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Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 14 Biomolecules Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 14 Biomolecules

PSEB 12th Class Chemistry Guide Biomolecules InText Questions and Answers

Question 1.
What are monosaccharides?
Answer:
Monosaccharides are carbohydrates which cannot be further hydrolysed to simpler molecules. The general formula of monosaccharides is (CH₂O)_n where n = 3-7.

Question 2.
What are reducing sugars?
Answer:
Carbohydrates which reduce Fehling’s solution to red precipitate of Cu₂O or Tollen’s reagent to metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars.

Question 3.
Write two main functions of carbohydrates in plants.
Answer:
Two main functions of carbohydrates in plants are:

1. Polysaccharides such as starch serve as storage molecules.
2. Cellulose, a polysaccharide, is used to build the cell wall.

Question 4.
Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose
Answer:
Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
Disaccharides: Maltose, lactose.

Question 5.
What do you understand by the term glycosidic linkage?
Answer:
Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule. For example, in a sucrose molecule, two monosaccharide units, α-D-glucose and β-D-fructose, are joined together by a glycosidic linkage.

Question 6.
What is glycogen? How is it different from starch?
Answer:
Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components-amylose (15 – 20%) and amylopectin (80 – 85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin. Glycogen chain consists of 10-14 glucose units whereas amylopectin chains consist of 20-25 glucose units.

Question 7.
What are the hydrolysis products of (i) sucrose and (ii) lactose?
Answer:
(i) On hydrolysis, sucrose gives one molecule of α-D glucose and one molecule of β-D-fructose.

(ii) On hydrolysis, lactose gives β-D-galactose and β-D-glucose.

Question 8.
What is the basic structural difference between starch and cellulose?
Answer:
Starch consists of two components-amylose and amylopectin. Amylose is a long linear chain of ≅α-D-(+)-glucose units joined by C₁ -C₄ glycosidic linkage (α-link).

Amylopectin is a branched-chain polymer of []α-D-glucose units, in which the chain is formed by C₁ -C₄ glycosidic linkage and the branching occurs by C₁ – C₆ glycosidic linkage.

On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C₁– C₄ glycosidic linkage (β-link).

Question 9.
What happens when D-glucose is treated with the following reagents?
(i) HI
(ii) Bromine water
(iii) HNO₃
Answer:
(i) HI s On prolonged heating with HI, glucose forms n-hexane.

(ii) Br₂ water: Glucose gets oxidised (gluconic acid) on reaction with a mild oxidising agent like bromine water.

(iii) HNO₃: On oxidation with nitric acid, glucose yields a dicarboxylic acid, that is saccharic acid.

Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open-chain structure.
Answer:

• Aldehydes give 2, 4-DNP test, Schiff’s test, and react with NaHSO₄ to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
• The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free -CHO group is absent from glucose.
• Glucose exists in two crystalline forms-α and β. The α-form (m.p. = 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p = 423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.

Question 11.
What are essential and non-essential amino acids? Give two examples of each type.
Answer:
Essential amino acids are required by the human body, but they cannot be synthesised in the body. They must be taken through food, e.g., valine and leucine. Non-essential amino acids are also required by the human body, but they can be synthesised in the body e.g., glycine and alanine.

Question 12.
Define the following as related to proteins
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) Peptide linkage: The amide formed between —COOH group of one molecule of an amino acid and -NH₂ group of another molecule of the amino acid by the elimination of a water molecule is called a peptide linkage.


(ii) Primary structure: The primary structure of protein refers to the specific sequence in which various amino acids are present in it, i.e., the sequence of linkages between amino acids in a polypeptide chain. The sequence in which amino acids are arranged is different in each protein. A change in the sequence creates a different protein.

(iii) Denaturation: In a biological system, a protein is found to have a unique 3-dimensional structure and a unique biological activity. In such a situation, the protein is called native protein. However, when the native protein is subjected to physical changes such as change in temperature or chemical changes such as change in pH, its H-bonds are disturbed.

This disturbance unfolds the globules and uncoils the helix. As a result, the protein loses its biological activity. This loss of biological activity by the protein is called denaturation. During denaturation, the secondary and the tertiary structures of the protein get destroyed, but the primary structure remains unaltered. One of the examples of denaturation of proteins is the coagulation of egg white when an egg is boiled.

Question 13.
What are the common types of secondary structures of proteins?
Answer:
There are two common types of secondary structure of proteins:
(i) α- helix structure: In this structure, the -NH group of an amino acid residue forms H-bond with the img group of the adjacent turn of the right-handed screw (α-helix).

(ii) β-pleated sheet structure: This structure is called so because it looks like the pleated folds of drapery. In this structure, all the peptide chains are stretched out to nearly the maximum extension and then laid side by side. These peptide chains are held together by intermolecular hydrogen bonds.

Question 14.
What type of bonding helps in stabilising the α-helix structure of proteins?
Answer:
The α-helix structure of proteins is stabilised by intramolecular H-bonding between C-O of one amino acid residue and the N-H of the fourth amino acid residue in the chain.

Question 15.
Differentiate between globular and fibrous proteins.
Answer:

Globular proteins	Fibrous proteins
1. These are water-soluble proteins.	These are water-insoluble proteins.
2. These are spherical in shape.	These are linear in shape.
3. Globular proteins are highly unstable.	Fibrous proteins are stable to moderate changes in temperature and pH.

Question 16.
How do you explain the amphoteric behaviour of amino acids?
Answer:
In aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as zwitterion.

Therefore, in zwitterionic form, the amino acid can act both as an acid and as a base.

Thus, amino acids show amphoteric behaviour.

Question 17.
What are enzymes?
Answer:
Enzymes are proteins that catalyse biological reactions. They are very specific in nature and catalyse only a particular reaction for a particular substrate. Enzymes are usually named after the particular substrate or class of substrate and sometimes after the particular reaction. For example, the enzyme used to catalyse the hydrolysis of maltose into glucose is named as maltase.

Again, the enzymes used to catalyse the oxidation of one substrate with the simultaneous reduction of another substrate are named as oxidoreductase enzymes. The name of an enzyme ends with ‘ – ase’.

Question 18.
What is the effect of denaturation on the structure of proteins?
Answer:
During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. Due to denaturation, the globular proteins (soluble in H20) are converted into fibrous proteins (insoluble in H₂O) and their biological activity is lost. For example, boiled egg which contains coagulated proteins cannot be hatched.

Question 19.
How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Answer:
Vitamins are classified into two groups depending upon their solubility in water or fat.

1. Water-soluble vitamins: These include vitamin B-complex (B₁, B₂, B₅, i.e., nicotinic acid, B₆, B₁₂, pantothenic acid, biotin, i.e., vitamin H and folic acid) and vitamin C.
2. Fat-soluble vitamins: These include vitamins A, D, E and K. These are stored in liver and adipose tissues (fat-storing tissues). Vitamin K is responsible for coagulation of blood.

Question 20.
Why are vitamin A and vitamin C essential to us? Give their important sources.
Answer:
Vitamin A: Vitamin A is essential for us because its deficiency can cause xerophthalmia (hardening of cornea of eye) and night blindness. Sources: Carrots, fish liver oil, butter and milk.
Vitamin C: Vitamin C is essential for us because its deficiency causes scurvy (bleeding gums) and pyorrhea (loosening and bleeding of teeth).
Sources: Amla, citrus fruits and green leafy vegetables.

Question 21.
What are nucleic acids? Mention their two important functions.
Answer:
Nucleic acids are biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins or chromosomes (proteins containing nuclei acids as the prosthetic group).

These are of two types: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).
The two main functions of nucleic acids are :

1. DNA is responsible for transmission of hereditary effects from one generation to another. This is because of the unique property of replication during cell division and the transfer of two identical DNA strands to the daughter cells.
2. DNA and RNA are responsible for synthesis of all proteins essential for the growth and maintenance of our body. Actually, the proteins are synthesised by various RNA molecules (rRNA, mRNA and tRNA) in the cell but the message for the synthesis of a particular protein is present in DNA.

Question 22.
What is the difference between a nucleoside and a nucleotide?
Answer:
A nucleoside is formed when 1-position of pyrimidine (cytosine, thymine or uracil) or 9-position of purine (guanine or adenine) base is connected to C-1 of sugar (ribose or deoxyribose) by a β-linkage. Hence, in general, nucleosides may be represented as: Sugar-Base.

A nucleotide contains all the three basic compounds of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. These are obtained by esterification of C’₅– OH group of the pentose sugar by phosphoric acid.

