Yearly Archives: 2021

NCERT Exercises

Question 1.
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Solution.
Hydrogen has electronic configuration 1s¹. Its electronic configuration is similar to the outer electronic configuration (ns¹) of alkali metals, which belong to the first group of the periodic table.

Question 2.
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Solution.
Hydrogen has three isotopes :

The mass ratio of hydrogen isotopes is 1 : 2 : 3.

Question 3.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Solution.
In monoatomic form hydrogen atom has only one electron in K-shell (1s¹) while in diatomic form, the K-shell is complete (1s²). This means that in diatomic form, hydrogen (H₂) has acquired the configuration of nearest noble gas, helium. It is therefore, quite stable.

Question 4.
How can the production of dihydrogen, obtained from ‘coal gasification’, be increased?
Solution.
The process of producing ‘syngas’ from coal is called “coal gasification”.

The production of dihydrogen can be increased by reacting carbon monoxide of syngas mixtures with steam in the presence of iron chromate as catalyst.

This is called water-gas shift reaction.

Question 5.
Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?
Solution.
Dihydrogen is prepared by the electrolysis of water using platinum electrodes in the presence of a small amount of acid or alkali.

During electrolysis, the reactions that take place are :

Question 6.
Complete the following reactions.

Solution.

Question 7.
Discuss the consequences of high enthalpy of H — H bond in terms of chemical reactivity of dihydrogen.
Solution.
Dihydrogen is quite stable and dissociates into hydrogen atoms only when heated at about 2000 K.

Above 2000 K, the dissociation of hydrogen into atoms is only 0.081% which increases to 95.5% at 5000 K. Its bond dissociation energy is very high.

Due to its high bond dissociation energy it is not very reactive.

Question 8.
What do you understand by

1.  electron- deficient,
2.  electron-precise and
3.  electron- rich compounds of hydrogen?

Provide justification with suitable examples.
Solution.

1.  Electron-deficient hydrides : The hydrides of group 13 elements have an incomplete octet and hence are electron deficient molecules. They act as Lewis acids e.g., B₂H₆
2.  Electron-precise hydrides : These compounds have the required number of electrons to write their conventional Lewis structures. All elements of group 14 form such compounds (e.g., CH₄) which have tetrahedral structure.
3.  Electron rich compounds of hydrogen : These compounds have excess electrons which are present as lone pairs, elements of group 15-17 form such compounds, e.g., NH₃ has one lone pair, H₂O has two lone pairs and HF has three lone pairs.

Question 9.
What characteristics do you expect from an electron-deficient hydride with respect to its structure and chemical reactions?
Solution.
It is expected to be a Lewis acid. For example, diborane B₂H₆ (dimer of BH₃) forms complex with LiH which gives H^– ion (Lewis base).

Question 10.
Do you expect the carbon hydrides of the type (C_nH_{2n + 2}) to act as Lewis acid or base? Justify your answer.
Solution.
The type (C_nH_{2n + 2}) is not a Lewis acid or a Lewis base because carbon atoms have a complete octet, therefore, hydrides behave as normal covalent hydrides.

Question 11.
What do you understand by the term ‘non-stoichiometric hydrides’? Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer.
Solution.
Hydrides of d-block (Group No. 3, 4, 5 and 6) and f-block are metallic or interstitial hydrides because in these hydrides, hydrogen occupies some interstitial sites in the metal lattice producing distortion. These hydrides are also called non-stoichiometric hydrides as, the law of constant composition does not hold good for them, e.g., LaH_2.87, YbH_2.55, TiH_1.5-1.8, ZrH_1.3-1.75, vH_0.56, NiH_0.6-0.7, PdH_0.6-0.8 etc. Alkali metals do not form these types of hydrides because alkali metals donate lone pair of electrons to hydrogen which forms H^– ion. As H^– ion is formed by complete transfer of electrons so, alkali metal and H atom are in fixed stoichiometric ratio.

Question 12.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Solution.
Some metals like palladium (Pd), platinum (Pt) have a tendency to adsorb very large volume of hydrogen and therefore, can be used as its storage media. This property has high potential for hydrogen storage and as a source of energy.

Question 13.
How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.
Solution.
Atomic hydrogen torch : When molecular hydrogen is passed through tungsten electric arc ~ 3000°C, at lowr pressure, it dissociates to form atoms of hydrogen known as atomic hydrogen.

When the atoms of hydrogen produced as above are made to fall at some metal surface, they combine to form molecular hydrogen again along with the liberation of large amount of heat energy. The temperature rises to 4000-5000°C. This is the principle of atomic hydrogen torch which is used for welding purposes.
Oxy-hydrogen torch : When hydrogen is burnt in oxygen, the reaction is highly exothermic in nature. A temperature ranging between 2800°C to 4000°C is generated. This temperature can be employed for welding purpose in the form of oxy-hydrogen torch.

Question 14.
Among NH₃, H₂O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?
Solution.
The strength of H-bonding depends upon the magnitude of the polarity of the bond. Since H-F bond has maximum polarity (F is the most electronegative element) hydrogen bonding is maximum in HF molecules.

Question 15.
Saline hydrides are known to react with water violently producing fire. Can CO₂, a well-known fire extinguisher, be used in this case? Explain.
Solution.
When saline hydrides react with water, the reaction is highly exothermic, the evolved hydrogen catches fire, e.g.,

Question 16.
Arrange the following :
(i) CaH₂, BeH₂ and TiH₂ in order of increasing electrical conductance.
(ii) LiH, NaH and CsH in order of increasing ionic character.
(iii) H – H, D – D and F – F in order of increasing bond dissociation enthalpy.
(iv) NaH, MgH₂ and H₂O in order of increasing reducing property.
Solution.
(i) BeH₂ is significantly covalent, CaH₂ is ionic and TiH₂ is metallic hydride. Hence, increasing electrical conductance : BeH₂<CaH₂<TiH₂
(ii) Electronegativity decreases as Li > Na > Cs. Thus, increasing ionic character :
LiH < NaH < CsH.
(iii) Due to lone pairs of F, bond pairs experience repulsion, hence, F-F has low bond dissociation energy. In D-D, due to higher nuclear attraction bond dissociation energy is geater than H- H. Increasing bond dissociation enthalpy : F-F < H-H < D-D.
(iv) NaH is ionic hydride. MgH₂ and H₂O are covalent hydrides but OH bond in H₂O is more stronger. Hence, increasing reducing power :
H₂O < MgH₂ < NaH.

