of the solution.
Answer:
Given: l = 2.1 cm; a = 4.2 sq.cm; R = 50 ohm

Question 14.
The specific conductivity of 0.02 M KCl solution at 25° C is 2.768 × 10^-3 ohm^-1 cm^-1. The resistance of this solution at 25° C, when measured with a particular cell, was 250.2 ohm. The resistance of 0.01 M CuSO₄ solution at 25° C measured with the same cell was 8331 ohm. Calculate the molar conductivity of the copper sulfate solution.
Answer:
Cell constant = Specific conductivity of KCl × Resistance = 2.768 × 10^-3 × 250.2
For 0.01 M CuSO₄

Question 15.
At 291 K, the molar conductivities at infinite dilution of NH₄Cl, NH₄OH, and NaCl are 129.8, 217.8, and 108.9 S cm² respectively. If the molar conductivity of a cent normal solution of NH₄OH is 9.33 S cm², what is the percentage dissociation of NH₄OH at this dilution? Also, calculate the dissociation constant of NH₄OH.
Answer:

∴ degree of dissociation (α)

Calculation of dissociation constant

Question 16.
The conductivity of a saturated solution of AgCl at 288 K. is found to be 1.382 × 10^-6 Scm^-1. Find its solubility. Given ionic conductances of Ag⁺ and Cl^– ion at infinite dilution are 61.9 S cm² mol^-1 and 76.3 S cm² mol^-1 respectively.
Answer:

Question 17.
Define (i) electrode potential and (ii) emf of a cell.
Answer:
(i) Electrode potential: The potential difference set up between the metal and its ions in the solution is called the electrode potential, (or) electrode potential may be defined as the tendency of an electrode to lose or gain electrons when it is in contact with a solution of its own ions.
(ii) emf of a cell; The difference between the electrode potentials of the two half cells in a galvanic cell is known as the cell potential (or) It is the driving forces that permit the flow of electrons from anode to cathode.

Question 18.
Explain the terms (i) oxidation potential and (ii) reduction potential of an electrode.
Answer:
(i) Oxidation potential: when an electrode is negatively charged with respect to the solution, i.e., it acts as an anode, the potential difference is called oxidation potential.
M ⇌ Mⁿ⁺ + ne
It refers to the tendency for oxidation to occur at the electrode.
(ii) Reduction potential: When an electrode is positively charged with respect to solution, i.e., it acts as the cathode, the potential difference developed is called reduction potential
Mⁿ⁺ + ne ⇌ M
It refers to the tendency for reduction to occur at the electrode.

Question 19.
What is a reference electrode? Mention its use.
Answer:
A reference electrode is an electrode whose electrode potential value is known. It is used to determine, electrode potentials of various electrodes.

Question 20.
Mention the characteristics of electrode chemical series.
Answer:

1. The negative sign of the standard reduction potential indicates that the electrode when joined with SHE, acts as an anode and oxidation occurs at this electrode. Similarly, the sign of standard reduction potential indicates that the electrode when joined with SHE acts as and reduction occurs on this electrode.
2. Greater the value of standard reduction potential move easily the substance (element or ions) is reduced or in other words, stronger oxidizing agent it is. Similarly, the lower the value of standard reduction potential of the substance greater is the extent of oxidation half-reaction and the substance acts as a reducing agent.
3. Reactivity of metals: A metal that has a high negative value (or small positive value) of standard reduction potential loses electrons is said to be chemically active. The chemical reactivity decreases from top to bottom of the series.

Question 21.
Given the standard reduction potentials

Arrange these metals in their increasing order of reducing character.
Answer:
Ag < Hg < Cr < Mg < K.
The lesser the reduction potentials greater is their reducing character.

Question 22.
Write down the electrode reactions and the net cell reaction for the following cells. Which electrode would be the positive terminal in each cell.

Answer:

Question 23.
The standard EMF of the cell:
Ni | N₁²⁺ || Cu²⁺ | Cu is 0.59 V. The standard reduction potential of the copper electrode is 0.34 V. Calculate the standard reduction potential of the nickel electrode.
Answer:

Question 24.
A cell is set up between copper and silver electrodes as Cu | Cu²⁺ || Ag⁺ | Ag.
Given: . Calculate the emf of the cell.
Answer:

Question 25.
Two half cells are Al³⁺ | Al and Mg²⁺ | Mg. The reduction potentials of these half cells are -1.66 and -2.36 V. Construct the galvanic cell, write the cell reaction and calculate the emf of the cell.
Answer:
For the construction of a galvanic cell,
= (Electrode with lesser value of reduction potential – Electrode with higher value of reduction potential )

Question 26.
A cell is prepared by dipping a copper rod in 1M CuSO₄ solution and a nickel rod in 1M NiSO₄ solution. The standard reduction potentials of copper and nickel electrodes are 0.34 V and – 0.25 V respectively.
(a) How will you represent the cell?
(b) What will be the cell reaction?
(c) What will be the standard emf of the cell?
(d) Which electrode will be positive?
Answer:
The cell is represented as

(d) The cathode is a copper electrode. Because reduction occurs at this electrode.

Question 27.
Predict whether zinc and silver react with 1M H₂SO₄ to give hydrogen gas or not. Given the standard reduction potentials of zinc and silver electrodes are -0.76 and 0.80 V respectively.

Answer:
The cell is

The emf of the cell is positive. Hence the cell reaction is spontaneous. Thus, zinc will liberate hydrogen gas from 1M H₂SO₄.
Similarly,

The cell is

The emf of the cell is negative. Hence the reaction will not occur, i.e., Ag will not displace hydrogen from 1M H₂SO₄.

Question 28.
Can a nickel sulphate be used to stir a solution of copper sulphate. Support your answer with a reason.
[OR]
Can a solution of 1M copper sulphate be stored in a vessel made of nickel.
Given:

Answer:
The reaction,
Ni + CuSO₄ → NiSO₄ + Cu
should not occur, if CuSO₄ solution to be stored in a nickel vessel.
The cell would be Ni | Ni²⁺ || Cu²⁺ | Cu

The emf is positive. This means copper sulfate reacts with nickel, Hence CuSO₄ cannot be stored in a nickel vessel.

Question 29.
Give the relationship between the electrode potential and the concentration of the electrode.
Answer:
This relationship is given by Nernst equation. For an electrode reaction,
Mⁿ⁺ + ne → M
the electrode potential is given by the relationship,

Question 30.
For general cell reactions

Write the Nernst equation.
Answer:

where is the number of electrons involved in the cell reaction?

Question 31.
Calculate the electrode potential of copper electrode, when a copper wire is dipped in 0.1 M CuSO₄ solution E_Cu²⁺/Cu⁰ = 0.34.
Answer:
The electrode reaction is Cu²⁺ + 2e → Cu
Applying Nernst equation

[Note: Decreasing concentration of the electrolyte decreases the electrode potential]

Question 32.
The following reaction takes place in a galvanic cell.

Calculate the emf of the cell.
Answer:

Question 33.
Give the relationship between the free energy charge of the cell reaction and the emf of the cell.
Answer:
ΔG° = -n FE_cell

Question 34.
Give the relationship between the free energy change for the cell reaction and the equilibrium constant.
Answer:
ΔG° = – 2.303 RT log K_c

Question 35.
(a) Calculate the standard free energy change and maximum work obtainable by a galvanic cell, where the cell reaction

(b) Also calculate the equilibrium constant for the reaction.
Answer:
(a) The cell is

Thus the maximum work that can be obtainable from the cell = 212.3 kJ mol^-1

Question 36.
Describe the nature of anode, cathode, the cell reaction, and salient features of the Leclanche cell.
Answer:
Anode: Zinc container
Cathode: Graphite rod in contact with Mn02
Electrolyte: ammonium chloride and zinc chloride in water.
Emf of the cell is about 1.5 V.
Cell reaction:
Oxidation at anode:
Zn(s) → Zn²⁺(aq) + 2e^– … (1)
Reduction at cathode:
2NH₄⁺(aq) + 2e^– → 2NH₃(aq) + H₂(g) … (2)
The hydrogen gas is oxidised to water by MnO₂
H₂(g) + 2MnO₂(s) → Mn₂O₃(s) + H₂O (l) …(3)
Equation (1) + (2) + (3) gives the overall redox reaction
Zn(s) + 2NH₄⁺ (aq) + 2MnO₂(s) → Zn²⁺(aq) + Mn₂O₃(s) + H₂O (l) + 2NH₃ ….. (4)
The ammonia produced at the cathode combines with Zn²⁺ to form a complex ion [Zn(NH₃)₄]²⁺(aq). As the reaction proceeds, the concentration of NH₄⁺ will decrease and the aqueous NH₃ will increase which lead to the decrease in the emf of cell.

Question 37.
Describe the nature of anode, cathode the cell reaction, and salient features of the mercury button cell.
Answer:
Anode: Zinc amalgamated with mercury.
Cathode: HgO mixed with graphite.
Electrolyte: Paste of KOH and ZnO.

Cell emf: about 1.35V.
Uses: It has a higher capacity and longer life. Used in pacemakers, electronic watches, cameras etc…

Question 38.
Explain the functioning of lead storage batteries.
[OR]
Explain the chemical reaction involved in the process of charging and recharging in a lead storage battery.
Answer:
Anode: Spongy lead
Cathode: Lead plate bearing PbO₂
Electrolyte: 38% by mass of H₂SO₄ with density 1.2g / mL.
Oxidation occurs at the anode:
Pb(s) → Pb²⁺(aq) + 2e^– … (1)
The Pb²⁺ ions combine with SO₄^-2 to from PbSO₄ precipitate.
Pb²⁺(aq) + SO₄² (aq) → PbSO₄(s) …….. (2)
Reduction occurs at the cathode:
PbO₂(s) + 4H⁺(aq) + 2e^– → Pb²⁺(aq) + 2H₂O (l) ……. (3)
The Pb²⁺ ions also combine with SO₄^2- ions from sulphuric acid to form PbS04 precipitate.
Pb²⁺(aq) + SO₄^2-(aq) → PbSO …… (4)
Overall reaction:
Equation (1) + (2) + (3) + (4) ⇒
Pb(s) + PbO₂ (s) + 4H⁺(aq) + 2SO₄^2-(aq) → 2PbSO₄(s) + 2H₂O (l)
The emf of a single cell is about 2 V. Usually six such cells are combined in series to produced 12 volt.
The emf of the cell depends on the concentration of H₂SO₄. AS the cell reaction uses SO₂² ions, the concentration H₂SO₄ decreases. When the cell potential falls to about 1.8V, the cell has to be recharged.

