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Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 5 Surface. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 5 Important Extra Questions Surface Chemistry

Surface Chemistry Important Extra Questions Very Short Answer Type

Question 1.
Give an example of a shape selective catalyst. (CBSE Delhi 2010, 2011)
Answer:
Zeolite catalyst known as ZSM-5.

Question 2.
Why are medicines more effective in colloidal state? (CBSE Delhi 2019)
Answer:
Medicines are most effective in colloidal state because colloids have large surface area and hence the medicines can be easily adsorbed and assimilated.

Question 3.
What is difference between an emulsion and a gel? (CBSE Delhi 2019)
Answer:
Emulsions are colloidal solutions in which both the dispersed phase and the dispersion medium are liquids. Gel is a colloidal system in which a liquid is dispersed in a solid.

Question 4.
Define ‘peptization’. (CBSE2012)
Answer:
The process of converting a freshly prepared precipitate into colloidal form by addition of a suitable electrolyte is called peptization.

Question 5.
What is meant by ‘shape-selective catalysis? (CBSE 2012)
Answer:
The catalytic reaction which depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysis.

Question 6.
What happens when a freshly precipitated
Fe(OH)₃ is shaken with little amount of dilute solution of FeCl₃? (CBSE 2013)
Answer:
A reddish brown colloidal solution of Fe(OH)₃ is obtained. This process is called peptization.

Question 7.
What is especially observed when a beam of light is passed through a colloidal solution? (CBSE 2013)
Answer:
Tyndall effect

Question 8.
Give one example each of sol and gel. (CBSE Delhi 2014)
Answer:
Sol: As₂S₃ sol.
Gel: Cheese.

Question 9.
Give one example each of ‘oil in water’ and ‘water in oil’ emulsion. (CBSE 2014)
Answer:
Oil in water emulsion:
Milk Water in oil emulsion: Butter

Question 10.
Write a method by which lyophobic colloids can be coagulated. (CBSE 2015)
Answer:
By electrophoresis lyophobic colloids can be coagulated. Other methods are by mixing two oppositely charged sols or by boiling or by addition of electrolyte.

Question 11.
Write the main reason for the stability of colloidal sols. (CBSE Delhi 2016)
Answer:
The stability of colloidal sols is because of the presence of electrical charge on the particles which prevents the particles to come close together.

Question 12.
What are lyophobic colloids? Give one example for them. (CBSE Delhi 2011)
Answer:
The colloidal solutions in which the particles of the dispersed phase have no affinity for the dispersion medium are called lyophobic colloids. For example, metal sulphides like As₂S₃.

Question 13.
Write one similarity between physisorption and chemisorption. (CBSE Delhi 2017, CBSE AI 2017)
Answer:
Both are surface phenomena and increase with increase in surface area.

Question 14.
Why are lyophilic colloidal sols more stable than lyophobic colloidal sols? (CBSE Delhi 2015)
Answer:
The lyophilic colloidal sols are more stable because they are highly hydrated in solution.

Question 15.
What is the difference between a sol and a gel? (CBSE Delhi 2017)
Answer:
In a sol, dispersion medium is liquid and dispersed phase is solid. On the other hand, in a gel, dispersion medium is solid and dispersed phase is liquid.

Question 16.
What type of colloid is formed when a liquid is dispersed in a solid? Give an example. (CBSE AI 2017)
Answer:
Gel
For example: Cheese

Question 17.
Which of the following is most effective electrolyte in the coagulation of Fe₂O₃.x H₂O/Fe³⁺ sol?
KCl, AlCl₃, MgCl₂, K₄[Fe(CN)₆] (CBSE Sample Paper 2011)
Answer:
Since Fe(OH)₃ sol is positively charged, the anion having highest charge will be most effective, i.e. [Fe(CN)₆]⁴.

Question 18.
Why is colloidal gold used for intramuscular injection? (CBSE Sample Paper 2011)
Answer:
Colloidal gold is more effective because of larger surface area and therefore, is easily assimilated with blood which is colloidal.

Question 19.
What is colloidion? (CBSE Sample Paper 2011)
Answer:
4% solution of nitrocellulose in a mixture of alcohol and ether.

Question 20.
Leather gets hardened after tanning. Why? (CBSE Delhi 2015)
Answer:
Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tannin which has negatively charged colloidal particles, it results in mutual coagulation. This results in the hardening of leather.

Question 21.
It is necessary to remove CO when ammonia is prepared by Haber’s process. Explain. (CBSE Delhi 2015)
Answer:
Carbon monoxide acts as a poison for the catalyst in Haber’s process and therefore, it will lower the activity of the catalyst. Thus, CO must be removed when ammonia is obtained by Haber’s process.

Question 22.
Addition of alum purifies water. Why? (CBSE AI 2015)
Answer:
Alum coagulates the impurities present in water by neutralising the charge.

Question 23.
Out of sulphur sol and proteins, which one forms multimolecular colloids? (CBSE Delhi 2016)
Answer:
Proteins are macromolecules which cannot form multimolecular colloids while sulphur sol has smaller S₈ molecules which can coagulate to form multimolecular colloids.

Question 24.
Write the chemical method by which Fe(OH)₃ sol is prepared from FeCl₃. (CBSE AI 2017)
Answer:

Question 25.
Out of MgCl₂ and AlCl₃ which one is more effective in causing coagulation of negatively charged sol and why? (CBSE Delhi 2016)
Answer:
According to Hardy-Schulze rule, for negatively charged sol, greater the valency of the positive ion of the electrolyte added, greater is its coagulating power. Thus, AlCl₃ (Al³⁺ ions) is more effective in causing coagulation of negatively charged sol than MgCl₂ (Mg²⁺ ions).

Question 26.
Give one example each of lyophobic sol and lyophilic sol. (CBSE Delhi 2014)
Answer:
Lyophobic : As₂S₃, Lyophilic : Gelatin

Question 27.
Out of BaCl₂ and KCl which one is more effective in causing coagulation of negatively charged colloidal sol. Give reason. (CBSE Delhi 2015)
Answer:
BaCl₂ because greater the valency of the coagulating ion (positive ion), greater is its tendency to coagulate.

Question 28.
What are the dispersed phase and . dispersion medium in milk? (CBSE AI 2014)
Answer:
Liquid (dispersed phase), Liquid (dispersion medium)

Question 29.
What type of colloid is formed when a solid is dispersed in liquid? Give an example. (CBSE AI 2017)
Answer:
Sol, paints

Question 30.
CO (g) and H₂(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions? (CBSE AI 2018)
Answer:
Selectivity of a catalyst. It is the ability of a catalyst to selectively form a particular product.

Surface Chemistry Important Extra Questions Short Answer Type

Question 1.
Describe a conspicuous change observed when
(i) a solution of NaCl is added to a sol of hydrated ferric oxide.
(ii) a beam of light is passed through a solution of NaCl and then through a sol. (C.B.S.E. Delhi 2012)
Answer:
1. Hydrated ferric oxide is positively charged. When an electrolyte such as NaCl is added, the excess Cl^– ions neutralise positive charge and cause coagulation of the sol.

2. When a beam of light is passed through a colloidal solution, the path of the light becomes visible when viewed from the direction at right angle to that of incident beam. This phenomenon is called Tyndall effect. This is because of scattering of light by colloidal particles. The particles first absorb the incident light and then a part of it gets scattered by them and therefore the part becomes visible when seen at right angle to the direction of incident beam.

Question 2.
Name the two groups into which phenomenon of catalysis can be divided. Give an example of each group with the chemical equation involved. (CBSE Delhi 2012)
Answer:
Catalysis can be divided into two groups
(i) Homogeneous catalysis
(ii) Heterogeneous catalysis

(i) Homogenous catalysis: When the catalyst is present in the same phase as the reactants and products, it is called homogeneous catalysis. For example,

SO₂(g) is oxidised to SO₃(g) in the presence of gaseous nitric oxide as catalyst.

(ii) Heterogeneous catalysis: When the catalyst is in different phase than the
reactants, it is called heterogeneous catalysis. For example,

Question 3.
Write the dispersed phase and dispersion medium of the following colloidal systems: (CBSE Delhi 2013)
(i) Smoke
(ii) Milk
Answer:
Colloidal system: Dispersed phase/Dispersion medium
(i) Smoke: Solid/Gas
(ii) Milk: Liquid/Liquid

Question 4.
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes? (CBSE Delhi 2013)
Answer:
The colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium are called lyophilic sols. For example, glue, gelatin, starch, proteins, etc.

The colloidal solutions in which there is no affinity between the particles of the dispersed phase and the dispersion medium are called lyophobic sols. For example, solutions of . metals like Ag, Au, Al(OH)₃, Fe(OH)₃, etc.

Lyophobic sols can be easily coagulated on the addition of small amounts of electrolytes.

Question 5.
What is the difference between oil/water (O/W) type and water/oil (W/O) type emulsions? Give an example of each type. (CBSE Delhi 2013)
Answer:
1. Oil-in-water emulsion (o/w). In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

2. Water-in-oil emulsion (w/o). In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

Question 6.
Peptizing agent is added to convert precipitate into colloidal solution. Explain. (CBSE Sample Paper 2011)
Answer:
Peptization is a process of converting a freshly prepared precipitate into colloidal form by the addition of an electrolyte called peptizing agent. The suitable ions from the peptizing agent (electrolyte) are adsorbed by the particles of the precipitate giving it positive or negative charge. The charged particles repel one another and break up the precipitate into smaller particles of the size of the colloid. Therefore, it results into the formation of colloid.