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Answer:
In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms hydrogen bond with guanine, while adenine forms hydrogen bond with thymine. As a result, the two strands are complementary to each other.

Question 24.
Write the important structural and functional differences between DNA and RNA.
Answer:
Structural differences

DNA	RNA
(i) The sugar present in DNA is 2-deoxy-D-(-)-ribose.	The sugar present in RNA is D-(-)-ribose.
(ii) DNA contains cytosine and thymine as pyrimidine bases.	RNA contains cytosine and uracil as pyrimidine bases.
(iii) DNA has a double-stranded a-helix structure.	RNA has a single-stranded a-helix structure.
(iv) DNA molecules are very large; their molecular mass may vary from 6 x 106 -16 x 106 u.	RNA molecules are much smaller with molecular mass ranging from 20,000 to 40,000 u.

Functional differences:

(i) DNA has unique property of replication.	RNA usually does not replicate.
(ii) DNA controls the transmission of hereditary effects.	RNA controls the synthesis of proteins.

Question 25.
What are the different types of RNA found in the cell?
Answer:
Three types of RNA present in the cell are :

1. Messenger RNA (m-RNA)
2. Ribosomal RNA (r-RNA)
3. Transfer RNA (t-RNA)

Chemistry Guide for Class 12 PSEB Biomolecules Textbook Questions and Answers

Question 1.
Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer:
Glucose and sucrose have hydroxyl (-OH) groups (5 in glucose and 8 in sucrose) which form strong hydrogen bonds with water molecules and hence these are soluble in water. Cyclohexane and benzene are non-polar compounds and do not have hydroxyl groups and hence they are not able to form hydrogen bonding with water molecules and, therefore, these are not soluble in water.

Question 2.
What are the expected products of hydrolysis of lactose?
Answer:
Lactose is composed of β-D galactose and β-D glucose. Thus, on hydrolysis, it gives D-(+)- galactose and D-(+)- glucose.

Question 3.
How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Answer:
D-glucose reacts with hydroxylamine (NH₂OH) to form an oxime because of the presence of aldehydic (-CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open-chain structure in an aqueous medium, which then reacts with NH₂OH to give an oxime.

But pentaacetate of D-glucose does not react with NH₂OH. This is because pentaacetate does not form an open-chain structure.

Question 4.
The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
Answer:
Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept a proton, thus giving rise to a dipolar ion known as a zwitterion.

Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. But halo-acids do not exhibit such dipolar behaviour. For this reason, the melting points and the solubility of amino acids in water is higher than those of the corresponding halo-acids.

Question 5.
Where does the water present in the egg go after boiling the egg?
Answer:
When an egg is boiled, the proteins present inside the egg get denatured and coagulated. After boiling the egg, the water present in it is absorbed by the coagulated protein through H-bonding.

Question 6.
Why cannot vitamin C be stored in our body?
Answer:
Because vitamin C are water-soluble vitamins and so these are readily excreted through urine and hence cannot be stored in our body.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer:
When a nucleotide from the DNA containing thymine is hydrolysed, thymine β-D-2-deoxyribose and phosphoric acid are obtained as products.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer:
On complete hydrolysis of RNA six compounds are isolated. These compounds are adenine, guanine, cytosine, uracil, D-ribose and phosphoric acid. There is, no relationship among the quantities of different bases obtained on hydrolysis of RNA suggests that RNA is single-stranded. If it would have been double-stranded like DNA the complementary pairing of bases would have given equal proportion of complementary bases.

Punjab State Board Syllabus PSEB 12th Class History Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 12.

PSEB 12th Class History Guide | History Guide for Class 12 PSEB

History Guide for Class 12 PSEB | PSEB 12th Class History Book Solutions

PSEB 12th Class History Book Solutions in English Medium

• Chapter 1 Physical Features of the Punjab and their influence on its History
• Chapter 2 Sources of the History of the Punjab
• Chapter 3 Political, Social and Economic Conditions of the Punjab in the beginning of the 16th Century
• Chapter 4 Guru Nanak Dev Ji’s Life and His Teachings
• Chapter 5 Development of Sikhism Under Guru Angad DevJi, Guru Amar Das Ji and Guru Ram Das Ji
• Chapter 6 Guru Arjan Dev Ji and His Martyrdom
• Chapter 7 Guru Hargobind Ji and Transformation of Sikhism
• Chapter 8 Guru Har Rai Ji and Guru Har Krishan Ji
• Chapter 9 Guru Tegh Bahadur Ji and His Martyrdom
• Chapter 10 Guru Gobind Singh Ji: The Foundation of Khalsa, His Battles and Personality
• Chapter 11 Banda Singh Bahadur
• Chapter 12 Abdus Samad Khan, Zakariya Khan and Mir Mannu: Their Relations with the Sikhs
• Chapter 13 Rise of the Dal Khalsa and its Mode of Fighting
• Chapter 14 Social and Economic Conditions of the Punjab under the Mughals
• Chapter 15 Invasions of Ahmad Shah Abdali and Disintegration of Mughal Rule in the Punjab
• Chapter 16 Origin and Growth of the Sikh Misls and their Nature of Organization
• Chapter 17 Maharaja Ranjit Singh’s Career and Conquests
• Chapter 18 Anglo-Sikh Relations 1800-1839 A.D.
• Chapter 19 Ranjit Singh’s relations with Afghanistan and his N.W.F. Policy
• Chapter 20 Civil and Military Administration of Ranjit Singh
• Chapter 21 Character and Personality of Maharaja Ranjit Singh
• Chapter 22 First Anglo-Sikh War: Causes and Results
• Chapter 23 Second Anglo-Sikh War, Causes, Results and Annexation of the Punjab
• PSEB 12th Class History Map Questions
• PSEB 12th Class History Source Based Questions

PSEB 12th Class History Book Solutions in Hindi Medium

• Chapter 1 पंजाब की भौतिक विशेषताएँ तथा उनका इसके इतिहास पर प्रभाव
• Chapter 2 पंजाब के ऐतिहासिक स्रोत
• Chapter 3 16वीं शताब्दी के आरंभ में पंजाब की राजनीतिक, सामाजिक और आर्थिक दशा
• Chapter 4 गुरु नानक देव जी का जीवन और उनकी शिक्षाएँ
• Chapter 5 गुरु अंगद देव जी, गुरु अमरदास जी और गुरु रामदास जी के अधीन सिख धर्म का विकास
• Chapter 6 गुरु अर्जन देव जी और उनका बलिदान
• Chapter 7 गुरु हरगोबिंद जी और सिख पंथ का रूपांतरण
• Chapter 8 गुरु हर राय जी और गुरु हर कृष्ण जी
• Chapter 9 गुरु तेग़ बहादुर जी और उनका बलिदान
• Chapter 10 गुरु गोबिंद सिंह जी : खालसा पंथ की स्थापना, उनकी लड़ाइयाँ एवं व्यक्तित्व
• Chapter 11 बंदा सिंह बहादुर
• Chapter 12 अब्दुस समद खाँ, जकरिया खाँ और मीर मन्नू-उनके सिखों के साथ संबंध
• Chapter 13 दल खालसा का उत्थान और इसकी युद्ध प्रणाली
• Chapter 14 मुगलों के अधीन पंजाब की सामाजिक और आर्थिक स्थिति
• Chapter 15 अहमद शाह अब्दाली के आक्रमण एवं मुग़ल शासन का विखँडन
• Chapter 16 सिख मिसलों की उत्पत्ति एवं विकास तथा उनके संगठन का स्वरूप
• Chapter 17 महाराजा रणजीत सिंह का जीवन और विजयें
• Chapter 18 ऐंग्लो-सिख संबंध : 1800-1839
• Chapter 19 महाराजा रणजीत सिंह के अफ़गानिस्तान के साथ संबंध तथा उसकी उत्तर-पश्चिमी सीमा नीति
• Chapter 20 महाराजा रणजीत सिंह का नागरिक एवं सैनिक प्रशासन
• Chapter 21 महाराजा रणजीत सिंह का आचरण और व्यक्तित्व
• Chapter 22 प्रथम ऐंग्लो-सिख युद्ध : कारण एवं परिणाम
• Chapter 23 द्वितीय एंग्लो-सिख युद्ध : कारण, परिणाम तथा पंजाब का विलय