Question 39.
Discuss the nature of the anode, cathode, and the cell reaction of the Lithium-ion battery. How does it act as a secondary cell?
Answer:
Anode: Porus graphite.
Cathode: transition metal oxide such as CoO₂. Electrolyte: Lithium salt in an organic solvent At the anode oxidation occurs:
Li(s) → Li⁺(aq) + e^–
At the cathode reduction occurs:
Li⁺ + CoO₂ (s) + e^– → Li CoO₂ (s)
Overall reactions:
Li(s) + CoO₂ → Li CoO₂ (s)
Both electrodes allow Li⁺ ions to move in and out of their structures.
During discharge, the Li⁺ ions produced at the anode move towards the cathode through the non-aqueous electrolyte. When a potential greater than the emf produced by the cell is applied across the electrode, the cell reaction is reversed and now the Li⁺ ions move from cathode to anode where they become embedded on the porous electrode. This is known as intercalation.
Uses: Used in cellular phones, laptop computer digital camera, etc…

Question 40.
Mention the various methods of protecting metals from corrosion.
Answer:
This can be achieved by the following methods.

1. Coating metal surface by paint.
2. Galvanizing by coating with another metal such as zinc, zinc is a stronger oxidizing agent than iron and hence it can be more easily corroded than iron, i.e., instead of iron, the zinc is oxidized.
3. Catholic protection: In this technique, unlike galvanizing the entire surface of the metal to be protected need not be covered with a protecting metal instead, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded.

Passivation: The metal is treated with strong oxidizing agents such as concentrated HNO₃. As a result, a protective oxide layer is formed on the surface of the metal.
Alloy formation: The oxidizing tendency of iron can be reduced by forming its alloy with other more anodic metals.
eg: stainless steel – an alloy of Fe and Cr.

Choose the correct answer:

1. In the electrolytic cell, the flow of electrons is from:
(a) cathode to anode in the solution
(b) cathode to the anode through external supply
(c) cathode to the anode through the internal supply
(d) anode to the cathode through the internal supply
Answer:
(b)
Hint: In electrolytic cells, electrons enter the cathode and leave at the anode. Thus in the external supply, electrons flow from cathode to anode. In the solution, only ions flow and not the electrons.

2. Which of the statements about solutions of electrolytes is not correct?
(a) the conductivity of the solution depends upon the size of the ions.
(b) conductivity depends upon the viscosity of the solution.
(c) conductivity does not depend upon the solvation of ions present in the solution.
(d) the conductivity of the solution increases with temperature.
Answer:
(c)
Hint: Greater the solvation of ions, lesser is the conductivity. Hence, (c) is an incorrect statement.

3. A dilute solution of Na₂SO₄ is electrolyzed using platinum electrodes. The products at the cathode and anode are:
(a) O₂ and H₂
(b) SO₂, Na
(c) O₂, Na
(d) S₂O₈^2-, H₂
Answer:
(a)
Hint: H20 is more readily reduced than Na+. It is more readily oxidised than S04 2. Hence, the electrode reactions are:
at cathode: 2H₂O + 2e → H₂ + 2OH^–
at anode: 2H₂O → O₂ + 4H⁺ + 4e.

4. The quantity of charge required to obtain one mole of aluminum from Al₂O₃ is:
(a) 1F
(b) 6F
(c) 3F
(d) 2F
Answer:
(c)
Hint: Al₂O₃ → 2Al.
i.e., 2Al₂O₃ + 6e → 2Al or
Al₂O₃ + 3e → Al.
Hence, to obtain one mol of Al, the charge required is 3F.

5. Electrolysis of dilute aqueous solution of NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mole of H₂ gas at the cathode is (1 Faraday = 96500 C mol^-1):
(a) 9.65 × 10⁴ sec
(b) 19.3 × 10⁴ sec
(c) 28.95 × 10⁴ sec
(d) 38.6 × 10⁴ sec
Answer:
(b)
Hint:

Thus,
0.5 mol of H₂ is liberated by IF = 96500 C 0.01 mol of H₂ will be liberated by charge

6. Al₂O₃ is reduced by electrolysis at low potentials and high currents. If 4.0 × 10⁴ amperes of current is passed through molten Al₂O₃ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency. At mass of Al = 27 g mol^-1):
(a) 8.1 × 10⁴ g
(b) 2.4 × 10⁵ g
(c) 1.3 × 10⁴ g
(d) 9.0 × 10³ g
Answer:
(a)
Hint: Al³⁺ + 3e → Al
1 mol of Al = 27 g of Al is deposited by 3F or 3 × 96500 C of electricity.
Quantity of electricity passed

7. What current is to be passed for 0.02 A for deposition of a certain weight of metal which is equal to electrochemical equivalent?
(a) 4 A
(b) 100 A
(c) 200 A
(d) 2 A
Answer:
(a)
Hint: Electrochemical equivalent is the weight deposited by 1 coulomb.
Q = I × t
1 = I × 0.25 or I = 4A

8. Acertain current liberates 0.504 g of hydrogen in 2 hours. How many grams of oxygen can be liberated by the same current in the same time?
(a) 2.0 g
(b) 0.4 g
(c) 4.0 g
(d) 8.0 g
Answer:
(c)
Hint: H₂O → H₂ + \(\frac{1}{2}\) O₂
Thus if one mole of H₂ is liberated, O₂ liberated = \(\frac{1}{2}\) mole

9. Resistance of 0.2 M solution of an electrolyte is 50Ω. The specific conductance of the solution is 1.3 Sm^-1. If the resistance of 0.4M solution of the same electrolyte is 260Ω the molar conductivity is:
(a) 6250 Sm² mol^-1
(b) 6.25 × 10^-4 Sm² mol^-1
(c) 625 × 10^-4 Sm² mol^-1
(d) 62.5 Sm² mol^-1
Answer:
(b)
Hint:
specific conductance = observed conductance × cell constant
K = \(\frac{1}{R}\) × cell constant
or cell constant = K × R
= 1.3 Sm^-1 × 50Ω
= 65 m^-1
For 0.4 M solution, specific conductance

10. An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to:
(a) increase in the number of ions
(b) increase in ionic mobility
(c) 100% dissociation of electrolyte at normal dilution
(d) increase in both i.e., number of ions and ionic mobility of ion.
Answer:
(b)
Hint: A strong electrolyte is completely ionized at all concentrations. Hence, the number of ions remains the same. However, on dilution, interionic forces decrease, and hence ionic mobility increases. Therefore, equivalent conductance increases.

11. In the electrolysis of which solution, OH^– ions are discharged in preference to Cl^– ions?
(a) dilute NaCl
(b) very dilute NaCl
(c) fused NaCl
(d) solid NaCl
Answer:
(b)
Hint: In very dilute NaCl solution, the following reactions take place on electrolysis.

12. The charge in coulombs on lg ion of N^-3 is:
(a) 28.9 × 10⁵ C
(b) 2.89 × 10⁵ C
(c) 2890 C
(d) 28900 C
Answer:
(b)
Hint: Charge on 1 gm ion of N^-3
= 3 × 1.6 × 10^-19 C
One g ion = 6.023 × 10²³ ion
Charge on 1 gm ion of N^-3
= 3 × 1.6 × 10^-19 × 6.023 × 10²³
= 2.89 × 10⁵ C

13. A solution of copper sulfate is electrolyzed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at the cathode is: (atomic mass of Cu = 63.5 g mol^-1)
(a) 29.6 g
(b) 0.296 g
(c) 2.96 g
(d) 1.564 g
Answer:
(b)
Hint: Current = 1.5 amperes
Time = 10 minutes
= 10 × 60 = 600 s
Quantity of electricity passed = i × t
= 1.5 × 600 = 900 C
The reaction occurring at cathode is
Cu²⁺ + 2e → Cu
i.e., 2F or 2 × 96500 C deposit
= 1 mol = 63.5 g of Cu
900 C will deposit = \(\frac{63.5}{2 \times 96500}\) × 900 = 0.296 g

14. The limiting molar conductance of HCl, CH₃COONa, and NaCl are respectively 425, 90, and 125 mho cm² mol^-1 at 25° C. The molar conductivity of 0.1 M CH₃COOH solution is 7.8 mho cm² mol^-1 at the same temperature. The degree of dissociation of 0.1 M acetic acid at this temperature is:
(a) 0.1
(b) 0.02
(c) 0.15
(d) 0.03
Answer:
(b)

15. Degree of dissociation of pure water is 1.9 × 10^-9. Molar conductances of H⁺ and OH^– ions at infinite dilution are 200 S cm² mol^-1 and 350 Scm² mol^-1 respectively.
(a) 3.8 × 10^-7 Scm² mol^-1
(b) 5.7 × 10^-7 Scm² mol^-1
(c) 9.5 × 10^-7 Scm² mol^-1
(d) 1.045 × 10^-6 Scm² mol^-1
Answer:
(b)
Hint:

16. When measured against a standard calomel electrode, an electrode is found to have a standard reduction potential of 0.100 V. If the standard reduction potential of the calomel electrode is 0.244 V, the standard reduction potential of the same electrode against the standard hydrogen electrode will be:
(a) – 0.144 V
(b) + 0.100 V
(c) – 0.344 V
(d) – 0.100 V
Answer:
(b)
Hint: Standard electrode potential is a fixed quantity for a given electrode. Hence, its standard electrode potential against SHE will remain the same. Viz., (0.100 V). (emf of the cell formed will change)

17. On the basis of the following E° values, the strongest oxidising agent is:


Answer:
(c)
Hint: The strongest oxidizing agent is a substance that is reduced move easily, i.e., it has the highest reduction potential. Reduction potentials of [Fe(CN)₆]^-3 and Fe³⁺ are 0.355 V and 0.77 V. Hence, Fe³⁺ is reduced more easily and hence is the strongest oxidizing agent.