For example, on treating a precipitate of iron (III) oxide with a small amount of FeCl₃ solution gives a reddish brown coloured colloidal solution.

Question 7.
Cottrell’s smoke precipitator is fitted at the mouth of chimney used in factories. Give reasons. (CBSE Sample Paper 2011)
Answer:
Smoke coming out of chimney of a factory is a colloidal solution of solid carbon particles which are charged in nature.

The mouth of the chimneys used in factories is fitted with Cottrell smoke precipitator. In this method, the smoke is allowed to pass through a chamber having a series of plates charged to very high potential (20,000 to 70,000 V).

Charged particles of smoke get attracted by charged plates, get precipitated and the gases coming out of chimney become free of charged carbon and dust particles.

Question 8.
Differentiate between peptization and coagulation.
(CBSE Sample Paper 2011, CBSE AI 2017)
Answer:
Peptization is the process of converting a freshly prepared precipitate into collodial form by the the addition of a suitable electrolyte. The electrolytes used for the purpose are called peptizing agents. On the other hand, coagulation is the phenomenon of precipitation of a collodial solution by the addition of excess of an electrolyte.

Question 9.
Explain:
(i) Sky appears blue in colour.
(ii) A freshly formed precipitate of ferric hydroxide can be converted to a colloidal sol by shaking it with a small quantity of ferric chloride. (CBSE Sample Paper 2011)
Answer:
1. Dust particles along with water suspended in air scatter blue light which reaches our eyes and therefore, sky looks blue to us.

2. When we add FeCl₃ to a freshly formed precipitate of Fe(OH)₃, peptization occurs. The Fe³⁺ ions are adsorbed on the surface of the precipitate which ‘ ultimately breaks down into smaller particles of colloidal size.

Question 10.
What happens when
(i) a freshly prepared precipitate of Fe(OH)₃ is shaken with a small amount of FeCl₃ solution?
(ii) persistent dialysis of a colloidal solution is carried out?
(iii) an emulsion is centrifuged? (CBSE 2018)
Answer:
(i) Peptization occurs / Colloidal solution of Fe(OH)₃ is formed.
(ii) Coagulation occurs
(iii) Demulsification or breaks into constituent liquids

Surface Chemistry Important Extra Questions Long Answer Type

Question 1.
Classify colloids where the dispersion medium is water. State their characteristics and write one example of each of these classes.
OR
Explain what is observed when
(i) an electric current is passed through a sol.
(ii) a beam of light is passed through a sol.
(iii) an electrolyte (say NaCl) is added to ferric hydroxide sol. (CBSE 2011)
Answer:
These colloids are of two types:
(i) Hydrophilic
(ii) Hydrophobic

Characteristics: In lyophilic colloids, there is affinity between dispersed phase and the dispersion medium while in the lyophobic colloids, there is no affinity rather hatred between the phases.

The main points of distinction between lyophilic and lyophobic sols are summarized below:

Example: Hydrophilic: Starch,
Hydrophobic : Metal sulphide

OR

(i) On passing electric current through colloidal solution, the colloidal particles move towards the oppositely charged electrodes, where they lose their charge and get coagulated. This is electrophoresis process.

(ii) When a beam of light is passed through a colloidal solution, the path of the light becomes visible when viewed from the direction at right angle to that of incident beam. This phenomenon is called Tyndall effect. This is because of scattering of tight by colloidal particles. The particles first absorb the incident light and then a part of it gets scattered by them and therefore the part becomes visible when seen at right angle to the direction of incident beam.

(iii) Hydrated ferric oxide is positively charged. When an electrolyte such as NaCl is added, the excess Cl^– ions neutralise positive charge and cause coagulation of the sol.

Question 2.
Explain how the phenomenon of adsorption finds application in each of the following processes:
(i) Production of vacuum
(ii) Heterogeneous catalysis
(iii) Froth floatation process (CBSE Delhi 2011)
Answer:
(i) Production of vacuum: The adsorption of air in liquid air helps to create a high vacuum in a vessel. This process is used in high vacuum instruments such as Dewar flask for storage of liquid air or liquid hydrogen. The remaining traces of air can be adsorbed by charcoal from the vessel evacuated by a vacuum pump to give a very high vacuum.

(ii) Heterogeneous catalysis: The phenomenon of adsorption is useful in the heterogeneous catalysis. Adsorption of reactants on the solid surface of catalysts increases the rate of reaction. The metals such as Fe, Ni, Pt, Pd, etc. are used in the manufacturing processes. Manufacture of ammonia using iron as catalyst (Haber process), manufacture of sulphuric acid by Contact process and use of finely divided nickel in the hydrogenation of oils are excellent examples of heterogeneous catalysis. Its use is based upon the phenomenon of adsorption.

(iii) In froth floatation process: A low grade sulphide ore is concentrated by separating it from silica and other earthly matter by adsorption using pine oil and frothing agent.

Question 3.
What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out. (CBSE 2012)
Answer:
The phenomenon of precipitation of a colloidal sol by the addition of excess of an electrolyte is called coagulation of colloidal sol. Coagulation of lyophobic sols can be carried out by the following methods:

1. By addition of an electrolyte: When an electrolyte is added to the sol, colloidal particles take up ions carrying opposite charge from the electrolyte and get neutralised resulting in coagulation.

2. By mutual precipitation: When two oppositely charged sols are mixed in equimolar proportions, they mutually neutralise their charge and both get coagulated.

3. By electrophoresis: When electrophoresis is carried out for a long time, the particles of dispersed phase move towards oppositely charged electrodes, lose their charge and get coagulated.

Question 4.
Explain the following terms giving a suitable example for each:
(i) Aerosol
(ii) Emulsion
(iii) Micelle (CBSE 2012)
Answer:
(i) Aerosol: It is a colloidal dispersion of a liquid in a gas. For example, fog, mist, cloud.

(ii) Emulsion. Emulsions are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium. Types of emulsion. These are of two types:

(a) Oil-in-water emulsion. In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

(b) Water-in-oil emulsion. In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

(iii) Micelle: The particles of colloidal size formed due to aggregation of several ions or molecules with lyophobic as well as lyophilic particles are called micelles. The common micelles system is soap. The micelles behave as normal electrolytes at low concentrations but as colloids at higher concentrations.

Question 5.
Write three distinct features of chemisorption which are not found in physisorption. (CBSE 2012)
Answer:
Characteristics of Chemisorption:

1. The forces between the adsorbate molecules and the adsorbent are strong chemical forces similar to chemical bonds. But no chemical bonds exist in physisorption.
2. Chemisorption is irreversible in nature but physisorption is reversible.
3. Chemisorption forms mono – molecular layers, but in physisorption multimolecular layers are present.

Question 6.
What are the characteristics of the following colloids? Give one example of each.
(i) Multimolecular colloids
(ii) Lyophobic sols
(iii) Emulsions (CBSE 2013)
Answer:
(i) The colloidal solutions in which atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimensions. The particles thus formed are called multimolecular colloids. For example, sols of gold atoms and sulphur (S₈) molecules.

(ii) The colloidal solutions in which there in no affinity (or love rather they have hatred) between the particles of the dispersed phase and the dispersion medium are called lyophobic sols.
For example, metal sulphides like As₂S₃.

(iii) The colloidal solutions in which both the dispersed phase and the dispersion medium are liquids are called emulsions. For example, milk.

Question 7.
Write the dispersed phase and dispersion medium of the following colloidal systems:
(i) Smoke
(ii) Milk
OR
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes? (CBSE Delhi 2013)
Answer:

OR
The colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium are called lyophilic sols. For example, glue, gelatin, starch, proteins, etc.

The colloidal solutions in which there is no affinity between the particles of the dispersed phase and the dispersion medium are called lyophobic sols. For example, ‘ solutions of metals like Ag, Au, Al(OH)₃, Fe(OH)₃, etc.
Lyophobic sols can be easily coagulated on the addition of small amounts of electrolytes.

Question 8.
What are emulsions? What are their different types? Give one example of each type. (CBSE 2014)
Answer:
Emulsions are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium.

Types of emulsions. These are of two types:
1. Oil-in-water emulsion. In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

(ii) Water-in-oil emulsion. In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

Question 9.
(i) In reference to Freundlich adsorption isotherm, write the expression for adsorption of gases on solids in the form of an equation.
(ii) Write an important characteristic of lyophilic sols.
(iff) Based on type of particles of dispersed phase, give one example each of associated colloid and multimolecular colloid. (CBSE Delhi 2014)
Answer:
(i) \(\frac{x}{m}\) = kp^1/n
where \(\frac{x}{m}\) is extent of adsorption, x is the mass of adsorbate, m is the mass of adsorbent, p is pressure at a particular temperature, k is a constant and n takes any whole number value.

(ii) in lyophilic sols, the particles of dispersed phase have a great affinity for the dispersion medium and are reversible in nature.

(iii) Associated colloids: Soap (e.g. sodium stearate).
Multimolecular colloids: Sol of gold (atoms).

Question 10.
Give reasons for the following observations:
(i) Leather gets hardened after tanning.
(ii) Lyophilic sol is more stable than lyophobic sol.
(iii) It is necessary to remove CO when ammonia is prepared by Haber’s process. (CBSE Delhi 2015)
Answer:
(i) Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tannin which has negatively charged colloidal particles, it results in mutual coagulation. This results in the hardening of leather.

(ii) Lyophilic sol is more stable than lyophobic sol because it is highly hydrated or solvated in solution.