PSEB 12th Class History Book Solutions in Punjabi Medium

• Chapter 1 ਪੰਜਾਬ ਦੀਆਂ ਭੌਤਿਕ ਵਿਸ਼ੇਸ਼ਤਾਵਾਂ ਅਤੇ ਉਨ੍ਹਾਂ ਦਾ ਇਸ ਦੇ ਇਤਿਹਾਸ ‘ਤੇ ਪ੍ਰਭਾਵ
• Chapter 2 ਪੰਜਾਬ ਦੇ ਇਤਿਹਾਸਿਕ ਸੋਮੇ
• Chapter 3 16ਵੀਂ ਸਦੀ ਦੇ ਆਰੰਭ ਵਿੱਚ ਪੰਜਾਬ ਦੀ ਰਾਜਨੀਤਿਕ, ਸਮਾਜਿਕ ਅਤੇ ਆਰਥਿਕ ਦਸ਼ਾ
• Chapter 4 ਗੁਰੂ ਨਾਨਕ ਦੇਵ ਜੀ ਦਾ ਜੀਵਨ ਅਤੇ ਉਨ੍ਹਾਂ ਦੀਆਂ ਸਿੱਖਿਆਵਾਂ
• Chapter 5 ਗੁਰੂ ਅੰਗਦ ਦੇਵ ਜੀ ਗੁਰੂ ਅਮਰਦਾਸ ਜੀ ਅਤੇ ਗੁਰੂ ਰਾਮਦਾਸ ਜੀ ਦੇ ਸਮੇਂ ਦੌਰਾਨ ਸਿੱਖ ਧਰਮ ਦਾ ਵਿਕਾਸ
• Chapter 6 ਗੁਰੂ ਅਰਜਨ ਦੇਵ ਜੀ ਅਤੇ ਉਨ੍ਹਾਂ ਦੀ ਸ਼ਹੀਦੀ
• Chapter 7 ਗੁਰੂ ਹਰਿਗੋਬਿੰਦ ਜੀ ਅਤੇ ਸਿੱਖ ਪੰਥ ਦਾ ਰੂਪਾਂਤਰਣ
• Chapter 8 ਗੁਰੂ ਹਰਿ ਰਾਏ ਜੀ ਅਤੇ ਗੁਰੂ ਹਰਿ ਕ੍ਰਿਸ਼ਨ ਜੀ
• Chapter 9 ਗੁਰੂ ਤੇਗ ਬਹਾਦਰ ਜੀ ਅਤੇ ਉਨਾਂ ਦੀ ਸ਼ਹੀਦੀ
• Chapter 10 ਗੁਰੂ ਗੋਬਿੰਦ ਸਿੰਘ ਜੀ : ਖ਼ਾਲਸਾ ਪੰਥ ਦੀ ਸਿਰਜਨਾ, ਉਨ੍ਹਾਂ ਦੀਆਂ ਲੜਾਈਆਂ ਅਤੇ ਸ਼ਖ਼ਸੀਅਤ
• Chapter 11 ਬੰਦਾ ਸਿੰਘ ਬਹਾਦਰ
• Chapter 12 ਅਬਦੁਸ ਸਮਦ ਖਾਂ, ਜ਼ਕਰੀਆ ਖਾਂ ਅਤੇ ਮੀਰ ਮੰਨੂੰ – ਉਨ੍ਹਾਂ ਦੇ ਸਿੱਖਾਂ ਨਾਲ ਸੰਬੰਧ
• Chapter 13 ਦਲ ਖਾਲਸਾ ਦਾ ਉੱਥਾਨ ਅਤੇ ਇਸ ਦੀ ਯੁੱਧ ਪ੍ਰਣਾਲੀ
• Chapter 14 ਮੁਗ਼ਲਾਂ ਅਧੀਨ ਪੰਜਾਬ ਦੀ ਸਮਾਜਿਕ ਅਤੇ ਆਰਥਿਕ ਅਵਸਥਾ
• Chapter 15 ਅਹਿਮਦ ਸ਼ਾਹ ਅਬਦਾਲੀ ਦੇ ਹਮਲੇ ਅਤੇ ਪੰਜਾਬ ਵਿੱਚ ਮੁਗਲ ਸ਼ਾਸਨ ਦਾ ਪਤਨ
• Chapter 16 ਸਿੱਖ ਮਿਸਲਾਂ ਦੀ ਉਤਪੱਤੀ ਤੇ ਵਿਕਾਸ ਅਤੇ ਉਨ੍ਹਾਂ ਦੇ ਸੰਗਠਨ ਦਾ ਸਰੂਪ
• Chapter 17 ਮਹਾਰਾਜਾ ਰਣਜੀਤ ਸਿੰਘ ਦਾ ਜੀਵਨ ਅਤੇ ਜਿੱਤਾਂ
• Chapter 18 ਐਂਗਲੋ-ਸਿੱਖ ਸੰਬੰਧ : 1800-1839
• Chapter 19 ਮਹਾਰਾਜਾ ਰਣਜੀਤ ਸਿੰਘ ਦੇ ਅਫ਼ਗਾਨਿਸਤਾਨ ਨਾਲ ਸੰਬੰਧ ਅਤੇ ਉਸ ਦੀ ਉੱਤਰ-ਪੱਛਮੀ ਸੀਮਾ ਨੀਤੀ
• Chapter 20 ਮਹਾਰਾਜਾ ਰਣਜੀਤ ਸਿੰਘ ਦਾ ਸਿਵਿਲ ਅਤੇ ਸੈਨਿਕ ਪ੍ਰਸ਼ਾਸਨ
• Chapter 21 ਮਹਾਰਾਜਾ ਰਣਜੀਤ ਸਿੰਘ ਦਾ ਆਚਰਣ ਅਤੇ ਸ਼ਖ਼ਸੀਅਤ
• Chapter 22 ਪਹਿਲਾ ਐਂਗਲੋ-ਸਿੱਖ ਯੁੱਧ : ਕਾਰਨ ਅਤੇ ਸਿੱਟੇ
• Chapter 23 ਦੂਸਰਾ ਐਂਗਲੋ-ਸਿੱਖ ਯੁੱਧ : ਕਾਰਨ, ਸਿੱਟੇ ਅਤੇ ਪੰਜਾਬ ਦਾ ਮਿਲਾਉਣਾ
• PSEB 12th Class History Map Questions

Map Question Topics
In this part there will be one question on the map. The question on the map will be compulsory. The question of Map will be set out of the following Map of Punjab.
1. Battles of Guru Gobind Singh Ji.
2. Important Battles of Banda Singh Bahadur.
3. Ranjit Singh’s Kingdom.
4. First Anglo-Sikh War.
5. Second Anglo-Sikh War.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Very Short Answer Type Questions

Question 1.
Out of and , which is an example of allylic halide?
X
Answer:
is an example of allylic halide.

Question 2.
Which of the following reactions is S_N1 type ?

Answer:
Reaction (ii) is S_N1 reaction.

Question 3.
Which one of the following compounds is more easily hydrolysed by KOH and why?
CH₃CHClCH₂CH₃ or CH₃CH₂CH₂CH₂Cl
Answer:
Due to +1 effect of alkyl groups the 2° carbonium ion CH₃—CH—CH₂—CH₃ derived from sec-butyl chloride is more stable than the 1° carbonium ion \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\stackrel{+}{\mathrm{C}} \mathrm{H}_{2}\) derived from n-propyl chloride. Therefore sec-butyl chloride gets hydrolysed more easily than n-propyl chloride under S_N1 conditions.

Question 4.
What is known as a racemic mixture ? Give an example.
Answer:
equimolar mixture of a pair of enantiomers is called racemic mixture. A racemic mixture is optically inactive due to external compensation.

Question 5.
Consider the three types of replacement of group X by group Y as shown here.

This can result in giving compound (A) or (B) or both. What is the process called if
(A) is the only compound obtained ?
(B) is the only compound obtained ?
(A) and (B) are formed in equal proportions ?
Answer:
(i) Retention
(ii) Inversion
(iii) Racemisation.

Question 6.
What is an asymmetric carbon?
Answer:
A carbon which is attached to four different atoms/groups is called asymmetric carbon. For example, the carbon atom in BrCHClI.

Question 7.
What is plane polarized light?
Answer:
A beam of light which has vibration in only one plane is called plane polarized light.

Question 8.
Why iodoform has appreciable antiseptic property ?
Answer:
Iodoform liberate I₂ when it comes in contact with skin. Antiseptic property of iodine is due to the liberation of I₂ not because of iodoform itself.
(responsible for antiseptic property)

Question 9.
How does the ordinary light differ from the plane polarized light?
Answer:
Ordinary light has oscillations in all the directions perpendicular to the path of propagation whereas plane polarised light has all oscillations in the same plane.

