CBSE Class 11

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 9 Hydrogen. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 9 Important Extra Questions Hydrogen

Hydrogen Important Extra Questions Very Short Answer Type

Question 1.
Which gaseous compound on treatment with dihydrogen produces methanol?
Answer:
Carbon monoxide (CO).

Question 2.
What are the constituents of water gas?
Answer:
Carbon monoxide and hydrogen.

Question 3.
Arrange H₂, D₂, T₂ in the decreasing order of their
(i) Boiling point
Answer:
T₂ > D₂ > H₂

(ii) Heat of fusion.
Answer:
T₂ > D₂ > H₂.

Question 4.
Which isotope of hydrogen
(i) does not contain neutron
Answer:
Protium

(ii) is radioactive?
Answer:
Tritium.

Question 5.
Out of the following metals which can be used to liberate H2 gas on reaction with dil. hydrochloric acid?
(i) Cu,
(ii) Zn,
(iii) Iron,
(iv) Silver,
(v) Magnesium.
Answer:
Only Zn, Fe, Mg.

Question 6.
Name one compound each of hydrogen in which it exists in:
(i) Positive oxidation state
Answer:
HCl

(ii) Negative oxidation state.
Answer:
NaH.

Question 7.
What is the importance of heavy water in nuclear power generation?
Answer:
It is used as a moderator in nuclear reactions to slow down the speed of fast-moving neutrons.

Question 8.
State two properties in which hydrogen resembles alkali metals.
Answer:

1. Both form unipositive ion
2. Both have one electron in their s orbital (ns1).

Question 9.
Give an example of each anionic and covalent hydride.
Answer:
Ionic Hydride NaH Covalent hydride NH₃.

Question 10.
Why is H₂O₂ concentrated at low pressure?
Answer:
Because it decomposes at ordinary pressure or on heating.

Question 11.
What is the mass of 1 mole of deuterium oxide and tritium oxide?
Answer:
D₂O = 20g, T₂O = 22g.

Question 12.
Arrange H₂O, H₂S, H₂Se, H₂Te in the decreasing order of boiling point.
Answer:
H₂O > H₂S > H₂Se > H₂Te.

Question 13.
Give one example of a zeolite used in softening hard water.
Answer:
Sodium aluminium silicate Na₂Al₂Si₂Og. xH₂O.

Question 14.
Name the compounds which retard the decomposition of H₂O₂.
Answer:
Acetanilide, glycerol.

Question 15.
Which is heavier out of ice and water?
Answer:
Water.

Question 16.
What is the trade name of hydrogen peroxide used as an antiseptic?
Answer:
Perhydrol.

Question 17.
What is the significance of H₂O₂ labelled as 30 volumes’?
Answer:
“30 volume” labelled hydrogen peroxide means that 1 mL of this sample of solution gives 30 mL of oxygen gas at STP.

Question 18.
What happens when water is added to calcium carbide.
Answer:
CaC₂ + O → Ca(OH)₂ + C₂H₂
Acetylene gas is produced.

Question 19.
What is the cause of the temporary hardness of water?
Answer:
Presence of Ca(HCO₃)₂ and Mg(HCO₃)₂ in water.

Question 20.
How is the temporary hardness of water removed?
Answer:
By boiling and filtering.

Question 21.
How is heavy water produced from ordinary water?
Answer:
By repeated electrolysis of ordinary water containing 3 % of NaOH in it.

Question 22.
Which of the two dihydrogen (H₂) or deuterium (D₂) undergoes reactions more readily?
Answer:
Dihydrogen (H₂).

Question 23.
Define hard water.
Answer:
Hard water is one that does not produce lather with soap solution readily.

Question 24.
Which isotope of hydrogen contains an equal number of protons and neutrons?
Answer:
Deuterium.

Question 25.
What happens when ethylene reacts with hydrogen peroxide?
Answer:
H₂C = CH₂ + H₂O₂ → HO-CH₂-CH₂-OH
Ethylene glycol

Question 26.
Name the phenomenon of absorption of hydrogen by palladium?
Answer:
Occlusion.

Question 27.
H₂O₂ is a better oxidant than H₂O. Explain.
Answer:
H₂O₂ is easily reduced to form H₂O and O.
H₂O₂ → H₂O + O.

Question 28.
Name one example of a reaction in which hydrogen acts as an oxidizing agent.
Answer:
The reaction of hydrogen with active metals
2Na + H₂ → 2NaH .

Question 29.
Concentrated H₂SO₄ cannot be used to dry moist H₂ gas. Why?
Answer:
Cone. H₂SO₄ on absorbing H₂O from moist H₂ gas produces so much heat that H₂ gas catches fire.

Question 30.
Give an example where water acts as an oxidizing agent.
Answer:
2Na + 2H₂O → 2NaOH + H₂

Question 31.
Name the element which when reacted with dil. H₂SO₄ gives pure hydrogen.
Answer:
Magnesium.

Question 32.
Does hydrogen support combustion?
Answer:
No.

Question 33.
Why is dihydrogen not preferred in balloons these days?
Answer:
Dihydrogen is a highly combustible gas and hence is likely to catch fire in presence of excess air.

Question 34.
Why is sodium less soluble in heavy water than in ordinary water?
Answer:
Due to the lower dielectric constant of D₂O over H₂O, NaCl is less soluble in heavy water.

Question 35.
Although D₂O resembles H₂O chemically, yet it is toxic. Explain.
Answer:
D₂O is toxic since D⁺ ions react at a much slower rate than H⁺ in enzyme catalysed reactions.

Question 36.
Is it correct to say that hydrogen can behave as a metal? State the conditions under which such behaviour can be possible.
Answer:
Yes. H2 can act as a metal under very high pressures.

Question 37.
Give two advantages of using hydrogen as a fuel over gasoline.
Answer:
The high heat of combustion and production of no pollutants like SO₂, NO₂, CO₂, etc.

Question 38.
What happens when chloroform is treated with heavy water in presence of an alkali?
Answer:

Question 39.
What is hydrolith? How is it prepared?
Answer:
Hydrolith is a calcium hydride (CaH₂). It can be prepared by

Question 40.
Write the structures of two complex metal hydrides which are used as reducing agent in organic synthesis.
Answer:
Lithium aluminium hydride (LiAlH₄) and sodium borohydride (NaBH₄).

Question 41.
What type of elements form interstitial hydrides.
Answer:
d and f-block elements.

Question 42.
Explain why beryllium forms a covalent hydride while calcium forms an ionic hydride?
Answer:
Because of higher electronegativity (= 1.5). Be forms covalent hydride while due to lower electronegativity (= 1.0), calcium forms ionic hydride.

Question 43.
Write two uses of interstitial hydrides.
Answer:

1. For storing Hg and
2. Catalysts of hydrogenation reactions.

Question 44.
Explain why electrolysis of ordinary water occurs faster than heavy water.
Answer:
Due to lower bond dissociation energy of H-O-H bonds in water than D-O-D bonds in D₂O, electrolysis of H₂O is much faster than of D₂O.

Question 45.
Can marine species live in distilled water?
Answer:
No, because distilled H₂O does not contain dissolved oxygen.

Question 46.
Can distilled water be called deionised water?
Answer:
Yes. Distilled water does not contain any cations and anions.

Question 47.
Which isotope of hydrogen is used as a tracer in organic reactions?
Answer:
Out of the three isotopes of hydrogen H, D, T both D and T can be used. But due to the radioactive nature of T, it is the only D that is used as a tracer in understanding the mechanism of organic reactions.

Question 48.
Which salts present in water make it permanent hard?
Answer:
Calcium and magnesium chlorides and sulphates.

Question 49.
Name a process that can remove both temporary and permanent hardness of the water.
Answer:
Pemiutit process.

Question 50.
Complete the reaction:
Fe(s) + H₂O(g) →?
Answer:
3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

Hydrogen Important Extra Questions Short Answer Type

Question 1.
Hydrogen forms three types of bonds in its compounds. Describe each type of bonding using suitable examples.
Answer:
Hydrogen forms compounds in three different ways:
1. By loss of electrons as in the reactions of H₂ with CuO

2. By gain of electrons as in reactions of H₂ with metals.

3. By sharing of electrons as in the reactions of H₂ with halogens

Question 2.
Name one example of a reaction in which dihydrogen acts as
(i) an oxidizing agent
Answer:
As an oxidizing agent

Here Na has been oxidized to Na while dihydrogen has been reduced to H⁺ ion.

(ii) a reducing agent.
Answer:

Here CuO has been reduced to copper and H₂ has been oxidized to H₂O.

Question 3.
The process \(\frac{1}{2}\) H₂(g) + e^– → H^– (g) is endothermic (DH = +151 kJ mol^-1), yet salt- like hydrides are known. How do you account for this?
Answer:
This is due to the reason that high lattice energy released (energy released during the formation of solid metal hydride from their corresponding gaseous ions, i.e., M⁺ and H⁺) more than compensates the energy, needed for the formation of H^– ions from H₂ gas.

Question 4.
Find the volume strength of 1.6 N H₂O₂ solution.
Answer:
Strength = Normality × EQuestion wt.
Eq. wt.of H₂O₂ = 17
∴ Strength of 1.6N H₂O₂ solution = 1.6 × 17g L^-1

Now 68g of H₂O₂ gives 22400 mL O₂ at NTP/STP
∴ 1.6 × 17g of H₂O₂ will give = \(\frac{22400}{68}\) × 1.6 × 17
= 8960 mL of O₂ at STP
But 1.6 × 17g of H₂O₂ are present in 1000 mL of H₂O₂ solution

Hence 1000 mL of H₂O₂ solution gives 8960 mL of O₂ at STP 1 mL of H₂O₂ will give = 8.96 mL of O₂ at STP.

Hence the volume strength of 1.6N H₂O₂ solution is = 8.96 volume

Question 5.
A sample of hard water is allowed to pass through an anion exchanger. Will it produce lather with soap easily?
Answer:
No. Ca²⁺ and Mg²⁺ ions are still present and these will react with soap to form curdy white ppt. Therefore it will not produce lather with soap solution easily.

Question 6.
Anhydrous Ba02 is not used for preparing H₂O₂ Why?
Answer:
BaSO₄ formed during the reaction of BaO₂ with H₂SO₄ forms a protective layer around unreacted BaO₂ and the reaction stops after some time.

Question 7.
Find the volume strength of the 2N H₂O₂ solution.
Answer:
Volume strength = 5.6 × Normality
Volume strength = 5.6 × 2
= 11.2 volumes.

Question 8.
Calculate the concentration in g. L-1 of a 20 volume H₂O₂ solution.
Answer:
1 L of 20 volume H₂O₂ solution on decomposition gives 20L of O₂ at STP.

Now 22.4 L of O₂ at STP is obtained from 68g of H₂O₂

20L of O₂ at STP is obtained from \(\frac{68}{22.4}\) × 20 = 60.7g

Thus the strength of 20 volume H₂O₂ solution = 60.7g L^-1.

Question 9.
Explain why an oxide ion is called a hard ion?
Answer:
Oxide ion (O^2-) is very small in size and thus cannot be easily polarised and hence it is called a hard oxide ion.

Question 10.
Statues coated with white lead on long exposure to the atmosphere turn black and the original colour can be restored on treatment with H₂O₂. Why?
Answer:
On long exposure to the atmosphere, white lead is converted into black PbS due to the action of H₂S present in the atmosphere.
PbO₂ + 2H₂S → PbS + 2H₂O

On the treatment of such blackened statues with H₂O_2, the black PbS gets oxidised to PbSO₄ and the colour is restored.
PbS + 4H₂O₂ → PbSO₄ + 4H₂O.

Question 11.
A mixture of hydrazine and H₂O₂ with Cu(II) catalyst is used as a rocket propellant. Why?
Answer:
The reaction between hydrazine (N₂H₄) and H₂O₂ is highly exothermic and is accompanied by a large increase in the volumes of the products and hence the mixture is used as a rocket propellant.

Question 12.
Calculate the volume of 10 volume H₂O₂ solution that will react with 200 ml. of 2N KMnO₄4 in an acidic medium.
Answer:
Normality 10 volume H₂O₂ = \(\frac{10 \times 68}{22.4 \times 17}=\frac{10}{5.6}\)N J
Applying normality equation

Question 13.
What is water gas? How is it prepared?
Answer:
An equimolar mixture of CO and H is called water gas. It is prepared by passing steam over a red hot iron.

Question 14.
The boiling point of H₂O is higher than that of H₂S. Explain.
Answer:
Due to intermolecular H-bonding in H₂O₂ extra energy in the form of heat is required to break these H-bonds due to which H₂O boils at a higher temperature than H₂S which does not have H-bonding.

Question 15.
Write two uses of interstitial hydrides.
Answer:
Two important uses to which interstitial hydrides are put are:

1. storing H₂ gas
2. Catalysts for hydrogenation reactions.

Question 16.
What is meant by auto protolysis of water?
Answer:
Auto-protolysis of water means that two molecules of water react with each other through proton transfer, i.e. one acts as the acid and the other acts as a base. The molecule which accepts a proton is converted into H₃O⁺ while the other which loses a proton is converted into OH^– ion

Question 17.
What is the difference between hydrolysis and hydration?
Answer:
Interaction of H⁺ ions and OH^– ions of H₂O with anions and cations of the salt respectively to give an acidic or a basic solution is called hydrolysis. For example:

Hydration,.on the other hand, means the addition of water to ions or molecules to form hydrated ions or hydrated salts. For examples:

Question 18.
Complete the following equations:
(i) Fe(s) + H₂O(g) →?
Answer:
3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

(ii) PbS(s) + H₂O₂(aq)→?
Answer:
PbS(s) + 4H₂O₂(aq) → PbSO₄(s) + 4H₂O

(iii) MnO₄^–(aq) + H₂O(g) →?
Answer:
MnO₄-(aq) + 5H₂O₂(aq) + 6H+ → 2MnO⁺²+ 8H₂O + 5O₂

Question 19.
Discuss the importance of heavy water in a nuclear reactor.
Answer:
Heavy water is used as a moderator in nuclear reactors because it slows down the fast-moving neutrons & therefore helps in controlling the process of nuclear fission. It has also been used as a tracer compound to study the mechanism of many chemical reactions.