(iii) Carbon monoxide acts as a poison for the catalyst in Haber’s process and therefore, it will lower the activity of the catalyst.
Thus, CO must be removed when ammonia is obtained by Haber’s process.

Question 11.
Give reasons for the following observations:
(i) Physisorption decreases with increase in temperature.
(ii) Addition of alum purifies the water.
(iii) Brownian movement provides stability to the colloidal solution. (CBSE 2015)
Answer:
(i) In physisorption, the attractive forces between adsorbent and adsorbate molecules are weak vander Waals forces. When temperature is increased, the kinetic energy of the molecules of the gas increases and they can easily leave the surface of adsorbent.
Therefore, physisorption decreases with increase in temperature.

(ii) Alum coagulates the impurities present in water by neutralising the charge.

(iii) The Brownian movement is due to the unbalanced bombardment of the particles by the molecules of the dispersion medium. This results in zigzag motion. Because of brownian movement, the particles do not settle down and hence is responsible for the stability of the sols.

Question 12.
Define the following terms:
(i) O/W Emulsion
(ii) Zeta potential
(iii) Multimolecular colloids (CBSE 2016)
Answer:
(i) O/W Emulsion: It is the emulsion in which oil is the dispersed phase and water is the dispersion medium. For example, milk is an emulsion of soluble fats in water.

(ii) Zeta potential: The potential difference between the fixed layer and the diffused layer of opposite charges is called Zeta potential.

(iii) Multimolecular colloids: When a dissolution, atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimension, the particles thus formed are called multimolecular colloids.

Question 13.
(i) Differentiate between adsorption and absorption.
(ii) Out of MgCl₂ and AlCl₂, which one is more effective in causing coagulation of negatively charged sol and why?
(iii) Out of sulphur sol and proteins, which one forms multimolecular colloids? (CBSE Delhi 2016)
Answer:
(i)

(ii) According to Hardy-Schulze rule, for negatively charged sol, greater the valency of the positive ion of the electrolyte added, greater is its coagulating power. Thus AlCl₃ (Al³⁺ ion) is more effective in causing coagulation of negatively charged sol than MgCl₂ (Mg²⁺ ions).

(iii) Proteins are macromolecules which cannot form multimolecular colloids while sulphur sol has smaller S₈ molecules which can coagulate to form multimolecular colloids.

Question 14.
What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out.
(CBSE Delhi 2013)
Answer:
The phenomenon of precipitation of a colloidal sol by the addition of excess of an electrolyte is called coagulation of colloidal sol.
Coagulation of lyophobic sols can be carried out by the following methods:
(i) By addition of an electrolyte: When an electrolyte is added to the sol, colloidal particles take up ions carrying opposite charge from the electrolyte and get neutralised resulting in coagulation.

(ii) By mutual precipitation: When two oppositely charged sols are mixed in equimolar proportions, they mutually neutralise their charge and both get coagulated.

(iii) By electrophoresis: When electrophoresis is carried out for a long time, the particles of dispersed phase move towards oppositely charged electrodes, lose their charge and get coagulated.

Question 15.
(i) What is the role of activated charcoal in gas mask?
(ii) A colloidal sol is prepared by the given method in figure. What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?

(iii) How does chemisorption vary with temperature? (CBSE Delhi 2019) Answer:
(i) Activated charcoalin gas mask adsorbs the poisonous or toxic gases from air. Therefore, these masks help to purify the air for breathing.

(ii) Negatively charged sol
Representation: Fe₂O₃. xH₂O/OH^–

(iii) In chemisorption, adsorption initially increases with rise in temperature and then decreases.

Question 16.
Define the following terms with an example in each case:
(i) Macromolecular sol
(ii) Peptization
(iii) Emulsion (CBSE Al 2013)
Answer:
(i) Macromolecular sol: In this type, the particles of the dispersed phase are sufficiently big in size (macro) to be of colloidal dimensions. In this case, a large number of small molecules are joined together through their primary valencies to form giant molecules. These molecules are called macromolecules and each macromolecule may consist of hundreds or thousands of simple molecules. The solutions of such molecules are called macromolecular solutions. For example, colloidal solution of starch, cellulose, etc.

(ii) Peptization: The process of converting a freshly prepared precipitate into colloidal form by addition of a suitable electrolyte is called peptization. The electrolytes used for the purpose are called peptizing agents.

(iii) Emulsions: They are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium. Normally, these are obtained by mixing an oil with water. Since the two do not mix well, the emulsion is generally unstable and is stabilised by adding a suitable reagent called emulsifier or emulsifying agent. The substances that are commonly employed for the purpose are gum, soap, glass powder, etc.

Question 17.
Define the following with a suitable example of each:
(a) Coagulation
(b) Multimolecular colloid
(c) Gel
Answer:
(a) Coagulation. The phenomenon of precipitation of a colloidal solution by the addition of excess of an . electrolyte is called coagulation. For
example, Fe(OH)₃ colloidal solution is coagulated by adding HCl.

(b) Multimolecular colloid. When on dilution, atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimensions, then the particles thus formed are called multimolecular colloids, for example, sols of gold atoms.

(c) Gel. A gel is a colloidal system in which a liquid is dispersed in a solid. For example, jellies, cheese.

OR

(a) Out of starch and ferric hydroxide sol, which one can easily be coagulated and why?
(b) What is observed when an emulsion is centrifuged?
(c) What is the role of promoters and poisons in catalysis? (CBSE Al 2019)
Answer:
(a) Ferric hydroxide sol is easier to coagulate because it is a lyophobic sol.
(b) De-emulsification occurs.
(c) Promoters increase the efficiency of catalyst whereas poisons inhibit the efficiency of catalyst.

Question 18.
Give reason for the following observations:
(i) When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed.
(ii) Finely divided substance is more effective as an adsorbent.
(iii) Lyophilic colloids are also called reversible sols.
Answer:
(i) The precipitated silver iodide adsorbs iodide ions from the dispersion medium resulting in the negatively charged colloidal solution.
(ii) Due to large surface area
(iii) If the dispersion medium is separated from the dispersed phase, the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols.

Question 19.
Answer the following questions:
(a) Which of the following electrolytes is most effective for the coagulation of Agl/Ag⁺ sol?
MgCl₂, K₂SO₄, K₄[Fe(CN)₆]
(b) What happens when a freshly precipitated Fe(OH)₃ is shaken with a little amount of dilute solution of FeCl₃?
(c) Out of sulphur sol and proteins, which one forms macromolecular colloids? (CBSE Sample Paper 2019)
Answer:
(a) K₄[Fe(CN)₆]
Hardy Schulze rule: The coagulation tendency of different electrolytes is different. It depends upon the valency of the active ion called flocculatins ion, which is the ion-carrying charge opposite to the charge on the colloidal particles. According to Hardy Schulze rule, greater the valency of the active ion or flocculating ion, greater will be its coagulating power. For example, to coagulate negative sol of AS₂S₃, the coagulating power of different cations has been found to decrease in the order as:
Al³⁺ > Mg²⁺ > Na⁺
Similarly, to coagulate a positive sol such as Fe(OH)₃, the coagulating power of different anions has been found to decrease in the order:
[Fe(CN)₆]4^– > PO₄^3- > SO₄^2- > Cl^–

(b) Due to the phenomenon of peptization, a reddish brown positively charged sol of ferric hydroxide is obtained when a freshly precipitated Fe(OH)₃ is shaken with a little amount of dilute solution of FeCl₃. It involves the adsorption of suitable ions from the electrolyte by . the particles of the precipitate. The charged particles repel one another and form a colloidal sol. Ferric ions from ferric chloride are adsorbed by Fe(OH)₃, precipitate and get converted into colloidal sol.

(c) Proteins are macromolecules which cannot form multi-molecular colloids while sulphur sol has smaller S₈ molecules which can coagulate to form multi-molecular colloids.

Question 20.
Answer the following:
(a) Why is ester hydrolysis slow in the beginning and then becomes faster after some time?
(b) In a solution of methylene blue, animal charcoal is added, the solution is then well shaken. What will be observed and why?
(c) Give an example of oil in water emulsion.
Answer:
(a) This is because of the process of autocatalysis. Ester on hydrolysis gives an acid which starts acting as a catalyst after sometime and therefore, the reaction becomes fast.

(b) Adsorption of a dye by charcoal. When animal charcoal is shaken with a solution of an organic dye such as methylene blue it is observed that the solution turns colourless. The discharge of the colour is due to the fact that the coloured component (generally an organic dye) gets adsorbed on the surface of animal charcoal. Therefore, animal charcoal is used for decolourising a number of organic substances in the form of their solutions.
(c) Milk, vanishing cream.

OR

Define the following:
(a) Associated colloids
(b) Electrophoresis
(c) Zeta potential (CBSE 2019C)
Answer:
(a) Associated colloids: These are the substances which when dissolved
in a medium behave as normal electrolytes at low concentration but behave as colloidal particles at higher concentration due to the formation of aggregated particles. The aggregate particles thus formed are called micelles. For example, in aqueous solution, soap (sodium stearate) ionises as:

In concentrated solution, these ions get associated to form an aggregate of colloidal size.

(b) The phenomenon of movement of colloidal particles under an applied electric field is called electrophoresis. If the particles accumulate near the negative electrode, the charge on the particles is positive. On the other hand, if the sol particles accumulate near the positive electrode, the charge on the particles is negative.

(c) The combination of two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer. The first layer of ions is firmly held and is termed fixed layer while the second layer is mobile which is termed diffused layer. The charges of opposite signs on the fixed and diffused parts of the double layer result in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charges is called the electrokinetic potential or zeta potential.