Question 10.
What do you .understand by the term optical activity of compounds?
Answer:
The property of certain compounds to rotate the plane of polarzed light in a characteristic way when it is passed through their solutions is called optical activity of compounds.

Short Answer Type Questions

Question 1.
Give reasons for the following:
(i) Ethyl iodide undergoes S_N2 reaction faster than ethyl bromide.
(ii) (±) 2-Butanol is optically inactive.
(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH₃ —X.
Answer:
(i) Since I^– ion is a better leaving group than Br^– ion, hence, CH3I reacts faster than CH₃Br in SN₂ reaction with OH^– ion.

(ii) (±) 2-Butanol is a racemic mixture, i.e., there are two enantiomers in equal proportions. The rotation by one enantiomer will be cancelled by the rotation due to the other isomer, making the mixture optically inactive.

(iii) In CH₃—X the carbon atom is sp²-hybridised while in halobenzene the carbon atom is sp³-hybridised. The sp²-hybridised carbon is more electronegative due to greater s-character and holds the electron pair of C—X bond more tightly than sp³-hybridised carbon with less s-character. Thus, C—X bond length in CH₃—X is bigger than C—X in halobenzene.

Question 2.
Give reasons for the following:
(i) Haloalkanes easily dissolve in organic solvents.
(ii) Halogen compounds used in industry as solvents are alkyl chlorides rather than bromides and iodides.
Answer:
(i) Haloalkanes dissolve in organic solvents because the new intermolecular attractions between haloalkanes and organic solvent molecules have much the same strength as ones being broken in the separate haloalkanes and solvent molecules.
(ii) Because alkyl chlorides are more stable and more volatile than bromides and iodides.

Question 3.
Write the mechanism of the following reaction

Answer:
S_N2 mechanism

Question 4.
Which would undergo S_N1 reaction faster in the following pairs and why ?

Answer:
(i)
Tertiary halide reacts faster than primary halide because of greater stability of 3°-carbocation.

(ii)
Because the secondary carbocation formed in the slowest step is more stable than the primary carbocation.

Question 5.
Give reasons:
(i) n-Butyl bromide has higher boiling point than f-butyl bromide. Racemic mixture is optically inactive.
(ii) The presence of nitro group (—NO₂) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.
Answer:
(i) n-butyl bromide being a straight chain alkyl halide has larger surface area than tert butyl bromide. Larger the surface area, larger the magnitude of the van der Waal’s forces and hence higher is the boiling point.

(ii) A racemic mixture contains the two enantiomers d and l in equal •proportions. As the rotation due to one enantiomer is cancelled by equal and opposite rotation of another enantiomer, therefore, it is optically inactive.

(iii) The presence of NO₂ group at o/p position in haloarenes helps in the stabilisation of resulting carbanion by -R and -I effects and hence increases the reactivity of haloarenes towards nucleophilic substitution reactions.

Question 6.
(i) Why are alkyl halides insoluble in water ?
(ii) Why is butan-l-ol optically inactive but butan-2-ol is optically active ?
(iii) Although chlorine is an electron withdrawing group, yet it is ortho, para directing in electrophilic aromatic substitution reactions. Why ?
Answer:
(i) This is due to the inability of alkyl halide molecule to form intermolecular hydrogen bonds with water molecules.
(ii)
due to presence of a chiral carbon butan-2-ol is an optically active compound.

(iii) As the weaker resonance (+R) effect of Cl which stabilise the carbocation formed tends to oppose the stronger inductive (-I) effect of Cl which destabilise the carbocation at ortho and para positions and makes deactivation less for ortho and para position.

Question 7.
(i) Allyl chloride can be distinguished from vinyl chloride by NaOH and silver nitrate test. Comment.
(ii) Alkyl halide reacts with lithium aluminium hydride to give alkane. Name the attacking reagent which will bring out this change.
Answer:
(i) Vinyl chloride does not respond to NaOH and silver nitrate test because of partial double bond character due to resonance.
(ii) Hydride ion (H )

Question 8.
Give the ITJPAC name of the product formed when :
(i) 2-Methyl-l-bromopropane is treated with sodium in the presence of dry ether.
(ii) 1-Methyl cyclohexene is treated with HI.
(iii) Chloroethane is treated with silver nitrite.
Answer:
(i) 2, 5-dimethylhexane
(ii) 1 -Methyl-1 -iodocyclohexane
(iii) Nitroethane

Long Answer Type Questions

Question 1.
Some alkyl halides undergo substitution whereas some undergo elimination reaction on treatment with base. Discuss the structural features of alkyl halides with the help of examples which are responsible for this difference.
Answer:
Primary alkyl halides follow S_N2 mechanism in which a nucleophile attacks at 180° to the halogen atom. A transition state is formed in which carbon is bonded to two nucleophiles and finally halogen atom is pushed out. In S_N2 mechanism, substitution of nucleophile takes place as follows :

Thus, in S_N2 mechanism, substitution takes place. Tertiary alkyl halides follow? S_N1 mechanism, In this case, tert-alkyl halides form 3° carbocation. Now, if the reagent used is a weak base then substitution occur while if it is a strong base then instead of substitution elimination occur.

As alc. KOH is a strong base, so elimination competes over substitution and alkene is formed.

Question 2.
Why are aryl halides less reactive towards nucleophilic substitution reactions than alkyl halides ? How can we enhance the reactivity of aryl halides ?
Answer:
Aryl halides are less reactive towards nucleophilic substitution reaction due to the following reasons :
(i) In haloarenes, the lone pair of electron on halogen are in resonance with benzene ring. So, C—Cl bond acquires partial double bond character which strengthen C—Cl bond. Therefore, they are less reactive towards nucleophilic substitution reaction.

(ii) In haloarenes, the carbon atom attached to halogen is sp²-hybridised. The sp²-hybridised carbon is more electronegative than sp³-hybridised carbon. This sp ²-hybridised carbon in haloarenes can hold the electron pair of C—X bond more tightly and make this C—Cl bond shorter than C—Cl bond haloalkanes.

(iii) Since, it is difficult to break a shorter bond than a longer bond therefore haloarenes are less reactive than haloarenes.
In haloarenes, the phenyl cation will not be stabilised by resonance therefore S_N1 mechanism ruled out.

(iv) Because of the repulsion between the nucleophile and electron rich arenes, aryl halides are less reactive than alkyl halides.

The reactivity of aryl halides can be increased by the presence of an electron withdrawing group (—NO₂) at ortho and para positions. However, no effect on reactivity of haloarenes is observed by the presence of electron withdrawing group at meta-position. Mechanism of the reaction is as depicted with OH^– ion.



From the above resonance, it is very clear that electron density is rich at ortho and para positions. So, presence of EWG will facilitate nucleophilic at ortho and para positions not on meta position.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 8 The d-and f-Block Elements Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 8 The d-and f-Block Elements

Very Short Answer Type Questions

Question 1.
Why first ionisation enthalpy of Cr is lower than that of Zn?
Answer:
Ionisation enthalpy of Cr is less than that of Zn configuration. In case of zinc, electron comes out from completely filled 4s-orbital. So, removal of electron from zinc requires more energy as compared to the chromium.

Question 2.
Zn, Cd and Hg are soft metals. Why?
Answer:
Because they have one or more typical metallic structures at normal temperatures.

Question 3.
Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?
Answer:
Oxygen can form multiple bonds with metals, while fluorine can’t form multiple bond with metals. Hence, oxygen has more ability to stabilise higher oxidation state rather than fluorine.

Question 4.
Mn shows the highest oxidation state of + 7 with oxygen but with fluorine it shows the highest oxidation state of +4. Why?
Answer:
This is due to ability of oxygen to form pπ – dπ bond.

Question 5.
Mn₂O₇ is acidic whereas MnO is basic.
Answer:
Mn has +7 oxidation state in Mn₂O₇ and +2 in MnO. In low oxidation state of the metal, some of the valence electrons of the metal atom are not involved in bonding. Hence, it can donate electrons and behave as a base. On the other hand, in higher oxidation state of the metal, valence electrons are involved on bonding and are not available. Instead effective nuclear charge is high and hence it can accept electrons and behave as an acid.

Question 6.
Copper atom has completely filled d-orbitals in its ground state but it is a transition element. Why?
Answer:
Copper exhibits +2 oxidation state wherein it has incompletely filled d orbitals (3d⁹ 4s⁰) hence a transition elements.