18. Standard electrode potentials for Sn⁴⁺ / Sn²⁺ couple is +0.15 V and that for Cr³⁺/Cr couple is – 0.74 V. These two couples in their standard states are connected to make a cell. The cell potential will be:
(a) +1.83 V
(b) +1.19 V
(c) +0.89 V
(d) +0.18 V
Answer:
(c)
Hint:

19. Cr₂O₇^2- + I^– → I₂ + Cr³⁺;

(a) 0.54 V
(b) – 0.54 V
(c) +0.18 V
(d) -0.18 V
Answer:
(a)
Hint: In the given reaction, I^– is oxidized to I₂ and Cr₂O₇²⁺ ions have been reduced to Cr³⁺. The cell is

20. Given the standard electrode potentials K⁺/K = -2.93 V; Ag⁺ / Ag = 0.80 V; Mg₂²⁺ / Hg = 0.79; Mg²⁺/Mg = -2.37 V; Cr²⁺/Cr = -0.74 V.
Arrange these metals in their increasing order of reducing power.
(a) Ag < Hg < Cr < Mg < K
(b) K < Mg < Cr < Hg < Ag
(c) Hg < Cr < Mg < K < Ag
(d) Mg < Cr < Hg < Ag < K
Answer:
(a)
Hint: Lower the reduction potential, more easily it is oxidized and hence greater is its reducing power.

21. The reduction potential of hydrogen half cell will be negative if:
(a) pH₂ = 1 atm and [H⁺] = 1.0 M
(b) pH₂ = 2 atm and [H⁺] = 1.0 M
(c) pH₂ = 2 atm and [H⁺] = 2.0 M
(d) pH₂ = 1 atm and [H⁺] = 2.0 M
Answer:
(b)
Hint: For hydrogen electrode,

22. 2Hg → Hg²⁺; E° = 0.855 V
Hg → Hg²⁺; E° = 0.799 V
Equilibrium constant for the reaction
Hg + Hg²⁺ ⇌ Hg₂²⁺ at 27° C is:
(a) 89
(b) 82.3
(c) 79
(d) none of these
Answer:
(c)
Hint:
E° for the reaction = 0.855 – 0.799 = 0.056 V

23. Standard free energies of formation of (in kJ mol^-1) at 298 K are -237.3, -394.2, and -8.2 for H₂O(l), CO₂(g), and pentane (g) respectively. The value of E°ell for the pentane-oxygen fuel cell is:
(a) 1.968 V
(b) 2.0968 V
(c) 1.0968 V
(d) 0.968 V
Answer:
(c)
Hint: The balanced equation for pentane oxygen cell reaction will be:

Students get through the TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Answer the following questions.

Question 1.
State the formula and the name of the conjugate base of each of the following acids.
(a) H₃O⁺,
(b) HSO₄^–,
(c) HF,
(d) NH₄⁺,
(e) CH₃NH₃⁺,
(f) CH₃PO₄,
(g) CH₃COOH,
(h) H₂PO₄^–,
(i) HS^–,
(j) HCO₃^–.
Answer:
(a) H₂O – water
(b) SO₄^-2 – sulphate ion
(c) F^– – fluoride ion
(d) NH₃ – ammonia
(e) CH₃NH₂ – methyl amine
(f) H₂PO₄^– – dihydrogen phosphate ion
(g) CH₃COO^– – acetate ion
(h) HPO₄^-2 – hydrogen phosphate ion
(i) S^-2 – sulphide ion
(j) CO₃^-2 – carbonate ion

Question 2.
State the formula and name of the conjugate acid of each of the following:
(a) OH^–,
(b) CH₃COOH,
(c) HPO₄^-2,
(d) CO₃^-2,
(e) NH₃,
(f) CH₃NH₂,
(g) HS^–,
(h) CH₃COO^–
(i) H₂PO₄^–,
(j) Cl^–.
Answer:
(a) H₃O – water
(b) CH₃COOH₂⁺ – protonated acetic acid
(c) H₂PO₄^– – dihydrogen phosphate ion
(d) HCO₃^– – bicarbonate ion
(e) NH₄⁺ – ammonium ion
(f) CH₃NH₃⁺ – protonated methyl amine
(g) H₂S – hydrogen sulphide
(h) CH₃COO – acetic acid
(i) H₃PO₄ – phosphoric acid
(j) HCl – hydrochloric acid

Question 3.
Which of the following behave both as Bronsted acids as well as Bronsted bases?
NH₃, HSO₄^–, H₂SO₄, HCO₃^–; H₃PO₄, HS^–, H₂O.
Answer:
NH₃, HSO₄^– , HCO₃^– ; HS^– and H₂O behave both as Bronsted acids and Bronsted bases.

Question 4.
Classify Lewis acids and bases from the following:
H₃O⁺, CH₃NH₂, BF₃, OH^–, Cu²⁺, S^2-, CH₃COO^–.
Answer:
Lewis acids: H₃O⁺, BF₃, Cu²⁺.
Lewis bases: CH₃NH₂, OH^–, S^2-, CH₃COO^–.

Question 5.
Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base.
(a) OH^–, (b) F^–, (c) H⁺, (d) BCl₃
Answer:
(a) OH^– is a Lewis base because it acts as an electron-pair donor.
(b) F^– is also a Lewis base.
(c) H⁺ is a Lewis acid because it acts as an electron pair acceptor.
(d) BCl₃ is an electron-deficient compound electron pair acceptor, hence Lewis acid.

Question 6.
Explain how the [H⁺] concentration or [OH^–] concentration, decides a solution as acidic, neutral, or alkaline.
Answer:

Question 7.
Calculate the hydrogen and hydroxyl ion concentrations in (i) 0.01M HNO₃, (ii) 0.001M KOH solution at 298 K.
Answer:
(i) HNO₃ is a strong acid. It ionizes as

(ii) KOH is a strong base. It ionizes as

Question 8.
Assuming complete dissociation, calculate the pH of the following solutions:
(i) 0.003M HCl,
(ii) 0.005M NaOH,
(iii) 0.002M HBr,
(iv) 0.002M KOH.
Answer:
(i) HCl → H⁺ + Cl^–
[H⁺] = [HCl] = 0.003M
pH = – log [H⁺]
= – log [0.003] = 2.52

(ii) NaOH → Na⁺ + OH^–
[OH^–] = [NaOH] = 0.005M
pOH = – log[OH^–]
= – log [0.005] = 2.301
pH = 14 – pOH
= 14 – 2.301 = 11.699

(iii) HBr → H⁺ + Br^–
[H⁺] = [HBr]
= 0.002M
pH = – log [H⁺] (or)
pH = – log [0.002] = 2.6989

(iv) KOH → K⁺ + OH^–
[OH^–] = [KOH]
= 0.002M
pOH = – log [OH ]
pH = – log (0.002) = 2.6929
pH = 14 – pOH (or)
= 14 – 2.6929 = 11.3011

Question 9.
Calculate the pH of the resulting mixtures: 10 ml of 0.02M Ca(OH)₂ + 25 ml of 0.1M HCl.
Answer:

Number of moles of OH^– ion = 2 x 0.002 = 0.004
Number of moles of H⁺ ion = 0.0025
Number of moles of OH^– remaining after neutralisation = 0.004 – 0.0023 = 0.0015
Molarity of OH^– ions

Question 10.
The concentration of H⁺ ions in 0.1 M solution of a weak acid is 1.0 x 10^-5 mol L^-1. Calculate the dissociation constant of the acid.
Answer:

[HA] can be taken as 0.1 as 1 x 10^-5 is very small. Hence,

Question 11.
Find whether the resulting solution is acidic, neutral, or basic when
(i) a strong acid is mixed with a strong base
(ii) a strong acid is mixed with a weak base
(iii) a weak acid is mixed with a strong base
(iv) a weak acid is mixed with a weak base.
Answer:
(i) A strong acid and a strong base give a neutral solution because both are completely ionized and the reaction goes on completion.
eg:

(ii) A strong acid and a weak base give an acidic solution, as the weak acid is hot completely ionized. The reaction does not go for completion and there is an excess of hydrogen ions in the solution.
eg:

The solution is therefore is acidic.
(iii) A weak acid and a strong base give a basic solution, as the weak acid is not completely ionized. The reaction does not go for completion and there is an excess of hydroxyl ions in the solution.

Hence the solution is basic.
(iv) A weak acid and a weak base give an acidic, neutral, or basic depending on the relative strength of acid or base. If both weak acid and weak base have equal strength, the resulting solution is neutral.

Question 12.
State Ostwald’s dilution law.
Answer:
When dilution increases, the degree of dissociation of weak electrolytes also increases.
The mathematical form is
α = \(\sqrt{\frac{\mathrm{K}_{a}}{\mathrm{C}}}\)
where ‘α’ is the degree of dissociation, K_a is the dissociation constant of the acid and ‘C’ is its concentration.

Question 13.
Explain the term ‘common ion effect’ with an example.
Answer:
The suppression of the dissociation of a weak electrolyte by the addition of an ion in common with that of the weak electrolyte is known as the common ion effect.
Consider the dissociation of acetic acid in the presence of sodium acetate or HCl.
In the absence of any added salt, the dissociation of acetic acid is represented as

The addition of acetate ions in the form of sodium acetate increases the concentration of acetate ions. This increases the value of K_a. But K_a is a constant at a given temperature. Thus to maintain the K_a value constant, H⁺ ions combine with acetate ion to form undissociated acetic acid. i.e., The degree of dissociation of acetic acid in the presence of acetate ion is less than in the absence of sodium acetate. The same happens in the presence of HCl.

Question 14.
What are buffer solutions? Give example.
Answer:

1. Buffer solutions resist changes in pH by the addition of small quantities of an acid or a base. This ability is known as buffer action.
2. Buffer solutions consist of a mixture of a weak acid and a salt of a weak acid and a strong base.
   eg: CH₃COOH + CH₃COONa. This type of buffer is known as an acid buffer.
3. Buffer solution which consists of a mixture of a weak base and a salt of a weak base and strong acid is known as a basic buffer.
   eg: NH₄OH and NH₄Cl.