Question 20.
Distinguish clearly between:
(a) Hard & soft water.
Answer:

(b) Temporary hardness & permanent hardness of the water.
Answer:

Question 21.
Explain the correct context in which the following terms are used:
(a) diprotium
Answer:
Diprotium: It is used for the correct term for Hr

(b) dihydrogen
Answer:
Dihydrogen: The term dihydrogen is used for the H2 molecule while referring to the isotopic mixture with natural abundance for H & D.

(c) proton
Answer:
Proton: It is used for H⁺.

(d) hydrogen
Answer:
Hydrogen: It is used in relation to the isotope.

Question 22.
Hydrogen forms three types of bonds in its compound. Describe each type of bonding using suitable examples.
Answer:
(a) Ionic bond: The binary hydrides of alkali metals like (LiH, KH & NaH otc.) form an ionic bond. They on electrolysis give hydride ion (H^–)
NaH(s) + H₂O(aq) → H₂(g) + NaOH(aq)
Na⁺ + H^– → NaH

(b) Metallic bond: d & f block elements (Metals) form metallic bonds.
Example: SeH₂, YH₂ & LaH₃ etc.

(c) Covalent bond: With p-block elements hydrogen form a covalent bond. For example,

Question 23.
Why do lakes freeze from the top to the bottom?
Answer:
There are intermolecular hydrogen bonding in H₂O molecule. The density of water is greater than ice. It may be noted that at 4°C water has maximum density. In the severe cold, the upper layer of the seawater freezes & the heavier water (density more than that of ice) is present below the surface of the ice. Due to this sea animals can live safely in the water.

Question 24.
Why does elemental hydrogen react with other substances only slowly at room temperature?
Answer:
The high bond energy of di-hydrogen (436 kJ mol^-1) makes it a very stable molecule & therefore its reaction with other elements are slow at room temperature.

Question 25.
Why is water an excellent solvent for ionic or polar substances?
Answer:
Water molecules are highly polar. When water molecules interact with ions of an ionic compound, a large amount of hydration energy is released. The hydration energy is more than the energy needed to overcome the interionic attractions of the ionic compounds as well as the energy needed to break the hydrogen-bonded, association of water molecules. Therefore, the ionic compounds dissolve readily in water.

When water molecule interacts with strongly polar substances, the hydration energy released is sufficient to break the molecules of polar substances into ions as well as to break the hydrogen-bonded, association structure of water molecules. Therefore, water is an excellent solvent for polar compounds.

Question 26.
How is hydrogen obtained from:
(i) Nitric acid
Answer:
By action of Mg on cold water & dill. HNO₃
Mg + 2HNO₃(aq) → Mg(NO₃)₂ + H₂

(ii) Alcohol
Answer:
By action of Na or K on alcohol
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂

(iii) Ammonia?
Write a chemical equation in each case.
Answer:
By passing NH₃ over heated Na or K 2Na + 2NH₃ → 2NaNH₂ + H₂

Question 27.
Explain how is dil. solution of hydrogen peroxide concentrated?
Answer:
Hydrogen peroxide is concentrated by its distillation under reduced pressure at a temperature below 333K & the absence of heat metal impurities. 90% H₂O₂ is obtained by this method. For further concentration, the solution is cooled to about 263K when crystals enriched in hydrogen peroxide separate out. This process of fractional crystallisation is repeated to get 100% H₂O₂

Question 28.
Describe how is strength of hydrogen peroxide expressed?
Answer:
Strength of H₂O₂ as a percentage: H₂O₂ in solution (W/V), 70%. H₂O₂ means 70g of H₂O₂ is present in 100 g of solution.

Volume strength: Strength of hydrogen peroxide is expressed as the volume of oxygen liberated at S.T.P. given by ml. of a sample of hydrogen peroxide on decomposition into water & oxygen, for example, 30 volumes H₂O₂ means that 30 ml O₂ is obtained at S.T.P. by decomposing 1 ml of H₂O₂2 solution.

Question 29.
What happens when:
(i) Barium peroxide is treated with cold dilute H₂SO₄.
Answer:
H2O2 is obtained:
BaO₂. 8H₂O + H₂SO₄ → BaSO₄ ↑ + H₂O₂ + 8H₂O

(ii) Sodium peroxide is treated with cold dilute H₂SO₄ & the resulting mixture is cooled below 273 K.
Answer:
H₂O₂ is obtained:

(iii) Barium peroxide is treated with Phosphoric acid.
Answer:
H₂O₂ is obtained:
3BaO₂ + 2H₃PO₄ → Ba₃(PO₄)₂ + 3H₂O₂

Question 30.
Why is water an excellent solvent for ionic or polar substances?
Answer:
Water is a polar solvent with a high dielectric constant. Due to the high dielectric constant of water, the coelomic force of attraction between cations & anions gets weakened. Thus, a water molecule is able to remove ions from the lattice site using ion dipole forces easily. The dissolution of ionic/polar substances in water is further favoured by the hydrations of ions by the water molecule.

Hydrogen Important Extra Questions Long Answer Type

Question 1.
(a) Compare atomic hydrogen with nascent hydrogen.
Answer:
Comparison of atomic and nascent hydrogen

The main point of differences are:

1. Nascent hydrogen can be produced even at room temperature but atomic hydrogen is produced only at very high temperature.
2. Nascent hydrogen can never be isolated, but atomic hydrogen can be isolated.
3. The reducing power of atomic hydrogen is much greater than that of nascent hydrogen.

In general reactivity of the three forms of hydrogen increases in order.
Molecular hydrogen (H₂) < Nascent hydrogen < Atomic hydrogen.

(b) What is (i) active hydrogen
Answer:
Active Hydrogen: It is obtained by subjecting a stream of molecular hydrogen at ordinary temperature to silent electric discharge at about 30,000 volts. It is very reactive in nature (half-life = 0.33 second, and combines directly at ordinary temperatures with Pb and S forming their hydrides

(ii) heavy hydrogen? How are they formed?
Answer:
Heavy hydrogen: It is manufactured by the electrolysis of heavy water containing a little of IT SO, or NaOH to make the solution conducting.

In the laboratory, it can be prepared by the action of heavy water on sodium metal.
2D₂O(l) + 2Na(s) → 2NaOD(aq) + D₂(g).

Question 2.
How is the solution of H₂O₂ concentrated?
Answer:
The concentration of hydrogen peroxide: Hydrogen peroxide obtained by any method is always in the form of a dilute solution. Great care is to be taken for concentrating its solution because it is unstable and decomposes on heating.
2H₂O₂ → 2H₂O + O₂

The decomposition of H2O2 is catalysed by the ions of heavy metals present as impurities.

The solution of H2O2 is concentrated by the following methods.
1. By careful evaporation on a water bath: A dilute solution of H₂O₂ is taken in a shallow evaporating dish and is heated at 313K – 323 K. Water evaporates slowly and a hydrogen-peroxide solution of about 15 – 50% strength is obtained.

2. By dehydration in a vacuum desiccator: The dilute (50 %) solution of H₂O₂ obtained as above, is further concentrated by placing the same in a vacuum desiccator containing concentrated H₂SO₄ as a dehydrating agent. Here, water vapours are absorbed by concentrated sulphuric acid. This is shown in the diagram

(Concentration of H₂O₂ in vacuum desiccator)

3. By distillation under reduced pressure: The solution of hydrogen peroxide is further concentrated by subjecting it to distillation under reduced pressure. The solution is distilled at 308 – 313 K under a reduced pressure of 15 mm Hg. Water present in the solution distils over leaving behind about 98 – 99% concentrated solution of hydrogen peroxide.

4. By crystallisation: The last traces of water present in H₂O are removed by freezing it in a freezing mixture of solid CO₂ and others. The crystals of hydrogen peroxide separate out. These crystals are removed, dried and then remitted to obtain 100% pure hydrogen peroxide.

5. Storage of hydrogen peroxide: In order to check the decomposition of hydrogen peroxide, a small amount of acetanilide (i.e. negative catalyst) is added to it before storing the hydrogen peroxide.

Hydrogen peroxide cannot be concentrated by distillation at ordinary pressure because it undergoes decomposition into water and oxygen as it is a highly unstable liquid. It decomposes even on long-standing or on heating.

Question 3.
What are the different methods used for the softening of hard water? Explain the principle of each method.
Answer:
Hard water can be softened by the following methods depending upon the nature of hardness.
(a) Temporary hardness:
1. By boiling: It can be removed by merely boiling the water. Boiling decomposes the bicarbonates to give carbon dioxide and insoluble carbonates, which can be removed by filtration.

2. Clark’s process: Temporary hardness can be removed by the addition of a calculated amount of lime, whereupon magnesium and/or calcium carbonates is precipitated.
Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ + 2H₂O
Mg(HCO₃)₂ + Ca(OH)₂ → CaCO3₃ + MgCO₃ + 2H₂O

(b) Permanent hardness:
1. With sodium carbonate: On treatment with washing soda, Ca²⁺ and Mg²⁺ in hard water are precipitated. The precipitate of the insoluble carbonates thus formed is removed by filtration.

2. Ion-exchange method: The common substance used for this process is zeolite which is hydrated sodium aluminium silicate, NaAl(SiO)₂, The exchange occurs when passing over the zeolite bed, sodium ions from zeolite are replaced by calcium and magnesium ions. Thus
Na(Ze) + Mg²⁺ → (Ze)₂Ca + 2Na⁺
2NaZe + Mg²⁺ → (Ze)₂Mg + 2Na⁺

when all the sodium ions of the zeolite have been replaced, the zeolite is said to be exhausted. It can be regenerated by treatment with a strong solution of sodium chloride.
2Na + (Ze)₂Ca → 2ZeNa + Ca²⁺.

Question 4.
Show how hydrogen peroxide can function both as an oxidising and a reducing agent.
Answer:
Oxidising properties: H₂O₂ has a tendency to accept electrons in chemical reactions and thus behaves as an oxidising agent in both acidic and alkaline medium.
H₂O₂ → H₂O + O
In acidic medium
H₂O₂ + 2H⁺ + 2e^– → 2H₂O

In alkaline medium
H₂O₂ + OH^– + 2e^– → 3OH^–
Example:
(a) In acidic medium:
2Fe²⁺ + 2H⁺ + H₂O₂ → 2Fe³⁺ + 2H₂O
(b) In alkaline medium:
3Cr3+ + 4H₂O₂ + 100H^– → 3CrO₄^2- + 8H₂O

Reducing properties: H₂O₂ can give electrons in a few reactions and thus behaves as a reducing agent.
In acidic medium
H₂O₂ → O₂ + 2H+ + 2e^–

In alkaline medium
H₂O₂ + 2OH^– → 2H₂O + O₂ + 2e^–

Reducing property in acidic medium:
2MnO₄^2- + 6H+ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂

Reducing property in basic medium:
2Fe³⁺ + H₂O₂ + 2OH^– → 2Fe²⁺ + O₂ + 2H₂O

Hydrogen Important Extra Questions Numerical Problems

Question 1.
Calculate the percentage strength & strength in g/L of 10 volume hydrogen peroxide solution.
Answer:
H₂O₂ decomposes on heating according to the equation:

From the equation
22.4L of O₂ at N.T.P are obtained from 68g of H202

∴ 10 ml of O₂ at N.T.P will be obtained from \(\frac{68}{22400}\) × 10g of H₂O₂
But 10 ml of O₂ at N.T.P are produced from 1 ml. of 10 volume H₂O₂ solution.

Thus 1 ml of 10 volume H₂O₂ solution contains \(\frac{68}{22400}\) × 10 g of H₂O₂
∴ 100 ml. of 10 volume H₂O₂ solution will contain
\(\frac{68}{22400}\) × \(\frac{10}{1}\) × 100 = 3.036g .

Thus a 10 volume H₂O₂ solution is approx. 3%
Alternatively, 1000 ml of 10 volume of H₂O₂ will contain H₂O₂
\(\frac{68}{22400}\) × 10 × 1000 = 30.36g

Therefore, strength of H₂O₂ in 10 volume H₂O₂ is 30.36 g/L

Question 2.
Calculate the normality of 20 volume hydrogen peroxide solution.
Answer:
Step-I: To calculate the strength in g/L of 20 volume H₂O₂ solution.
By definition, 1L of 20 volume H₂O₂ solution on decomposition gives 20 litres of oxygen at N.T.P. consider the chemical equation.

Now 22.4 litres of 02 at N.T.P. will be obtained from H₂O₂
\(\frac{68}{22400}\) × 20g = 60.7g

Thus, the strength of 20 volume H₂O₂ solution is 60.7 g/l

Step-II: To calculate the equivalent wt. of H₂O₂ consider the chemical equation,

From the above equation, 32 parts by wt. of oxygen are obtained from 68 parts by wt. of H₂O₂
∴ 8 parts by wt. of oxygen will be obtained from
\(\frac{68}{32}\) × 8 = 17 parts by wt. of H₂O₂
∴ Equivalent wt. of H₂O₂ = 17

Step-III: To calculate the normality = \(\frac{\text { Strength }}{\text { Eq. wt. }}=\frac{60.7}{17}\) = 3.57
Hence normality of 20 volume H₂O₂ solution = 3.57N

Question 3.
Find the volume strength of the 1.6N H₂O₂ solution.
Answer:
We know that strength = Normality × Eq. wt. of H₂O₂

∴ Strength of 1.6N H₂O₂ solution = 1.6N × 17
Now 68g of H₂O₂ gives 22400 ml O₂ at N.T.P.