Question 17.
Compare the structures of H₂O and H₂O₂.
Solution.
In H₂O molecule, the oxygen is sp³-hybridised. Two half-filled sp³-orbitals form O – H σ-bonds, while the other two contain lone pairs of electrons. The expected ∠HOH is 109.5°, but the experimental value is 104.5°. This is because lp-lp repulsion > bp-bp repulsion.

On the other hand, H₂O₂ is a non-planar molecule. The dihedral angle between two planes is 111.5° and ∠OOH is 94.8°. (Gas phase)

Question 18.
What do you understand by the term ‘auto-protolysis’ of water? What is its significance?
Solution.
The reaction in which two water molecules react to give ions with proton transfer is called auto-protolysis of water or k self ionization of water. A proton from one H₂O molecule is transferred to another water molecule leaving behind OH^– ion and forming a H₃O⁺ ion.
H₂O_(l) + H₂O(l) ⇌ H₃O₊(aq) + OH^–(aq)
Significance : Auto-protolysis proves that water is amphoteric in nature.

Question 19.
Consider the reaction of water with F₂ and suggest, in terms of oxidation and reduction, which species are oxidised/reduced?
Solution.
2F_2(g) + 2H₂O_(l) ➝ 4H⁺(aq) + 4F ^–_(aq) + O_2(g)
In this reaction, H₂O is oxidised to oxygen by F₂. Fluorine has been reduced to F^– ion.

Question 20.
Complete the following chemical reactions.

Classify the above into
(a) hydrolysis,
(b) redox and
(c) hydration reactions.
Solution.

Question 21.
Describe the structure of the common form of ice.
Solution.
Water molecule in solid form is in a perfect tetrahedral shape when each oxygen atom is tetrahedrally (sp³ hybridization) surrounded by four hydrogen atoms, two of which are covalently joint with oxygen and rest of the two form hydrogen bonds with oxygen. This gives ice an open cage-like structure in which each oxygen atom is in contact with four hydrogen atoms and each hydrogen atom is attached to two oxygen atoms (covalently with one and through hydrogen bonding by the other).

Question 22.
What causes temporary and permanent hardness of water?
Solution.
Temporary hardness of water is due to the presence of magnesium and calcium hydrogen carbonates. It can be very easily removed by simply boiling the hard water for sometime. Permanent hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in water. It cannot be easily removed.

Question 23.
Discuss the principle and method of softening of hard water by synthetic ion-exchange resins.
Solution.
Hard water is made soft by a very common ion exchange resin method. In this the hard water is allowed to pass over a zeolite bed [zeolites are sodium aluminium silicates (Na₂Al₂Si₂O₈.xH₂O)] during which the Na⁺ ions from zeolite are replaced by Ca²⁺ and Mg²⁺ ions.
Na₂Z + Ca²⁺➝ CaZ + 2Na⁺
Na₂Z + Mg²⁺ ➝ MgZ + 2Na⁺
When whole of the Na⁺ ions of the zeolite have been exchanged, the zeolite is regenerated by treating it with strong (or saturated) solution of NaCl.
MZ + 2NaCl ➝ Na₂Z + MCl₂ (M = Mg, Ca)

Question 24.
Write chemical reactions to show the amphoteric nature of water.
Solution.
Water is amphoteric in nature because it can behaves both as an acid and base.

Question 25.
Write chemical reactions to justify that
hydrogen peroxide can function as an oxidising as well as reducing agent.
Solution.
H₂O₂ acts as oxidising and reducing agent in both acidic as well as in basic medium.

Question 26.
What is meant by ‘demineralised’ water and how can it be obtained?
Solution.

Question 27.
Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful?
Solution.
No, demineralised water is not always useful for drinking purposes. It is usually tasteless. Moreover, some ions such as Na⁺, K⁺ and Mg²⁺, etc. are essential to the body.
In order to make demineralised water more useful, proper amount of additional salts of sodium and potassium etc. must be dissolved in it.

Question 28.
Describe the usefulness of water in biosphere and biological systems.
Solution.
A major part of all living organisms is made up of water. It is a crucial compound for the survival of all forms of life. It is a solvent of great importance.
Liquid water is found in water bodies, such as ocean, sea, lake, river, stream, canal, pond, or puddle. The majority of water on the Earth is ocean water. Water is important in both chemical and physical weathering processes at the Earth’s surface. Water exists as vapours in the atmosphere.
All known forms of life depend on water. Water is vital both as a solvent in which many of the body’s solutes dissolve and as an essential part of many metabolic processes within the body.

Question 29.
What properties of water make it useful as a solvent? What types of compound can it
(i) dissolve, and
(ii) hydrolyse?
Solution.
Water is useful as a solvent rather excellent solvent due to the following properties :
(a) It is a liquid over a wide range of temperature (0° to 100°C).
(b) It has high enthalpv of vapourisation and heat capacitv.
(c) It is polar in nature and has a high dielectric constant (78.39).
(i) It can dissolve polar substances and also some organic compounds due to hydrogen bonding.
(ii) Water can hydrolyse oxides, halides, phosphides, nitrides, etc due to interionic attraction between them.

Question 30.
Knowing the properties of H₂O and D₂O, do you think that D2O can be used for drinking purposes?
Solution.
No, heavy water (D₂O) cannot be used for drinking purposes because it is injurious to health due to presence of D⁺ ions. Heavy water of high concentration retards the growth of plants and animals. Heavy water has germicide and bactericide properties.

Question 31.
What is the difference between the term ‘hydrolysis’ and ‘hydration’?
Solution.
When H⁺ and OH^– ions of H₂O interacts with anion and cation of the salt respectively of give the original acid and base then the reaction is called hydrolysis. It results in the change of pH of solution.
When H₂O is added to ion or molecule to give hydrated ion or compound then the reaction is called hydration.

Question 32.
How can saline hydrides remove traces of water from organic compounds?
Solution.
Saline hydrides such as NaH release H^– ions which act as strong Bronsted base whereas H₂O is a weak Bronsted acid. They combine with water to liberate hydrogen gas.
NaH + H₂O ➝NaOH + H₂
As a result, these hydrides can be used to remove traces of water from organic compounds.