Question 7.
Why is zinc not regarded as a transition element?
Answer:
As zinc atom has completely filled d-orbitals (3d10) in its ground state as well as oxidised state, therefore, it is not regarded as transition element.

Question 8.
Zn²⁺ salts are white while Cu²⁺ salts are coloured. Why?
Answer:
Cu²⁺(3d⁹4s⁰) has one unpaired electron in d-subshell which absorbs radiation in visible region resulting in d-d transition and hence Cu²⁺ salts are coloured. Zn²⁺(3d¹⁰4s⁰) has completely filled d-orbitals. No radiation is absorbed for d-d transition and hence Zn²⁺ salts are colourless.

Question 9.
The second and third row of transition elements resemble each other much more than they resemble the first row. Explain, why?
Answer:
Due to lanthanoid contraction, the atomic radii of the second and third row transition elements is almost same. So, they resemble each other much more as compared to first row elements and show similar character.

Question 10.
Why does copper not replace hydrogen from acids?
Answer:
Because Cu shows E^⊖ positive value.

Short Answer Type Questions

Question 1.
Why do transition elements show variable oxidation states? How is the variability in oxidation states of d-block different from that of the p-block elements?
Answer:
In transition elements, the energies of (n – 1)d orbitals and ns orbitals are nearly same. Therefore, electrons from both can participate in bond formation and hence show variable oxidation states.

In transition elements, the oxidation states differ from each other by unity e.g., Fe²⁺ and Fe³⁺, Cu⁺ and Cu²⁺ etc., while in p-block elements the oxidation state differ by units of two, e.g., Sn²⁺ and Sn⁴⁺, Pb²⁺ and Pb⁴⁺ etc. In transition elements, the higher oxidation states are more stable for heavier elements in a group e.g., Mo(VI) and W(VI) are more stable than Cr(VI) in group 6 whereas in p-block, elements the lower oxidation states are more stable for heavier elements due to the inert pair effect, e.g., Pb(II) is more stable than Pb(IV) in group 16.

Question 2.
When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallised from the solution. When compound (C) is treated with KC1, orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.
Answer:
K₂Cr₂O₇ is an orange compound. It is formed when Na₂Cr₂O₇ reacts with KCl. In acidic medium, yellow coloured \(\mathrm{CrO}_{4}^{2-}\) (chromate ion) changes into dichromate.
The given process is the preparation method of potassium dichromate from chromite ore.
A = FeCr₂O₄; B = Na₂CrO₄; C = Na₂Cr₂O₇; D = K₂Cr₂O₇

Question 3.
Mention the type of compounds formed when small atoms like H, C and N get trapped inside the crystal lattice of transition metals. Also give physical and chemical characteristics of these compounds.
Answer:
When small atoms like H, C and N get trapped inside the crystal lattice of transition metals.
(a) Such compounds are called interstitial compounds.
(b) Their characteristic properties are :

1. They have high melting point, higher than those of pure metals.
2. They are very hard.
3. They retain metallic conductivity.
4. They are chemically inert.

Question 4.
On the basis of lanthanoid contraction, explain the following:
(i) Nature of bonding in La₂O₃ and Lu₂O₃.
(ii) Trends in the stability of oxosalts of lanthanoids from La to Lu.
(iii) Stability of the complexes of lanthanoids.
(iv) Radii of 4d and 5d block elements.
(v) Trends in acidic character of lanthanoid oxides.
Answer:
(i) As the size decreases covalent character increases. Therefore, La₂O₃ is more ionic and Lu₂O₃ is more covalent.
(ii) As the size decreases from La to Lu, stability of oxosalts also decreases.
(iii) Stability of the complexes increases as the size of lanthanoids decreases.
(iv) Radii of 4d and 5d block elements will be almost same.
(v) Acidic character of oxides increases from La to Lu.

Question 5.
A solution of KMnO₄ on reduction yields either a colourless solution of a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?
Answer:
Oxidising behaviour of KMnO₄ depends on pH of the solution.
In acidic medium (pH < 7),

Question 6.
Identify the following:
(i) Oxoanion of chromium which is stable in acidic medium.
(ii) The lanthanoid element that exhibits + 4 oxidation state.
Answer:
(i) Cr₂O₇
(ii) Cerium

Question 7.
The magnetic moments of few transition metal ions are given below:

Metal ion	Metal ion
Sc3+	0.00
Cr2+	4.90
Ni2+	2.84
Ti3+	1.73

(at no. Sc = 21, Ti = 22, Cr = 24, Ni = 28)
Which of the given metal ions :
(i) has the maximum number of unpaired electrons?
(ii) forms colourless aqueous solution?
(iii) exhibits the most stable + 3 oxidation state?
Answer:
(i) Cr²⁺
(ii) Sc³⁺
(iii) Sc³⁺

Question 8.
Consider the standard electrode potential values (M²⁺/M) of the elements of the first transition series.

Explain:
(i) E⁰ value for copper is positive.
(ii) E⁰ value of Mn is more negative as expected from the trend.
(iii) Cr²⁺ is a stronger reducing agent than Fe²⁺.
Answer:
(i) E⁰ value for copper is positive because the high energy to transform Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.
(ii) E⁰ value of Mn is more negative as expected from the trend because Mn²⁺ has d⁵ configuration i. e., stable half-filled configuration.
(iii) Cr²⁺ is a stronger reducing agent than Fe²⁺ because d⁴ to d³ occurs in case of Cr²⁺ to Cr³⁺ (more stable \(t_{2 g}^{3}\)) while it changes from d⁶ to d⁵ in case of Fe²⁺ to Fe³⁺.

Long Answer Type Questions

Question 1.
Write similarities and differences between the chemistry of lanthanoids and that of actinoids.
Answer:
Similarities between lanthanoids and actinoids :

1. Both lanthanoids and actinoids mainly show an oxidation state of +3.
2. Actinoids show actinoid contraction like lanthanoid contraction is exhibited by lanthanoids.
3. Both lanthanoids and actinoids are electropositive.

Differences between lanthanoids and actinoids :

1. The members of lanthanoid exhibit less number of oxidation states than the corresponding members of actinoid series.
2. Lanthanoid contraction is smaller than the actinoid contraction.
3. Lanthanoids except promethium cure non-radioactive metals while actinoids are radioactive metals.

Question 2.
(a) Assign reasons for the following:
(i) Zr and Hf have almost identical radii.
(ii) The , value for copper is positive (+0.34 V).
(b) Although +3 oxidation state is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?
Answer:
(a) (i) This is due to filling of 4/ orbitals which have poor shielding effect (lanthanoid contraction).
(ii) This is because the sum of enthalpies of sublimation and ionisation is not balanced by hydration enthalpy.
(b) It is because after losing one more electron Ce acquires stable 4f⁰ electronic configuration.

Question 3.
(a) How do you prepare :
(i) K₂MnO₄ from MnO₂?
(ii) Na₂Cr₂O₇ from Na₂CrO₄?
(b) Account for the following :
(i) The enthalpy of atomisation is lowest for Zn in 3d series of the transition elements.
(ii) Actinoid elements show wide range of oxidation states.
Answer:
(a) (i) Pyrolusite is fused with KOH in the presence of atmospheric oxygen to give K₂MnO₄.

(b) (i) In the formation of metallic bonds, no electrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bonds. That is why the enthalpy of atomisation of zinc is the lowest in the series.
(ii) This is due to comparable energies of 5f, 6d and 7s orbitals.

Punjab State Board PSEB 12th Class Physics Book Solutions Chapter 6 Electromagnetic Induction Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

PSEB 12th Class Physics Guide Electromagnetic Induction Textbook Questions and Answers

Question 1.
Predict the direction of induced current in the situations described by the following Figs. 6.18 (a) to (f).


Answer:
(a) As the magnet moves towards the solenoid, the magnetic flux linked with the solenoid increases. According to Lenz’s law, the induced e.m.f. produced in the solenoid in such that it opposes the very cause producing it i. e., it opposes the motion of the magnet. Hence the face q of it becomes the south pole and p becomes north pole. Therefore, the current will flow along pqin the coili. e., along qrpqin this figurei. e., clockwise when seen from the side of the magnet according to clock rule.

(b) As the north pole moves away from xy coil, so the magnetic flux linked with this coil decreases. Thus according to Lenz’s law, the induced e.m.f. produced in the coil will oppose the motion of the magnet. Hence the face, X becomes S-pole, so the current will flow in the clockwise direction i.e., along yzx in the cone.