∴ 1.6 × 17g of H₂O₂ will give
\(\frac{22400}{68}\) × 1.6 × 17 = 8960 ml of O₂ at N.T.P.

But 1.67 × 17g of H₂O₂ are present in 1000 ml of H₂O₂ solution.
Hence 1000 ml of H₂O₂ solution gives 8960 ml of O₂ at N.T.P.

∴ 1 ml of H₂O₂ solution will give = \(\frac{8960}{1000}\)
= 8.96 ml of O₂ at N.T.P.
Hence the volume strength of 1.6N H₂O₂ solution = 8.96 volume.

Question 4.
Calculate the amount of H₂O₂ present in 10 ml of 25 volume H₂O₂ solution.
Answer:
ml. of 25 volume H₂O₂ liberate O₂
10 × 25 = 250 ml. at N.T.P

∴ Amount of H₂O₂ that will liberate 250 ml of O₂ at N.T.P.
= \(\frac{68 \times 250}{22400}\) = 0.759 g

Question 5.
10 ml of a given solution of H₂O₂ contains 0.91 g of H₂O₂ Express its strength in volume.
Answer:
68g of H₂O₂ produce O₂ = 22400 ml at N.T.P.

∴ 0.91g of H₂O₂ will produce O₂
\(\frac{22400 \times 0.91}{68}\) = 300 ml at N.T.P.

Volume strength = \(\frac{300}{10}\) = 30

Question 6.
Calculate the strength in volumes of a solution containing 30.36 g/litre of H₂O₂.
Answer:

68g of H₂O₂ Produces 22.4 L O₂ at N.T.P.

30.36g of K₂O₂ will produce \(\frac{22.4}{68}\) × 30.46
= 10 L O₂ at N.T.P.
The given solution of H₂O₂ produces 10 L of O₂ at N.T.P.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 8 Redox Reactions. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 8 Important Extra Questions Redox Reactions

Redox Reactions Important Extra Questions Very Short Answer Type

Question 1.
What are redox reactions? Give an example.
Answer:
Redox reaction is a reaction in which oxidation and reduction take place simultaneously, e.g.
Zn (s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Question 2.
Define oxidation and reduction in terms of electrons.
Answer:
Oxidation involves loss and reduction involves the gain of electrons.

Question 3.
Define an oxidizing agent. Name the best oxidizing agent.
Answer:
The oxidizing agent is a substance that can gain electrons easily. F₂ is the best oxidizing agent.

Question 4.
What is meant by reducing agent? Name the best reducing agent.
Answer:
The reducing agent is a substance that can lose electrons easily. Li is the best reducing agent.

Question 5.
In the reaction MnO₂ + 4HCl → MnCl₂ + H₂O which species is oxidized?
Answer:
HCl is oxidized to Cl₂

Question 6.
What is the oxidation state of Ni in Ni (CO)₄?
Answer:
Zero.

Question 7.
What is a redox couple?
Answer:
The redox couple consists of the oxidized and reduced form of the same substance taking part in an oxidation or reduction half-reaction, for example.
Zn²⁺(aq) / Zn, Cl₂ / Cl^–(aq) etc.

Question 8.
Define oxidation and reduction in the term of oxidation numbers.
Answer:
Oxidation involves an increase in oxidation number while reduction involves a decrease in oxidation number.
Sn²⁺ + 2Hg²⁺ → Sn⁴⁺ + Hg²⁺

Here Sn²⁺ gets oxidised while Hg²⁺ gets reduced.

Question 9.
What is the sum of oxidation numbers of all atoms in HIO₄4?
Answer:
Zero.

Question 10.
What is the oxidation number of N in (NH₄)₂ SO₂?
Answer:

The oxidation number of N is (NH₄)₂ SO₄ is – 3.

Question 11.
What is the oxidation number of O in
(i) OF₂
(ii) O₂F₂
Answer:
The oxidation number of oxygen in OF₂ is +2 whereas the oxidation number of oxygen in O₂F₂ is +1

Question 12.
At what concentration of M+(aq) will its electrode potential become equal to its standard electrode potential?
Answer:
At the concentration of IMol / L.

Question 13.
Select the oxidant in the following reaction.
H₂O₂ + O₃ → H₂O + 2O₂
Answer:
O₃ (zero to – 2)

Question 14.
What is the electrode potential of a standard hydrogen electrode?
Answer:
Zero.

Question 15.
Set up an electrochemical cell for the redox reaction: Ni²⁺(aq) + Fe(s) → Ni(s) + Fe²⁺(aq)
Answer:
Fe(s) / Fe²⁺(aq) || Ni²⁺(aq) / Ni(s)

Question 16.
What is the oxidation numbers of
(i) C in CH₂O
Answer:
Zero

(ii) pt in [pt(C₂H₄)Cl₃]
Answer:
2

Question 17.
What is the oxidation number of Mn in KMnO₄?
Answer:
The oxidation number of Mn in KMnO₄ is
KMnO₄ = 1 + x + 4 (- 2) = 0
x = + 7

Question 18.
What happens to the oxidation number of an element in oxidation?
Answer:
It increases.

Question 19.
Which reaction occurs at the cathode in a galvanic cell?
Answer:
Reduction.

Question 20.
Name one compound in which the oxidation number of Cl is +4.
Answer:
ClO₂.

Question 21.
If the reduction potential of an electrode is 1.28 v. What will be its oxidation potential?
Answer:
– 1.28 v.

Question 22.
Can we store copper sulfate in an iron vessel?
Answer:
No, because iron is more reactive than copper and thus holes will be developed in an iron vessel.
Cu²⁺ (aq) + Fe(s) → Fe²⁺(aq) + Cu(s)

Question 23.
The E° of Cu²⁺ / Cu is + 0.34 v. What does it signify?
Answer:
Copper lies below hydrogen in the activity series.

Question 24.
Calculate the oxidation number of underlined elements in the followings:
Na₂B₄O₇,O₃O₄
Answer:
Na₂B₄O₇ = 2 + 4x – 14 = 0
4x = 12
x = 3 .
O₃O₄ = x – 8 – 0
x = + 8

Question 25.
What is the oxidation number of alkali metals in its compounds?
Answer:
+1.

Question 26.
Determine the change in the oxidation number of S in H2S and S02 in the following reaction:
2H₂S(g) + SO₂(g) → 3S(s) + 2H₂S(g)
Answer:
The oxidation number of S changes from – 2 in H₂S and +4 in SO₂ to zero in elemental sulfur.

Question 27.
Can we use KCl as an electrolyte in the following cell?
Cu(s) | Cu²⁺(aq) || Ag⁺(aq)| Ag(s)
Answer:
KCl cannot be used as an electrolyte as a salt bridge because Cl^– ions will combine with Ag⁺ ions to form a precipitate of AgCl.

Question 28.
Define the EMF of the cell.
Answer:
EMF of the cell is defined as the difference in the electrode potential of the two half cells when the cell is not sending current through the circuit.

Question 29.
A solution of Na₂SO₄ was electrolyzed by using some inert molecules. What are the products at the electrodes?
Answer:
O₂ & H₂.

Question 30.
The reduction potentials are:
Cl₂ + 2e^– → 2Cl – E° = 1.36V
F₂ + 2e^– → 2F- E° = 2.87 V
Which is the better oxidising agent?
Answer:
F.

Question 31.
What is the relationship between standard oxidation potential and standard reduction potential?
Answer:
Both are equal in magnitude but opposite in sign.

Question 32.
Determine the oxidation number of nitrogen in HNO₃.
Answer:
HNO₃
1 + x +. 3(- 2) = 0
x = + 5

Question 33.
What is the oxidation number erf Fe in Na₄[Fe(CN)₆]?
Answer:

x – 1 × 6 = – 4 ,
x = +2

Question 34.
What is the maximum and minimum oxidation number of Nitrogen?
Answer:
The highest oxidation number of N is +5 and its minimum oxidation number is -3.

Question 35.
What is the basic principle of balancing the redox reactions by the ion-electron method?
Answer:
A number of electrons lost during oxidation = number of electrons gained during reduction.

Question 36.
What is the oxidation state of S in H₂SO₄ & H₂SO₃?
Answer:
H₂SO₄ = 2 ( + 1) + x + 4 (- 2) = 0
x = 8 – 2 = +6 ‘ ,
H₂SO₃ = 2 ( + 1) + x + 3 ( – 2) = 0
x = +4

Question 37.
What is the oxidation number of Xe in Ba₂XeO₂?
Answer:
Ba₂XeO₂ = 2 × 2 + x + 2( – 2) = 0
4 + x – 4 = 0
x = 0

Question 38.
Define half cell.
Answer:
The combination of an electrode and the solution in which it is dipped is called a half-cell.

Question 39.
Indicate the oxidizing and reducing agent in the following reaction:
2Cu²⁺ + 4I^– ⇌ 2CuI + I₂
Answer:
Cu²⁺ is an oxidizing agent and I^– is the reducing agent.

Question 40.
Write correctly the balanced half-reaction and the overall reaction for the following skeletal equation.
Fe(OH)₂ + H₂OZ ⇌ Fe(OH)₃ + H₂O (in basic medium)
Answer:
[Fe(OH)₂ + OH^– ⇌ Fe(OH)₃ + e^–] × 2
[H₂O₂ + 2e^– ⇌ 2OH^–]
2Fe(OH)₂ + H₂O₂ ⇌ 2Fe(OH)₃

Question 41.
Calculate the oxidation number of P in MgP₂O₇
Answer:
MgP₂O₇ = 2 × 2 + 2x + 7(-2) = 0
2x = 10
x = + 5

Question 42.
Arrange the following in order of increasing oxidation number of Mn:
MnCl₂, MnO₂, Mn(OH)₃, KMnO₄
Answer:

Question 43.
Arrange the followings in order of decreasing the oxidation number of oxygen:
HXO₄, HXO₃, HXO₂, HXO
Answer:

Question 44.
Arrange the following in order of increasing oxidation number of iodine:
I₂, HI, HIO₄, ICl
Answer:

Question 45.
How do you arrive at the oxidation number of Cr in Cr₂O₇^2-?
Answer:
(Cr₂O₇ )² = 2x – 14 = – 2
x = + 6

Question 46.
What is the charge on I mole of electrons?
Answer:
96500 coulombs.

Question 47.
Set up an electrochemical cell for the redox reaction.
Ni²⁺(aq) + Fe(s) ⇌ Ni(s) + Fe²⁺(aq)
Answer:
Fe(s) / Fe²⁺(aq) || Ni²⁺(aq) / Ni(s)

Question 48.
Define oxidation number?
Answer:
The oxidation number of an element may be defined as the charge which an atom of the element has in its ions. It is also known as the oxidation state.

Question 49.
Define electrode potential.
Answer:
The tendency, of an electrode to lose or gain electrons is called electrode potential.

Question 50.
Identify the strongest and weakest reducing agent from the metal.
Zn, Cu, Ag, Na, Sn
Answer:
Strongest reducing agent = Na Weakest reducing agent = Ag

Redox Reactions Important Extra Questions Short Answer Type

Question 1.
HNO3 acts only as an oxidant whereas HNOz acts both as an oxidant and reductant. Why?
Answer:
Ox. No. of N in HNO₃ = +5
Ox No. of N in HNO₂ = +3
Maximum oxidation numbers which N can show is = + 5
(∴ It has only 5 valance electrons 2S²2P³)

The Ox. No. of N in HNO₃ is maximum and it can only decrease. Therefore HNO₃ can act only as an oxidant. Minimum Ox. No. of N is -3.

Thus HNO₂ in which Ox. No. of N is +3 Can decrease as well as increase. Thus HNO₂ can act as an oxidant as well as a reductant.

Question 2.
Balance the following equation by the ion-electron method.
Zn(s) + NO₃^– → Zn²⁺(aq) + NH⁺ (aq) + H₂O(l) (In acid solution)
Answer:
Oxidation half reaction

Multiply (1) equation by 4 to equalise the no. of electron in both. Add both half reaction
4 Zn(s) → Zn²⁺(aq) + 8e^–
Zn(s) + NO₃^– (aq) + 10H⁺ (aq) → 4Zn²⁺ (aq) +NH⁺(aq)+ 3H₂2O(1)
– 1 + 10 = + 8 + 1
+ 9 = + 9

Question 3.
Balance the following equation in acidic medium by oxidation number method.

Answer:

Question 4.
Indicate the oxidising and reducing agent in the following reactions:
Answer:
(i) 2Mg + SO₂ → 2MgO + S
Mg = Reducing agent
SO₂ = Oxidising agent

(ii)2Cu²⁺ + 41 → 2CuI + I₂
Cu²⁺ = Oxidising agent
I- = Reducing agent

(iii)SO₂ + 2H₂S → 2H₂O + 3S
SO₂ = Oxidising ageiit
H₂S = Reducing agent

(iv)Sn²⁺ + 2Hg²⁺ → Hg₂²⁺ + Sn⁴⁺
Sn²⁺ = Reducing agent
Hg²⁺ = Oxidising agent

Question 5.
Which of the following redox reaction is oxidation & which is reduction?
Answer:
(i) Zn → Zn²⁺ + 2e^–
Oxidation
(ii) Cl2 + 2e^– → 2Cl-
Reduction
(iii) Fe → Fe²⁺ + 2e^–
Oxidation
(tv) Sn⁴⁺ + 2e^– → Sn²⁺
Reduction

Question 6.
What are the minimum and maximum oxidation numbers shown by sulfur?
Answer:
The minimum oxidation number shown by S is – 2 since it can acquire 2 more electrons to achieve the nearest inert gas [Ar] configuration.