Question 33.
What do you expect the nature of hydrides is, if formed by elements of atomic numbers 15, 19, 23 and 44 with dihydrogen? Compare their behaviour towards water.
Solution.
The elements with atomic numbers 15, 19, 23 and 44 are phosphorus (P), potassium (K), vanadium (V) and ruthenium (Ru) respectively. The nature of their respective hydrides is :
PH₃ (Molecular hydride), KH (Ionic), interstitial or metallic hydrides with vanadium (V). Ruthenium being a transition metal of group 8 does not form hydride. PH₃ acts as a weak base with water. KH reacts violenty with water to give KOH and H₂.

Question 34.
Do you expect different products in solution when aluminium(III) chloride and potassium chloride are treated separately with
(i) normal water
(ii) acidified water, and
(iii) alkaline water?
Write equations wherever necessary.
Solution.
Both the compounds are salts and they react differently with water.
Aluminium (III) chloride or A1C13 will react with water as follows :
AlCl₃ + 3H₂O ➝ Al(OH)₃ + 3HCl.
The reaction is known as hydrolysis.
(i) In normal water, both Al(OH)₃ and HCl will be present.

Question 35.
How does H₂O₂ behave as a bleaching agent?
Solution.
H₂O₂ acts as bleaching agent due to the release of nascent oxygen.
H₂O₂ ➝ H₂O + [O]
Thus, the bleaching action of hydrogen peroxide is permanent and is due to oxidation. It oxidises the colouring matter to a colourless product.

Question 36.
What do you understand by the terms:

1.  hydrogen economy
2.  hydrogenation
3.  syngas
4.  water-gas shift reaction
5.  fuel-cell ?

Solution.

1.  Hydrogen economy : The energy is transported and stored in the form of liquid or gaseous hydrogen is the principle due to which hydrogen is considered to be a possible source of clean energy. This proposal to use hydrogen as fuel is called hydrogen economy.
2.  Hydrogenation : The process of addition of hydrogen to unsaturated hydrocarbons is known as hydrogenation.
   
3.  Syngas : The mixture of CO and H₂ is used for the synthesis of methanol and a number of hydrocarbons, it is also called synthesis gas or syngas. Nowadays, syngas is produced from sewage, saw dust, scrap wood, newspapers, etc.
4.  Water-gas shift reaction: The production of dihydrogen can be increased by reacting CO of syngas mixtures with steam in the presence of iron chromate as catalyst.
   
   This is called water-gas shift reaction.
5.  Fuel cell: It is a commercial cell in which the chemical energy produced during the combustion of fuel is converted directly into electricity, e.g., H₂ – O₂ fuel cell. A fuel cell is used as a source of electrical energy in the space vehicles.

NCERT Exercises

Question 1.
Assign oxidation number to the underlined elements in each of the following species :
(a) NaH₂PO₄
(b) NaHSO₄
(c) H₄P₂O₇
(d) K₂MnO₄
(e) CaO₂
(f) NaBH₄
(g) H₂S₂O₇
(h) KAI(SO₄)₂.12H₂O
Solution.
Let the oxidation no. of underlined element in all the given compounds = x

Question 2.
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
(a) KI₃
(b) H₂S₄O₆
(c) Fe₃O₄
(d) CH₃CH₂OH
(e) CH₃COOH
Solution.
(a) In KI₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. In the structure, K⁺[I – I <— I]^–, a coordinate bond is formed between I2 molecule and I ion. The oxidation number of two iodine atoms forming the I2 molecule is zero while that of iodine ion forming the coordinate bond is -1. Thus, the O.N. of three iodine atoms in KI₃ are 0, 0 and -1 respectively.

Question 3.
Justify that the following reactions are redox reactions:
(a) CuO(s) + H_2(g) ➝ Cu₍₅₎ + H₂O(g)
(b) Fe₂O_3(s) + 3CO_(g) ➝ 2Fe_(s) + 3CO_2(g)
(c) 4BCI_3(g) + 3LiAIH_4(s) ➝ 2B₂H_6(g) + 3LiCI_(s) + 3AICI_3(s)
(d) 2K_(s) + F_2(g) ➝ 2K⁺F^–_(s)
(e) 4NH₃(g) + 5O_2(g) ➝ 4NO_(g) + 6H₂O_(g)
Solution.

Question 4.
Fluorine reacts with ice and results in the change:
H₂O_(s) + F_2(g) ➝ HF_(g) + HOF_(g)
Justify that this reaction is a redox reaction.
Solution.

Question 5.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇^2- and NO₃^–. Suggest structure of these compounds. Count for the fallacy.
Solution.

Question 6.
Write formulas for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin(IV) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate
(f) Chromium(III) oxide
Solution.
(a) HgCl₂
(b) NiSO₄
(c) SnO₂
(d) Tl₂SO₄
(e) Fe₂(SO₄)₃
(f) Cr₂O₃

Question 7.
Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
Solution.

Question 8.
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Solution.
The oxidation state of sulphur in sulphur dioxide is +4. It can be oxidised to +6 oxidation state or reduced to +2. Therefore, sulphur dioxide acts as a reducing agent as well as oxidising agent. Similarly, the oxidation state of oxygen in hydrogen peroxide is -1. It can be oxidised to O₂ (zero oxidation state) or reduced to H₂O or OH^– (-2 oxidation state) and therefore, acts as reducing as well as oxidising agents.
However, both ozone and nitric acid can only decrease their oxidation number and therefore, act only as oxidising agents.

Question 9.
Consider the reactions :
(a) 6CO₂(g) + 6H₂O_(l) ➝ C₆H₁₂O_6(ag) + 6O_2(g)
(b) O_3(g) + H₂O_2(l) ➝ H₂O_(l) + 2O_2(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Solution.

Question 10.
The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?
Solution.
In AgF₂ oxidation state of Ag is +2 which is very unstable. Therefore, it quickly, accepts an electron to form the more stable +1 oxidation state.
Ag²⁺ + e^– ➝ Ag⁺
Therefore, AgF₂, if formed, will act as a strong oxidising agent.

Question 11.
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
Solution.
(i) C is a reducing agent while O₂ is an oxidising agent. If excess of carbon is burnt in a limited supply of O₂, CO is formed in which oxidation state of C is +2 but when O₂ is in excess CO formed gets oxidised to CO₂ in which oxidation state of C is + 4.

Question 12.
How do you count for the following observations ?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added toaninorganicmixture containing chloride, we get colourless pungent smelling gas HCI, but if the mixture contains bromide then we get red vapour of bromine. Why?
Solution.
(a) In neutral medium, KMnO₄ acts as an oxidant as follows :
MnO₄^– + 2H₂O + 3e^– ➝ MnO₂ + 40H^–
In laboratory, alkaline KMnO₄ is used to oxidise toluene to benzoic acid.