Question 21.
Write the differences between physisorption and chemisorption With respect to the following:
(i) Specificity
(ii) Temperature dependence
(iii) Reversibility and
(iv) Enthalpy change (CBSE Delhi 2013)
Answer:

Question 22.
Describe some features of catalysis by zeolites.
Answer:
Zeolites are microporous aluminosilicates of the general formula M_x/n [(AlO₂)_x (SiO₂)y] mH₂O. These are most important oxide catalysts. These are used in petrochemical industries for cracking of hydrocarbons and isomerisation. The reactions in zeolites depend upon the size of the cavities (cages) or pores (apertures) present in them.

The most remarkable feature of zeolite catalysis is the shape selectivity. Therefore, the selectivity of catalyst depends on the pores structure. It has been observed that the pore size in zeolites generally varies between 260 pm and 740 pm. Depending upon the size of the reactants and products compared to the size of the cages or pores of zeolite, reactions proceed in specific manner.

A zeolite catalyst called ZSM-5 converts alcohols to gasoline, by first dehydrating the alcohol by loss of water.

Question 23.
What do you understand by adsorption isotherm? Discuss briefly Freundlich adsorption isotherm.
Answer:
Adsorption isotherm. A graph between amount of adsorption and gas pressure keeping temperature constant is called an adsorption isotherm.

Freundlich adsorption isotherm. The behaviour of adsorption with pressure can be expressed by adsorption isotherm as shown in figure.

The inspection of the graph reveals the following facts:
(i) At low pressure, the graph is almost straight line which indicates that x/m is directly proportional to the pressure. This may be expressed as:
\(\frac{x}{m}\) ∝ P or \(\frac{x}{m}\) = kP ……. (i)
where k is constant.

(ii) At high pressure. The graph becomes almost constant which means that x/m becomes independent of pressure. This may be expressed as:

(iii) Thus, in the intermediate range of pressure, x/m will depend upon the power of pressure which lies between 0 and 1, i.e. fractional power of pressure. This may be expressed as:

where n can take any value between 1 and a larger integer depending upon the pressure. The above relationship is also called Freundlich’s adsorption isotherm.

Question 24.
Explain the following in connection with colloids:
(i) Hardy-Schulze rule
(ii) Dialysis.
Answer:
(i) Hardy-Schulze rule: This rule states that greater the valency of the active ion or flocculating ion, greater will be its coagulating power. According to this rule:

(a) The ions carrying the charge opposite to that of sol particles are effective in causing coagulation of the sol.

(b) Coagulation power of an electrolyte is directly proportional to the valency of the active ions.
For example, for negative As₂S₃ sol, the coagulating power of cations decreases as Al³⁺ > Mg²⁺ > K⁺
Similarly, for positively charged sol such as Fe (OH)₃, the coagulating power decreases as
[Fe(CN)₃]^4- > PO₄^3- > SO₄^2- > Cr^–

(ii) Dialysis. The method used to separate the impurities from the colloidal solution is called dialysis. Its principle is based upon the fact that colloidal particles cannot pass through a parchment or cellophane membrane while the ions of the electrolyte can pass through it. The colloidal solution is taken in a bag made of cellophane or parchment. The bag is suspended in fresh water. The impurities slowly diffuse out of the bag leaving behind pure colloidal solution. Dialysis can be used for removing HCl from the ferric hydroxide sol. The method is shown in figure.

The ordinary process of dialysis is slow. To increase the process of purification, the dialysis is carried out by applying electric field. This process is called electrodialysis.

Question 25.
Explain the following terms:
(i) Electrodialysis
(ii) Tyndall effect.
Answer:
(i) Electrodialysis. It is the process of separating the impurities from the colloidal solution by applying electric field. It is based on the fact that colloidal particles cannot pass through a parchment or cellophane membrane while the ions of the electrolyte can pass through it. The colloidal solution is taken in a bag made of cellophane or parchment and is suspended in a fresh water and electric current is applied. The impurities slowly diffuse out of the bag leaving behind pure colloidal solution.

(ii) Tyndall effect. When a strong beam of light is passed through a true solution placed in a beaker, in a dark room, the path of the light does not become visible. However, if the light is passed through a sol, placed in the same room, the path of the light becomes visible when viewed from a direction at right angle to that of the incident beam.

This phenomenon was studied for the first time by Tyndall and therefore it is called Tyndall effect. The cause to Tyndall effect is the scattering of light by colloidal particles, i.e. these particles first absorb the incident light and then a part of it gets scattered by them. Since the intensity of the scattered light is at right angle to the plane of the incident light, the path becomes visible only when seen in that direction. The particles in true solution are too small in size to cause any scattering, i.e. the Tyndall effect is not noticed in true solutions.

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 5 Magnetism and Matter. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 5 Important Extra Questions Magnetism and Matter

Magnetism and Matter Important Extra Questions Very Short Answer Type

Question 1.
A small magnetic needle pivoted at the center is free to rotate In a magnetic meridian. At what place will the needle
be vertical?
Answer:
At the potes

Question 2.
What is the angle of dip at a place where the horizontal and vertical components of the earth’s magnetic field are equal?
Answer:
450

Question 3.
How does the intensity of a paramagnetic sample vary with temperature?
Answer:
it decreases with the increase in temperature.

Question 4.
What should be the orientation of a magnetic dipole in a uniform magnetic field so that its potential energy is maximum?
Answer:
It should be anti-parallel to the applied magnetic field.

Question 5.
What is the value of angle of dip at a place on the surface of the earth where the ratio of the vertical component to the horizontal component of the earth’s magnetic field is 1/\(\sqrt{3}\)?
Answer:
Using the expression tan δ = \(\frac{B_{V}}{B_{H}}=\frac{1}{\sqrt{3}}\)
Therefore, δ = 30°

Question 6.
Where on the surface of the earth is the angle of dip 90°? (CBSE Al 2011)
Answer:
Poles.

Question 7.
Where on the surface of the earth is the angle dip zero? (CBSE Al 2011)
Answer:
Magnetic equator

Question 8.
What are permanent magnets? Give one example. (CBSE Delhi 2013)
Answer:
It is an arrangement that has a permanent dipole moment, e.g. bar magnet.

Question 9.
At a place, the horizontal component of the earth’s magnetic field is B, and the angle of dip is 60°. What is the value of the horizontal component of the earth’s magnetic field at the equator? (CBSE Delhi 2017)
Answer:
Zero.

Question 10.
Is the steady electric current the only source of the magnetic field? Justify your answer. (CBSE Delhi 2013C)
Answer:
No, the magnetic field is also produced by alternating current.

Question 11.
Write one important property of a paramagnetic material. (CBSEAI2019)
OR
Do the diamagnetic substances have a resultant magnetic moment in an atom in the absence of an external magnetic field?
Answer:
It moves from the weaker to the stronger regions of the magnetic field.
OR
No

Question 12.
Steel is preferred for making permanent magnets, whereas soft iron is preferred for making electromagnets. Give one reason.
Answer:

1. The Retentivity of steel is more than that of soft iron.
2. Soft iron has a smaller value of coercivity than steel.

Question 13.
Define angle of inclination at a given place due to the earth’s magnetic field. (CBSE 2019C)
Answer:
The angle of inclination is the angle at which the earth’s magnetic field at a given place makes with the horizontal line in the magnetic meridian.

Question 14.
Relative permeability of a material p = 0.5. Identify the nature of the magnetic material and write its relation to magnetic susceptibility. (CBSE Delhi 2014C)
Answer:
Paramagnetic.
Susceptibility is small and greater than one.

Magnetism and Matter Important Extra Questions Short Answer Type

Question 1.
(a) Define the term magnetic susceptibility and write its relation in terms of relative magnetic permeability.
Answer:
It refers to the ease with which a substance can be magnetized. It is defined as the ratio of the intensity of magnetization to the magnetizing field. The required relation is µ_r = 1 + χ_m

(b) Two magnetic materials A and B have relative magnetic permeabilities of 0. 96 and 500. Identify the magnetic materials A and B. (CBSE Al, Delhi 2018C)
Answer:
A: Paramagnetic,
B: Ferromagnetic

Question 2.
A magnetic needle free to rotate in a vertical position orients itself with its axis vertical at a certain place on the earth. What are the values of
(a) the angle of dip and
(b) the horizontal component of the earth’s magnetic field at this place? Where will this place be on the earth?
Answer:
The angle of dip is 90° and the horizontal component of the earth’s magnetic field is zero. This place is the magnetic pole of the earth.

Question 3.
Out of the two magnetic materials ‘A’ has relative permeability slightly greater than unity while ‘B’ has less than unity. Identify the nature of the materials ‘A’ and ‘B’. Will their susceptibilities be positive or negative? (CBSE Delhi 2014)
Answer:

• ‘A’ is paramagnetic and ‘B’ is diamagnetic.
• ‘A’ will have positive susceptibility while
• ‘B’ will have negative susceptibility.

Question 4.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its northern tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place. (CBSE Delhi 2011)
Answer:
Given δ = 30°, B_H = 0.4 G, B = ?
Using the expression
B_H = Bcos δ we have
B = \(\frac{B_{H}}{\cos \delta}=\frac{0.4}{\cos 30^{\circ}}=\frac{0.4}{\sqrt{3} / 2}=\frac{0.8}{\sqrt{3}}\) G

Question 5.
The susceptibility of a magnetic material is -0.085. Identify the type of magnetic material. A specimen of this material is kept in a non-uniform magnetic field. Draw the modified field pattern.
Answer:
The material is a diamagnetic material as diamagnetic materials have negative susceptibility. The modified field pattern is as shown below.