The maximum Ox. No. shown by S is +6 since it has 6 valance electrons. (3S² 3P⁴)

Question 7.
Arrange the molecule NH₃, NO₂, HN₃, NO₂ & N₂H₄ in the decreasing order of the oxidation state of nitrogen.
Answer:

Question 8.
Can the reaction.
Cr₂O₇ ^2- + H₂O ⇌ 2CrO₄^2- + 2H⁺ be regarded as redox reaction.
Answer:
Ox. No. of Cr in Cr₂O₇^2- = + 6
Since the ox. no if Cr in CrO₄^2- = + 6

There is no change in ox. no. in the reaction therefore this reaction cannot be regarded as a redox reaction.

Question 9.
Balance the oxidation—reduction reaction
FeS₂ + O₂ → Fe₂O₃ + SO₂
Answer:
Fe²⁺ → Fe³⁺ 1 ↑
S₂^-1 → 2S⁺⁴ 10 ↑
O₂ → 2O^2- 4↓ × 11
4[Fe⁺² + S₂^-1] + 11O₂ → 4Fe³⁺ + 8S⁺⁴ + 22O^-2
or
3FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

Question 10.
Write the following redox reactions using half cell reactions:
(i) Zn(s) + FeCl₂(aq) → ZnCl₂(aq) + Fe(s)
Answer:
(Zn(s) → Zn²⁺(aq) + 2e^–
Fe²⁺(aq) + 2e^– → Fe(s)

(ii) Mg(s) + Cl2(g) → MgCl₂(s)
Answer:
Mg(s) → Mg²⁺ + 2e^–
Cl₂(g) + 2e^– → 2Cl^– (aq)

(iii) Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)
Answer:
Mg(s) → Mg²⁺ + 2e^–
2H⁺(aq) + 2e^– → H₂(g)

Question 11.
How would you know whether a redox reaction is taking place in an acidic/alkaline or neutral medium?
Answer:
If H⁺ or any acid appears on either side of the chemical equation, the reactions take place in the acidic medium. If OH^– or any base appears on either side of a chemical equation, the solution is basic. If neither H⁺; OH^– nor any acid or base is present in the chemical equation, the solution is neutral.

Question 12.
Is it possible to store:
(i) Copper sulphate solution in a zinc vessel.
Answer:
we cannot place CuSO₄ solution in a zinc vessel, if the following redox reaction occur.
Zn + CuSO₄ → ZnSO₄ + Cu
or
Zn + Cu²⁺ → Zn²⁺ + Cu

By convention, the cell may be represented by
Zn / Zn²⁺ || Cu²⁺ / Cu
K°_cell = E°_cu²⁺ cu- E°_zn²⁺
Zn = 0.34 -(- 0.76) = + 1.10v.
Since EMF comes out positive, therefore CuSO₄ reacts with zinc. So CuSO₄ solution cannot be stored in a zinc vessel.

(ii) Copper sulphate solution in silver vessel.
Answer:
We cannot store CuSO₄ solution in a silver vessel if the following reaction occurs.
2Ag + Cu²⁺ → 2Ag⁺ + Cu

By convention, the cell of the above redox reaction may be represented by
Ag / Ag⁺ || Cu²⁺ / Cu and E°cell = E°_cu²⁺ , cu^–, E°_Ag⁺
Ag = 0.34 – 0.80 = – 0.56 v.

Since the EMF of the cell is -ve, therefore CuSO₄ does not react with silver or CuSO₄ solution cannot be stored in a silver vessel.

Question 14.
The electrode potential of four metallic elements (A, B, C, D) are +0.80, -0.76, +0.12 and +0.34v. respectively. Arrange them in order of decreasing electron positive character.
Answer:
Higher the electrode potential, (E°) lower is the tendency of the metal to lose electrons .and hence lower is the electropositive character. So the electrode potential increase in the order of

Question 15.
What are the maximum and minimum oxidation numbers of N?
Answer:
The highest ox. no. of N is +5 since it has five electrons in the valence shell (2S2 2P3) and its minimum o.no. is -3 since it can accept three more electrons to acquire the nearest inert gas (Ne) configuration.

Question 16.
Find out the ox. no. of Cl in HCl, HClO, ClO;, CaOCl₂ & ClO₂.
Answer:
ox. no. of Cl = -1 in HCl, +1 in HClO, + 7 in ClO₄^– 0, in CaOCl₂ and + 4 in ClO₂.

Question 17.
What will happen if a nickel spatula is used to stir a solution of copper sulfate?
Answer:
Since E° of Ni (-0.25v) is lower than that of copper (+ 0.34v), therefore nickel has a higher tendency to lose electrons & copper has a higher tendency to gain electrons. Cu will deposit over the nickel spatula.

Question 18. Justify the reaction:
2Cu₂O(S) + Cu₂S (s) → 6Cu (s) + SO₂ (g) is a redox reaction. Identify the species oxidized, reduced, which acts as an oxidant and which acts as reductant.
Answer:

In this reaction copper is reduced from + 1 oxidation state to zero oxidation State and sulfur is oxidized from – 2 to +4 state. Therefore the reaction is a redox reaction further Cu₂O helps sulfur in Cu₂S itself and Cu₂O to decrease its oxidation number hence sulfur of Cu₂S is a reducing agent.

Question 19.
Write two information about the reaction-
Mg + H₂SO₄ → MgSO₄ + H₂
Answer:

1. Magnesium reacts with sulphuric acid to form magnesium sulfate and hydrogen.
2. 24 grams of magnesium reacts with 98 gm of sulphuric acid-producing (24 + 32 + 64) = 120 gm of magnesium sulfate and 2 gm of hydrogen.

Question 20.
Calculate the oxidation number of N in H – C = N
Answer:
As nitrogen is more electronegative than carbon, therefore, each covalent bond contributes one unit negative value to the oxidation number to nitrogen. Thus o. no. of nitrogen in HCN = -3

Question 21.
I₂ & Br₂ are added to a solution containing Br- and I- ions. What reaction will occur if,
I₂ + 2e^– → 21, E° = 0.54 v
& Br₂ + 2e^– → 2Br^–, E° = + 1.09 v?
Answer:
Since E° of Br₂ is higher than that of I₂, therefore Br₂ has a higher tendency to accept electrons than I₂. Conversely I- ions have a higher tendency to lose electrons than Br^– ions. Therefore, the following reaction will occur.

In other words, I^– ions will be oxidized to I₂ while Br₂ will be reduced to Br^– ions.

Question 22.
Is it possible to store copper sulfate solution in a gold vessel?
Answer:
We can store CuSO₄ solution in a gold vessel if the following redox reaction occur
2Au + 3Cu²⁺ → 2Au³⁺ + 3Cu
The cell corresponding to the above redox reaction may be represented as:
Au / Au³⁺ || Cu²⁺ / Cu and E°cell = E°_Cu²⁺ , Cu E°_Au³⁺
Au = 0.34 – 1.50 = 1.16 v

since the EMF of the above cell reaction is – ve, therefore, CuSO₄ solution does not react with gold in other words, the CuSO₄ solution can be stored in a gold vessel.

Question 23.
When magnesium ribbon burns in air two products are formed, magnesium oxide and magnesium nitride point out the oxidizing and reducing agent.
Answer:

Question 24.
What is the basic difference between electrochemical and electrolytic cells?
Answer:
In electrochemical cells electrical energy is generated as a result of the redox reaction which occurs in an electrolytic cell, the chemical reaction takes place as a result of electrical energy supplied.

Question 25.
Why are articles made of iron coated with zinc to check their rusting?
Answer:
Coating a layer of zinc on iron articles is called galvanization articles made up of iron are coated with zinc since zinc forms a protective coating on iron. When such articles are exposed to air. Zinc is oxidized to Zn²⁺ ions in preference to iron. Therefore zinc sacrifices itself for the sake of iron. The rusting of iron is prevented.

Question 26.
Identify the oxidizing agent, reducing agent, and the substance undergoing oxidation & reduction in the following reaction.
H₂SO₄ + 2HBr → SO₂ + Br₂ + 2H₂O
Answer:

Oxidizing agent – H₂SO₄
Reducing agent – HBr
Substance oxidized – HBr
Substance reduced – H₂SO₄

Question 27.
Consider the following galvanic cell.
Cd / Cd²⁺ (1M) || H⁺(1M) / H₂(g / atm) pt
(i) Write the overall cell reaction
Answer:
The anodic reaction is
Cd(s) → Cd²⁺(aq) + 2e^–

The Cathodic reaction is
2H⁺(aq) + 2e^– → H₂(g)

The overall reaction is
Cd(s) + 2H⁺(aq) → Cd²⁺(aq) + H₂(g)

(ii) What do the double vertical lines denote?
Answer:
The double vertical lines denote the salt bridge which connects the oxidation & reduction half cells.

Question 28.
Write the following redox reaction in the oxidation & reduction half-reaction.
(a) 2K(s) + Cl₂(g) → 2KCl(s)
Answer:
K(s) → K⁺(aq) + e^– (Oxidation)
Cl₂(g) + 2e^– → 2Cl^– (Reduction)

(b) 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)
Answer:
Al(s) → Al³⁺ (aq) + 3e^– (Oxidation)
Cu²⁺ + 2e^– → Cu(s) (Reduction)

Question 29.
Find out the oxidation number of iron in
[Fe(H₂O)₅ (NO)+] SO₄^—
Answer:
[Fe(H₂O)₅(NO)⁺]²⁺
x + 0 + 1 = 2
x = 2 – 1
= +1

Question 30.
Calculate the oxidation number of all the atoms in- CrO₄^2-.
Answer:
If x is the ox. no. of Cr in CrO₄^2- ion then
x + 4 × (- 2) x = – 2
x = 8 – 2 = 6
x = 6
For O → 6 + 4(x) = – 2
4x = – 8
x = \(\frac{-8}{4}\) = – 2

Redox Reactions Important Extra Questions Long Answer Type

Question 1.
Give the rules on the basis of which oxidation numbers are assigned to various elements.
Answer:
Various atoms are assigned oxidation number on the basis of the following rules:

1. An element in the free state has an oxidation number equal to zero, e.g. H₂, He, K, Ag all have zero ox. no.
2. In a binary compound of a metal and a non-metal, the oxidation number of metal is positive while that of non-metal is negative. In NaCl the ox. no. of sodium +1 and ox. n. of chlorine is  -1.
3. In a covalent compound, the atom with higher electronegativity has a negative oxidation number while another atom has a positive oxidation number.
4. The oxidation number of the radical or ions is equal to the electrical charge on it. for e.g. the ox. no. of Na+ is +1.
5. In neutral molecules, the algebraic sum of the oxidation number of all the atoms is zero.

Question 2.
Starting with the correctly balanced half-reaction, write the overall net ionic equation for the following change:
Chloride ion is oxidised to Cl₂ by MnO₄^– (in acid solution)
Answer:
The skeletal equation is
MnO₄^–+ H⁺ + Cl^–(aq) → Mn²⁺ + Cl₂(g) + H₂O (l)

Ox. no. of Mn change from +7 in MnO₄^– to +2
Whereas ox. no. of chlorine change from -1 in Cl^– ions to 0 in Cl₂.

Question 3.
Write the method used for balancing redox reaction by oxidation number method.
Answer:
The following steps are used for balancing the reactions by this methods:

1. Writing the skeletal equation for all the reactants and products of the reaction.
2. Assignment of the oxidation number of all atoms in each compound in the skeletal equation. Identify the atoms undergoing a change in their oxidation number.
3. Calculating the increase or decrease in oxidation number per atom and then for the whole molecule in which it occurs. If these are not equal then multiplying by suitable coefficients such that these become equal.
4. Now balancing the chemical reaction with respect to all atoms except H & O.
5. Finally balancing with respect to H & O atom for balancing oxygen atoms add H20 molecules to the side deficient in it.

Redox Reactions Important Extra Questions Numerical Problems

Question 1.
Determine the oxidation number of O in the following: OF2, Na₂O₂ & CH₃COOH
Answer:
(i) OF₂
Let the ox. no. of O = x
The ox. no. of each F = — 1
x – 2 = 0
x = +2

(ii) Na₂O₂
Let the o. no. of O = x
ox. no. of each Na = + 1
2 + 2x = 0
2x = – 2
x = – 1

(iii) CH₃COOH
Let the ox. no. of O = x
The ox. no. of each carbon atom = — 1
The ox. no. of hydrogen = +1
– 2 + 4 + 2x = 0
2x + 2 = 0
x = – 1

Question 2.
Determine the volume 6f M/8 KMnO₄ solution required to react completely with 25.0 cm3 of M/4 FeSO₄ solution in an acidic medium.
Answer:
The balanced ionic equation for the reaction is
MnO₄^– + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
from the balanced equation, it is evident that-
1 mole of KMnO₄ = 5 moles of FeSO₄

Applying the molarity equation to the balanced redox equation.

Thus the volume of M/8 KMnO₄ solution required = 10.0 ml.