Question 13.
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:
Solution.

Question 14.
Consider the reactions:

Why does the same reductant, thiosulphate react differently with iodine and bromine?
Solution.

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Solution.
The halogens (X₂) have strong electron accepting tendency and have positive standard oxidation potential values. They are therefore, powerful oxidising agents. The decreasing order of oxidising powers of halogens is :

Question 16.
Why does the following reaction occur ?

Solution.

Question 17.
Consider the reactions :

Solution.

Question 18.
Balance the following redox reactions by ion – electron method:
Solution.

Question 19.
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

Solution.

Question 20.
What sorts of informations can you draw from the following reaction?
(CN)_2(g) + 2OH^–_(aq) ➝ CN^–_(aq) + CNO^–_(aq) + H₂O_(l)
Solution.

Question 21.
The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂, and H⁺ ion. Write a balanced ionic equation for the reaction.
Solution.

Question 22.
Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.
Solution.
(a) F: Fluorine is the most electronegative element and shows only -1 oxidation state.
(b) Cs : Because of the presence of single electron in the valence shell, (alkali metals) Cs exhibits an oxidation state of +1 only.
(c) I: Because of the presence of seven electrons in the valence shell, I shows an oxidation state of-1 in compounds with more electropositive elements (such as H, Na, K, Ca, etc.) and oxidation states of +3, +5, +7 in compounds of I with more electronegative elements (such as, O, F, etc.)
(d) Ne: It is an inert gas (with high ionization enthalpy and highly positive electron gain enthalpy) and hence, it exhibits neither negative nor positive oxidation states.

Question 23.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Solution.

Question 24.
Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.
Solution.

Question 25.
In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen?
Solution.
The reaction involved in the manufacturing process is :

Question 26.
Using the standard electrode potentials, predict if the reaction between the following is feasible:

Solution.

Question 27.
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes.
(iii) A dilute solution of H₂SO₄ with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes.
Solution.

(1) An aqueous solution of AgNO₃ using platinum electrodes :     (P.I.S.A. Based)
Both AgNO₃ and water will ionise in aqueous solution

At cathode : Ag⁺ ions with less discharge potential are reduced in preference to H⁺ ions which will remain in solution. As a result, silver will be deposited at cathode.
Ag⁺ (aq) + e^– → Ag (deposited)
At anode : An equivalent amount of silver will be oxidised to Ag⁺ ions by releasing electrons.

As result of electrolysis, Ag from silver anode dissolves as Ag⁺(aq) ions while an equivalent amount of Ag⁺(aq) ions from the aqueous AgNO₃ solution get deposited on the cathode.

(2) An aqueous solution of AgNO₃ using platinum electrodes :

In the case, the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes.
At cathode : Ag⁺ ions will be reduced to Ag which will get deposited at the cathode.
At anode : Both NO^–₃ and OH^– ions will migrate. But OH^– ions with less discharge potential will be oxidised in preference to NO₃ ions which will remain in solution.
OH- (aq) → OH + e^–; 4OH → 2H₂O(l) + O₂ (g)
Thus, as a result of electrolysis, silver is deposited on the cathode while O₂ is evolved at the anode. The solution will be acidic due to the presence of HNO₃.

(3) A dilute solution of H₂SO₄ using platinum electrodes :
On passing current, both acid and water will ionise as follows :

At cathode :
H⁺ (aq) ions will migrate to the cathode and will be reduced to H₂.
H+ (aq) + e-→ H ; H + H→ H₂ (g)
Thus, H₂ (g) will evolve at cathode.
At anode : OH ions will be released in preference to SO^2-₄ ions because their discharge potential is less. They will be oxidised as follows :
OH^– (aq) → OH + e^– ; 4OH → 2H₂O(l) + O₂ (g)
Thus, O₂ (g) will be evolved at anode. The solution will be acidic and will contain H₂SO₄.

(4) An aqueous solution of CuCl₂ using platinum electrodes :

The electrolysis proceeds in the same manner as discussed in the case of AgNO₃ solution. Both CuCl₂ and H₂O will ionise as follows :

At cathode :  Cu²⁺ ions will be reduced in preference to H⁺ ions and copper will be deposited at cathode.
Cu²⁺ (aq) + 2e^– → Cu (deposited)
At anode : Cl^– ions will be discharged in preference to OH^– ions which will remain in solution.
Cl^–→Cl^–+ e^– ; Cl + Cl → Cl₂ (g) (evolved)
Thus, Cl₂ will evolve at anode.

Question 28.
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
Solution.

Since a metal with lower electrode potential is a stronger reducing agent, therefore, Mg can displace all the above metals from their aqueous solutions, Al can displace all metals except Mg from the aqueous solutions of their salts. Zn can displace all metals except Mg and Al from the aqueous solutions of their salts while Fe can displace only Cu from the aqueous solution of its salts. Thus, the order in which they can displace each other from the solution of their salts is Mg, Al, Zn, Fe, Cu.

Question 29.
Given the standard electrode potentials,

Solution.

Question 30.
Depict the galvanic cell in which the reaction,
Zn_(s) + 2Ag⁺_(aq) ➝ zn²⁺_(aq) + 2Ag_(s) takes place. Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of current in the cell and
(iii) individual reaction at each electrode.
Solution.
The given redox reaction is

NCERT Exercises

Question 1.
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The
volume of the container is suddenly increased.
(a) What is the initial effect of the change on vapour pressure?
(b) How do rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Solution.
(a) Vapour pressure decreases since equilibrium is disturbed and the rate of condensation falls considerably.
(b) Rate of evaporation increases while that of condensation falls initially.
(c) On restoration of the equilibrium again, the vapour pressure becomes the same.

Question 2.
What is K_c for the following equilibrium when the equilibrium concentration of each substance is : [SO₂] = 0.60 M, [O₂] = 0.82 M and [SO₂] = 1.90 M
2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
Solution.
2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
Applying law of chemical equilibrium,

Question 3.
At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms. I_2(g) ⇌ 2l_(g)Calculate Kp for the equilibrium.
Solution.

Question 4.
Write the expression for the equilibrium constant, K_c for each of the following reactions:

Solution.

Question 5.
Find out the value of K_cof each of the following equilibria from the value of Kp:

Solution.