For coil pq, the south pole of the magnet moves towards end q and thus this end will acquire south polarity so as to oppose the motion of the magnet, hence the current will flow along prq in the coil.

(c) The induced current will be in the anticlockwise direction i.e., along yzx.

(d) The induced current will be in the clockwise direction i.e., along zyx.

(e) The battery current in the left coil will be from right to left, so by mutual induction, the induced current in the right coil will be in the opposite direction i.e., from left to right or along xry.

(f) In this case, there is no change in magnetic flux linked with the wire, so no current will flow through the wire since there is no induced current as the field lines lie in the plane of the loop.

Question 2.
Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19.
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.

(a) When a wire of irregular shape turns into a circular loop, the magnetic flux linked with the loop increases due to increase in area. The circular loop has greater area than the loop of irregular shape. The induced e.m.f. will cause current to flow in such a direction so that the wire forming the loop is pulled inward from all sides i.e., current must flow in the direction adcba as shown in Fig. (a) i.e., in anticlock-wise direction so that the magnetic field produced by the current ((directed out of the paper) opposes the applied field.

In Fig. (b), a circular loop deforms into a narrow straight wire i.e., upper side of loop should move downwards and lower end should move upwards to oppose the motion of the circular loop, thus its area decreases as a result of which the magnetic flux linked with it decreases. To oppose the decrease in magnetic flux, the induced current should flow anti clockwise in the loop i. e., along a’d’ d b’ a’. Due to the flow of anti-clockwise current, the magnetic field produced will be out of the page and hence the applied field is supplemented.

Question 3.
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried hy the solenoid changes steadily from 2.0 A to
4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Answer:
Number of turns per unit length of the solenoid, n = 15 turns/cm = 1500 turns/m
The solenoid has a small loop of area, A = 2.0 cm² = 2 × 10^-4 m²
Current carried by the solenoid changes from 2 A to 4 A.
.-. Change in current in the solenoid, dI = 4 – 2 = 2A
Change in time, dt = 0.1 s
We know that the magnetic field produced inside the solenoid is given by
B = μ₀nI
If Φ be the magnetic flux linked with the loop, then
Φ = BA = μ₀nI A
Induced emf in the solenoid is given by Faraday’s law as
e = –\(\frac{d \phi}{d t}\)
e = – \(\frac{d}{d t}\) (Φ) = –\(\frac{d}{d t}\) μ₀nI A
μ₀n A \(\frac{d I}{d t}\)
∴ Magnitude of e is given by
= A μ₀n × (\(\frac{d I}{d t}\))
= 2 × 10^-4 × 4π × 10^-7 × 500 × \(\frac{2}{0.1}\)
7.54 × 10 ^-6 V
Hence, the induced voltage in the loop is = 7.54 × 10 ^-6 V

Question 4.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^-1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Answer:
Length of the rectangular wire, l = 8 cm = 0.08 m
Width of the rectangular wire, b = 2 cm = 0.02 m
Hence, area of the rectangular loop A = lb
= 0.08 × 0.02
= 16 × 10^-4 m²
Magnetic field strength, B = 0.3 T
Velocity of the loop, v = 1 cm/s = 0.01 m / s

(a) Emf developed in the loop is given as
e = Blv
= 0.3 × 0.08 × 0.01 = 2.4 × 10^-4 V

= \(\frac{b}{v}\) = \(\frac{0.02}{0.01}\) = 2 s
Hence, the induced voltage is 2.4 × 10^-4 V which lasts for 2s.

(b) Emf developed,
e = Bbv = 0.3 × 0.02 × 0.01 = 0.6 × 10^-4 V

\(\frac{l}{v}\) = \(\frac{0.08}{0.01}\) 8s
Hence, the induced voltage is 0.6 × 10^-4 V which lasts for 8 s.

Question 5.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Answer:
Length of the rod, l = 1m
Angular frequency, ω = 400 rad/s
Magnetic field strength, B = 0.5 T
One end of the rod has zero linear velocity, while the other end has a linear velocity of l ω.
Average linear velocity of the rod, v = \(\frac{l \omega+0}{2}=\frac{l \omega}{2}\)
Emf developed between the centre and the ring,
e = Blv = Bl(\(\frac{l \omega}{2}\)) = \(\frac{B l^{2} \omega}{2}\)
= \(\frac{0.5 \times(1)^{2} \times 400}{2}\) = 100V
Hence, the emf developed between the centre and the ring is 100 V.

Question 6.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s^-1 in a uniform horizontal magnetic field of magnitude 3.0 × 10^-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Answer:
Here, n = number of turns in the coil = 20
r = radius ofcoil = 8.0 cm = 8 × 10^-2 m
ω = angular speed of the coil = 50 rad s^-1.
B = magnetic field = 3.0 × 10^-2 T
Let e₀ be the maximum e.m.f. in the coil = ?
and e_av be the average e.m.f. in the coil = ?
We know that the instantaneous e.m.f. produced in a coil is given by
e = BA ω sinωt.
for e to be maximum e_max, sin ωt = 1.
∴ e_max = B A n ω = B.πr² nω
where A = πr² is the area of the coil

i.e., e_av is zero as the average value of sincot for one complete cycle is always zero.
Now R = resistance of the closed loop formed by the coil = 10 Ω
Let I_max = maximum current in the coil = ?
∴ Using the relation,
I_max = \(\frac{e_{\max }}{R}\), we get
I_max = \(\frac{0.603}{10}\) = 0.0603 A
Let P_av be the average power loss due to Joule heating = ?
∴ P_av = \(\frac{e_{\max } \cdot I_{\max }}{2}\) = \(\frac{0.603 \times 0.0603}{2}\)
= 0.018 Watt
The induced current causes a torque opposing the rotation of the coil. An external agent must supply torque and do work to counter this torque in order to keep the coil rotating uniformly. Thus the source of the power dissipated as heat in the coil is the external agent i. e., rotor.

Question 7.
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s^-1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10^-4 Wb m^-2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Answer:
Length of the wire, l = 10 m
Falling speed of the wire, v = 5.0 m/s
Magnetic field strength, B = 0.3 × 10^-4 Wb m^-2

(a) emf induced in the wire,
e = Blv = 0.3 × 10^-4 × 5 × 10
= 1.5 × 10^-3 V

(b) Using Fleming’s right hand rule, it can be inferred that the direction of the induced emf is from west to east.

(c) The eastern end of the wire is at a higher electrical potential.

Question 8.
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Initial current, I₁ = 5.0 A
Final current, I₂ = 0.0 A
Change in current, dl = I₁ – I₂ = 5 – 0 = 5 A
Time taken for the change, dt = 0.1 s
Average emf, e = 200 V
For self-inductance (I) of the circuit, we have the relation for average emf as
e = L\(\frac{d I}{d t}\)
L = \(\frac{e}{\left(\frac{d I}{d t}\right)}\)
= \(\frac{200}{\frac{5}{0.1}}=\frac{200 \times 0.1}{5}\) 4H
Hence, the self induction of the circuit is 4 H.

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Answer:
Mutual inductance of the pair of coils, μ = 1.5 H
Initial current, I₁ = 0 A
Final current, I₂ – 20 A
Change in current, dI = I₂ – I₁ = 20 – 0 = 20 A
Time taken for the change, dt = 0.5 s
Induced emf, e = \(\frac{d \phi}{d t}\) ………… (1)

Where d Φ is the change in the flux linkage with the coil.
Emf is related with mutual inductance as
e = μ\(\frac{d I}{d t}\) ……………. (2)
Equating equations (1) and (2), we get
\(\frac{d \phi}{d t}\) = μ\(\frac{d I}{d t}\)
or dΦ = μdI
∴ dΦ = 1.5 × (20) = 30 Wb
Hence, the change in the flux linkage is 30 Wb.

Question 10.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wings having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10^-4 T and the dip angle is 30°.
Answer:
Speed of the jet plane, v = 1800 km/h = 1800 × \(\frac{5}{18}\) = 500 m/s
Wing span of the jet plane, l = 25 m
Earth’s magnetic field strength, B = 5.0 × 10^-4 T
Angle of dip, δ = 30°
Vertical component of Earth’s magnetic field,
B_V = B sinδ
= 5 × 10^-4 × sin30°
= 5 × 10^-4 × \(\frac{1}{2}\) = 2.5 × 10^-4 T
Voltage difference between the ends of the wing can be calculated as
e = (B_V) × l × v
= 2.5 × 10^-4 × 25 × 500 = 3.125 V
Hence, the voltage difference developed between the ends of the wings is 3.125 V.