Question 3.
How many grams of K₂Cr₂O₇ is required to oxidize Fe²⁺ present in 15.2 gm of FeSO₄ to Fe³⁺ if the reaction is carried out in an acidic medium.
Answer:
The balanced chemical equation for the redox reaction is
K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O
from the balanced equation, it is clear that 6 moles of FeSO₄ = 1

a mole of K₂Cr₂O₇ or 6 × 152 gm of FeSO₄ are oxidized by
K₂Cr₂O₇ = 294 gm
or
15.2 gm of FeSO₄ are oxidized by K₂Cr₂O₇

Question 4.
15.0 cm3 of 0.12 M KMnO₄ solution are required to oxidize 20 ml of FeSO₄ solution in an acidic medium. What is the concentration of FeSO₄ solution?
Answer:
The balanced chemical equation for the redox reaction is
2KMnO₄ + 10FeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8 H₂O

Applying molarity equation to the above redox reaction

Question 5.
16.6 gm of pure KI was dissolved in water and the solution was made up to one liter, cm³ of this solution was acidified with 20 cm³ of 2 MHCl the resulting solution required 10 cm³ of decinormal KIO₃ for complete oxidation of I- ions to ICl. Find out the value of v.
Answer:
The chemical equation for the redox reaction is
IO- + 2I^– + 6HCl → 3ICl + 3Cl^– + 3H₂O

Molarity of KI solution = \(\frac{16.6}{166}\) = 0.1 m

Applying molarity equation
\(\frac{0.1 \times v}{2}\) (KI) = \(\frac{10 \times 0.1}{1}\)(KIO3)
v = 20 cm³

Question 6.
Calculate the cone of hypo (Na₂S₂O₃ 5H₂O) solution in g dm-3 if 10.0 of this solution decolorized 15 ml of M/40 iodine solution.
Answer:
The balanced equation for the redox reaction is
2S₂O₃^2- + I₂ → 2I^– + S₄O₆^2-
from the balanced equation, it is evident that
2 moles of Na₂S₂O₃ = 1 mole of I₂

Applying the molarity equation we have,

Thus, the molarity of the hypo solution = 3/40 M
mol. mass of Na₂S₂O₃.5H₂O = 248 g mol^-1
cone, of Na₂S₂O₃.5H₂O = \(\frac{248 \times 3}{40}\)
= 18.6 gdm^-3

Question 7.
How many millimoles of potassium dichromate is required to oxidize 24 cm³ of 0.5 M mohr’s salt solution in an acidic medium.
Answer:
No. of millimoles of K₂Cr₂O₇ present in 24 cm³ of 0.5 m solution = 24 × 0.5 = 12
The balanced chemical equation for the redox reaction is
K₂Cr₂O₇ + 6(NH₄)₂SO₄.FeSO₄.6H₂O + 7H₂SO₄ → K₂SO₄ + 6(NH₄)₂SO₄ + 3Fe₂(SO₄)₃ + Cr₂(SO₄)₃ + 43H₂O from the balanced equations.

6 moles mohr’s salt are oxidised by K₂Cr₂O₇ = 1 moles
∴ 12 millimoles of mohr’s salt will be oxidised by
K₂Cr₂O₇ = \(\frac{1}{6}\) × 12 = 2 millimoies.

Question 8.
2.48 gm of Na₂S₂O₃.xH₂O was dissolved per liter of the solution. 20 cm³ of this solution required 10 cm³ of 0.01 M iodine solution. Find out the value of x?
Answer:
The balanced equation for the redox reaction is
2Na₂S₂O₃ + I2 → Na₂S₄O₆ + 2NaI
Let the molarity of Na₂S₂O₃ .xH₂O solution = M₁.

Applying molarity equation to the above redox reaction, we have
\(\frac{\mathrm{M}_{1} \times 20}{2}\)(Na2S2O3) = \(\frac{10 \times .01}{1}\)(I2)
∴ M₁ = 0.01 M

mol wt. of Na₂S₂O₃.xH₂O
= 2 × 23 + 2 × 32 + 3 × 16 + x × 18
= 158 + 18x

Amount of Na₂S₂O₃.xH₂O present per litre
= (158 + 18x) × 0.01 g

But the actual amount dissolved = 2.48 g. equating these values, we have
(158 + 18x) × 0.01 = 2.48
or
x = 5.

Question 9.
The half cell reactions with their oxidation potentials are:
Pb(s) → Pb²⁺(aq) + 2e^–, E°_oxi = + 0.13 v.
Ag(s) → Ag⁺(aq) + e^–, E°_oxi = – 0.80 v.
Write the cell reaction and calculate its EMF.
Answer:
The equations are as
Pb²⁺(aq) + 2e^– → Pb(s), E°_oxi = – 0.13 v
Ag⁺(aq) + e^– → Ag(s), (E° = + 0.80 v) …(2)

To obtain the equation for the cell reaction, multiply equation (2) by 2 and subtract equation (1) from it we get
Pb(s) + 2Ag⁺(aq) → Pb²⁺(aq) + 2Ag(s)
E°_cell = + 0.80 – ( – 0.13) = + 0.93 v

Question 10.
Predict whether zinc or silver reacts with 1 M H₂SO₄ to give out hydrogen or not. Given that the standard potential of zinc & silver are – 0.76 v & + 0.80 v respectively.
Answer:
(a) To predict the reaction of zinc with H₂SO₄ If Zn reacts, the following reactions should take place.
Zn + H₂SO₄ → ZnSO₄ + H₂
i.e. Zn + 2H⁺ → Zn²⁺ + H₂

By conventions, the cell will be represented as-
Zn / Zn²⁺ ∥ H⁺ / H₂

standard EMF of the cell,
E°cell = E°_H⁺/H2 – E°_zn²⁺/zn
= 0 – (- 0.76) = + 0.76 v

Thus the EMF of the cells comes out to be positive. Hence the reaction takes place.

(b) To predict the reaction, of silver with H₂SO₄.
If Ag reacts, the following reactions should take place.
2Ag + H₂SO₄ → Ag₂SO₄ + H₂
2Ag + 2H⁺ → 2Ag⁺ + H₂

By convention, the cell may be written as
Ag / Ag⁺ ∥ H⁺ / H₂
E°_cell = E°_H⁺,H2 – E°_Ag⁺,Ag
= 0 – 0.80 = – 0.80 v.
The EMF of the cell is negative
Hence, this reaction does not take place.

Question 11.
Calculate the oxidation number of Sb atoms in Sb₂O₅
Answer:
Sb₂O₅: Oxygen in common oxide has an oxidation state of -2. Therefore if x is the oxidation number of Sb in Sb₂O₅ then.
2 × x × 5 × (-2) = 0
x = \(\frac{10}{2}\) = 5
The ox. no. of Sb in Sb₂O₅ = +5

Question 12.
Calculate the ox. no. of sulphur is Na₂S₂O₃
Answer:
Ox. no. of various atom in Na₂S₂O₃
Na = +1, S = x, 0 = – 2
2( + l) + 2x + 3(- 2) = 0
2 + 2x – 6 = 0
2x = 4
x = + 2

Question 13.
Determine the oxidation number of O. in CH₃COOH.
Answer:
Let the ox. no. of O = x
The oxidation no. of each carbon atom = – 1
– 2 + 4 + 2x = 0
2x + 2 = 0
x = – 1

Question 14.
Determine the ox. no. of all the atoms in KClO₄.
Answer:
The ox.no. of K = +1
The ox. no. of Cl = x
The ox. no. of O = — 2
1 + x – 8 = 0
x – 7 = 0 .
x = 7
(K = +1, O = -2, Cl = 7)

Question 15.
Calculate the ox. no. of metal atom in Fe((CN)₆)^3-
Answer:
Fe((CN)₆)^3-
The ox. no. of CN- = – 1

Let the ox. no. of Fe be x
x + 6 (- 1) = – 3
x = + 3

Question 16.
Find out the ox. no. of iron in
[Fe(H₂O)₅(NO)⁺]SO₄
Answer:
[Fe(H₂O)₅(NO)⁺]²⁺
x + 0 + 1 = 2
x = 2 – 1 = +1

Question 17.
Find out the ox. no. of chlorine in HCl & HClO.
Answer:
(i) Cl in HCl is
+1 + x = 0
x = – 1

(ii) Cl in HCIO is
+ 1 + x – 2 = 0
x = +1

Question 18.
Find the ox. state of S in S₂O₄^2- or HSO₃^–
Answer:
(i) S₂O₄^2-
2x – 8 = – 2
2x = 6
x = +3

(ii) HSO₃^–
+ l + x – 6 = – 1
x – 5 = – 1
x = + 4

Question 19.
Calculate the emf of the cell.
M₁ = 1.00 M, M₂ = 0.40 M, PH₂ = 1.00 atm
Answer:
[Pb²⁺(aq)]
LOOM (H⁺ (aq)) = 0.40 M
emf 0.00 – (- 0.13) + \(\frac{0.059}{2}\) log \(\left(\frac{0.40}{1.00}\right)^{2}\)
= 0.13 – 0.0295 × 0.7959
= 0.13 – 0.023 = 0.107 v

Question 20.
Calculate the number of coulombs required to deposit 40.5g of aluminum when the electrode potential is
Answer:
The atomic mass of Al is 27.
The number of coulombs required to deposit 27g of Al is
3 × faradays
or
3 × 96500
Thus, the charge required to deposit 27 g of Al
= 3 × 96500 coulombs

Charge required to deposit 40.5 g of Al
= 3 × \(\frac{96500}{27}\) × 40.5 = 434250 c.

Question 21.
Calculate the ox. number of oxygen in OF₂
Answer:
OF₂
x – 2 = 0
x = +2

Question 22.
Calculate the ox. no. of Pb in Pb₃O₄
Answer:
3x + 4( -2) = 0
3x = 8; x = \(\frac{8}{3}\)

Question 23.
Calculate the ox. no. of S in H₂SO₄.
Answer:
H = +1, S = x, O = – 2
2 × 1 + x + 4 (- 2) = 0
2 + x – 8 = 0
x = + 6

Question 24.
Calculate the ox. no. of carbon atom in C₆H₁₂O₆
Answer:
If x is the oxidation number of c then,
6 × x + 12 × (+1) + 6 × (-2) = 0
6x = 0
x = 0

Question 25.
Calculate the ox. no. of Cr atom in CrO₄^2-.
Answer:
CrO₄^2- if x is the ox. no. of Cr in CrO₄^2- then
x + 4 × (-2) = – 2
x = 8 – 2 = 6
x = + 6

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 7 Equilibrium. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 7 Important Extra Questions Equilibrium

Equilibrium Important Extra Questions Very Short Answer Type

Question 1.
Write the expression for the equilibrium constant K_p for the reaction 3Fe (s) + 4H₂O (g) ⇌ Fe₃O₄ (s) + 4H₂ (g)
Answer:
K_p = \(\frac{p_{\mathrm{H}_{2}}^{4}}{p_{\mathrm{H}_{2} \mathrm{O}}^{4}}=\frac{p_{\mathrm{H}_{2}}}{p_{\mathrm{H}_{2} \mathrm{O}}}\)

Question 2.
How are K_c and K_p related to each other in the reaction
N₂ (g) + O₂ (g) ⇌ 2NO (g)
Answer:
K_p = K_c.

Question 3.
What is the equilibrium constant expression for the reaction
Al (s) + 3H⁺ (aq) ⇌ Al³⁺ (aq) + \(\frac{3}{2}\)H₂ (g)
Answer:
K_c = [Al³⁺ (aq)][H₂ (g)^3/2/[H+ (aq)]³.

Question 4.
What happens to the equilibrium
PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g) if nitrogen is added to it
(i) at constant volume
Answer:
The state of equilibrium remains unaffected.

(ii) at constant pressure?
Answer:
Dissociation increases, i.e., the equilibrium shifts forward.

Question 5.
What does the equilibrium K < I indicate? ,
Answer:
The reaction does not proceed much in the forward direction.

Question 6.
For an exothermic reaction, what happens to the equilibrium constant if the temperature is increased?
Answer:
K = K/K_b.
K_b increases much more than when the temperature is increased in an exothermic reaction. Hence K decreases.

Question 7.
Under what conditions, a reversible process becomes irreversible?
Answer:
If one of the products (gaseous) is allowed to escape out (i.e., in the open vessel).

Question 8.
What is the effect of increasing pressure on the equilibrium?
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)?
Answer:
Equilibrium will shift in the forward direction forming more ammonia.

Question 9.
What is- the effect of increasing pressure on the equilibrium.
CaCO₃ (s) ⇌ CaO (s) + CO₂ (g)
Answer:
K_c = [CaO (s)][CO₂ (g)J/[CaCO₃ (s)]
K_c = [CO₂ (g)]
Similarly, K_p = P_CO2.

Question 10.
The equilibrium constant for the reaction SO₃ (g) ⇌ SO₂ (g) + \(\frac{1}{2}\) O₂ (g) is 0.18 at 900 K. What will be the equilibrium
constant for the reaction SO₂ (g) + \(\frac{1}{2}\) O₂ (g) ⇌ SO₃ (g)?
Answer:
K = \(\frac{1}{0.18}\) = 5.55.

Question 11.
For which of the following cases does the reaction go farthest to completion: K = 1, K = 1010, K = 10-10.
Answer:
The reaction having K = 1010 will go farthest to completion because the ratio (product)/(reactants) is maximum in this case.

Question 12.
Under what conditions ice water system is in equilibrium?
(a) at 273 K
(b) below 273 K
(c) above 273 K.
Answer:
(a) At 273 K.

Question 13.
The equilibrium constant for the reaction
N₂ (g) + 3H₂ (g) ⇌ 2NH^3-(g) is K. How is the equilibrium constant for the reaction
NH₃ (g) ⇌ \(\frac{1}{2}\)N₂ (g) + \(\frac{3}{2}\) H₂ (g) related to K?
Answer:
If K’ is the equilibrium constant for the reaction
NH₃3(g) = \(\frac{1}{2}\)N2(g)+ \(\frac{3}{2}\)H₂(g)
Then, K’ = \(\frac{1}{K}\).

Question 14.
Write the relation between Kc and Kp for the reaction
PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)
Answer:
Δn (g) = 2 – 1 = 1
Hence K_p = K_c × (RT)^Δn = K × RT.

Question 15.
The equilibrium constant of a reaction is 2 × 10^-3 at 25°C and 2 × 10^-2 at 50°C. Is the reaction exothermic or endothermic? ,
Answer:
It is endothermic as the equilibrium constant has increased with temperature.