Question 6.
For the following equilibrium, K_c= 6.3 x 10¹⁴ at 1000 K. NO_(g) + O_3(g) ⇌ NO_{2(g) + O2(g)} the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?
Solution.
NO_(g) + O_3(g) ⇌ NO_2(g) + O_2(g)

For the reverse reaction,

Question 7.
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.
Solution.
For the concentration of pure solid or pure liquid,


Since density of pure solid or liquid is constant at constant temperature and molar mass is also constant therefore, their molar concentrations are constant and are included in the equilibrium constant.

Question 8.
Reaction between N₂ and O₂ takes place as follows:
2N_2(g) + O_2(g) ⇌ 2N₂O_(g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 x 10^-37, determine the composition of equilibrium mixture.
Solution.

Question 9.
Nitric oxide reacts with Br2 and gives nitrosyl bromide as reaction given below:
2NO_(g) + Br_2(g) ⇌ 2NOBr_(g) When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.
Solution.

Question 10.
At 450 K, K_p = 2.0 x 10¹⁰.bar for the given reaction at equilibrium.
2SO_2(g)+ O_2(g) ⇌ 2SO_3(g)
What is K_c at this temperature?
Solution.

Question 11.
A sample of HI_(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI_(g) is 0.04 atm. What is K_p for the given equilibrium ?
2HI_(g)  H_2(g) + I_2(g)
Solution.

Question 12..
AA mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant K_c for the reaction given as follows :
N_2(g) + 3H_2(g)  2NH_3(g) is 1.7 × 10^-2.
Is this reaction at equilibrium ? If not, what is the direction of net reaction ?
Solution.

Question 13.
The equilibrium constant expression for a gas reaction is,

Write the balanced chemical equation corresponding to this expression.
Solution.
Balanced chemical equation for the reaction is
4 NO_(g) + 6 H₂O_(g)  4NH_3(g) + 5 O_2(g)

Question 14.
If 1 mole of H₂O and 1 mole of CO are taken in a 10 litre vessel and heated to 725 K, at equilibrium point 40 percent of water (by mass) reacts with carbon monoxide according to equation,
H₂O_(g) + CO_(g)  H_2(g) + CO_2(g)
Calculate the equilibrium constant for the reaction.
Solution.

Question 15.
At 700 K, the equilibrium constant for the reaction H_2(g) + I_2(g)  2HI_(g) is 54.8. If 0.5 mol L^-1 of HI_(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I_2(g) ? Assume that we initially started with HI_(g) and allowed it to reach equilibrium at 700 K.
Solution.

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ?
2ICl_(g)  I_2(g) + Cl_2(g) ; K_c = 0.14
Solution.

Question 17.
K_p = 0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4 atm pressure and allowed to come to equilibrium.
C₂H_6(g)  C₂H_4(g) + H_2(g)
Solution.

Question 18.
The ester, ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as :
CH₃COOH_l + C₂H₅OH_l  CH₃COOC₂H_5l + H₂O_l
(i) Write the concentration ratio (concentration quotient) Q for this reaction. Note that water is not in excess and is not a solvent in this reaction.
(ii) At 293 K, if one starts with 1.000 mol of acetic acid and 0.180 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.50 mol of ethanol and 1.000 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate
is found after some time. Has equilibrium been reached ?
Solution.

Question 19.
A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was reached, the concentration of PCl₅ was found to be 0.5 x 10^-1 mol L^-1. If K_c is 8.3 × 10^-3, what are the concentrations of PCl₃ and Cl₂ at the equilibrium ?

Solution.

Question 20.
One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂
FeO_(s) + CO_(g)  Fe_(s) + CO_2(g); K_p = 0.265 atm at 1050 K
What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial pressures are :Pco = 1.4 atm and Pco₂ = 0.80 atm ?
Solution.

Question 21.
Equilibrium constant Kc for the reaction, N_2(g) + 3H_2(g)  2NH_3(g) at 500 K is 0.061. At particular time, the analysis shows that the composition of the reaction mixture is : 3.0 mol L^-1 of N₂ , 2.0 mol L^-1 of H₂,0.50 mol L^-1 of NH₃. Is the reaction at equilibrium ? If not, in which direction does the reaction tend to proceed to reach the equilibrium ?
Solution.

Question 22.
Bromine monochloride (BrCl) decomposes into bromine and chlorine and reaches the equilibrium :
2BrCl_(g)  Br_2(g) + Cl_2(g)
For which K_c is 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3× 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium ?
Solution.

Question 23.
At 1127 K and 1 atmosphere pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% by mass.
C_(s) + CO_2(g)  2CO_(g)
Calculate K_c for the reaction at the above temperature.
Solution.

Question 24.

Solution.

Question 25.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume ?
(a) PCl_5(g)  PCl_3(g) + Cl₂
(b) CaO_(s) + CO_2(g)  CaCO_3(s)
(c) 3Fe_(s) + 4H₂O_(g)  Fe₃O_4(s) + 4H_2(g)
Solution.
Applying Le Chatelier’s principle, on decreasing the pressure, equilibrium shifts to the direction in which pressure increases, i.e., number of moles of gaseous substances is more. Thus, moles of reaction products will increase in reaction (a), decrease in reaction (b) and remain same (n_p, = n_r gaseous) in reaction (c).

Question 26.
Which of the following reactions will get affected by increase in pressure ? Also mention whether the change will cause the reaction to go in the forward or backward direction.

Solution.

Question 27.
The equilibrium constant for the following reaction is 1.6 x 10⁵ at 1024 K.
H_2(g) + Br_2(g)  2HBr_(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
Solution.

Question 28.
Dihydrogen gas is obtained from the natural gas by partial oxidation with steam as per following endothermic reaction :
CH_4(g) + H₂O_(g)  CO_(g) + 3H_2(g)
(a) Write the expression for K_p for the above reaction
(b) How will the value of K_p and composition of equilibrium mixture be affected by :
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst ?
Solution.

(b)

1. By Le Chatelier’s principle, on increasing pressure, equilibrium will shift in the backward direction where number of moles decreases.
2. As the given reaction is endothermic, by Le Chatelier ‘sprinciple, equilibrium will shift in the forward direction with increasing temperature.
3. Equilibrium composition will not be disturbed by the presence of catalyst but equilibrium will be attained quickly.