Question 6.
A uniform magnetic field gets modified as shown below when two specimens X and Y are placed in it.

(a) Identify the two specimens X and Y.
Answer:
X is a diamagnetic substance and Y is a paramagnetic substance.

(b) State the reason for the behavior of the field lines in X and Y.
Answer:
This is because the permeability of a diamagnetic substance is less than one and that of a paramagnetic substance is greater than one.

Question 7.
Three identical specimens of magnetic materials nickel, antimony, and aluminum are kept in a non-uniform magnetic field. Draw the modification in the field lines in each case. Justify your answer.
Answer:
Nickel is ferromagnetic, antimony is diamagnetic and aluminium is paramagnetic. Therefore, they will show the behavior as shown in the following figures.

Question 8.
Define neutral point. Draw lines of force when two identical magnets are placed at a finite distance apart with their N-poles facing each other. Locate the neutral points.
Answer:
It is a point near a magnet where the magnetic field of the earth is completely balanced by the magnetic field of the magnet. The figure is as shown below.

The cross indicates the neutral point.

Question 9.
The susceptibility of a magnetic material is -2.6 x 10^-5. Identify the type of magnetic material and state its two properties. (CBSE Delhi 2012)
Answer:
Diamagnetic:

1. Very small and negative susceptibility
2. Permeability is less than one.

Magnetism and Matter Important Extra Questions Long Answer Type

Question 1.
Write the expression for the magnetic dipole moment for a closed current loop. Give its SI unit. Derive an expression for the torque experienced by a magnetic dipole in a uniform magnetic field.
Answer:
The required expression is m = nIA.
It is measured in A m².

Consider a uniform magnetic field of strength B. Let a magnetic dipole be suspended in it such that its axis makes an angle 6 with the field as shown in the figure below. If ‘m’ is the strength of each pole, the two poles experience two equal and opposite force ‘B’ each. These forces constitute a couple that tends to rotate the dipole. Suppose the couple exerts a torque of magnitude τ.

Then

τ = either force × arm of the couple
= mB × AN = mB × 2 L sin θ
or
Since m × 2L is the magnetic dipole moment of the magnet.
Therefore τ = MB sin θ in vector form

we have \(\vec{τ}\) = \(\vec{M}\) × \(\vec{B}\)

Question 2.
(a) State Gauss’s law for magnetism. Explain Its significance.
Answer:
(a) Gauss’s Law for magnetism states that “The total flux of the magnetic field, through any closed surface, is aLways
zero, i.e. ∮\(\vec{B}\).\(\vec{dL}\) = 0

This law implies that magnetic monopoles do not exist” or magnetic field lines form closed loops.

(b) Write the four Important properties of the magnetic field lines due to a bar magnet. (CBSE Delhi 2019)
Answer:
Four properties of magnetic field lines are as follows:

1. Magnetic field lines always form continuous closed loops.
2. The tangent to the magnetic field line at a given point represents the direction of the net magnetic field at that point.
3. The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field.
4. Magnetic field lines do not intersect.

Question 3.
A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm.
(a) What is the magnetic moment of the magnet?
Answer:
(a) From equation τ = mB sin θ,
For θ = 30°, sin θ = 0.5

Thus, 0.016 = m × 800 × 10^-4 × 0.5 or m = 160 × 2/800 = 0.40 Am²

(b) What is the work done in moving it from its most stable to a most unstable position?
Answer:
From equation U= -mB cos θ, the most stable position is for θ = 0° and the most unstable position is for θ = 180°.

Work done is given by
W = U_m (θ = 180°) – Um for (θ = 0°)
= 2 mB = 2 × 0.40 × 800 × 10^-4
= 0.064 J

(c) The bar magnet is replaced by a solenoid of cross-sectional area 2 × 10^-4 m² and 1000 turns but of the same magnetic moment. Determine the current flowing through the solenoid. (NCERT Delhi 2005C)
Answer:
From part (i), m = 0.40 Am² Using equation m = NIA.
0.40 = 1000 × l × 2 × 10^-4
l = 2 A

Question 4.
(a) What happens if a bar magnet is cut into two pieces: (i) transverse to its length and (ii) along its length?
Answer:
(a) In either case one gets two magnets each with a north pole and a south pole.

(b) A magnetized needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to torque. Why?
Answer:
The needle experiences no net force because the field is uniform.

The iron nail experiences a non¬uniform field due to the bar magnet. There is an induced magnetic moment in the nail; therefore, it experiences both force and torque.

The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of the magnet than the induced north pole.

Question 5.
Name the elements of the earth’s magnetic field at a place. Explain their meaning.
Answer:
The elements required to completely specify the earth’s magnetic field are called magnetic elements. These elements are:
(a) Declination (θ)
(b) Dip or inclination (δ)
(c) Horizontal component of the earth’s magnetic field (B_H).

(a) Declination (θ): Declination at a place is defined as the acute angle between the magnetic meridian and the geographic meridian.
In the figure above ∠BAB’ = θ is called the angle of declination. There are several periodic variations in the value of declination for a given location.

(b) Dip (δ): Dip at a place is defined as the angle between the direction of the total intensity of the earth’s magnetic field and a horizontal line in the magnetic meridian at that place. It is denoted by 8.

Suppose a freely suspended magnet aligns itself along the direction AB as shown in the figure, then this gives the direction of the earth’s total magnetic field at that place. Then the angle ∠BAO = δ is called the angle of dip or simply dip at that point.

(c) Horizontal component of the earth’s magnetic field (BH): It is the component of the total intensity of the earth’s magnetic field along a horizontal line in the magnetic meridian. It is denoted by BH or H.

Question 6.
Name the three types of magnetic materials which behave differently when placed in a non-uniform magnetic field. Give two properties for each of them.
Answer:
The three types of magnetic materials are
(a) Diamagnetic
(b) Paramagnetic and
(c) Ferromagnetic

Question 7.
The susceptibility of a magnetic material is 0-9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field. (CBSEAI, Delhi 2018)
Answer:
We know that the susceptibility of a paramagnetic substance is positive but small. Hence, the given magnetic material is paramagnetic in nature.

When a specimen of a paramagnetic substance is placed in a magnetizing field, the magnetic field lines prefer to pass through the specimen rather than through air. Thus, magnetic induction 6 inside the sample is more than the magnetic induction B0 outside the sample.

Question 8.
(a) Draw the magnetic field lines due to a circular loop of area \(\vec{A}\) carrying current l. Show that it acts as a bar magnet of magnetic moment \(\vec{m}\) = I \(\vec{A}\).
Answer:
(a) The magnetic field lines are as shown.

Magnetic field due to circular loop on its axis at far off points B = \(\frac{\mu_{0}}{4 \pi} \frac{2 l A}{x^{3}}\)

Magnetic field due to a bar magnet at its axial point is B = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{x^{3}}\)
Comparing the above two we have m = IA

(b) Derive the expression for the magnetic field due to a solenoid of length ‘21’, radius ‘a’ having ‘n’ number of turns per unit length and carrying a steady current T at a point on the axial line, distant ‘r’ from the center of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of the magnetic moment ‘m’? (CBSEAI 2015)
Answer:
(b) By the-Biot-Savarts law, the magnetic field at point P on the axis of the solenoid due to a small element of length dx at a distance ‘x’ from the center of the solenoid is

dB = \(\frac{\mu_{0} n d x / a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}\)

Hence total magnetic field is given by
B = ∫dB = \(\frac{\mu_{0} / a^{2} n^{+L}}{2} \int_{-L}^{d x} \frac{d x}{\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}\)

For r >> a (R >> L)
We have
B = ∫dB = \(\frac{\mu_{0} / a^{2} n^{+L}}{2 r^{3}} \int_{-L}^{+L} d x=\frac{\mu_{0} \ln }{2} \frac{2 L a^{2}}{r^{3}}\)

Magnetic moment of a solenoid
m = (n × 2L) I (πa²)

Therefore,
B = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\) as that of a bar magnet.

Numerical Problems :

• Torque on a magnetic dipole τ = MB sin θ
• Potential energy of a magnetic dipole U = – MB cos θ
• \(B_{H}^{2}+B_{v}^{2}=B^{2}\)
• tan θ = \(\frac{B_{V}}{B_{H}}\)
• H = B cos δ
• V = B sin δ

Question 1.
A bar magnet of the magnetic moment 6 J T¹ is aligned at 60° with a uniform external magnetic field of 0 – 44 T. Calculate
(a) the work is done in turning the magnet to align its magnetic moment (i) normal to the magnetic field and (ii) opposite to the magnetic field and
Answer:

(b) the torque on the magnet in the final orientation in case (ii). (CBSEAI, Delhi 2018)
Answer:
τ = M × B sin θ₂
= 6 × 0.44 × sin 180° = 0

Question 2.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 x 10^-2 J. What is the magnitude of the magnetic moment of the magnet?
Answer:
Given θ = 30°, B = 0.25 T, τ = 4.5 × 10^-2 J,M = ?
Using the expression τ = MB sin θ we have
M = \(\frac{\tau}{B \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}\) = 0.36 J T^-1

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 4 Chemical Kinetics. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 4 Important Extra Questions Chemical Kinetics

Chemical Kinetics Important Extra Questions Very Short Answer Type

Question 1.
Define rate of a reaction. (CBSE Delhi 2010)
Answer:
The rate of a reaction is the change of concentration in any one of the reactants or products per unit time.