Question 16.
What is the thermodynamic criterion for the state of equilibrium?
Answer:
At equilibrium, (ΔG)_TP = 0.

Question 17.
Which measurable property becomes constant in water ⇌ water vapor equilibrium at constant temperature?
Answer:
Vapour pressure.

Question 18.
What happens to the dissociation of PCl₅ in a closed vessel if helium gas is introduced into it at the same temperature?
Answer:
No effect.

Question 19.
What are the conditions for getting NH3₃ by Haber’s process?
N₂ (g) + 3H₂ (g) 2NH₃ (g); ΔH = – Q.
Answer:
High concentration of N₂ and H₂, low temperature, and high pressure.

Question 20.
What happens if ferric salt is added to the equilibrium of the reaction between Fe³⁺ and SCN̅ ions?
Answer:
The red color deepens. Equilibrium shifts in the forward direction.

Question 21.
Does the value of equilibrium constant change on adding, a catalyst?
Answer:
No.

Question 22.
Name the factors which affect the equilibrium state.
Answer:
Temperature, pressure, and concentration.

Question 23.
What happens to the ionic product of water if some acid is added into the water?
Answer:
It remains unchanged.

Question 24.
What is the p^H of 0.1 M HCl?
Answer:
p^H = – log [H⁺] = – log 10^-1 = 1.

Question 25.
Which conjugate base is stronger: CN̅ or F̅?
Answer:
CN̅ is a stronger base than F̅

Question 26.
The dimethyl ammonium ion (CH₃)₂ NH₂⁺ is a weak acid. What is its conjugate base?
Answer:
(CH₃)₂NH.

Question 27.
Write the solubility product expression for:
Ag₂CrO₄ (s) ⇌ 2Ag+ (aq) + CrO₄^2- (aq)
Answer:
K_sp = [Ag+]²[CrO₄^2-].

Question 28.
When does salt get precipitated in solution?
Answer:
When the Ionic product of its ions exceeds its solubility product.

Question 29.
The K_sp value of salt is high. What does it indicate?
Answer:
This means that the salt is highly soluble in water.

Question 30.
Select the Lewis acid and Lewis base in
SnCl₄ + 2Cl^– → [SnCl6₆]^2-.
Answer:
SnCl₄ is acid and Cl is a base.

Question 31.
Out of pure water and 0.1. KCl solution in which AgCl will dissolve more?
Answer:
Pure water.

Question 32.
Write the conjugate acid and conjugate base of water?
Answer:
Conjugate acid H₃O⁺ and conjugate base OH̅.

Question 33.
Select Lewis acids and Lewis bases from the following: Cu²⁺, H₂O, BF₃, OH.
Answer:
Lewis acids: Cu²⁺, BF₃
Lewis base: H₂O, OH.

Question 34.
Give two examples of cations that can act as Lewis acids.
Answer:
Ag⁺, H⁺.

Question 35.
What is the difference between a conjugate acid and a conjugate base?
Answer:
A conjugate acid and its conjugate base differ from each other by a proton.

Question 36.
What is the active mass of water?
Answer:
55.5 mol L^-1

Question 37.
Whether the p^H value of an aqueous solution of sodium acetate will be 7 or greater than 7?
Answer:
It is greater than 7.

Question 38.
An old sample of an aqueous solution of CuSO₄ is acidic. Why?
Answer:
On keeping CuS04 solution undergoes hydrolysis to give H₂SO₄ a strong acid and a weak base Cu(OH)₂

Question 39.
What happens to the p^H of ammonium acetate solution when a few drops of acid are added to it?
Answer:
p^H will remain unchanged as ammonium acetate (CH₃COONH₄) solution is a buffer.

Question 40.
What happens when HCl gas is passed through NaCl solution?
Answer:
NaCl will precipitate out.

Question 41.
What is the value of H₃O⁺ ions and OH ions in water at 298 K?
Answer:
[H₃O⁺] = [OH̅] 1.0 × 10^-7 mL^-1.

Question 42.
Will the ionic products of water increase or decrease on increasing the temperature?
Answer:
It increases with the increase in temperature.

Question 43.
Give one example of
(i) acidic buffer
Answer:
Acidic buffer: CH₃COOH + CH₃COONa

(ii) basic buffer.
Answer:
Basic buffer: NH₄OH + NH₄Cl.

Question 44.
Write down the conjugate base of [Al(H₂O)₆]³⁺.
Answer:
[Al(0H)(H₂O)₅]²⁺ is the conjugate base of [Al(H₂O)₆]³⁺.

Question 45.
The K_sp of CuS, Ag₂S, HgS is 10^-31, 10^-44, 10^-54 respectively. What is the order of the solubility of these sulfides?
Answer:
CuS > Ag₂S > HgS.

Question 46.
K_sp for HgSO4 is 6.4 × 10^-5. What is the solubility of the salt?
Answer:
S = \(\sqrt{\mathrm{K}_{s p}}=\sqrt{6.4 \times 10^{-5}}\)= 8 × 10^-3 mol L^-1.

Question 47.
The solubility product (K ) of silver chloride is 1.8 × 10^-10 at 298 K. What is the solubility of AgO in 0.01 M HCl in mol L^-1?
Answer:
[Ag+] = \(\frac{\mathrm{K}_{s p}}{\left|\mathrm{Cl}^{-}\right|}=\frac{1.8 \times 10^{-10}}{0.01}\) = 1.8 × 10^-8 mol L^-1
∴ Solubility of AgCl = 1.8 × 10^-8 Mol L^-1.

Question 48.
How does dilution with water affect the p^H of a buffer solution?
Answer:
Dilution with water of a buffer solution has no effect on the p^H of a buffer solution.

Question 49.
What is the p^H of our blood? Why does it not change in spite of the variety of foods and Spices we eat?
Answer:
the p^H of our blood is about 7.4. It remains constant because blood is a buffer.

Question 50.
Why does boric acid act as Lewis acid?
Answer:
Boric a.cid acts as a Lewis acid by accepting electrons from hydroxyl ions.
B(OH)₃ + 2HOH → B(OH)^–₄ + H₃⁺O.

Question 51.
When a precipitate formed when solutions of BaCl₂ and Na₂SO₄ are mixed?
Answer:
When in the final solution after mixing, the ionic product.
[Ba²⁺] [SO₄^2- ] exceeds Ksp for BaSO₄.

Question 52.
What is the common ion effect?
Answer:
The suppression of the dissociation of a weak electrolyte by the addition of a strong electrolyte having a common ion.

Question 53.
Arrange them in increasing order of the extent of hydrolysis.
CCl₄, MgCl₂, AlCl₃ PCl₅, SiCl₄.
Answer:
The increasing order of the extent of hydrolysis is CCl₄ < MgCl₂ < AlCl₃ < PCl₅ < SiCl₄.

Question 54.
Mention two different ways of drawing the following equilibrium towards the right

[W.B. JEE 2003]
Answer:

1. By adding more of CH₃COOH or CH₃CH₂OH.
2. By removing the ester or water formed.

Question 55.
Give one example of everyday life in which there is gas ⇌ solution equilibrium.
Answer:
Soda-water bottle.

Question 56.
What is the relationship between p^K_a and p^K_b values where K_a, K_b represent ionization constants of the acid and its conjugate base respectively?
Answer:
p^K_a + p^K_b = p^K_w =14.

Question 57.
What is the relationship between p^H and p^OH?
Answer:
p^H + p^OH= p^K_w = 14.

Question 58.
What is the function of adding NH₄OH in group V?
Answer:
It converts any NH₄HCO₃ present into (NH₄)2CO₃.

Question 59.
What happens to the solubility of AgCl in water if NaCl solution is added to it?
Answer:
Solubility of AgCl decreases due to the common ion effect.

Question 60.
Write the expression for comparison of relative strengths of two weak acids in terms of their ionization constants.
Answer:
\(\frac{\text { Strength of acid }_{1}}{\text { Strength of acid }_{2}}=\sqrt{\frac{\mathrm{K}_{a_{1}}}{\mathrm{~K}_{a_{2}}}}\)

Equilibrium Important Extra Questions Short Answer Type

Question 1.
Justify the statement that water behaves like acid as well as a base on the basis of the protonic concept.
Answer:
Water ionizes as H₂O + H₂O ⇌ H₃O⁺ + OH^–
With strong acids, water behaves as a base by accepting a proton from an acid.
HCl + H₂O ⇌ H₃O⁺ (aq) + Cl^– (aq)
While with bases, water behaves as an acid by liberating a proton
NH₃ + H₂O ⇌ NH₄⁺ (aq) + OH (aq).

Question 2.
What is p^OH? What is its value for pure water at 298 K?
Answer:
p^OH = – log [OH^–]
p^H + p^OH = 14 for pure water at 298 K
p^H = 7
or
p^OH of water at 298 = 7.

Question 3.
Calculate the p^H of a buffer solution containing 0.1 moles of acetic acid and 0.15 mole of sodium acetate. The ionization constant for acetic acid is 1.75 × 10^-5.
Answer:

Question 4.
An aqueous solution of CuSO₄ is acidic while that of Na₂SO₄ is neutral. Explain.
Answer:
CuSO₄ + 2H₂O ⇌ Cu(OH)₂ + H₂SO₄ (weak base strong acid)
CuS04 is the salt of weak base Cu(OH)₂ and a strong acid H₂SO₄.
Thus the solution will have free H⁺ ions and will, therefore, be acidic.

Na₂SO₄, being the salt of a strong acid H₂SO₄ and a strong base.
NaOH does not undergo hydrolysis. The solution is, therefore, neutral.

Question 5.
The dissociation constants of HCN, CH₃COOH, and HF are 7.2 × 10^-10, 1.8 × 10^-5, and 6.7 × 10^-4 respectively. Arrange them in increasing order of acid strength.
Answer:
More the value of Ka, the stronger the acid
Their Ka1S are 6.7 × 10^-4 > 1.8 × 10^-5 > 7.2 × 10^-10
∴ HCN < CH₃COOH < HF.

Question 6.
The dissociation of PCl₅ decreases in presence of Cl₂. Why?
Answer:
For PCl₅ ⇌ PCl₃ + Cl₂.
According to Le Chatelier’s principle, an increase in the concentration of Cl₂ (one of the products) at equilibrium will favor the backward reaction, and thus the dissociation of PCl₅ into PCl₃ and Cl₂ decreases.

Question 7.
The dissociation of HI is independent of pressure while dissociation of PCl₅ depends upon the pressure applied. Why?•
Answer:
For 2HI ⇌ H₂ + I₂
K_c = \(\frac{x^{2}}{4(1-x)^{2}}\)
where x = degree of dissociation

For PCl₅ ⇌ PCl₃ + Cl₂
K = \(\frac{x^{2}}{V(1-x)^{2}}\); V = Volume of container.

K_c for HI does not have a volume factor- and dissociation is independent of volume and hence pressure.
K_c for PCl₅ has volume in the denominator and hence dissociation of PCl₅ depends upon the volume and consequently pressure.

Question 8.
The reaction between ethyl acetate and water attains a state of equilibrium in an open vessel, but not the decomposition of CaCO3. Explain.
Answer:
CH₂ COOC₂H₅ (l) + H₂O (l) ⇌ CH₃COOH (l) + C₂H₅OH (l)
Here both reactants and products are liquids and they will not escape from the vessel even if it is open. Therefore equilibrium is attained.
CaSO₃ (s) ⇌ CaO (s) + CO₂ (g)
Here CO₂ is a gas. It will escape from the vessel if it is open and so backward reaction cannot take place. Therefore equilibrium is attained.

Question 9.
The degree of dissociation of N₂O₄ is α according to the reaction
N₂O₄ (g) ⇌ 2NO₂ (g) at temperature T and total pressure P.
Find the expression for the equilibrium constant of this reaction.
Answer:

Total moles at eqbm. = 1 – α + 2α = 1 + α
If P is total pressure

Question 10.
Why NH₄Cl is added in precipitating III group hydroxides before the addition of NH4OH?
Answer:
To prevent precipitation of IV group hydroxides (especially Mn) along with III group hydroxides. NH₄Cl decreases dissociation of NH₄OH and thus limited OH ions are present in solution to precipitate III group cations only
NH₄OH ⇌ NH₄ (aq) + OH (aq) to a small extent
NH₄Cl (aq) ⇌ NH₄ (aq) + Cl (aq) to a large extent
Due to common ion affect the degree of dissociation of NH₄OH decreases leaving only a small no. of OH̅ ions.

Question 11.
If concentrations are expressed in moles L^-1 and pressures in atmospheres. What is the ratio of K_p to K_c for the
2SO₂ + O₂ (g) ⇌ 2SO₃ (g)
Answer:

Question 12.
The equilibrium constants for the reactions
N₂ + O₂ ⇌ 2NOand
2NO + O₂ ⇌ 2NO₂ are K₁ and K₂ respectively, then what would be the equilibrium constant for the reactions.
N₂ + 2O₂ ⇌ 2NO₂?
Answer:

Question 13.
What qualitative information can be obtained from the magnitude of the equilibrium constant?
Answer:

1. Large values of equilibrium constant (> 10³) show that the forward direction is favored i.e. concentration of products is much larger than that of the reactants of equilibrium.
2. Intermediate values of K (10^-3 to 10³) show that the concentrations of the reactants and products are comparable.
3. The low value of K (< 10^-3) shows that the backward reaction is favored, i.e., the concentration of reactants is much large than that of. products.

Question 14.
The following reaction has attained equilibrium
CO (g) + 2H₂ (g) ⇌ CH₃OH (g); ΔH° = – 92.0 kJ mol^-1.
What will happen if
(i) the volume of the vessel is suddenly reduced to half?
Answer:
K_c = [CH₃OH]/[CO][H₂]², K_p = PCH₃OH/PCO × PH₂
When the volume of the vessel is reduced to half, the concentration of each reactant or product becomes double. Thus
Q_c = 2[CH₃OH]/2[CO] × {2[H₂]}² = i K..
As Q_c < K_p, equilibrium will shift in the forward direction.