Question 11.
Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that.the field decreases from its initial value of 0.3 T at the rate of 0.02 Ts^-1. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?
Answer:
Sides of the rectangular wire loop are 8 cm and 2 cm.
Hence, area of the rectangular wire loop,
A = length × width = 8 × 2 = 16 cm
= 16 × 10^-4 m²
Initial value of the magnetic field, B = 0.3 T
Rate of decrease of the magnetic field, \(\frac{d B}{d t}\) = 0.02 T/s
emf developed in the loop is given as
e = \(\frac{d \phi}{d t}\)
where, Φ = Change in flux through the loop area
= AB
∴ e = \(\frac{d(A B)}{d t}=\frac{A d B}{d t}\)
= 16 × 10^-4 × 0.02 =0.32 × 10^-4 V
= 3.2 × 10^-5 V
Resistance of the loop, R = 1.6 Ω
The current induced in the loop is given as
i = \(\frac{e}{R}\)
= \(\frac{0.32 \times 10^{-4}}{1.6}\) = 2 × 10^-5A
Power dissipated in the loop in the form of heat is given as
P = i²R
= (2 × 10^-5)² × 1.6
= 6.4 × 10^-10 W
The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.

Question 12.
A square loop of side 12 cm with its sides parallel to X and F axes is moved with a velocity of 8 cm s^-1 in the positive x-direction in an environment containing a magnetic field in the positive 2-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10^-3 T cm^-1 along the negative jtr-direction (that is it increases by 10^-3 T cm^-1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10-3 T s1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
Answer:
Here, a = side of the square loop = 12 cm = 12 × 10^-2 m
\(\vec{v}\) = velocity of loop parallel to x-axis = 8 cms^-1
= 8 × 10^-2 ms^-1.
Let B = variable magnetic field acting away from us ⊥ ar to the XY plane along z axis i. e., plane of paper represented by x.
\(\) = 10^-3 Tcm^-1
= 10^-3 × 10² Tm^-1
= 0.1 Tm^-1
= field gradient along – ve x direction.

\(\frac{d B}{d t}\) = rate of variation with me
= 10^-3 Ts^-1
R = resistance of the loop = 4.5 mΩ = 4.5 × 10^-3 Ω
Let I = induced current = ? and its direction = ?
∴ A = area of loop = a² = (12 × 10^-2)² m² = 144 × 10^-4 m².
The magnetic flux changes (i) due, to the variation of B with time and
(ii) due to motion of the loop in non-uniform \(\vec{B}\).
Thus if Φ be the total magnetic flux of the loop, then Φ is calculated as Area of shaded part = adx
Let dΦ = magnetic flux linked with shaded part = B(x,t)adx

∴ From (3), \(\) = 144 × 10^-7 + 1152 × 10^-7
= 1296 × 10^-7 Wbs^-1
Clearly the two effect add up as these cause a decrease in flux along the + z direction.
∴ If e be the induced e.m.f. produced, then
e = –\(\frac{d \phi}{d t}\) = -1296 × 10^-7 V
= -12.96 × 10^-5 V
∴ e = 12.96 × 10^-5 V
∴ I = \(\frac{e}{R}\) = \(\frac{12.96 \times 10^{-5}}{4.5 \times 10^{-3}}\) 2.88 × 10^-2 A.
The direction of induced current is such as to increase the flux through the loop along +z-direction. Thus if for the observer, the loop moves to the right, the current will be seen to be anti-clockwise.

Question 13.
It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small fiat search coil of area 2 cm² with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction. The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Q. Estimate the field strength of magnet.
Answer:
Area of the small flat search coil, A = 2cm² = 2 × 10^-4m²
Number of turns on the coil, N = 25
Total charge flown in the coil, Q = 7.5 mC = 7.5 × 10 ^-3 C
Total resistance of the coil and galvanometer, R = 0.50 Ω
Induced current in the coil,
I = \(\frac{\text { Induced emf }(e)}{R}\) ………….. (1)
Induced emf is given us
e = -N\(\frac{d \phi}{d t}\) ……………… (2)
Combining equations (1) and (2), we get


Hence, the field strength of the magnet is 0.75 T.

Question 14.
Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic Held are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mfl. Assume the field to be uniform.

(a) Suppose K is open and the rod is moved with a speed of 12 cm s^-1 in me airection snown. dive me polarity ana magnitude of the induced emf.

(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?

(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.

(d) What is the retarding force on the rod when K is closed?

(e) How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s^-1) when K is closed? How much power is required when K is open?

(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?

(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Answer:
Here, B = 0.50 T
l = length of the rod = 15 cm = 15 × 10^-2 m
R = resistance of the closed loop containing the rod = 9.0 mΩ
= 9 × 10^-3 Ω.

(a) v = speed of the rod = 12 cms^-1 = 12 × 10^-2 ms^-1.
The magnitude of the induced e.m.f. is
E = Blv = 0.50 × 15 × 10^-2 × 12 × 10^-12 = 9 × 10^-3 V
According to Fleming’s left hand rule, the direction of Lorentz force —^ ^ ^
\(\vec{F}\) = -e(\(\vec{V} \times \vec{B}\)) on electrons in PQ is from P to Q. So the end P of the rod will acquire positive charge and Q will acquire negative charge,

(b) Yes. When the switch K is open, the electrons collect at the end Q, so excess charge is built up at the end Q. But when the switch K is closed, the accumulated charge at the end Q is maintained by the continuous flow of current.

(c) This is because the presence of excess charge at the ends P and Q of the rod sets up an electric field \(\vec{E}\). The force due to the electric field (q\(\vec{E}\)) balances the Lorentz magnetic force q(\(\vec{V} \times \vec{B}\)). Hence the net force on the electrons is zero.

(d) When the key K is closed, current flows through the rod. The retarding force experienced by the rod is
F = BIl = B(\(\frac{E}{R}\)) l
where, I = \(\) is the induced current. R
F = \(\frac{0.50 \times 9 \times 10^{-3} \times 15 \times 10^{-2}}{9 \times 10^{-3}}\)
= 7.5 × 10^-2 N.

(e) The power required by the external agent against the above retarding force to keep the rod moving uniformly at speed 12 cms^-1 (= 12 × 10^-2 m/s) when K is closed is given by
p = FV = 7. 5 × 10^-2 × 12 × 10^-2
= 90 × 10^-4 W
= 9 × 10^-3 W

(f) Power dissipated as heat is given by
P = I²R = (\(\frac{E}{R}\))² R = \(\frac{E^{2}}{R}\)
= \(\frac{\left(9 \times 10^{-3}\right)^{2}}{9 \times 10^{-3}}\)
= 9 × 10^-3 W.
The source of this power is the power provided by the external agent calculated in (e).

Zero. This is because when the magnetic field is parallel to the rails, θ = 0°, so induced e.m.f. E = Blv sinθ = Blv sin 0 = 0. In this situation, the moving rod does not cut the field lines, so there is no change in the magnetic flux, hence E = 0.

Question 15.
An air-cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^-3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic Held near the ends of the solenoid.
Answer:
Length of the solenoid, l = 30 cm = 0.3 m
Area of cross-section, A = 25 cm² = 25 x 10^-4 m²
Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5 A
Current flows for time, t = 10^-3 s
Average back emf, e = \(\frac{d \phi}{d t}\) ……………. (1)
where,
dΦ = NAB ………….. (2)
and B = μ₀ \(\frac{N I}{l}\) …………. (3)
Using equations (2) and (3) in equation (1), we get
e = \(\frac{\mu_{0} N^{2} I A}{l t}\)
\(=\frac{4 \pi \times 10^{-7} \times(500)^{2} \times 2.5 \times 25 \times 10^{-4}}{0.3 \times 10^{-3}}\)
= 6.5 V
Hence, the average back emf induced in the solenoid is 6.5 V.

Question 16.
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Figure 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s.
Calculate the induced emf in the loop at the instant when x = 0.2 m.
Take a = 0.1 m and assume that the loop has a large resistance.
Answer:
(a) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).
Magnetic flux associated with element dy, dΦ = BdA

= where,
dA = Area of element dy = a dy
B = Magnetic field at distance y = \(\frac{\mu_{0} I}{2 \pi y}\)
I = Current in the wire

Question 17.
A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis as shown in Fig. 6.22. A uniform magnetic field extends over a circular region within the rim. It is given by,
B = -Bk (r ≤ a; a < R)
= 0 (otherwise)
What is the angular velocity of the wheel after the field is suddenly switched off?