Question 15.
Give the characteristics of a buffer solution.
Answer:

1. It should have a definite pH. i.e., it has reserve acidity or alkalinity.
2. Its pH does not change on standing.
3. Its pH does not change on dilution.
4. Its pH is only slightly changed by the addition of a small quantity of acid or base.

Question 16.
Give examples of acidic and basic buffers.
Answer:
Acidic buffer:

1. CH₃COOH + CH₃COONa
2. Boric acid and borax.
3. Pthalic acid and potassium acid pthalate.

Basic buffer:

1. NH₄OH + NH₄Cl
2. Glycine and glycine hydrochloride.
3. Aniline and aniline hydrochloride.

Question 17.
Explain buffer action with a suitable example.
Answer:
The buffer action in a solution containing CH₃COOH and CH₃COONa is explained as follows:
The dissociation of the buffer components occurs as below.


If an acid is added to this mixture, it will be consumed by the conjugate base CH₃COO^– to form the undissociated weak acid i.e., the increase in the concentration of H⁺ does not reduce the pH significantly.

If a base is added, it will be neutralized by H₃O⁺, and the acetic acid is dissociated to maintain the equilibrium. Hence the pH is not significantly altered.

Question 18.
Explain the terms ‘buffer capacity” and ‘buffer index”.
Answer:
The buffer capacity is the ability of the buffer to resist changes in pH by the addition of small quantities of acid or base. It is measured in terms of buffer capacity. The quantitative measure of buffer capacity is the buffer index. It is defined as the number of equivalents of acid or base to one liter of the buffer solution to change its pH by unity.
β = \(\frac{d \mathrm{~B}}{d(\mathrm{pH})}\)
where P = buffer index, dB = number of gram equivalent of acid/base added to one liter of buffer solution, d(pH) = The change in pH after the addition of an acid or base.

Question 19.
Derive Henderson equation for the determination of pH of an acid buffer.
Answer:
An acid buffer consists of a mixture of a weak acid and salt with strong acid. The ionization of weak acid, HA is given by
HA ⇌ H⁺ + A^–
The dissociation constant of the weak acid is given by,

It can be assumed that the concentration of [A^–] ions from the complete ionization of the salt compared to the ionization of the weak acid. Hence, [A^–] concentration may be considered as the concentration of the salt.
So,

Taking logarithms on both sides and reversing the sign,

This is known as Henderson’s equation.

Question 20.
Write Henderson’s equation for a basic buffer.
Answer:

Question 21.
Suppose it is required to make a buffer solution of pH = 4. Using acetic acid and sodium acetate. How much sodium acetate to be added to 1 liter of N/10 acetic acid? The dissociation constant of acetic acid is 1.8 × 10^-5.
Answer:
Applying Henderson’s equation,
The molecular mass of CH₃COONa = 82
Amount of salt = 0.018 × 82 = 1.476 g.

Question 22.
Calculate the pH of a buffer solution containing 0.15 moles of NH₄OH and 0.25 moles of NH₄Cl. (K_b for NH₄OH = 1.8 × 10^-5)
Answer:

Question 23.
Calculate the pH of 0.1M ammonia solution. Calculate the pH after 50 ml of this solution is treated with 25 ml of 0.1M HCl. The dissociation constant of ammonia, K_b = 1.77 × 10^-5.
Answer:
For NH₄OH

When NH₃ is added, neutralization occurs

(M_NH4Cl) number of mole of NH₄Cl = 0.0025
(M_NH4OH) Remaining mole of NH₄OH = 0.005 – 0.0025 = 0.0025
According to Henderson’s equation,

Question 24.
Define hydrolysis.
Answer:
The reaction between cation or anion of the salt with water to produce the acid and base from which the salt formed is known as hydrolysis.
salt + H₂O = acid + base

Question 25.
Define the term degree of hydrolysis.
Answer:
It is the fraction of one mole of the salt that undergoes hydrolysis is known as the degree of hydrolysis.

Question 26.
Define the term hydrolysis constant (K_h), with an example.
Answer:
The equilibrium constant for the reaction, salt + H₂O ⇌ acid + base, is known as the hydrolysis constant (K_h). Consider the hydrolysis of acetate ion.

Question 27.
The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of the ammonium acetate solution.
Answer:
Ammonium acetate is the salt of a weak acid and weak base.

Question 28.
The ionization constant of nitrous acid is 4.5 × 10^-4. Calculate the pH of 0.04 M sodium nitrite solution and its degree of hydrolysis.
Answer:
Sodium nitrite is a salt of a weak acid and a strong base.

Substituting these values in equation (1)
pH = \(\frac{1}{2}\) [14 + 3.346 – 1.3979]
= 7.974
Degree of hydrolysis (h)

Question 29.
Calculate the degree of hydrolysis and pH of a 0.1M sodium acetate solution. The hydrolysis constant of sodium acetate is 5.6 × 10^-10.
Answer:

Question 30.
Calculate the pH of an aqueous solution of 1.0M ammonium formate assuming complete dissociation. pK_a for formic acid is 3.8 and pK_b of ammonia = 4.8.
Answer:
Ammonium formate is a salt of a weak acid and weak base.

Question 31.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer:
Pyridinium hydrochloride is a salt of a weak base and strong acid.

Question 32.
Define solubility product.
Answer:
It is defined as the product of the concentration of ions in a saturated solution of an electrolyte at a given temperature.

Question 33.
Give expressions for the solubility product of the following: (i) BaSO₄, (ii) H₂S, (iii) Sb₂S₃, (iv) AlI₃.
Answer:

Question 34.
Mention the criteria of precipitation of an electrolyte.
Answer:
The ionic product should exceed the value of the solubility product, for precipitation to occur.

Question 35.
How will you decide whether a solution is saturated or unsaturated, from the ionic product values?
Answer:

1. When the ionic product is less than the solubility product, the solution is unsaturated.
2. When an ionic product is equal to the solubility product, the solution is saturated.

Question 36.
Obtain a relationship between solubility and solubility product of the following: (i) AgCl, (ii) Ag₂CO₃, (iii) CaF₂, (iv) Cr(OH)₃.
Answer:

Question 37.
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to the precipitation of copper iodate? K_s for cupric iodate = 7.4 x 10^-8.
Answer:
Given, [NaIO₃] = 0.002 M
[Cu(ClO₃)₂] = 0.002 M
∴ [IO₃^–] = 0.002 M
[Cu⁺²] = 0.002 M
When equal volumes of these solutions are mixed, their concentrations will be halved.
i.e., [IO₃^–] = 0.001 M
[Cu⁺²] = 0.001 M
Cu(IO₃)₂ ⇌ Cu⁺² + 2IO₃^–
K_s = [Cu⁺²][IO₃^–]²
= 0.001 x (0.001)2 = 10^-9
Since IP is less than solubility products precipitation will not occur.

Question 38.
What is the minimum volume of water required to dissolve 1 gm of calcium sulfate at 298 K. The K_s for calcium sulfate is 9.1 x 10^-6.
Answer:
Let the solubility of CaSO₄ at 298 K be ‘s’ mol L^-1.

Minimum volume of water required to dissolve
1 g of CaSO₄ = \(\frac{1000}{0.41}\)
= 2439 mL = 2.439 L

Question 39.
The solubility product of BaS04 is 1.5 x 10^-9 Find its solubility (i) in pure water and (ii) in 0.1M BaCl₂, solution.
Answer:
(i)

Let ‘s’ be the solubility in mol L^-1. Then
K_s = s²
(or) 1.5 x 10⁷ = s²
s = \(\sqrt{1.5 \times 10^{-9}}\)
= 3.87 x 10^-5 mol L⁷

(ii) Let (s’) be the solubility of BaSO₄ in 0.1M BaCl₂ solution.
Total [Ba²⁺] = [Ba²⁺] from BaSO₄ from [Ba²⁺] from BaCl₂.

Question 40.
Calculate the pH at which Mg(OH)₂ begins to precipitate from a solution containing 0.10 Mg²⁺ ions.
Answer:

Choose the correct answer:

1. What is the conjugate base of OH?
(a) O₂
(b) H₂O
(c) O^–
(d) O^2-
Answer:
(d)
Hint:

2. C₂H₅ONa is …………. of C₂H₅OH.
(a) strong acid
(b) weak acid
(c) strong base
(d) weak base
Answer:
(c)
Hint: C₂H₅OH is a strong base. a weak acid, C₂H₅O^– is a strong base

3. Among the following, the one which can act as Bronsted acid as well as Bronsted base is:
(a) H₃PO₄
(b) AlCl₃
(c) CH₃COO^–
(d) H₂O
Answer:
(d)
Hint:

4. At 25° C, the dissociation constant of a base, BOH is 1.0 x 10^-12. The concentration of hydroxyl ions in 0.01M aqueous solution of the base would be:
(a) 1.0 x 10^-6 mol lit^-1
(b) 1.0 x 10^-7 mol lit^-1
(c) 2.0 x 10^-6 mol lit^-1
(d) 1.0 x 10^-5 mol lit^-1
Answer:
(b)
Hint:

5. Which of the following is the correct statement?
(a) HCO₃^– is the conjugate base of CO₃^2-
(b) NH₂^– is the conjugate acid of NH₃
(c) H₂SO₄ is the conjugate acid of HSO₄^–
(d) NH₃ is the conjugate base of NH₂^–
Answer:
(c)
Hint:

6. Three reactions involving H₂PO₄^– are given below:

In which of the above H₂PO₄^– act as an acid?
(a) (iii) only
(b) (i) only
(c) (ii) only
(d) (i) and (ii)
Answer:
(c)

7. Which one of the following will decrease the pH of 50 ml of 0.01M hydrochloric acid?
(a) Addition of 50 ml of 0.01 M HCl
(b) Addition of 50 ml of 0.002 M HCl
(c) Addition of 150 ml of 0.002 M HCl
(d) Addition of 5 ml of 1 M HCl
Answer:
(d)
Hint: For 0.001 M HCl, [H⁺] = 10^-2M; pH = 2

8. If pK_a for fluoride ion at 25° C, is 10.83, the ionisation constant of hydrofluoric acid at this temperature is:
(a) 1.74 x 10^-5
(b) 3.52 x 10^-3
(c) 6.75 x 10^-4
(d) 5.38 x 10^-2
Answer:
(c)
Hint:
pK_a + pK_b = 14
pK_a = 14 – 10.83 = 3.17
pK_a= – log Ka = – log 3.17
log K_a= -3.17 = 4.83
(or) K_a= 6.76 x 10^-4