(ii) The partial pressure of hydrogen is suddenly doubled (ii) an inert gas is added to the system.
Answer:
As volume remains constant, molar concentration will not change. Hence there is no effect on the state of equilibrium.

Question 15.
How does the degree of ionization of a weak electrolyte vary with concentration? Give exact relationship. What is this law called?
Answer:
α = \(\sqrt{\mathrm{K}_{i} / c}\) . It is called Ostwald’s dilution law.
(K_i is ionization constant and c is the molar concentration).

Question 16.
The ionization constant for formic acid and acetic acid is 17.7 × 10^-5 and 1.77 × 10^-5. Which acid is stronger and how many times the other if equimolar concentrations of the two are taken?
Answer:
K_a for HCOOH > K_a for CH₃COOH. Hence formic acid is stronger.
Further \(\frac{\text { Strength of } \mathrm{HCOOH}}{\text { Strength of } \mathrm{CH}_{3} \mathrm{COOH}}=\sqrt{\frac{\mathrm{K}_{\mathrm{HCOOH}}}{\mathrm{K}_{\mathrm{CH}_{3} \mathrm{COOH}}}}\)
= \(\sqrt{10}\) = 3.16 times.

Question 17.
Out of CH₃COO̅ and OH̅ which is the stronger base and why?
Answer:
OH̅ ions can combine with H⁺ ions more readily than CH₃COO̅ ions can do. Hence OH̅ is a stronger base.

Question 18.
What is p^K_w? What is it? value at 25°C?
Answer:
p^K_w = – log ^K_w = – log 10^-14 = 14.

Question 19.
What are p^H and p^OH values of a neutral solution at a temperature at which = 10-13?
Answer:
p^K_w = p^H + p^OH, But p^K_w =13
Hence p^H = p^OH = 6.5.

Question 20.
The ionization constants of HF = 6.8 × 10^-4. Calculate the ionization constant of the corresponding conjugate base.
Answer:
K_b = \(\frac{\mathrm{K}_{w}}{\mathrm{~K}_{c}}=\frac{10^{-14}}{6.8 \times 10^{-4}}\) = 1.47 × 10^-11.

Question 21.
What is the difference between an ionic product a rich solubility product?
Answer:
Solubility product is the product of the molar concentrations of the ions in a saturated solution, but the ionic product is for any solution.

Question 22.
Why common salt is added to precipitate soap from the solution during its manufacture?
Answer:
Soap is the sodium salt of higher fatty acid [RCOONa].
On adding common salt, Na+ ion concentration increases.
Hence the equilibrium RCOONa (s) ⇌ RCOO̅ + Na⁺ shifts in the backward direction, i.e., soap precipitates out.

Question 23.
Through a solution containing Cu²⁺ and Ni²⁺, H₂S gas is passed after adding dil. HCl, which will precipitate out and why?
Answer:
Cu²⁺ ions will precipitate out because in the acidic medium only ionic product [Cu²⁺][S^2-] exceeds the solubility product of CuS.

Question 24.
Why in Group V of qualitative analysis sufficient NH₄OH solution should be added before adding (NH₄)₂CO₃ solution?
Answer:
This is done to convert NH₄HCO₃ usually present in large amounts to (NH₄)₂CO₃.
NH₄HCO₃ + NH₄OH → (NH₄)₂CO₃ + H₂O.

Question 25.
The p^H of an enzyme-catalyzed reaction has to be maintained between 7 and 8. What indicator should be used to monitor and control the p^H?
Answer:
Bromothymol blue or phenol red or cresol red.

Question 26.
The ionization constant of formic acid is 1.8 × 10^-4. Around what p^H will its mixture with sodium formate give buffer solution of highest capacity?
Answer:
Buffer solution of highest capacity is formed at which
p^H = P^K_a = – l0g 1.8 × 10^-4 = 3.74.

Question 27.
Why PO₄ ion is not amphiprotic?
Answer:
An amphiprotic ion is one that can donate proton as well as accept a proton. PO₄^3- ion can accept a proton(s) but cannot donate any proton. Hence, PO₄^3- is not amphiprotic.

Question 28.
In the reaction between BF₃ and C₂H₅OC₂H₅ which one of them will act as an acid. Justify your answer.
Answer:
The reaction between BF₃ and C₂H₅OC₂H₅ is

As BF₃ is electron-deficient and accepts a pair of electrons from C₂H₅OC₂H₅, hence BF₃ is the Lewis acid.

Question 29.
Will the water be the same at 4°C and 25°C? Explain. [IIT 2003]
Answer:
No. the p^H of water is not the same at 4°C and 25°C. This is because with an increase in temperature dissociation of H20 molecules increases. Hence concentration of [H⁺] ions will increase, i.e., p^H will decrease. Thus p^H of H₂O at 4°C will be more than at 25°C.

Question 30.
What type of salts are Na₂HPO₃ and NaHS? [W.B. /EE 2003]
Answer:
Na₂HPO₃ is obtained by the reaction between NaOH and H₃PO₃ a dibasic acid.

Both displaceable hydrogens are displaced by Na. No acidic hydrogen is left. Hence Na₂HPO₃ is a normal salt. NaHS is obtained by the replacement of one acidic hydrogen of H₂S by Na (on reaction with NaOH)’. Hence NaHS is an acidic salt.

Question 31.
Explain why the p^H of 0.1 molar solution of acetic acid will be higher than that of 0.1 molar solution of HCl.
Answer:
Acetic acid is a weak electrolyte. It is not completely ionized and hence gives less H⁺ ion concentration. HCl is a strong acid. It is completely ionized giving more H⁺ ions concentration. As p^H = – log [H⁺], the lesser the no. of H⁺ ions, the more the value of p^H. Therefore p^H of 0.1 M acetic acid is more than that of 0.1 M HCl.

Question 32.
Benzoic acid is a monobasic acid. When 1.22 of its pure sample are dissolved in water and titrated against base 50 mL of 0.2 M NaOH are used up. Calculate the molar mass of benzoic acid.
Answer:
1000 ml of 1.0 M NaOH will neutralize acid
= \(\frac{1.22}{50 \times 0.2}\) × 1000 = 122 g.

But 1000 ml of 1.0 M NaOH contains 1 mole of NaOH and will neutralize 1 mole of monobasic acid. Hence the molar mass of benzoic acid is 122 g mol^-1

Question 33.
Explain why ammonium chloride is acidic in liquid ammonia solvent.
Answer:
When NH₄Cl is present in liquid ammonia, the following reaction takes place:
NH₄ + NH₃ ⇌ NH₃ + NH₄
Thus NH₄Cl gives protons to liquid ammonia solvent.
Hence it is acidic.

Question 34.
Arrange the following in order of their increasing basicity.
H₂O, OH̅, CH₃OH, CH₃ O̅.
Answer:
H₂O < CH₃OH < OH̅ < CH₃ O̅.

Question 35.
The following can act both as Bronsted acid and Bronsted base. Write the formula in each case (of the product).
(i) HCO3-
Answer:
CO₃^2-, H₂CO₃

(ii) H₂PO4-
Answer:
HPO₄^–, H₃PO₄

(iii) NH₃
Answer:
NH̅₂, NH₄⁺

(iv) HS̅.
Answer:
S^2-, H₂S.

Question 36.
NaCl solution is added to a saturated solution of PbCl₂. What will happen to the concentration of Pb²⁺ ions?
Answer:
Pb²⁺ ion concentration will decrease to keep K constant.

Question 37.
Which is a stronger base in each of the following pairs and why?
(i) H₂O, Cl^–
Answer:
H₂O is a stronger base being the conjugate base of a weak acid H₃O⁺. Cl^– ion is the conjugate base of a strong acid and so is weak.

(ii) CH₃COO, OH.
Answer:
OH is the stronger base, being the conjugate base of H₂O a very weak acid.

Question 38.
For an aqueous solution of NH₄Cl, prove that [H₃O]⁺ = \(\sqrt{\mathbf{K}_{h} c}\) [CBSE PMT 2004]
Answer:
For salt of a strong acid weak base

Question 39.
What are the conjugate bases of the following?
CH₃OH, HN₃, [Al(H₂O)6]³⁺.
Answer:
CH₃O (methoxide ion), N₃ (azide ion), [Al(H₂O)S(OH)]²⁺.

Question 40.
Write reaction for autoprotolysis of water. How is the ionic product of water-related to ionization constant of water? Derive the relationship.
Answer:
Autoprotolysis of H₂O takes place as follows:

Question 41.
Glycine is an a-aminoacid that exists in the form of Zwitter ion as NH₃CH₂OO̅. Write the formula of its conjugate acid and conjugate base.
Answer:

Equilibrium Important Extra Questions Long Answer Type

Question 1.
Explain chemical equilibrium with th^ help of an example of formation and decomposition of hydrogen iodide.
Answer:
Consider the reaction between hydrogen and iodide at a constant temperature of 720 K in a closed vessel. The reaction involved is:
H₂(g) + I₂(g) → 2HI(g)

Accordingly, the effective collision amongst the reactant molecules will result in the production of HI. Since the product molecules are not permitted to leave the vessel (i.e., the reaction is carried out in a closed vessel), they will also collide amongst themselves leading to the formation of reactant molecules. Under these conditions, the reaction takes place in both directions. Hence, it is called a reversible reaction.

Graphical representation of the change of reaction rates with time for the formation and decomposition of hydrogen iodide

Forward reaction: H₂ (g) + I₂ (g) → 2HI (g)
Backward reaction: 2HI (g) → H₂ (g) + I₂ (g)
Reversible reaction: H₂ (g) + I₂ (g) ⇌ 2HI (g).

To begin with, with the concentration of the reactants being higher in comparison to the product molecules, the rate of the forward reaction will be high as compared to the backward reaction. As the reaction proceeds further, the molar concentration of the reactants will gradually decrease while that of the product will gradually increase.

Apparently, the rate of forwarding reaction goes on decreasing while that of the backward reaction. This state is the reversible chemical reaction is called a chemical equilibrium state.

Question 2.
Name and explain the factors which influence the equilibrium state.
Answer:
The various factors which influence the equilibrium state are:
1. Concentration: Concentration change influences the equilibrium state. If the concentration of the reactants is increased, the equilibrium will shift in such a direction in which more to the products are formed and vice-versa.
On the other hand, if the concentration of the products is increased, the equilibrium will shift in such a direction in which more of the reactants are formed.

2. Temperature: Like concentration, the temperature change also affects the equilibrium state. An increase in temperature of the system will shift the equilibrium in such a direction in which heat is absorbed (i.e. rate of endothermic reaction will increase).

On the other hand, a decrease in temperature of the system will shift the equilibrium in such a direction in which heat is evolved (i.e., rate of exothermic reaction will increase).

3. Pressure: Like concentration and temperature, the pressure also influences the equilibrium state only when the reaction proceeds with a change in volume. An increase in pressure of the system will shift the equilibrium in such a direction in which the volume of the system decreases.

On the other hand, a decrease in pressure of the system will shift the equilibrium in such a direction in which the volume of the system increases.

To explain the effect of temperature, pressure, and concentration on the equilibrium state, consider the combination of N₂ and H₂ to form NH₃
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g); ΔH = – 93.6 kJ

The reaction is reversible, exothermic, and accompanied by a decrease in volume.
Effect of temperature: According to Le-Chatelier’s principle, an increase in temperature shifts the equilibrium in the direction in which heat is absorbed, and a decrease in temperature shifts the equilibrium in the direction in which heat is evolved. Since the formation of ammonia is accompanied by the evolution of heat, it is favored by a decrease in temperature.

Effect of pressure: According to Le-Chatelier’s principle, an increase of pressure on a system in equilibrium, favors the direction which is accompanied by a decrease in volume and vice-versa. While going from, left to right in the above reaction, there is a decrease in the number of moles or say volume, the formation of ammonia is favored by an increase in pressure.

Effect of concentration: According to Le-Chatelier’s principle, an increase of concentration of any of the substances in the system shifts the equilibrium in the direction in which the concentration of that substance is reduced. Thus, the addition of N₂ or H₂ favors the formation of ammonia.

Question 3.
What is salt hydrolysis? Explain hydrolysis of salts of
(i) strong acids and strong bases
(ii) strong acids and weak bases
(iii) strong bases and weak acids
(iv) strong acids and weak bases.
Answer:
Salt hydrolysis: Hydrolysis is a process in which a salt reacts with water to form acid and base.
Salt + Water ⇌ Acid + Base
B A + H₂O ⇌ HA + BOH

That is the interaction of the cations of the salt with OH ions furnished by water and anions of the salt with H+ ions furnished by water to form an acidic or basic solution is called salt hydrolysis.
(i) Salts of strong acids and strong bases like NaCl, KCl, KNO₃ NaNO₃, Na₂SO₄, K2SO₄ do not undergo hydrolysis because the acids an.d bases furnished by them in aqueous solutions are strong acids and strong bases which are completely dissociated.
NaCl + H₂O ⇌ NaOH + HCl
NaOH (aq) ⇌ Na+ + OH^–
HCl (Aq) ⇌ H+ + Cl^–

Since [H⁺] = [OH^–] the resulting solution is neutral and its pH = 7.

(iii) Hydrolysis of salts of strong acids and weak bases:
The salts belonging to this type are NH₄NO₃, NH₄Cl, (NH₄)₂SO₄, CuSO₄, AlCl₃, Ca(NO₃)_2, etc.