Question 29.
Describe the effect of :
(a) addition of H₂
(b) addition of CH₃OH
(c) removal of CO
(d) removal of CH₃OH on the equilibrium of the reaction:
2H_2(g) + CO_(g)  CH₃OH ?
Solution.
2H_2(g) + CO_(g)  CH₃OH Effect of
(a) addition of H₂ : The equilibrium will shift in the forward direction.
(b) addition of CH₃OH : The equilibrium will shift in the backward direction.
(c) removal of CO : The equilibrium will shift in the backward direction.
(d) removal of CH₃OH : The equilibrium will shift in the forward direction.

Question 30.
At 473 K, the equilibrium constant Kc for the decomposition of phosphorus pentachloride (PCl₅) is 8.3 × 10^-3. If decomposition proceeds as :
PCl_5(g)  PCl_3(g) + Cl_2(g) ; ∆H = + 124.0 kJ mol^-1
(a) Write an expression for K_c for the reaction.
(b) What is the value of K_c for the reverse reaction at the same temperature ?
What would be the effect on K_c if
(i) more PCl₅ is added
(ii) pressure is increased
(iii) the temperature is increased?
Solution.

Question 31.
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
CO_(g) + H₂O_(g)  CO_2(g) + H_2(g)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam so that pco = PH₂O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium ? K_p = 0.1 at 400°C.
Solution.

Question 32.
Predict which of the following will have appreciable concentration of reactants and products :
(a) Cl_2(g)  2Cl_(g) ; K_c = 5 ×10^-39
(b) Cl_2(g) + 2NO_(g)  2NOCl_(g) ; K_c = 3.7 × 10⁸
(c) Cl_2(g) + 2NO₂ (g)  2NO₂Cl_(g) ; K_c = 1.8.
Solution.
(a) Since the value of K_c is very small
∴ There will be appreciable concentration of the reactants.
(b) Since K_c is very large.
∴ There will be appreciable concentration of the products.
(c) Since K_c = 1.8, the concentration of reactants and products will be in comparable amounts.

Question 33.
The value of K_c for the reaction 3O_2(g)  2O_3(g) is 2.0 × 1 x 10^-50 at 25°C. If equilibrium concentration of O₂ in air at 25°C is 1.6 × 10^-2, what is the concentration of O₂ ?
Solution.

Question 34.
The reaction CO_(g) + 3H_2(g)  CH_4(g) + H₂O_(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.
Solution.

Question 35.
What is meant by conjugate acid-base pair ? Find the conjugate acid/base for the following species :
HNO₂,CN^–,HClO₄,OH^–,CO₃^2-,S^2-.
Solution.

Question 36.
Which of the following are Lewis Acids ?
H₂O, BF₃, H⁺ and NH₄⁺
Solution.
BF₃, H⁺ and NH₄⁺ are Lewis acids because they can accept a lone pair of electrons.

Question 37.
What will be the conjugate bases for the Bronsted acids ? HF, H₂SO₄ and  ?
Solution.

Question 38.
Write the conjugate acids for the following Bronsted acids.
, NH₃ and HCOO^–
Solution.

Question 39.
The species H₂O, HCO₃^–, HSO₄^– and NH₃ can act both as Bronsted acid and base. For each case, give the corresponding conjugate acid and base.
Solution.

Question 40.
Classify the following species into Lewis acids and Lewis bases and show how these can act as Lewis acid/Lewis base ?
(a) OH^–
(b) F^–
(c) H⁺
(d) BCl₃
Solution.
(a) OH^– : OH^– is a Lewis base because it can donate lone pair of electrons.
(b) F^– : F^– is a Lewis base because it can donate lone pair of electrons.
(c) H⁺ : H⁺is a Lewis acid because it can accept lone pair of electrons.
(d) BCl₃ : is a Lewis acid because it is electron deficient and can accept a lone pair of electrons.

Question 41.
The concentration of hydrogen ions in a sample of soft drink is 3.8 × 10^-3 M. What is its pH value ?
Solution.

Question 42.
The pH of soft drink is 3.76. Calculate the concentration of hydrogen ions in it.
Solution.

Question 43.
The ionisation constants of HF, HCOOH and HCN at 298 K are 6.8 x 10^-4, 1-8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionisation constant of the corresponding congugate bases.
Solution.

Question 44.
The ionisation constant of phenol is 1.0 x 10^-10. What is the concentration of phenate ion in 0.05 M solution of phenol and pH of solution ? What will be the degree of ionisation if the solution is also 0.01 M in sodium phenate ?
Solution.

Question 45.
The first dissociation constant H₂S is 9.1 × 10^-8. Calculate the concentration of HS^– ions in its 0.1 M solution and how much will this concentration be affected if the solution is 0.1 M in HCl also. If the second dissociation constant of H₂S is 1.2 × 10^-13, calculate the concentration of S^2- ions under both the conditions.
Solution.

Question 46.
The ionization constant of acetic acid is 1.74 × 10^-5. Calculate the degree of dissociation of acetic acid in 0.05 M solution. Calculate the concentration of acetate ions in solution and the pH of the solution.
Solution.

Question 47.
It has been found that the pH of 0.01 M solution of organic acid is 4.15. Calculate the concentration of the anion, ionization constant of the acid and the pK_a.
Solution.

Question 48.
Assuming complete dissociation, calculate the pH of the following solutions :
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH.
Solution.

Question 49.
Calculate the pH of the following solutions :
(a) 2 g of TIOH dissolved in water to give 2 litre of solution.
(b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(d) 1 mL of 13.6 M HCl diluted with water to give 1 litre of solution.
Solution.

Question 50.
The degree of ionization of 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and pK_a of bromoacetic acid.
Solution.

Question 51.
The pH of 0-005 M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionizationconstant and pK_b.
Solution.

Question 52.
What is the pH of 0.001 M aniline solution ? The ionization constant of aniline is 4.27 × 10^-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Solution.

Question 53.
Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74. How is degree of dissociation affected when the solution contains (a) 0.01 M HCl (b) 0.1 M HCl ?
Solution.

Question 54.
The ionization constant of dimethylamine is 5.4 × 10^-4. Calculate the degree of ionization of its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.01 M NaOH ?
Solution.

Question 55.
Calculate the hydrogen ion concentration in the following biological fluids whose pH values are as follows :
(a) Human muscles fluid 6.83
(b) Human stomach fluid 1.2.
(c) Human blood 7.38
(d) Human saliva 6.4
Solution.

Question 56.
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each case.
Solution.

Question 57.
0.561 g KOH is dissolved in water to give 200 mL of solution at 298K. Calculate the concentration of potassium, hydrogen and hydroxyl ions. What is the pH of the solution ?
Solution.

Question 58.
The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Solution.