9. The pH of a solution obtained by mixing 100 ml of a solution of pH = 3 with 400 ml of a solution of pH = 4 is:
(a) 3 – log 2.8
(b) 1 – log 2.8
(c) 4 – log 2.8
(d) 5 – log 2.8
Answer:
(c)
Hint:

10. The pH of 0.1M aqueous solution of a weak acid, HA is 3. What is its degree of dissociation?
(a) 1%
(b) 10%
(c) 50%
(d) 25%
Answer:
(a)
Hint: HA ⇌ H⁺ + A^–
[H⁺] = Cα
0.1 x α = 10^-3
α = 10^-2 i.e., 1%

11. The pK_a of acetic acid is 4.74. The concentration of acetic acid is 0.01 M. The pH of acetic acid is:
(a) 3.37
(b) 4.37
(c) 4.74
(d) 0.474
Answer:
(a)

12. How many times 1M CH₃COOH solution should be diluted so that pH of the solution is doubled?
(a) 20 times
(b) 200 times
(c) 5.55 x 10² times
(d) 5.55 x 10⁴ times
Answer:
(d)
Hint: pH of a weak acid

13. 40 ml of 0.1M ammonia is mixed with 20 ml 0.1M HCl. What is the pH of the mixture? (pK_b for ammonia is 4.74).
(a) 4.74
(b) 2.26
(c) 9.26
(d) 5.00
Answer:
(c)
Hint: 40 ml of 0.1M NH₃ solution
= 40 x 0.1 milli = 4 millimol
20 ml of 0.1 M HCl = 20 x 0.1 = 2 millimol
2 millimol of HCl neutralise 2 millimol of NH₄OH, to form 2 millimol of NH₄OH.
NH₄OH left = 2 millimol
Total volume = 60 ml

14. The pH of a solution formed on mixing 20 ml of 0.05M H₂SO₄ with 5 ml of 0.45 M NaOH at 298 K is:
(a) 6
(b) 2
(c) 12
(d) 1
Answer:
(c)
Hint:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
20 ml of 0.05M H₂SO₄
= 20 ml of 0.05 millimol
= 1.0 millimol

5 ml of 0.45M NaOH
= 5 x 0.45 millimol
= 2.25 millimol

2 millimol of NaOH will react with 1 millimol of H₂SO₄.
∴ NaOH left in solution = 0.5 millimol Volume of the solution = 25 ml
∴ [OH^–] = \(\frac{0.5}{25}\) = 0.01 M = 10^-2 M
[H⁺] = 10^-12;
pH = 12

15. Which of the following salts will have the highest pH in water?
(a) KCl
(b) NaCl
(c) Na₂CO₃
(d) CuSO₄
Answer:
(c)
Hint: KCl and NaCl, being salts of strong acid and strong base do not get hydrolysed. Their aqueous solutions are neutral and hence their pH = 7.
Na₂CO₃ being a salt of weak acid and a strong base, gives NaOH as a product of hydrolysis. Hence, its aqueous solution is alkaline and has pH >7.
CuSO₄ being a salt of strong acid and weak base, gives H₂SO₄, as a product of hydrolysis H₂SO₄ is a strong acid and its pH < 7.

16. The H₃O⁺ ion concentration of a solution of pH 6.58 is:
(a) antilog (- 6.58)
(c) antilog (- 5.58)
Answer:
(a)
Hint: pH = – log [H₃O⁺];
log [H₃O⁺] = -pH = – 6.58
[H₃O⁺] = antilog (-6.58)
= 2.63 x 10^-7 g ion/lit

17. 30 CC of \(\frac{M}{4}\) HCl, 20 CC of \(\frac{M}{2}\) HNO₃, and 40 CC of \(\frac{M}{4}\) NaOH solutions are mixed and the volume is made upto 1 dm³. The pH of the resulting solution is:
(a) 2
(b) 1
(c) 3
(d) 8
Answer:
(a)
Hint:
Total millimoles of H^–
= (30 x \(\frac{1}{3}\)) + (20 x \(\frac{1}{2}\))
= 10 + 10 = 20
Total millimol of OH^–
= 40 x \(\frac{1}{4}\) = 10
∴ H⁺ ions left after the neutralisation = 10 millimoles Volume of the solution
= 1 dm3
∴ Molarity of H⁺ ions
= \(\frac{10}{1000}\) = 10^-2M
pH = – log [H⁺]
= – log [10^-2] = 2

18. The pH of 0.1 M solutions of the following salts increases in the order:
(a) NaCl < NH₄Cl < NaCN < HCl
(b) HCl < NH₄Cl < NaCl < NaCN
(c) NaCN < NH₄Cl < NaCl < HCl
(d) HCl < NaCl < NaCN < NH₄Cl
Answer:
(b)
Hint: NaCl = neutral (pH = 7); NH₄Cl = slightly acidic (pH < 7); NaCN = basic (pH > 7); HCl = strongly acidic (pH << 7).

19. A weak acid HX has the dissociation constant 1 x 10^-5. It forms NaX on reaction with alkali. The degree of hydrolysis of 0.1 M solution of NaX is:
(a) 0.0001%
(b) 0.01 %
(c) 0.1%
(d) 0.15 %
Answer:
(b)
Hint: Hydrolysis reaction is
X^– + H₂O ⇌ HX + OH^–.
For a salt of weak acid with strong base,

20. Among the following hydroxides, the one which has the lowest value of K_s (solubility product) at ordinary temperature is:
(a) Mg(OH)₂
(b) Ca(OH)₂
(c) Ba(OH)₂
(d) Be(OH)₂
Answer:
(d)
Hint: The solubility increases down the group due to an increase in the size of the ion and a decrease in lattice energy. Lower the solubility, lower is the K_s.

21. On adding 0.1M solution each of Ag⁺, Ba²⁺, Ca²⁺ ions in a Na₂SO₄ solution, the species first precipitated is (K_s(CaSO₄) = 10^-6, K_s(BaSO₄) = 10^-11, K_s(Ag₂SO₄) = 10^-5:
(a) Ag₂SO₄
(b) BaSO₄
(c) CaSO₄
(d) All of these
Answer:
(b)

The minimum [SO₄^2-] concentration required for precipitation in for BaSO₄.

22. The K_s of Ag₂CrO₄, AgCl, AgBr, and AgI are respectively 1.1 x 10^-12, 1.8 x 10^-6, 5.0 x 10^-13 and 8.3 x 10^-17. Which of the following salts will precipitate last of AgNO₃ solution containing equal moles of NaCl, NaBr, Nal and Na₂CrO₄.
(a) AgBr
(b) Ag₂CrO₄
(c) AgI
(d) AgCl
Answer:
(b)

As the solubility of Ag₂CrO₄ is highest, it will be precipitated last at all.

23. The Ks of Ag₂CrO₄ is 1.1 x 10 12 at 298 K.
The solubility in mol per litre of Ag₂CrO₄ in 0.1M. AgNO₃ solution is:
(a) 1.1 x 10^-11
(b) 1.1 x 10^-10
(c) 1.1 x 10^-12
(d) 1.1 x 10^-9
Answer:
(b)

24. Using Gibb’s free energy change, AG° = +63.3 kJ for the following reaction:
Ag₂CO₃ (s) ⇌ 2Ag⁺(aq) + CO₃^2-(aq)
the K_s for Ag₂CO₃ in water at 25° C is (R = 8.314 JK^-1 mol^-1)
(a) 3.2 x 10^-26
(b) 8.0 x 10^-12
(c) 2.9 x 10^-3
(d) 7.9 x 10^-2
Answer:
(b)
Hint:

25. A buffer solution is prepared in which the concentration of NH₃ is 0.30M and the concentration of NH₄⁺ is 0.20M. If the equilibrium constant, K_b for NH₃ equals 1.8 x 10^-5, what is the pH of the solution?
(a) 8.73
(b) 9.08
(c) 9.43
(d) 11.72
Answer:
(b)

= 4.74 – 0.176 = 4.56
pH = 14 – pOH = 14 – 4.56 = 9.44

26. What is the [H⁺] in mol/L of a solution that is 0.20M in CH₃COONa and 0.10M in CH₃COOH? K_b for CH₃COOH = 1.8 x 10^-5:
(a) 9.0 x 10⁶
(b) 3.5 x 10^-6
(b) 1.1 x 10^-5
(d) 1.8 x 10^-5
Answer:
(a)

27. In what volume ratio of NH₄Cl and NH₄OH solution (each 1M) should be mixed to get a buffer solution of pH, 9.80 (pK_b for NH₄OH is 4.74):
(a) 1 : 2.5
(b) 2.5 : 1
(c) 1 : 3.5
(d) 3.5 : 1
Answer:
(c)
Hint:
Suppose ‘V₁’ ml of 1M NH₄Cl is mixed with ‘V₂’ ml of 1M NH₄OH solution:
V₁ ml of 1M NH₄Cl = V₁ millimole
V₂ ml of 1M NH₄OH = V₂ millimole
pH = 9.80, Hence pOH = 14 – 9.80 = 4.20

28. Which one of the following pairs of solutions is not an acidic buffer?
(a) H₂CO₃ + Na₂CO₃
(b) H₃PO₄ + Na₃PO₄
(c) HClO₄ + NaClO₄
(d) CH₃COOH + CH₃COONa
Answer:
(c)
Hint: HClO₄ is a strong acid. Hence, it cannot be used to make an acidic buffer.

Students get through the TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Answer the following questions.
Question 1.
Express the rate of the following reaction in terms of disappearance of the reactant and appearance or formation of the product: A → B
Answer:
Rate of disappearance of the reactant = \(-\frac{\Delta[\mathrm{A}]}{\Delta t}\)
where the negative sign indicates the concentration of the reactant decreases with time.
Rate of formation of the product = \(+\frac{d[\mathrm{B}]}{d t}\)
The positive sign indicates that the concentration of the product increases with time.

Question 2.
Mention the unit of rate of reaction.
Answer:

For a gaseous reaction, a unit of rate = atm sec^-1 since concentration is expressed in partial pressures.