Question 2.
Define order of a reaction. (CBSE Delhi 2010, CBSE 2011)
Answer:
The sum of the powers of the concentration terms of the reactants in the rate law expression is called the order of that chemical reaction. For example, if rate law expression is
rate = K[A]^p[B]^q[C]^r
Order of reaction = p + q + r

Question 3.
For a chemical reaction R → P, the variation in the concentration (R) vs. time (t) plot is given as

(i) Predict the order of the reaction.
(ii) What is the slope of the curve? (CBSE 2014)
Answer:
(i) Zero order reaction
(ii) Slope = – k

Question 4.
Define ‘activation energy’ of a reaction. (CBSE Delhi 2011)
Answer:
Activation energy is the minimum energy that the reactants must possess for the conversion into the products during their collisions.
It is equal to difference between the threshold energy needed for the reaction and the average kinetic energy of the molecules.

Question 5.
Identify the order of a reaction if the units of its rate constant are:
(i) L^-1 mol s^-1
(ii) L mol^-1 s^-1 (CBSE All India 2011)
Answer:
(i) Zero order
(ii) Second order

Question 6.
For a reaction: A + B → P, the rate law is given by, r = k [A]^1/2 [B]² .
What is the order of reaction? (CBSE All India 2013)
Answer:
Order = \(\frac{1}{2}\) + 2 = 2\(\frac{1}{2}\) or 2.5.

Question 7.
In the Arrhenius equation, what does the factor e^-E₀/RT correspond to?
(CBSE Sample Paper 2017-18)
Answer:
e^-E₀/RT corresponds to the fraction of molecules having kinetic energy greater than E_a.

Chemical Kinetics Important Extra Questions Short Answer Type

Question 1.
For a reaction,
C₂H₄(g) + H₂(g) → C₂H₆(g),
rate = 5.5 × 10^-14 [C₂H₄].
(a) Write the unit of rate constant.
(b) Calculate its half-life (t_1/2). (CBSE 2019)
Answer:
(a) Hydrogenation of ethene is an example of first order reaction so the unit of rate constant is s^-1 or time^-1.

(b) t_1/2 = \(\frac{0.693}{k}=\frac{0.693}{5.5 \times 10^{-14}}\)
= 1.26 × 10¹³s
(or any other unit of time)

Question 2.
For a reaction

The proposed mechanism is as given below:
(1) H₂O₂ + I^– → H₂O + IO^– (slow)
(2) H₂O₂ + IO^– → H₂O + I^– + O₂ (fast)
(i) Write rate law for the reaction.
(ii) Write the overall order of reaction.
(iii) Out of steps (1) and (2), which one is rate-determining step? (CBSE Delhi 2019)
Answer:
(i) Rate = k[H₂O₂][I^–]
(ii) 2
(iii) Step 1 is rate determining.

Question 3.
A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when
(i) the concentration of B alone is increased to three times?
(ii) the concentrations of A as well as B are doubled? (CBSE Delhi 2010)
Answer:
The rate law may be expressed as:
Rate = k[A] [B]²
(i) Rate becomes nine times
(ii) Rate becomes eight times.

Question 4.
What do you understand by the rate law and rate constant of a reaction? Identify the order of reaction if the units of its rate constant are:
(i) L^-1 mol s^-1
(ii) L mol^-1 s^-1 (CBSE 2011)
Answer:
Rate law is the representation of rate of reaction in terms of concentration of the reactants.
Rate constant is the rate of reaction when the concentration of reactants is unity. Order of reaction from rate constants
(i) Zero order
(ii) Second order

Question 5.
A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half? What is the unit of rate constant for such a reaction? (CBSE 2011)
Answer:
For a second order reaction,
rate = K[A]²
If the concentration of A is reduced to half, rate will become 1 /4 of the original value.
Units of k.

Question 6.
Distinguish between rate expression and rate constant of a reaction. (CBSE 2011)
Answer:
Rate expression is a mathematical expression which denotes the rate of a reaction in terms of molar concentration of reactants.

Rate constant is the rate of the reaction when the concentration of reactants is unity. It is proportionality constant in the rate law and is independent of initial concentrations of the reactants.

Question 7.
What do you understand by the ‘order of reaction’? Identify the reaction order from each of the following units of reaction rate constant:
(i) L^-1 mol s^-1
(ii) L mol^-1 s^-1 (CBSE Delhi 2012)
Answer:
The sum of the powers of the concentrations of reactants in the rate law expression is called order of a reaction. For example, if rate law expression for a reaction is:
rate = K[A]^a[B]^a[C]c
then order = a + b + c
(i) Zero order
(ii) Second order

Question 8.
(i) For a reaction A + B → P, the rate law is given by,
r = k[A]^1/2 [B]²
What is the order of this reaction?
(ii) A first order reaction is found to have a rate constant k = 5.5 × 10^-14 s^-1. Find the half life of the reaction. (CBSE 2013)
Answer:
(i) Order of reaction 1
= \(\frac{1}{2}\) + 2 = 2.5

(ii) t_1/2 = \(\frac{0.693}{k}=\frac{0.693}{5.5 \times 10^{-14}}\)
= 1.26 × 10¹³s

Question 9.
Write two differences between ‘order of reaction’ and ‘molecularity of reaction. (CBSE Delhi 2014)
Answer:

Question 10.
The rate constant for a reaction of zero order in A is 0.0030 mol L^-1s^-1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M? (CBSE 2010)
Answer:
For a zero order reaction,
k = 0.0030 mol L^-1s^-1
[A]₀ = 0.10 M, [A] = 0.075M

Answer:

Question 11.
The rate of a reaction depends upon the temperature and is quantitatively expressed as k = \(A e^{-E_{a} / R T}\)
(i) If a graph Is plotted between log k and 1/T, write the expression for the slope of the reaction.
(ii) If at under different conditions, E_a1 and E_a2 are the activation energy of two reactions. If E_a1 = 40 J / mol and E_a2 = 80 J / mol which of the two has a larger value of the rate constant? (CBSE Sample Paper 2019)
Answer:
(ii) Arrhenius proposed a quantitative relationship between rate constant
and temperature as k = \(A e^{-E_{a} / R T}\) Taking logarithm,
In k = In A – \(\frac{E_{a}}{R} \times \frac{1}{T}\)
Converting to common logarithm (In X = 2.303 log X), we get
2.303 log k = 2.303 log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\).
Dividing each side by 2.303, we get
log k = log A – \(\frac{E_{a}}{2.303 R T}\)
So, it is clear that increasing the temperature or decreasing the activation energy will result in an increase in the rate of reaction.

(i) Slope = – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)

Plot of log k versus 1/T to calculate activation energy.

(ii) k₁ > k₂

Question 12.
The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce the concentration of the reactant to 1/10th of its initial value?
Answer:
For a first order reaction,

Question 13.
Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
log k = log A – \(\frac{E_{a}}{2.303 R}\left(\frac{1}{T}\right)\)
where E_a is the activation energy. When a graph is plotted for log k vs. \(\frac{1}{T}\), a straight line with a slope of – 4250 K is obtained. Calculate ‘E_a’ for the reaction (R = 8.314 JK^-1mol^-1). (CBSE Delhi 2013)
Answer:
Slope = –\(\frac{E_{a}}{2.303 R}\) = -4250K
E_a = – 2.303 × R × Slope
= – 2.303 × 8.314 JK^-1 mol^-1 × (-4250 K)
= 81375.3 J mol^-1 or 81.375 kJ mol^-1.

Question 14.
For the reaction
C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆
write:
(i) Rate of reaction expression,
(ii) Rate law equation,
(iii) Molecularity,
(iv) Order of reaction (CBSE Sample paper 2011)
Answer:
(i)

(ii) Rate law equation:
Rate = k[C₁₂H₂₂O₁₁]
(iii) Molecularity = 2
(iv) Order = 1

Question 15.
Consider the decomposition of hydrogen peroxide in alkaline medium which is catalysed by iodide ions.

This reaction takes place in two steps as given below:
Step -1 H₂O₂ + I^– → H₂O + IO^– (slow)
Step – II H₂O₂ + IO^– → H₂O + I^– (fast)
(i) Write the rate law expression and determine the order of reaction w.r.t. H₂O₂.
(ii) What is the molecularity of each individual step? (C8SE Sample paper 2011)
Answer:
(i) Rate = k[H₂O₂] [I^–]
Order w.r.t. H₂O₂ = 1
(ii) Molecularity: Step I = 2, Step II = 2

Question 16.
For a reaction: A + B → P, the rate law is given as: Rate = k[A][B]²
(i) How is the rate of reaction affected when the concentration of B is doubled?
(ii) What is the order of reaction if A is present in large excess? (CBSE All India 2015)
Answer:
(i) Rate becomes four times
(ii) Order = 2

Question 17.
What is the effect of adding a catalyst on
(i) Activation energy (E_a) and
(ii) Gibbs free energy (∆G) of a reaction? (CBSE All India 2017)
Answer:
(i) Activation energy decreases
(ii) No effect on ∆G because Gibbs free energies of products and reactants remain the same in the presence of catalyst.

Question 18.
For a reaction:

Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k. (CBSE Sample paper 2018)
Answer:

Rate = k
(i) Order of reaction: Zero order Molecularity = 2 (bi molecular)
(ii) Unit of k = mol L^-1 s^-1 or atm s^-1

Question 19.
Explain the following terms: (CBSE 2014)
(i) Rate constant (k).
(ii) Half life period of a reaction (t_1/2).
Answer:
(i) Rate constant is the rate of the reaction
when the molar concentration of each of the reactants is unity. For example, for a reaction,
A + B → AB
Rate of reaction is
Rate = k[A][B]
where k is the rate constant. When
[A] = 1 and [B] = 1
Rate of reaction = k × 1 × 1 = k
It is also catted specific reaction rate and is independent of the initial concentration of the reactants.