Question 59.
The ionisation constant of propionic acid is 1.32 × 10^-5. Calculate the degree of ionisation of acid in its 0.05 M solution and also its pH. What will be its degree of ionisation if the solution is 0.01 M in HCl also ?
Solution.

Question 60.
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate ionisation constant of the acid and also its degree of dissociation in the solution.
Solution.

Question 61.
The ionisation constant of nitrous acid is 4.5 × 10^-4. Calculate the pH value of 0.04 M NaNO₂ solution and also its degree of hydrolysis.
Solution.

Question 62.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionisation constant of pyridine.
Solution.

Question 63.
Predict if the solutions of the following salts are neutral, acidic or basic :
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂, and KF
Solution.

Question 64.
The ionisation constant of chloroacetic acid is 1.35 × 10^-3. What will be the pH of 0.1 M acid solution and of its 0.1 M sodium salt solution ?
Solution.

Question 65.
The ionic product of water at 310 K is 2.7 × 10^-14. What is the pH value of neutral water at this temperature ?
Solution.

Question 66.
Calculate the pH of the resultant mixtures:
(a) 10 mL. of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HCl
(b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂
(c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH.
Solution.

Question 67.
Determine the solubilites of silver chromate, barium chromate and ferric hydroxide at 289 K from their solubility product constants. Determine also the molarities of the individual ions.

Solution.

Question 68.
The solubility product constants of Ag₂CrO₄ and AgBr are 1.1 × 10^-12 and 5.0 × 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Solution.

Question 69.
Equal volumes of 0.002 M solutions of sodium iodate and copper chlorate are mixed together. Will it lead to the precipitation of copper iodate ? (For copper iodate K_sp = 7.4 × 10^-8).
Solution.

Question 70.
The ionization constant of benzoic acid is 6.46 × 10^-5 and K_sp for silver benzoate is 2.5 × 10^-13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water ?
Solution.

Question 71.
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide ? (For iron sulphide, K_sp = 6.3 × 10^-18).
Solution.

Question 72.
What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K. For calcium sulphate K_sp = 9.1 × 10^-6?
Solution.

Question 73.
The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is 1.0 × 10^-19 M. If 10 mL of this solution is added to 5 mL of 0.04 M solution of FeSO₄, MnCl₂, ZnCl₂ and CaCl₂, in which solutions precipitation will take place ?
Given K_sp for
FeS = 6.3 × 10^-18,
MnS = 2.5 × 10^-13,
ZnS = 1.6 × 10^-24 and
CdS = 8.0 × 10^-27.
Solution.

NCERT Exercises

Question 1.
Choose the correct answer. A thermodynamic state function is a quantity
(a) used to determine heat changes
(b) whose value is independent of path
(c) used to determine pressure volume work
(d) whose value depends on temperature only.
Answer.
(b): State function is a property of the system whose value depends only upon the state of the system and is independent of the path or the manner by which the state is reached.

Question 2.
For the process to occur under adiabatic conditions, the correct condition is
(a) ΔT = 0
(b) Δp = 0
(c) g = 0
(d) w=0
Answer..
(c): Adiabatic system does not exchange heat with the surroundings.

Question 3.
The enthalpies of all elements in their standard states are
(a) unity
(b) zero
(c) <0
(d) different for each element.
Answer.
(b) : By convention, the standard enthalpy of formation of every element in its standard state is zero.

Question 4.
ΔU° of combustion of methane is – X kJ mol^-1. The value of ΔH° is
(a) = ΔU°
(b) >ΔU°
(c) <ΔU°
(d) =0
Answer.
(c) : CH_4(g)+2O₂ → CO_{2(g)+2H2Ol
Δn = 1 – 3 = -2. ‘
ΔH° = ΔU° + ΔnRT = -X- 2RT}

Question 5.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol^-1,-393.5 kJ mol^-1 and-285.8 kJ mol^-1 respectively. Enthalpy of formation of CH_4(g) will be
(a) -74.8 kJ mol^-1
(b) -52.27 kJ mol^-1
(c) +74.8 kJ mol^-1
(d) +52.26 kJ mol^-1
Solution.

Question 6.
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(a) possible at high temperature
(b) possible only at low temperature
(c) not possible at any temperature
(d) possible at any temperature
Solution.
(d): A + B → C + D + q, ΔS = +ve
Here, ΔH = -ve
ΔG = ΔH – TΔS
For reaction to be spontaneous, ΔG should be -ve. As ΔH = -ve and ΔS is +ve, ΔG will be -ve at any temperature.

Question 7.
In a process, 701 J of heat is absorbed by a system and 394 S of work is done by the system. What is the change in internal energy for the process?
Solution.
Heat absorbed by the system (q) = 701 J
Work done by the system (w) = -394 J
According to first law of thermodynamics,
ΔU = q + w = 701 + (-394) = 701 – 394 = 307 J

Question 8.
The reaction of cyanamide, NH₂CN_(g), with dioxygen was carried out in a bomb calorimeter, and AU was found to be -742.7 kJ mol 1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH₂CN_(g) +  → N₂+CO_2(g)+H₂O_l
Solution.

Question 9.
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.
Solution.
Mass of Al = 60 g
Rise in temperature, ΔT = 55 – 35 = 20°C
Molar heat capacity of Al = 24 J mo^-1 K^-1
Specific heat capacity of Al = 
∴ Energy required = m x c x ΔT
= 
= 1.068kJ or 1.07kJ

Question 10.
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. Δ_fusH = 6.03 kJ mol^-1 at 0°C.
C_p[H₂O_(l)] = 75.3 J mol^-1 K^-1,
C_p[H₂O_(s)] = 36.8 J mol^-1 K^-1
Solution.

Question 11.
Enthalpy of combustion of carbon to CO₂ is -393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
Solution.
C + O₂ → CO₂; ΔT = -393.5 kJ
∵ When 44 g of COz is formed from carbon and dioxygen gas, heat released = 393.5 kj
∴ When 35.2 g of CO₂ is formed from carbon and dioxygen gas, heat released
= 
Thus, ΔH = -314.8 kJ

Question 12.
Enthalpies of formation of CO_(g), CO_2(g), N₂O_(g) and N₂O_4(g) are -110, – 393, 81 and 9.7 kJ mol ^-1 respectively. Find the value of Δ_rf for the reaction
N₂O_4(g) + 3CO_(g) → N₂O_(g) + 3CO_2(g)
Solution.