Question 3.
Explain how will you determine experimentally determine
(i) average rate of reaction
(ii) instantaneous rate of reaction and
(iii) initial rate of reaction.
Answer:
The changes in concentration of the reactants or products are experimentally determined by withdrawing a small portion of the reaction mixture (usually 2 or 5 ml) at suitable intervals of time and then freeze to about 0°C to stop the reaction. The concentration of the reactant or product is measured by a suitable method. A plot of concentration-time is made from the graph, the average rate, instantaneous and initial rates can be determined.
(i) Average rate: Two equivalent points are taken with respect to the time at which the average rate is to be determined. The concentrations corresponding to these points are noted from the graph. The difference in concentration is divided by the time interval between equidistant points. Two equidistant points with respect to T is selected. Let these be ‘t₁,’ and ‘t₂’ respectively. The corresponding concentrations are x₁, and x₂ respectively.

(ii) Instantaneous rate of reaction: For the determination of the instantaneous rate of reaction at any time ‘t’, a tangent is the direction to the curve at the point ‘P’ corresponding to that time. The slope of this tangent gives the instantaneous rate of reaction.

Where Q is the angle of a tangent with the time axis. In a similar manner, the rate of formation of the product can be determined by measuring the slope of the tangent at a particular instant in concentration to the time curve.
(iii) Initial rate of reaction:

The rate of reaction is maximum at the time of mixing the concentration. This cannot be determined experimentally. This is determined from the concentration is time curve. The curve is extrapolated to zero time and the slope of this curve at zero time gives the initial rate of the reaction.
\(\frac{\Delta x}{\Delta t}\) = initial rate.

Question 4.
In the reaction 2A → products the concentration of A decreases from 0.5 mol L¹ to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this period.
Answer:

Question 5.
The concentration of reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate using units of time both in minutes and in seconds.
Answer:

Question 6.
The decomposition of N₂O₅ is expressed by the equation.

If during a certain interval of time the rate of decomposition of N₂O₅ is 1.8 x 10^-3 mol liter 1 min^-1, what will be the rates of formation of NO₂ and O₂ during the same interval.
Answer:
The rate expression for the decomposition of N₂O₅ is

Question 7.
For each of the following reactions, express the given rate of change of concentration of the product or reactant in terms of rate of change of concentration of other reactants or products in that reaction.

Answer:
(a)

(b) The equality in this case is,


(c) The equality in this case is,

Question 8.
From the concentration of R at different times given below, calculate the average rate of the reaction.
Answer:
R → P during different intervals of time

Average rate of reaction between 0 → 5 sec

Average-rate of reaction between 5-10 sec

Similarly, average rate between 10 → 20 and 20 → 30 see can be calculated.
Note: What we understand from the above problem is that the rate of reaction is not a constant quantity. It decreases with time.

Question 9.
The decomposition of N₂O₅ is CCl₄ solution at 318K has been studied by monitoring the concentration of the N₂O₅ in the solution. Initially, the concentration of N₂O₅ in the solution is 2.33 M and after 184 minutes it is reduced to 2.08 M. The reaction takes place according to the equation
2N₂O₅ → 4NO₂ + O₂.
Calculate the average rate of the reaction in terms of hours, minutes and seconds. What is the rate of production during this period?
Answer:
Average rate of reaction (mol L^-1 min^-1)

For the given reaction

Question 10.
In hydrogenation reaction at 25°C it is observed that hydrogen gas pressure falls from 2 atm to 1.2 atm in 50 min. Calculate the rate of reaction in molarity per sec.
Answer:
R = 0.0821 lit mol^-1 deg^-1

Question 11.
Explain why the average rate cannot be used to produce the rate of reaction at any instant.
Answer:
Because the rate decreases with time as the reaction proceeds. It is not constant.

Question 12.
Bring out the difference between the rate and rate constant of a reaction?
Answer:

Question 13.
The decomposition of dimethyl ether leads to the formation of CH₄, H_2, and Co and the reaction rate in given by
rate = R [CH₃OCH₃]^3/2
The rate is followed by an increase in pressure in a closed vessel so that the rate is expressed in terms of partial pressure of dimethyl ether.
rate = R[P_CH3OCH3]^3/2
If the pressure is measured in the bar and the time in seconds then what are theTxmts of rate and rate constant?
Answer:
The rate law of the reaction

Question 14.
State the order with respect to each reactant, overall reaction, and the units of the rate constant in each of the following reactions.

Answer:
(a) Order with respect to NO and Br₂ are 2 and 1 respectively, over all order is 3.

(b) Order \(\frac{3}{2}\)

(c) Order with respect to H₂ and NO are 1 and 2 respectively.

= mol^-2 litre² sec^-1
(d) Order with respect to CO and Cl₂ are 2 and \(\frac{1}{2}\) respectively.
Over all order = 2\(\frac{1}{2}\) = \(\frac{5}{2}\)

(e) Order with respect to H₂O₂ and I^– are one respectively. Over all order 1 + 1 = 2

Question 15.
Identify the reaction order from the following rate constants.
(a) k = 3.1 × 10^-4 sec-1
(b) k = 1.12 × 10^-2
(c) k = 1.35 × 10^-2 mol^-2 let² s^-1
(d) k = 3.4 × 10^-3 mol^– let s^-1
Answer:
(a) Let the reaction be of nth order

i.e., reaction is of first order.
(b) (atm)^1-n × s^-1 = atm^-1 s^-1
1 – n = -1 : or n = 2
i.e., reaction is of second order.
(c)

i.e., reaction is of third order.
(d)

i.e., reaction is of second order.

Question 16.
Derive an expression for the rate constant for the first-order reaction.
Answer:
A reaction whose fate depends on the reactant concentration raised to the first power is called a first order reaction. Let us consider the following Cl₂ first-order reaction,
A → product
Rate law can be expressed as Rate = k [A]¹
Where, k is the first order rate constant.

Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial concentration [A₀] to [A] at the later time.

This equation is in a natural logarithm. To convert it into the usual logarithm with base 10, we have to multiply the term by 2.303.

Question 17.
How will you determine the first-order rate constant graphically?
Answer:
For a first order reaction, the rate constant k is given by

where [A₀] is the initial concentration of the reactant and [A] is the concentration of the reactant at time it\
Rearranging this equation in the form . y = mx + c, we get
log [A₀] – log [A] = \(\frac{kt}{2.303}\)
2.303 [log [A] – log [A]] = kt
log [A₀] – log [A] = \(\frac{k}{2.303}\) t.
Hence a plot of log [A] vs t gives a straight line with a negative slope having the volume of \(\frac{-k}{2.303}\)

Question 18.
Give example for first order reaction.
Answer:
(i) Decomposition of dinitrogen pentoxide
N₂O₅ → (g) 2NO₂ (g) + \(\frac{1}{2}\) O₂ (g)
(ii) Decomposition of thionylchloride;
SO₂Cl₂(l) → SO₂ (g) + Cl₂ (g)
(iii) Decomposition of the H₂O₂ in aqueous solution;
H₂O₂ (aq) → H₂O (l) + \(\frac{1}{2}\) O₂ (g)
(iv) Isomerisation of cyclopropane to propene.

Question 19.
Show that acid hydrolysis of an ester is a pseudo first order reaction.
Answer:
A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction. Let us consider the acid hydrolysis of an ester,

If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis, i.e., the concentration of water remains almost constant.
Now, we can define k [H₂O] = k’ ; Therefore the above rate equation becomes
Rate = k’ [CH₃COOCH₃]
Thus it follows first-order kinetics.

Question 20.
Give examples of zero-order reactions.
Answer:
(i) Photochemical reaction between H₂ and I₂

(ii) Decomposition of N₂O on a hot platinum surface

(iii) Iodination of acetone in an acid medium is zero-order with respect to iodine.

Question 21.
Give the general rate equation for nth-order reaction involving one reactant A.
Answer:
The general rate equation for an nth order reaction involving one reactant [A],
A → product

Consider the case in which n ≠ 1, integration of above equation between [A₀] and [A] at time t = 0 and t = t respectively gives

Question 22.
Derive an expression to calculate t_1/2 for a zero-order reaction.
Answer:
Let us calculate the half life period for a zero-order reaction.

The half-life of a zero-order reaction is directly proportional to the initial concentration of the reactant.

Question 23.
Give the general expression for t_1/2 of the nth order (n ≠ 1) reaction.
Answer:
Half-life for an nth order reaction involving reactant A and n ≠ 1

Question 24.
Give the characteristics of first order reaction.
Answer:

1. A change in concentration unit will not change the numerical value of ‘k’
   k = \(\frac{2.303}{t}\) log \(\frac{a}{a-x}\)
   Where ‘a’ and (a – x) are the initial concentration of the reactant and concentration of the reactant at time Y respectively. Thus for the first-order reaction, any quantity which is proportional to the concentration can be used in place of concentration to calculate ‘k’.
2. The time taken for the completion of the same fraction of change is independent of the initial concentration.
3. Time required to calculate nth fraction of reaction t_1/2 can be calculated as When x = \(\frac{a}{n}\) ; t = t_1/n
   Substituting these values in,
   
   The above equation can be used to calculate t_3/4 of a reaction.
   
4. In a first-order reaction, the amount of reactant remaining after V half-lives can be calculated as follows:
   where [A₀] – initial concentration of the reactant, [A] = concentration of the reactant remaining after ‘n’ half-lives.

Question 25.
Calculate the half life of a first order reaction from their rate constant given below:
(a) 200 sec^-1 (b) 2 min^-1 (c) 4 year^-1
Answer:

= 3.465 x 10~3 sec
= 0.1732 year.

Question 26.
The rate constant for a first-order reaction is 60 sec^–. How much time will it take to reduce the initial concentration to its 1/16th value?
Answer:

Question 27.
The thermal decomposition of a compound is of the first order. If 50% of the sample is decomposed in 120 minutes, how long will it take for 90% of the compound to decompose?
Answer:
Half life of the reaction = 120 min.
We know that

If a= 100, x = 90, a-x = 10
= 5.77 x 10^-3 min^-1
Applying first order rate equation.