(ii) Half life period of a reaction is the time during which the concentration of a reactant Is reduced to half of its initial concentration. In other words, it is the time during which half of the reaction Is completed. It is denoted as t_1/2.

Chemical Kinetics Important Extra Questions Long Answer Type

Question 1.
Show that time required for 99% completion is twice the time required for the completion of 90% reaction. (CBSE Delhi 2013)
Answer:
For the first order reaction,
t = \(\frac{2.303}{k} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]}\)
Let initial concentration, [A₀] = a
For 99% completion of reaction, [A]

For 90% completion of reaction,

Dividing equation (i) by (ii)
\(\frac{t_{(99 \%)}}{t_{(90 \%)}}\) = 2
or t_(99%) = 2 × t_(90%)

Question 2.
The decomposition of NH₃ on platinum surface is zero order reaction. If rate constant (k) Is 4 × 10^-3 M s^-1, how long will it take to reduce the initial concentration of NH₃ from 0.1 M to 0.064 M? (CBSE Delhi 2019)
Answer:
k = 4 × 10^-3M s^-1, [A]₀ = 0.1 M, [A] = 0.064 M

Question 3.
The thermal decomposition of HCO₂H is a first order reaction with a rate constant of 2.4 × 10^-3 s^-1 at a certain temperature. Calculate how long will it take for three fourth of initial quantity of HCO₂H to decompose. (log 4 = 0.6021) (CBSE 2011)
Answer:
We know that
t = \(\frac{2.303}{k} \log \frac{[A]_{0}}{[A]}\)
If initial conc. of A is a, then
[A] = a – \(\frac{3}{4}\)a = \(\frac{1}{4}\)a
k = 2.4 × 10^-3 s^-1

= \(\frac{2.303}{2.4 \times 10^{-3}}\) × 0.6021 = 578 s

Question 4.
For the reaction:
2N0(g) + Cl₂(g) → 2NOCl(g)
the following data were collected. All the measurements were taken at 263 K:

(i) Write the expression for rate law.
(ii) Calculate the value of rate constant and specify its units.
(iii) What is the initial rate of disappearance of Cl₂ In exp. 4? (CBSE 2012)
Answer:
(i) Rate = k[NO]²[Cl₂]
(ii) According to experiment 1,
[NO] = 0.15 M; [Cl₂] = 0.15 M
Rate = 0.6 M/min.
Rate = k[NO]²[Cl₂]
0.6 M/min = k(0.15M)² (0.15 M)
k = 177.8 M^-2 min^-1
or
= 177.8 mol^-2 L² min^-1
Therefore, units of k are M^-2 min^-1 or mol^-2 L² min^-1.

(iii) In experiment 4, [NO] = 0.25 M; [Cl₂] = 0.25 M and k = 177.8 M^-2 min^-1
Rate = k[NO]² [Cl₂]
= 177.8 M^-2 min^-1 × (0.25 M)² × 0.25 M
= 2.78 M min^-1.

Question 5.
The following data were obtained during the first order thermal decomposition of
SO₂Cl₂ at a constant volume:
SO₂Cl₂(g) → SO₂(g) + Cl₂(g)

Calculate the rate constant.
(Given: log 4 = 0.6021, log 2 = 0.3010) (CBSE Delhi 2014)
Answer:
SO₂Cl₂ → SO₂ + Cl₂
At t = 0 s     0.4 atm            0            0
At t = 100 s (0.4 – x) atm   x atm   x atm
P_t = 0.4 – x + x + x = 0.4 + x
0.7 = 0.4 + x      or    x = 0.3
k = \(\frac{2.303}{t}\) log \(\frac{p_{0}}{2 p_{0}-p_{t}}\)
t = 100 s
∴ k = \(\frac{2.303}{100}\) log \(\frac{0.4}{2 \times 0.4-0.7}\)
= \(\frac{2.303}{100}\) log \(\frac{0.4}{0.1}\)
∴ k = \(\frac{2.303}{100}\) × 0.6021 = 1.38 × 10^-2 s^-1

Question 6.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation:
log k = 14.2 – \(\frac{1.0 \times 10^{4} \mathrm{~K}}{\mathrm{~T}}\)
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) (CBSE Delhi 2016)
Answer:
According to Arrhenius equation,
log k = log A – \(\frac{E_{a}}{2.303 \mathrm{RT}}\) …….. (i)
The given equation is
log k = 14.2 – \(\frac{1.0 \times 10^{4} \mathrm{~K}}{\mathrm{~T}}\) ………. (ii)
Comparing eq. (1) and (ii)
\(\frac{E_{a}}{2.303 \mathrm{RT}}\) = 1.0 × 10⁴
∴ E_a = 1.0 × 10⁴ × 2.303 × 8.314
= 19.147 × 10⁴ J mol^-1
Now t_1/2 = \(\frac{0.693}{k}\)
or k = \(\frac{0.693}{t_{1 / 2}}\)
= \(\frac{0.693}{200 \mathrm{~min}}\) = 3.465 × 10^-3 min ^-1

Question 7.
The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the activation energy of the reaction assuming that it does not change with temperature (R = 8.3 14 JK^-1 mol^-1). (CBSE All India 2010)
Answer:
T₁ = 300 K, T₂ = 320 K, R = 8.314 JK^-1 mol^-1

Question 8.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours? (CBSE Sample Paper 2011)
Answer:
k = \(\frac{0.693}{t_{1 / 2}}\) = \(\frac{0.693}{3.0}\) = 0.231 hr^-1
Let initial cone, of sucrose = 1 M
Cone, after 8 hr. = (1 – x) where x is the amount of sucrose decomposed

Conc. of sucrose left after 8 hr = 1 – 0.842
= 0.158 M

Question 9.
The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_a) of the reaction assuming that it does not change with temperature.
[R = 8.314 J K^-1 mol^-1, log 4 = 0.6021] (CBSE All India 2013)
Answer:

Question 10.
The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at constant volume.
SO₂Cl₂(g) → SO₂(g) + Cl₂(g)

Calculate the rate of reaction when total pressure is 0.65 atm. (CBSE Sample Paper 2011)
Answer:
For the decomposition of SO₂Cl₂
SO₂Cl₂(g) → SO₂(g) + Cl₂(g)
At t = 0          p₀ atm              0        0
At time, t      (p₀ – x) atm       x        x
p_t = p₀ – x + x + x = p₀ + x
or x = p_t – p₀
Now at time t, p(SOCl₂) = p₀ – x
= P₀ – (P_t – P₀) = 2 p₀ – p_t
∴ k = \(\frac{2.303}{t}\) log \(\frac{p_{0}}{2 p_{0}-p_{t}}\)
when t = 100 s
k = \(\frac{2.303}{100}\) log \(\frac{0.5}{2 \times 0.5-0.6}\)
= \(\frac{2.303}{100}\) log 1.25 = 2.2316 × 10^-3 s^-1
When p_t = 0.65 atm, i.e. p₀ + p = 0.65 atm
p = 0.65 – p₀ = 0.65 – 0.50 = 0.15 atm
∴ Pressure of SO₂Cl₂ at time t = 0.50 – 0.15
= 0.35 atm
Rate = 2.2316 ×10^-3 × 0.35
= 7.8 × 10^-5 atm s^-1

Question 11.
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. {CBSE All India 2017)
Answer:
For the first order reaction:

Question 12.
A first order reaction takes 23.1 minutes for 50% completion. Calculate the time required for 75% completion of this reaction. (log 2 = 0.301, log3 0.4771, log4 = 0.6021) (CBSE All India 2015)
Answer:
For the first order reaction:

Question 13.
The decomposition of phosphine, PH₃, proceeds according to the following equation:
4PH₃(g) → P₄(g) + 6H₂(g)
It is found that the reaction follows the following rate equation:
Rate = k[PH₃].
The half-life of PH₃ is 37.9 sat 120 C.
(i) How much time is required for 3/4th of PH₃ to decompose?
(ii) What fraction of the original sample of PH₃ remains behind after 1 minute? (C8SE All India 2010)
Answer:

Question 14.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with t_1/2 = 3 hrs. Calculate the fraction of sucrose which remains after 8 hrs. (CBSE Sample Paper 2011)
Answer:

Question 15.
Starting from 10 g of a radioactive element, 0.25 g was left after 5 years. Calculate
(i) Rate constant for the decay of the radioactive element.
Answer:

(ii) The amount left after one year.
Answer:
Amount left after 1 year

(iii) The time required for half of the element to decay.
Answer:
Time for decay of half of the element
t_1/2 = \(\frac{0.693}{k}=\frac{0.693}{0.7379}\) = 0.9392 year

(iv) Average life of the element. (CBSE All India 2011)
Answer:
t = \(\frac{1}{k}=\frac{1}{0.7379}\) = 1.3552 year

Question 16.
In general, It is observed that the rate of a chemical reaction becomes double with every 1o°. rise in temperature. If this generalization holds for a reaction In the temperature range 298 to 308 K, what would be the value of activation energy for this reaction? (R = 8.3 14 JK^-1 mol^-1) (CBSE Delhi 2000)
Answer:
Arrhenius equation is:

Question 17.
A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.
(Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.3 14 Jk^-1 mol^-1) (CBSE All India 2018)
Answer:
Let us first calculate k₁ and k₁ at temperatures 300 K and 320 K. We know that