Question 13.
Given: N_2(g) + 3H_2(g) → 2NH_3(g); Δ_rH° = -92.4 kJ mol^-1 What is the standard enthalpy of formation of NH₃ gas?
Solution.
N_2(g) + 3H_2(g) → 2NH_3(g) ; Δ_rH° = -92.4 kJ mol^-1
∴ Standard enthalpy of formation of NH_3(g)
= 

Question 14.
Calculate the standard enthalpy of formation of CH₃OH_l from the following data :

Solution.

Question 15.
Calculate the enthalpy change for the process
CCl_4(g) → C_(g) + 4Cl_(g)
and calculate bond enthalpy of C – Cl in CCl_4(g).
Δ_vapH°(CCl₄) = 30.5 kJ mol^-1,
Δ_fH°(CCl₄) = -135.5 kJ mol^-1,
Δ_aH°(C) = 715.0 kJ mol^-1, where Δ_aH° is enthalpy of atomisation Δ_aH°(Cl₂) = 242 kJ mol^-1
Solution.

Question 16.
For an isolated system, ΔU = 0, what will be ΔS?
Solution.
When energy factor has no role to play, for the process to be spontaneous ΔS must be +ve i.e., ΔS > 0.

Question 17.
For the reaction at 298 K, 2A + B → C,
ΔH = 400 kJ mol^-1 and ΔS = 0.2 kJ K^-1 mol^-1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.
Solution.
According to Gibbs Helmholtz equation, ΔG = ΔH – TΔS.

Question 18.
For the reaction, 2Cl_(g) → Cl_2(g), what are the signs of ΔH and ΔS?
Solution.
ΔH is negative because bond energy is released and ΔS is negative because there is less randomness among the molecules than among the atoms.

Question 19.
For the reaction
2A_(g) + B_(g) → 2D_(g), ΔU° = -10.5 kJ and ΔS° = – 44.1 JK^-1.
Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously.
Solution.
ΔH° = ΔU° + Δn_(g)RT

Question 20.
The equilibrium constant for a reaction is 10. What will be the value of ΔG° ?
R = 8.314J K^-1 mol^-1, T= 300 K.
Solution.
We know that ΔG° = – 2.303RT logK
= – 2.303 x 8.314 x 300 x log10
= – 5744.14 J/mol

Question 21.
Comment on the thermodynamic stability of NO_(g), given

Solution.
Since Δ_(r)H° is +ve, i.e., enthalpy of formation of NO is positive, therefore NO is unstable. But Δ_(r)H° is negative for the formation of NO₂. So, NO₂ is stable.

Question 22.
Calculate the entropy change in surroundings when 1.00 mol of H₂O_l is formed under standard conditions.
Δ_(r)H° = – 286 kJ mol^-1.
Solution.

NCERT Exercises

Question 1.
What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30°C?
Solution.
According to Boyle’s law, at constant temperature,

Question 2.
A vessel of 120 mL capacity contains a certain amount of gas of 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C.What would be its pressure?
Solution.
Since temperature and amount of gas remain constant, therefore, Boyle’s law is applicable.

Question 3.
Using the equation of state PV = nRT; show that at a given temperature density of a gas is proportional to gas pressure P.
Solution.
According to ideal gas equation for ‘n’ moles of a gas,

Question 4.
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Solution.

Question 5.
Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
Solution.

Question 6.
The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts?
Solution.
The reaction between aluminium and caustic soda is

Question 7.
What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm³ flask at 27°C?
Solution.

Question 8.
What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C?
Solution.
Partial pressure of hydrogen gas

Question 9.
Density of a gas is found to be 5.46 g/dm³ at 27°C and at 2 bar pressure. What will be its density at STP?
Solution.
We know that PV = nRT

Question 10.
34.05 mL of phosphorus vapours weigh 0.0625 g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?
Solution.
Molar mass of phosphorus, M

Question 11.
A student forgot to add the reaction mixture to the round bottomed flask at 27°C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?
Solution.
Suppose the number of moles of gas present at 27°C in flask of volume V at pressure P is n₁ then assuming ideal gas behaviour,

Question 12.
Calculate the temperature of 4.0 mol of a gas occupying 5 dm³ at 3.32 bar.
(R – 0.083 bar dm³ K^-1 mol^-1)
Solution.

Question 13.
Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Solution.
Number of moles of dinitrogen

Question 14.
How much time would it take to distribute one Avogadro number of wheat grains, if 10¹⁰ grains are distributed each second?
Solution.
Time taken to distribute 10¹⁰ grains = 1 sec.
Time taken to distribute 6.023 x 10²³ grains

Question 15.
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm³ at 27°C.
R = 0.083 bar dm³ K^-1 mol^-1
Solution.
Partial pressure of oxygen gas,

Question 16.
Payload is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air =1.2 kg m^-3 and R = 0.083 bar dm³ K^-1 mol^-1)
Solution.

Question 17.
Calculate the volume occupied by 8.8 g of CO₂ at 31.1°C and 1 bar pressure.
(R = 0.083 bar L K^-1 mol^-1)
Solution.
According to ideal gas equation, PV= nRT

Question 18.
2.9 g of a gas at 95°C occupied the same volume as 0.184 g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Solution.
Let molar mass of gas = M

Question 19.
A mixture of dihydrogen and dioxygen at one bar pressurecontains 20% by weight of dihydgrogen. Calculate the partial pressure of dihydrogen.
Solution.
Let the total mass of the mixture be 100 g

Question 20.
What would be the SI unit for the quantity pV²T²/n?
Solution.

Question 21.
In terms of Charles’ law explain why -273°C is the lowest possible temperature.
Solution.
According to Charles’ law,
Volume of gas at 1°C,
 (∵ V₀ = Volume at 0°C)
Volume of gas at -273°C, 
Thus, -273°C is the lowest possible temperature because below this temperature, the volume will become negative, and that is meaningless. This lowest temperature is called absolute zero temperature.

Question 22.
Critical temperature for carbon dioxide and methane are 31.1°C and -81.9°C respectively. Which of these has stronger intermolecular forces and why?
Solution.
Higher the critical temperature, more easily the gas can be liquefied i.e. greater are the intermolecular forces of attraction, CO₂ has stronger intermolecular forces than methane.

Question 23.
Explain the physical significance of van der Waals parameters.
Solution.
van der Waals constant ‘a’ is a measure of the magnitude of the attractive forces among the molecules of a gas, while constant ‘b’ is a measure of effective size of the gas molecules.