Question 28.
Sucrose decomposes in acid solution into glucose and fructose according to the first-order rate law. With t_1/2 = 3.00 hrs. What fraction of sucrose remains after 8 hours?
Answer:

Therefore fraction of reactant remaining = 0.1576

Question 29.
Explain the pressure change method in determining a first-order reaction.
Answer:
If pressure is given in gaseous reactions then we use the following kinetic equation.

where p₀ is the pressure at the initial stage p₀-x is the pressure at time ‘t’ values of p₀ and x can be calculated using the following examples.

Case I: If total pressure of the reaction mixture is given in place of pressure of reactants.
p_t = (p₀ – x + x + x + x)
Where p_t is the pressure of reaction vessel at time ‘t’.
Case II: If the pressure of the reaction vessel after a long time or infinite time is given
p_t = p₀ + p₀ +p₀

Question 30.
The decomposition of Cl₂O₇ at 400K in the gas phase to Cl₂ and O₂ is a first order reaction.
(i) After 55 seconds at 400K, the pressure of Cl₂O₇ falls from 0.062 to 0.044 atm. Calculate the rate constant.
(ii) Calculate the pressure of Cl₂O₇ after 100 seconds of decomposition at this temperature.
Answer:
(i) As pressure ∝ concentration.

(ii) Applying the first-order rate equation.

Question 31.
For the decomposition of azo isopropanol to hexane and nitrogen at 543K, the following data are obtained.

Calculate the rate constant for the reaction.
Answer:
The reaction is

Total pressure – p₀ – x + x + x
p_t = p₀ + x
or x = p_t – p₀
a ∝ p₀
∴ (a – x) ∝ p₀ – (p_t – p₀)
= 2p₀ – p_t
For the first order reaction

Question 32.
Explain Oswald dilution method for determining the order of the reaction.
Answer:
It is applied for these reactions which involve more than one reactant. In this method, the concentration of all reactants are taken in large excess than one. The changes in the concentration of the reactant that is not taken in excess will influence the rate of reaction.
The concentrations of all other reactants practically remain same during the course of reaction. Consider the reaction.
m₁A + m₂B + m₃C → product
Rate = k[A]^α [B]^β [C]^γ
When B and C are taken in excess, the rate law will reduce to
rate = k’ [A]^α
where k’ = k[B]^β [C]^γ
Two sets are taken in which the concentration of A is [A₁] and [A₂] while B and C are present in large excess,
r₁ = [A₁]^α
r₂ = [A₂]^α

From this, the value of α, i.e., the order with respect to A is obtained. The process is repeated one by one and order with respect to others is determined.
Total order of reaction = α + β + γ

Question 33.
Show that if the concentration of a reactant is doubled, the rate of the reaction is also doubled for a first-order reaction, increases four times for a second-order reaction, increases by eight times for a third-order reaction, i.e., A → product.
Answer:
(i) If the concentration of A is doubled.

Hence, for the first-order reaction, if the concentration is doubled, the rate is also doubled.
(ii) For a second-order reaction.

i.e., for a second-order reaction, if the concentration of A is doubled, the rate of the reaction increases by 4 times.
(iii) For a third-order reaction,

Hence, for a third-order reaction, if the concentration of the reactant is doubled the rate increases 8 times.

Question 34.
Compounds A and B react according to the following chemical equation
A (g) +2B(g) → 2C(g)
Answer:
The concentration of either ‘A’ or ‘B’ was changed keeping the concentration of one of the reactants constant and rates were measured as a function of initial concentration following results were obtained. Find the order with respect to A and B and write the rate law for the reaction

From experiments 1 and 3,
[B] = constant; [A] = doubled
rate is doubled.
Hence rate ∝ [A] i.e., the order with respect to A is 1.
From experiments 1, and 2,
[A] = constant; [B] doubled rate is quadrupled.
Hence, rate ∝ [B]²
i.e., order with respect by B is 2.
Combining rate = k [A] [B]².

Question 35.
The initial rate of reactions
3A + 2B + C → products at different initial concentrations are given below.

Answer:
(i) Consider equation (1) and (2)
The concentration of A and B are constant. The rate remains the same for different concentrations of C. Hence, the order with respect to C is zero.
(ii) Consider equations (3) and (1)
The concentration of A and C are constant. The concentration of B is doubled, rate increases by 2 times. Hence order with respect to B is 1.
(iii) Consider equations (4) and (1), The concentration of B and C remain constant. When the concentration of A is doubled, the rate increases by 4 times. Hence order with respect to A is 2.
Alternate method:
Suppose order with respect to A, B and C are α, β and γ respectively. Then the rate law at different concentration of A, B and C are
5.0 × 10^-3 = k [0.010]^α [0.005]^β [0.010]^γ …(1)
5.0 × 10^-3 = k [0.010]^α [0.005]^β [0.015]^γ …(2)
1.0 × 10^-3 = k [0.010]^α [0.010]^β [0.010]^γ …(3)
1.25 × 10^-3 = k [0.005]^α [0.005]^β [0.010]^γ …(4)
Dividing (1) by (2)

i.e., order with respect to C is zero.
Dividing (3) by (2)

Dividing (1) by (4) ⇒ 4 = 2^α or α = 2
Order with respect to A, B, C are 2, 1 and 0.

Question 36.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected by increasing the concentration of B, three times, keeping the concentration of A constant?
(iii) How is the rate affected by when the concentration of both A and B are doubled?
Answer:
(i) A + B → product
rate = k [A] [B]²
(ii) If the concentration of B is tripled, the rate increases by 9 times i.e.,

(iii) If the concentration of both A and B are doubled, the rate increases by 8 times.

Question 37.
The following rate data were obtained at 303K for the following reaction.
2A + B → C + D.

What is the order with respect to each reactant and the overall order of the reaction? write the rate law. Also, calculate the rate constant for the reaction and the units of the rate constant.
Answer:
From experiments 1 and 4, it is seen that [B] the concentration of B is the same but [A] has been made 4 times, the rate of the reaction has also become 4 times. Hence the order with respect to A is 1. i.e., rate ∝ [A],
From experiments 2 and 3, it may be noted that [A] is kept the same but [B] is doubled. The rate of reaction increases by 4 times, i.e., the order with respect to B is 2 combining both,
rate = k[A] [B]². This is the rate law for the reaction.
Alternatively, suppose the order with respect to A is α and the order with respect to B is β, the general rate law for the reaction is
rate = k [A]^α [B]^β
From the data given

Now,

Unit of rate constant (k)

Calculation of rate constant: Substitute the values of α and β in any of the equations.

Hence k = 6.0 mol^-1 L² min^-1.

Question 38.
In a reaction between A and B, the initial rate of reaction was measured at a different initial concentration of A and B as given below.

What is the order with respect to A and B?
Answer:
The general rate law is rate = k [A]^α [B]^β where α, β are ordered with respect to A and B respectively.
The given data is


Thus order with respect to A is 0.5 and that of B is 0.

Question 39.
The decomposition of ammonium nitrate in an aqueous solution was studied by placing the apparatus in a thermostat maintained at a particular temperature. The volume of nitrogen gas collected at different intervals of time was as follows:

The above data prove that the reaction is of the first order.
Answer:
For a first-order reaction,

In the present case, V_∞ = 35.05 cm³. The value of R at each instant of time can be calculated as follows:
Values of ‘k’ come nearly constant and hence the reaction is first order.

Question 40.
What do understand by a fraction of effective collisions? Mention its significances.
Answer:
Fraction of effective collisions (f) is given by the following expression

To understand the magnitude of collision factor (f), Let us calculate the collision factor (f ) for a reaction having an activation energy of 100 kJ mol^-1 at 300K.

Thus, out of 10¹⁸ collisions, only four collisions are sufficiently energetic to convert reactants to products.

Question 41.
Explain the importance of the proper orientation of molecules in the collision theory.
Answer:
Even if the reactant possesses sufficient activation energy, their collisions need not result in the formation of products. The molecules should orient themselves in such a way that their collision becomes fruit fill. The fraction of effective collisions (f) having orientation is given by steric factor (p). Thus, the rate of reaction is given by
rate = p x f x collision rate
(or) rate = pz\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\)

Question 42.
Define activation energy of a reaction.
Answer:
Activation energy is the extra amount of energy that is supplied from outside so that colloiding molecules must produce effective collisions.

Question 43.
Arrhenius equation is given by k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\). Based on this equation answer the following questions.
(i) Can reactions have zero activation energy?
(ii) Can a reaction have negative activation energy?
Answer:
(i) k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\), when the expression becomes k = e° = A.
The rate constant k = collision frequency. This implies that every collision is an effective collision and leads to product formation which is not possible. Thus activation energy can not be zero.
(ii) When E_a is negative, the above equation

Rate constant having a higher value than collision frequency which is not possible.

Question 44.
Explain the effect of temperature on reaction rate based on Arrhenius’s theory.
Answer:
An increase in temperature increases the fraction of the total number of molecules having the necessary energy of activation which in terms increases the number of effective collisions and hence the rate.

Question 45.
Write Arrhenius equation and explain the terms? What is the significance of frequency factor A in the equation?
Answer:
The Arrhenius equation is k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\) Where ‘k’ is the rate constant for the reaction, A is called the pre-exponential factor, E_a is the energy of activation and R is the gas constant and T is the absolute temperature. The two quantities A and E_a are called the Arrhenius parameters.
The frequency factor gives the frequency of binary collision of the reacting molecules per liter per second. It practically remains constant and does not vary with temperature.

Question 46.
Explain how the energy of activation for a reaction is determined.
Answer:
The rate constant of a reaction is determined at two different temperatures. If R₁ is the rate constant at temperature T₁ arid R₂ is the rate constant at temperature T₂ thereby using a logarithmic form of Arrhenius equation

E_a can be calculated.