Question 18.
The rate constant of a first order reaction increases from 2 × 10^-2 to 6 × 10^-2 when the temperature changes from 300 K to 320 K. Calculate the energy of activation. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) (CBSE 2019 C)
Answer:

Question 19.
For the reaction A + B → products, the following initial rates were obtained at various given initial concentrations:

Determine the half-life period.
OR
A first order reaction is 50% complete In 50 minutes at 300 K and the same reaction is again 50% complete In 25 minutes at 350 K. Calculate activation energy of the reaction. (CBSE Sample Paper 2019)
Answer:
Rate = k [A]^x [B]^y
0.05 = k [0.1]^x [0.1]^y …… (i)
0.10 = k [0.2]^x [0.1]^y ……. (ii)
0.05 = k [0.1]^x [0.2]^y …….. (iii)
Dividing equation (ii) by (i)
\(\frac{0.10}{0.05}\) = (2)^x
x = 1
Dividing equation (iii) by (i)
\(\frac{0.05}{0.05}\) = (2)^y
y = 0
Rate = k [A]¹ [B]⁰

Question 20.
For a certain chemical reaction variation in concentration [A] vs. time (s) plot as shown in given figure:

(i) Predict the order of the given reaction?
(ii) What does the slope of the line and intercept indicate?
(iii) What is the unit of rate constant k? (CBSE Sample paper 2018-19)
Answer:
(i) Zero order reaction
(ii) Slope represents k; Intercept represents [R]₀
(iii) mol L^-1 s^-1

Question 21.
The following data were obtained during the first order thermal decomposition of N₂O₅(g) at a constant volume:
2N₂O₅(g) → 2N₂O₄(g) + O₂(g)

Calculate the rate constant.
OR
Two reactions of the same order have equal pre-exponential factors but their activation energies differ by 24.9 kJ mol^-1. Calculate the ratio between the rate constants of these reactions at 27°C. (Gas constant R 8.314 J K^-1 mol^-1) (CBSE 2018)
Answer:
2N₂O₅(g) → 2N₂O₄(g) + O₂(g)
At t = 0       0.5 atm       0 atm      0 atm
At time t   0.5-2x atm   2x atm    x atm
P_t = P_N2O5 + P_N2O5 + P_O2
= (0.5 – 2x) + 2x + x = 0.5 + x
x = p_t – 0.5
P_N2O5 = 0.5 – 2x
= 0.5 – 2(p_t – 0.5)
= 1.5 – 2p_t
At t = 100s; p_t = 0.512 atm
P_N2O5 = 1.5 – 2 × 0.512 = 0.47 atm

OR
The Arrhenius equation: k = Ae^-Ea/RT
Taking log on both sides:
log k = log A – \(\frac{E_{a}}{2.303 R T}\)
For reaction (i) log k₁ = log A – \(\frac{E_{o}(1)}{2.303 R T}\)
For reaction (ii) log k₂ = logA – \(\frac{E_{o}(2)}{2.303 R T}\)
Subtracting (i) from (ii)

Question 22.
(a) Define order of reaction. How does order of a reaction differ from molecularity for a complex reaction?
Answer:
(a) Order of reaction is the sum of the power to which the concentration terms are raised in the rate law equation. Order of a reaction is applicable for the complex reaction but molecularity has no meaning for the complex reaction. 0.693

(b) A first order reaction is 50% complete in 25 minutes. Calculate the time for 80% completion of the reaction.
Answer:

(a) The decomposition of a hydrocarbon has value of rate constant as 2.5 × 10⁴ s^-1 at 27 °C. At what temperature would rate constant be 7.5 × 10⁴ s^-1 if energy of activation is 19.147 × 10³ J mol^-1?
Answer:

0.1431 T₂ = T₂ – 300
0.1431 T₂ – T₂ = 300
0.8569T₂ = – 300
T₂ = \(\frac{300}{0.8569}\) = 350 k

(b) Write a condition under which a bimolecular reaction is kinetically first order. Give an example of such a reaction.
(Given: log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990) (CBSE AI 2019)
Answer:
A bimolecular reaction is kinetically of first order when one of the reactant is in excess. For example, hydrolysis of methyl acetate.
CH₃COOCH₃ + H₂O (excess) → CH₃COOH + CH₃OH

Question 23.
Nitrogen pentoxide decomposes according to equation:
2N₂O₅ (g) →4NO₂(g) + O₂(g)
The first order reaction was allowed to proceed at 40° C and the data below were collected:

(i) Calculate the rate constant. Include units with your answer.
Answer:
Since it is a first order reaction,
k = \(\frac{2.303}{t}\) log \(\frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]}\)
k can be calculated as:
at t = 20 min


The average value of rate constant,
k = 0.01624 min^-1 or = 1.624 × 10^-2 min^-1

(ii) What will be the concentration of N₂O₅ after 100 minutes?
Answer:
Concentration of [N₂O₅] at t = 100 min

(iii) Calculate the initial rate of reaction. (CBSE 2011)
Answer:
Initial rate of reaction,
rate = k[N₂O₅]
= 0.01624 × 0.40
= 6.496 × 10^-3

Question 24.
(i) Explain the following terms:
(a) Rate of a reaction
Answer:
The rate of reaction is the change of concentration in any one of the reactants or products per unit time. For a reaction, A → B rate may be expressed as
= \(\frac{-d[A]}{d t}\) or \(\frac{d[B]}{d t}\)

(b) Activation energy of a reaction
Answer:
Activation energy is the minimum energy which the reacting molecules should acquire so that they react to give the products. It is also defined as the energy required to form the products.

(ii) The decomposition of phosphine, PH₃, proceeds according to the following equation:
4PH₃(g) → P₄ (g) + 6H₂(g)
It Is found that the reaction follows the following rate equation:
Rate = k[PH₃]
The half life of PH₃ is 37.9 s at 120°C.
(a) How much time is required for 3/4th of PH₃ to decompose?
Answer:

(b) What fraction of the original sample of PH₃ remains behind after 1 minute? (CBSE 2010)
Answer:

Question 25.
Explain the following terms:
(a) Order of a reaction
Answer:
The sum of the powers of the concentration of reactants in the rate law expression is called the order of the reaction.
For example, if rate law expression for a reaction is:
Rate = k[A]^a[B]^b[C]^c
then order of reaction is
order = a + b + c

(b) Molecularity of a reaction
Answer:
Molecularity is the number of molecules taking part in a reaction. For example, for the reaction:
aA + bB → Product
Molecularity is a + b

(ii) The rate of a reaction Increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature (R = 8.314 JK^-1 mol^-1). (CBSE 2010)
Answer:
T₁ = 300 k, T₂ = 320 k,
R = 8.314 Jk^-1 mol^-1

Question 26.
(i) A reaction is second order in A and first order in B.
(a) Write the differential rate equation.
(b) How is the rate affected on Increasing the concentration of A three times?
(C) How is the rate affected when the concentrations of both A and B are doubled?
Answer:
(a) Rate = k[A]²[B]
(b) 9 times
(C) 8 times

(ii) A first order reaction takes 40 minutes for 30% decomposition. Calculate t_1/2 for this reaction. (Given log 1.428 = 0.1548)
Answer:
Let initial conc. = a
Conc. after 40 min

OR

(i) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer:
For a first order reaction,
t = \(\frac{2.303}{k}\) log \(\frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]}\)
Let intial concentration, [A]₀ = a
For 99% completion of reaction,

Dividing equation (i) by (ii)
\(\frac{t_{(99 \%)}}{t_{(90 \%)}}\) = 2 or t_(99%) = 2 × t_(90%)

(ii) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
log k = log A – \(\frac{E_{a}}{2.303 R}\left(\frac{1}{T}\right)\)
where E_a is the activation energy. When a graph is plotted for log k vs. \(\frac{1}{\mathrm{~T}}\), a straight line with a slope
of – 4250 K is obtained. Calculate ‘E_a’ for the reaction (R = 8.3 14 JK^-1 mol^-1). (CBSE Delhi 2013)
Answer:
Slope = – \(\frac{E_{a}}{2.303 R}\) = – 4250 k
∴ E_a = 2.303 × 8.314 Jk^-1 mol^-1 × 4250 k
= 81.375 kJ mol^-1

Question 27.
(i) What is rate of reaction? Write two factors that affect the rate of reaction.
Answer:
Rate of reaction is defined as the change in concentration of reactants or products per unit time. Factors that affect rate of a reaction
(a) Concentration of reactant
(b) Temperature

(ii) The rate constant of a first order reaction increases from 4 × 10^-2 to 8 × 10^-2 when the temperature changes from 27 °C to 37°C. Calculate the energy of activation (E_a).
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
Answer:
log \(\frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right]\)
k₁ = 4 × 10^-2,
k₂ = 8 × 10^-2,
T₁ = 273 + 27 = 300 K
T₂ = 273 + 37 = 310 K,
R = 8.314 JK^-1 mol^-1

= 53598 J mol^-1
= 53.598 KJ mol^-1

OR

(i) For a reaction A + B → P, the rate is given by.
Rate = k [A] [B]²
(a) How is the rate of reaction affected if the concentration of B is doubled?
(b) What is the overall order of reaction if A is present in large excess?
Answer:
Rate = K[A][B]²
(a) becomes four times
(b) Second order

(ii) A first order reaction takes 23.1 minutes for 50% completion. Calculate the time required for 75% completion of this reaction.
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021) (CBSE 2015)
Answer:
For a first order reaction,