Answer:
Let ω be the angular velocity of the wheel of mass M and radius R.
Let e = Induced e.m.f. produced.
The rotational K.E. of the rotating wheel = \(\frac{1}{2}\) Iω² ………… (1)
where, I = Moment of inertia of wheel
= \(\frac{1}{2}\) MR² …………… (2)

or Work done = eQ
Applying the work energy theorem, we get
Rotational K.E. = Work done
or RotationalK.E. = Q × e …………… (3)
We know that the e.m.f. of a rod rotating in a uniform magnetic field is
given by \(\frac{1}{2}\) Bωa² , since here the magnetic field is changing, we assume the average over the time span and thus average value of e.m.f. is given by

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 4 Chemical Kinetics Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 4 Chemical Kinetics

Very Short Answer Type Questions

Question 1.
For which type of reactions, order and molecularity have the same value?
Answer:
If the reaction is an elementary reaction, order is same as molecularity.

Question 2.
Why is the probability of reaction with molecularity higher than three very rare?
Answer:
The probability of more than three molecules colliding simultaneously is very small. Hence, possibility of molecularity being three is very low.

Question 3.
State a condition under which a bimolecular reaction is kinetically first order.
Answer:
A bimolecular reaction may become kinetically of first order if one of the reactants is in excess.

Question 4.
Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
Answer:
Thermodynamically the conversion of diamond to graphite is highly feasible but this reaction is very slow because its activation energy is high.

Question 5.
Why is it that instantaneous rate of reaction does not change when a part of the reacting solution is taken out?
Answer:
Instantaneous rate is measured over a very small interval of time, hence, it does not change when a part of solution is taken out.

Question 6.
A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction?
Solution:
As t_75% = 2t_50%
Therefore, it is a first order reaction.

Question 7.
Define threshold energy of a reaction.
Answer:
Threshold energy is the minimum energy which must be possessed by reacting molecules in order to undergo effective collision which leads to formation of product molecules.

Question 8.
Why does the rate of a reaction increase with rise in temperature?
Answer:
At higher temperatures, larger fraction of colliding particles can cross the energy barrier (i.e., the activation energy), which leads to faster rate.

Question 9.
What is the difference between rate law and law of mass action?
Answer:
Rate law is an experimental law. On the other hand, law of mass action is a theoretical law based on the balanced chemical reaction.

Question 10.
What do you understand by ‘Rate of reaction’?
Answer:
The change in the concentration of any one of the reactants or products per unit time is termed as the rate of reaction.

Question 11.
In the Arrhenius equation, what does the factor e a corresponds to?
Answer:
e^-E_a/RT corresponds to the fraction of molecules that have kinetic energy greater than E_a,

Short Answer Type Questions

Question 1
What do you understand by the ‘order of a reaction’? Identify the reaction order from each of the following units of reaction rate constant:
(i) L^-1mols^-1
(ii) Lmol^-1s^-1.
Solution:
The sum of powers of the concentration of the reactants in the rate law expression is called order of reaction.
For a general reaction: aA + bB → Products
If rate = k[A]^m [B]ⁿ; order of reaction = m + n

(i) General unit of rate constant, k = (mol L^-1 )^1-ns^-1
L^-1mol s^-1 = (mol L^-1 )^1-ns^-1
-1 = -1 + n ⇒ n = 0 ∴ Reaction order = 0

(ii) L mol^-1 s^-1 = (mol L^-1)^1-n s^-1
1 = -1 + n ⇒ n = 2 ∴ Reaction order = 2

Question 2.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation :
log k = 14.2 – \(\frac{1.0 \times 10^{4}}{T}\)K
Calculate E_a for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 JK^-1 mol^-1)
Solution:
Comparing the equation, log k = 14.2 – \(\frac{1.0 \times 10^{4}}{T}\) K with the equation,
log k = log A = \(\frac{E_{a}}{2.303 R T}\), we get
\(\frac{E_{a}}{2.303 R}\) = 1.0 × 10⁴ K or E_a = 1.0 × 10⁴ K × 2.303 × R
E_a = 1.0 × 10⁴ K × 2.303 × 8.314 JK^-1
= 19.1471 × 10⁴ Jmol^-1
= 191.47 kJ mol^-1
For a first order reaction, t_t/2 = \(\frac{0.693}{k}\) or k = \(\frac{0.693}{t_{1 / 2}}\)
k = \(\frac{0.693}{200 \mathrm{~min}}\) = 3.465 × 10 ^-3min^-1

Question 3.
The reaction, N₂(g) + O₂(g) ⇌ 2NO(g) contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500 K, equilibrium constant K for it is 1.0 × 10^-5. Suppose in a case [N₂] = 0.80 mol L^-1 and [O₂] = 0.20 mol L^-1 before any reaction occurs. Calculate the equilibrium concentrations of the reactants and the product after the mixture has been heated to 1500 K.
Solution:

[N₂] = 0.8 – 6.324 × 10⁴ mol L^-1
= 0.799 molL^-1
[O₂] = 0.2 – 6.324 × 10^-4 mol L^-1
= 0.199 mol L^-1

Question 4.
For a general reaction, A → B, plot of concentration of A vs time is given in figure. Answer the following questions on the basis of this graph.

(i) What is the order of the reaction?
(ii) What is the slope of the curve?
(iii) What are the units of rate constant?
Answer:
(i) Zero order
(ii) Slope = – k
(iii) Units of rate constant = mol L^-1 s^-1

Question 5.
For a reaction, A + B → products, the rate law is rate = k [A][B]a^3/2. Can the reaction be an elementary reaction? Explain.
Answer:
During an elementary reaction, the number of atoms or ions colliding to react is referred to as molecularity. Had this been an elementary reaction the order of reaction with respect to B would have been 1, but in the given rate law it is \(\frac{3}{2}\). This indicated that the reaction is not an elementary reaction.

Question 6.
The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce 1 g of the reactant to 0.0625 g?
Answer:
We know that, t = \(\frac{2.303}{k}\) log \(\frac{[R]_{0}}{[R]}\)
t = \(\frac{2.303}{60}\) log \(\frac{1}{0.0625}\)
t = 0.0462 s

Long Answer Type Questions

Question 1.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

t/s	0	30	60
[CH3COOCH3]/mol L-1	0.60	0.30	0.15

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
Solution:

As the value of k is same in both the cases, therefore, hydrolysis of methylacetate in aqueous solution follows pseudo first order reaction.

(ii) Average rate = \(-\frac{\Delta\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]}{\Delta t}\)
= \(\frac{-[0.15-0.30]}{60-30}\) = \(\frac{0.15}{30}\)
Average rate = 0.005 mol L^-1s^-1

Question 2.
Describe how does the enthalpy of reaction remain unchanged when a catalyst is used in the reaction?
Answer:
A catalyst is a substance which increases the speed of a reaction without itself undergoing any chemical change.
According to “intermediate complex formation theory” reactants first combine with the catalyst to form an intermediate complex which is short-lived and decomposes to form the products and regenerating the catalyst.

The intermediate formed has much lower potential energy than the intermediate complex formed between the reactants in the absence of the catalyst.

Thus, the presence of catalyst lowers the potential energy barrier and the reaction follows a new alternate pathway which require less activation energy.

We know that, lower the activation energy, faster is the reaction because more reactant molecules can cross the energy barrier and change into products.

Enthalpy, △H is a ‘state function. Enthalpy of reaction, i.e., difference in energy between reactants and product is constant, which is clear from potential energy diagram.
Potential energy diagram of catalysed reaction is given as:

Question 3.
All energetically effective collisions do not result in a chemical change. Explain with the help of an example.
Answer:
Only effective collision lead to the formation of products. It means that collisions in which molecules collide with sufficient kinetic energy (called threshold energy = activation energy + energy possessed by reacting species).

And proper orientation lead to a chemical change because it facilitates the breaking of old bonds between (reactant) molecules and formation of the new ones i.e., in products.
e.g., formation of methanol from bromomethane depends upon the orientation of the reactant molecules.

The proper orientation of reactant molecules leads to bond formation whereas improper orientation makes them simply back and no products are formed.

To account for effective collisions, another factor P (probability of steric factor) is introduced K = PZ_ABe^-Ea/RT.