Question 47.
Describe how the energy of activation of a reaction is determined graphically.
Answer:
The Arrhenius equation is written in the form,

This equation is of the form y = mx + C i.e., equation for a straight line. Thus, if a plot of log k is \(\frac{1}{T}\) is a straight line, the validity of the equation is confirmed.
Thus the values of k at different temperatures are found and a plot of log k vs \(\frac{1}{T}\) is made.
A straight line with a negative slope is obtained from the value of the slope, E_a can be calculated.
Slope = \(\frac{-\mathrm{E}_{a}}{2.303 \mathrm{R}}\)

Question 48.
The rate constant for a reaction is 1.2 x 10^-3 sec^– at 30°C and 2.1 x 10^-3 sec^-1 at 40°C. Calculate the energy of activation.
Answer:
Given that k₁ = 1.2 x 10^-3 sec^-1
T₁ = 30 + 273 = 303K
k₂ = 2.1 x 10^-3 sec^–
T₂ = 40 + 273 = 313K
Substituting these values in the equation
Solving E_a = 44126.3J = 44.13kJ

Question 49.
The rate of particular reaction doubles when temperature changes from 27°C to 37°C. Calculate the energy of activation.
Answer:
Given that T₁ = 27°C = 273 +21 = 300K
T₂ = 37°C = 273 + 37 = 310K
k₂ = 2k₁
Substituting these values in the equation.
This as solving for E_a
E_a = 53598.6 J mol^-1 or 53.6 kJ mol^-1

Question 50.
The activation energy of a reaction is 94.14 kJ mol^-1 and the value of the rate constant at 313K. is 1.8 x 10^-1 sec^-1 . Calculate the frequency factor (A).
Answer:
Given that, E_a = 94.14 kJ mol^-1 = 9414 J mol^-1
T = 313K,
k = 1.8 x 10^-1 sec^-1 Substituting these values in
= (log 1.8 – 5)+ 15.7082
= (0.2553 – 5)+ 15.708
= 10.963J
A = Anti log (10.963 J)
= 9.194 x 10^-10 sec

Question 51.
The first order rate constant for the decomposition of ethyl iodide by the reaction
C₂H₅I (g) → C₂H₄ (g) + HI (g)
at 600 K is 1.60 x 10^-1 s^-1. Its energy of activation is 209 kJ / mol. Calculate the rate constant of the reaction at 700K.
Answer:
Given T₁ = 600 K
k₁ = 1.60 x 10^-5 sec^-1
T₂ = 700K
k₂ = ?
E_a = 299 kJ mol^-1 = 299000 J mol^-1
Substituting these values in the equation.

Question 52.
Rate constant ‘k’ of a reaction varies with temperature according to the equation
where E_a is the energy of activation for the reaction. When a graph is plotted for log k vs \(\frac{1}{T}\) a straight line with slope – 6670 K is obtained.
Calculate the energy of activation for the reaction and mention is units.
(R = 8.314 J k^-1 mol^-1)
Answer:
Slope of the line = \(\frac{-\mathrm{E}_{a}}{2.303 \mathrm{R}}\) = -6670K
E_a = 2.303 x 8.314 x 6670 = 127711.4 J mol^-1

Choose the correct answers:

1. Which of the following statements is not correct about the order of reaction?
(a) The order of a reaction can be fractional.
(b) Order of reaction is an experimental quantity.
(c) The order of the reaction is always equal to the sum of the stoichiometric coefficients of reactants in a balanced chemical equation for the reaction.
(d) The order of a reaction is the sum of the powers of molar concentrations of the reactants in the rate law expression.
Answer:
(c)

2. Which of the following statements is correct?
(a) The rate of reaction decreases with the passage of time as the concentration of the reactant decreases.
(b) The rate of reaction is the same at any time during the reaction.
(c) The rate of reaction is independent of temperature change.
(d) The rate of reaction decreases with an increase in the concentration of reactants.
Answer:
(a)

3. Which of the following expression is correct for the rate of reaction given below:

Answer:
(c)

4. Time required for 100 percent completion of a zero-order reaction is:

Answer:
(c)

5. For the reaction aA+ bB → cC,

if , then a, b and c respectively are:
(a) 3,1,2
(b) 2,1,3
(c) 1,3,2
(d) 6, 2, 3
Answer:
(c)
Hint: Dividing throughout by 3, we get

This is so for the reacting A + 3B → 2C

6. The rate of a gaseous reaction is given by the expression k [A] [B]. If the volume of the reaction vessel is suddenly reduced to 1/4th of the initial volume, the reaction rate relating to the original rate will be:
(a) 1/10
(b) 1/8
(c) 8
(d) 16
Answer:
(d)
Hint: Rate = k ab. When volume is reduced to 1/4th, concentrations will become = 4 times.
New rate = k (4a) (4b) = 16 kab = 16 times.

7. In a reaction A → B, the rate of reaction increases two times on increasing the concentration of the reactant four times, then order of reaction is:
(a) 0
(b) 2
(c) 1/2
(d) 4
Answer:
(c)
Hint: (i) r = ka^α, (ii) 2r = k(4a)^α
Dividing (ii) by (i),
4^α = 2 or 2^2α = 2 or 2α = 1 or α = 1/2.

8. The rate of the reaction 2 NO + Cl₂ → 2 NOCl is given by the rate equation: rate = k[NO]²[Cl₂]. The value of the rate constant can be increased by:
(a) increasing the temperature
(b) increasing the concentration of NO
(c) increasing the concentration of Cl₂
(d) doing all of these
Answer:
(a)
Hint: The rate constant of a reaction depends only on temperature and does not depend upon concentrations of the reactants.

9. The unit of rate constant for a zero-order reaction is:
(a) mol L^-1 s^-1
(b) L mol^-1 s^-1
(c) L² mol^-2 s^-1
(d) s^-1
Answer:
(a)

10. Rate constant of a reaction (k) is 175 litre² mol^-2 sec^-1. What is the order of reaction?
(a) first
(b) second
(c) third
(d) zero
Answer:
(c)
Hint: On the basis of given units of k, the reaction is of 3rd order.

11. The reaction A → B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B?
(a) 1 hour
(b) 0.5 hour
(c) 0.25 hour
(d) 2 hours
Answer:
(a)
Hint: The fraction of A reacted in each case is same (0.2 / 0.8 = 1/4),
(0.9 – 0.675)/ 0.90 = 0.225 / 0.90 = 1/4. Hence, time taken is same.

12. 75% of the first-order reaction was completed in 32 min. 50% of the reaction was completed in:
(a) 24 min
(b) 8 min
(c) 16 min
(d) 4 min
Answer:
(c)
Hint: 75 % of reaction is completed in two half-lives, i.e., 2 x t_1/2 = 32 min or t_1/2 = 16 min.

13. 1/[A] vs time is a straight line. The order of the reaction is:
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(b)
Hint: log [A] vs time is linear for 1st order reactions, 1/[A] vs time is linear for 2nd order reactions; 1/[A]² vs time is linear for 3rd order reactions.

14. If a graph is plotted between in k and 1/T for the first-order reaction, the slope of the straight line so obtained is given by:

Answer:
(a)

15. 10g of a radioactive isotope is reduced to 1.25g in 12 years. Therefore, half-life period of the isotope is:
(a) 24 years
(b) 4 years
(c) 3 years
(d) 8 years
Answer:
(b)

16. The half-life period of a radioactive element is 20 days. What will be the remaining mass of 100 g of it after 60 days?
(a) 25 g
(b) 50 g
(c) 125 g
(d) 20 g
Answer:
(c)
Hint: 60 days = 3 half-lives, i.e, n = 3;

17. Activation energy of a chemical reaction can be determined by:
(a) determining the rate constant at standard temperature.
(b) determining the rate constants at two temperatures.
(c) determining the probability of collision.
(d) using catalyst.
Answer:
(b)
Hint:
Knowing k₁ and k₂ at T₁ and T₂, E_a can be determined. (Arrhenius equation)

18. According to Arrhenius equation, rate constant k is equal to A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\). Which of the following options represents the graph of k vs \(\frac{1}{T}\)?

Answer:
(a)
Hint:
k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\). In k = ln A – \(\frac{\mathrm{E}_{a}}{\mathrm{RT}}\). Hence, graph of In k vs \(\frac{1}{T}\) will be linear with negative slope and intercept = ln A.

19. Which of the following statement is incorrect about the collison theory of chemical reaction?
(a) It considers reacting molecules or atoms to be hard spheres and ignores their structural features.
(b) Number of effective collisions determines the rate of reaction.
(c) Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.
(d) Molecules should collide with sufficient threshold energy and proper orientation for the collision to be effective.
Answer:
(c)
Hint: (c) is incorrect because the formation of the product depends not only on energy but also on proper orientation at the time or collision.

20. A first-order reaction is 50% completed in 1.26 x 1014 s. How much time would it take for 100% completion?
(a) 1.26 x 1015 s
(b) 2.52 x 1014 s
(c) 2.52 x 1028 s
(d) infinite
Answer:
(d)
Hint: The whole of the substance never reacts because in every half-life, 50% of the substance reacts. Hence, the time taken for 100% completion of a reaction is infinite.

21. Compounds ‘A’ and ‘B’ react according to the following chemical equation:
A (g) + 2 B(g) → 2C (g)
The concentration of either ‘A’ or ‘B’ was changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.

(a) Rate = k [A]² [B]
(b) Rate = k [A] [B]²
(c) Rate = k [A] [B]
(d) Rate = k [A]² [B]⁰
Answer:
(b)
Hint: From expts.l and 3, [B] = constant, [A] = doubled, rate is doubled. Hence, Rate ∝ [A].
From expts. 1 and 2, [A] = constant, [B] = doubled, rate quadrupled. Hence, Rate ∝ [B]².
Combining, Rate = k [A][B]².

22. The rate constant of reaction A → B is 0.6 x 10³ mole per liter per second. If the concentration of A is 5 M, then the concentration of B after 20 minutes is:
(a) 0.36 M
(b) 0.72 M
(c) 1.08M
(d) 3.60 M
Answer:
(b)
Hint: The unit mol L^-1 s^-1 of the rate constant show that it is a reaction of zero order. For a zero order reaction, x = kt. Amount of A reached (x) = (0.6 x 10^-3 mol L^-1 s^-1 ) (20 x 60 s) = 0.72 mol L^-1 [B] formed = [A] reacted = 0.72 mol L^-1

23. The half-life of a substance in a certain enzyme-catalyzed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L^-1 to 0.04 mg L^-1, is:
(a) 414 s
(b) 552 s
(c) 690 s
(d) 276 s
Answer:
(c)