Let us take the case of NH₄NO₃
NH₄NO₃ + H₂O ⇌ NH₄OH + HNO₃
NH₄⁺ + NO₃^– + H₂O ⇌ NH₄OH + HNO₃
or
NH₄⁺ + H₂O ⇌ NH₄OH + H⁺

The resulting solution after hydrolysis is basic (p^H > 7). Since only the anions of the salt have taken place in the hydrolysis, it is called anionic hydrolysis.

(iv) Hydrolysis of salts of weak acids and weak bases:
The salts belonging to this type are:
CH₃COONH₄, (NH₄)₂CO₃, Ca₃(PO₄)2 etc.

Let us take the case of hydrolysis of CH₃COONH₄
CH₃COONH₄ + H₃O ⇌ CH₃COOH + NH₄OH
or
CH₃COO̅ + NH₄ + H₂O ⇌ CH₃COOH + NH₄OH

Since both the cations and anions of the salt have participated in the hydrolysis, it is known as cationic as well as anionic hydrolysis. The nature of the solution or p^H depends upon the relative strengths of the acid and base that are formed on hydrolysis.

Equilibrium Important Extra Questions Numerical Problems

Question 1.
Calculate the pH of \(\frac{N}{1000}\) sodium hydroxide solution assuming complete ionisation (K_w = 1.0 × 10^-14).
Answer:
Since NaOH is completely ionized
∴ [NaOH] = [OH^–] = 10^-3 N = 10^-3 M

Now [H₂O⁺][OH^–] = K_w = 10^-14 [Given]
∴ [H₃O⁺] = \(\frac{\mathrm{K}_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{10^{-14}}{10^{-3}}\) = 10^-11
p^H = – log [H₃O⁺] = – log 10^-11 = 11.

Question 2.
Calculate the p^H of a 0.01 N solution of acetic acid. K_a for acetic acid is 1.8 × 10^-5 at 25°C.
Answer:

Applying the law of chemical equilibrium
K_a = [CH₃COO̅][H₃O⁺]/[CH₃COOH]

Question 3.
Calculate the p^H value of a solution of 0.1 M NH₃ (Kb = 1.8 × 10^-5). ,
Answer:
NH₃ H₂O ⇌ NH₄⁺ + OH^–

Question 4.
Equal volumes of solutions with p^H = 4 and p^H = 10 are mixed. Calculate the p^H of the resulting solution?
Answer:
p^H = 4
∴ [H3O⁺] = 10^-4 M

p^H = 10
∴ [H3O⁺] = 10^-10 M
or [OH^–] = 10^-4 M
Thus, they will exactly neutralize each other, and the p^H of the resulting solution will be = 7.

Question 5.
An aqueous solution of 0.02 g of NaOH in 50 mL has been prepared. What is its p^H and p^OH.
Answer:
50 mL of NaOH contains 0.02 g of it
1000 mL of it contains = \(\frac{0.02}{50}\) × 1000 = 0.4 g

Strength of NaOH = 0.4 gL-1

Question 6.
A reaction mixture containing N₂ at 0.50 atm, H₂ at 3.0 atm and NH₃ at 0.50 atm is heated to 450°C, in which direction the reaction N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g) will go if K_p = 4.28 × 10^-5?
Answer:
Concentration quotient Q = \(\frac{p^{2} \mathrm{NH}_{3}}{p_{\mathrm{N}_{2}} \times p_{\mathrm{H}_{2}}^{3}}=\frac{(0.50)^{2}}{0.5 \times(3.0)^{3}}\) = 0.018

K_p = 4.28 × 10^-5
As Q > > K_p reaction will go in the backward direction.

Question 7.
Under what pressure must an equimolar mixture of PCl₃ and Cl₂ be placed at 250°C in order to obtain PCl₅ at 1 atm > K_p for dissociation of PCl₅ = 1.78.
Answer:
Let partial pressure of PCl₃ at eqbm. = p atm
Then partial pressure of Cl₂ at eqbm. = p atm
Partial pressure of PCl₅ at eqbm. = 1 atm

Then for PCl₅ ⇌ PCl₃ + Cl₂


∴ Total initial pressure of PCl₃ and Cl₂ = 2.33 + 2.33 = 4.66 atm.

Question 8.
At 773 K, the equilibrium constant K_c for the reaction N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g) is 6.02 × 10^-2 L² mol^-2. Calculate the value of K_p at the same temperature.
Answer:

Question 9.
K_p for the reaction N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g) is 49 at a certain temperature. Calculate the value of K_p at the same
temperature for the reaction NH₃ ⇌ \(\frac{1}{2}\)N₂ + \(\frac{3}{2}\)H₂ (g)
Answer:
Given for the reaction N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g), K_p = 49
∴ for the reverse reaction 2NH₃ (g) ⇌ N₂ (g) + 3H₂ (g)

Question 10.
In the equilibrium CaCO₃ (s) ⇌ CaO (s) + CO₂ (g) at 1073 K, the pressure of CO₂ is formed to be 2.5 × 10⁴ Pa. What is the equilibrium constant for this reaction at 1073 K.
Answer:
With reference to the standard state pressure of 1 bar which is = 10⁵ Pa

Question 11.
Determine the concentration of CO₂ which will be in equilibrium with 2.5 × 10^-2, mol L^-1 of CO at 100°C for the reaction FeO (s) + CO (g) ⇌ Fe (s) + CO₂ (g); Kc = 5.0.
Answer:
K_c = \(\frac{\left[\mathrm{CO}_{2}\right]}{[\mathrm{CO}]}\)
i.e., 5 = \(\frac{\left[\mathrm{CO}_{2}\right]}{2.5 \times 10^{-2}}\)
or
[CO₂] = 5 × 2.5 × 10^-2 = 12.5 × 10^-2 mol L^-1

Question 12.
In a reaction between hydrogen and iodine, 6.34 moles of hydrogen and 4.02 moles of iodine are found to be in equilibrium with 42.85 moles of hydrogen iodide at 350°C. Calculate the equilibrium constant.
Answer:

Question 13.
Prove that the pressure necessary to obtain 50% dissociation of PCl₅ at 500 K is numerically equal to the three times the value of the equilibrium constant, K_p.
Answer:

Total no. of moles = 1.5
If P is the total required pressure, then

Question 14.
The equilibrium constant for the reaction A₂ + B₂ ⇌ 2AB is K_p. What will be the equilibrium constant for the reaction AB ⇌ \(\frac{1}{2}\)A2 + \(\frac{1}{2}\) B₂2? [A, B and AB are all gases] [WB JEE 2004]
Answer:
For A₂ + B₂ ⇌ 2AB, eqbm. constant = K_p
For the reverse reaction 2AB ⇌ A₂ + B₂ eqmb. constant = \(\frac{1}{\mathbf{K}_{p}}\)

On dividing by 2, AB ⇌ \(\frac{1}{2}\)A₂ + \(\frac{1}{2}\)B₂
eqbm. constant = \(\sqrt{\frac{1}{K_{p}}}\)

Question 15.
The concentration of hydronium ions in a cup of black coffee is 1.3 × 10^-5 M. Find the pH of the coffee. Is this coffee acidic or alkaline?
Answer:
Here, given [H₃O⁺] = 1.3 × 10^-5 M
p^H = – log [H₃O⁺] = – log (1.3 × 10^-5-5)
= 5 – log 1.3 = 5 – 0.1139 = 4.8861
As p^H is less than 7, the black coffee is acidic.

Question 16.
A solution is found to contain 0.63 g of nitric acid in 100 mL of the solution. What is the pH of the solution.
Answer:
HNO₃ → H⁺ + NO₃ completely
∴ [HNO₃] = [H⁺] as nitric acid is a strong acid.
Cone, of HNO₃ = 0.63 g per 100 mL

Strength L^-1 = 6.3
[HNO₃] = \(\frac{6.3}{63}\) = 0.1 = 10^-1 M
∴ [H⁺] = 10^-1 M
∴ p^H = – log 10^-1 = 1.

Question 17.
The value of K_w is 9.55 × 10^-14 at a certain temperature. Calculate the p^H of water at this temperature.
Answer:
K_w = 9.55 × 10^-14
Now for water [H₃O⁺] = [OH^–]
∴ K_w = [H₃O⁺] × [OH^–] = [H₃O⁺]²
or
[H₃O⁺] = \(\sqrt{\mathrm{K}_{w}}=\sqrt{9.55 \times 10^{-14}}\) = 3.09 × 10^-7M
p^H = – log (3.09 × 10^-7) = – (- 7 – log 3.09) = 7 – log 3.09 = 7 – 0.49 = 6.51

Question 18.
Calculate the mass of HCl to be dissolved per litre of the solution so that its p^H = 1.301.
p^H = – log [H₃O⁺]
[H₃⁺O] = antulog (- 1.301) = antilog 2.699
= 5.0 × 10^-2 × 36.5 gL^-1 = 1.825 gL^-1

Question 19.
Calculate the pH of 0 001 N H₂SO₄ solution.
Answer:
H₂SO₄ is completely ionised as
H₂SO₄ + 2H₂O ⇌ 2H₃O+ + SO₄^2- as it is, a strong acid.
[H₃O⁺] = 2[H₂SO₄] as one molecule of H₂SO₄ gives 2 H₃O⁺ ions

But H₂SO₄ is given to be 0.001 N = 0.001 × 49 g L^-1 [Eq. wt of H₂SO₄ = 49]

Question 20.
What would be the p^H of a solution obtained by mixing 100 mL of 0.1 NHCl and 9.9 mL of 1.0 N NaOH solution?
Answer:
100 mL of 0.1 N HCl = 100 × 0.1 = 10 millieq.
9.9 mL of 1 N NaOH = 9.9 × 1 = 9.9 millieq.
∴ HCl left unneutralized = 10 – 9.9 = 0.1 millieq
Volume of solution = 100 + 9.9 = 109.9 mL

Normality of resulting HCl solution
= \(\frac{0.1}{109.9}=\frac{0.1}{110}\) = 0.09 × 10^-4 N

Molarity = 9.09 × 10^-4 M = [H⁺] = [HCl]
p^H = – log(9.09 × 10^-4) = 3.05

Question 21.
Calculate the pH of 10^-8 M HCl solution.
Answer:
From acid [H⁺] = 10^-8 M as HCl is completely ionised
Now the concentration of [H⁺] = 10^-7 M cannot be ignored as compared to [H⁺] = 10^-8 M from HCl as HCl is very dilute solution.

∴ Total [H⁺] = 10^-8 + 10^-7 = 10^-8(1 + 10)
= 11 × 10^-8 M
p^H = – log (11 × 10^-8) = 8 – log 11
= 8 – 1.04 = 6.96.

Question 22.
Calculate the hydrolysis constant and degree of hydrolysis and p^H of 0.10 M KCN solution at 25°C. Ka for HCN = 6.2 × 10^-10.
Answer:
KCN is the salt of a weak acid HCN and strong base KOH
∴ K_h = Hydrolysis constant

The hydrolysis reaction of KCN is

Question 23.
Calculate the p^H of 0.01 M solution of NH₄CN. Given that the K_a for HCN = 6.2 × 10-10 and K_b for NH₃ = 1.6 /10“5.
Answer:
Applying the formula
P^H = 7+ [p^K_a – p^K_b] = 7 + [- log K_a + log K_b]
= 7 + \(\frac{1}{2}\)[- log (6.2 × 10^-10) + log (1.6 × 10^-5)]
= 7+ \(\frac{1}{2}\)[(10- 0.7924) + (5 – 0.2041)]
p^H = 9,31

Question 24.
The solubility product of AgCl in water is 1.5 × 10^-10. Calculate its solubility in 0.01 M NaCl solution.
Answer:
As NaCl dissociates completely
∴ [NaCl] = [Cl] = 0.01 M

If solubility of AgCl in 0.01 M is s mol L^-1, then
[Ag⁺] = [Cl] = s mol L^-1

∴ Total [Cl] = 0.01 + s ≈ 0.01 M
Ksp for AgCl = [Ag+][Cl- ] = s × 0.01 = 0.01 s

∴ 0.01 s = 1.5 × 10^-10
s = 1.5 × 10^-8 M.

Question 25.
Given that the solubility product of BaSO₄ is 1 × 10^-10. Will a precipitate form when
(i) Equal volumes of 2 × 10^-3 M BaCl₂ solution and 2 × 10^-4 M NaSO₄ solutions are mixed?
Answer:
BaCl₂ ionizes completely in solution

Ionic product of BaSO4 = [Ba++][SO4^—]
= 10^-3 × 10^-4 = 10^-7

As Ionic product 10^-7 is > K_sp 1 × 10^-10
Hence a precipitate of BaSO₄ will be formed.

(ii) Equal volumes of 2 × 10^-8 M BaCl₂ solution and × 2 × 10^-3 M Na₂SO₄ solutions are mixed?
Answer:
Similarly [Ba⁺⁺] = \(\frac{2 \times 10^{-8}}{2}\) = 10^-8 M
[SO₄^—] = \(\frac{2 \times 10^{-3}}{2}\) = 10^-3

Ionic product of BaSO₄ = 108 × 10 = 10^-11
It is less than the solubility product K_sp × 1 × 10^-10
Hence, no precipitate will be formed in this case.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 6 Thermodynamics. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 6 Important Extra Questions Thermodynamics

Thermodynamics Important Extra Questions Very Short Answer Type

Question 1.
Under what conditions the heat evolved or absorbed is equal to the internal energy change?
Answer:
At constant volume.

Question 2.
What is the sign of AH for endothermic reactions and why?
Answer:
AH is positive as ΔH = H_p – H_r and H_r < H_p.

Question 3.
What is the relationship between the standard enthalpy of formation and the enthalpy of a compound?
Answer:
They are equal.

Question 4.
Why enthalpy of neutralization of HF is greater than 57.1 kJ mol^-1?
Answer:
This is due to the high hydration energy of fluoride ions.

Question 5.
What are the specific heat capacity and molar heat capacity for water?
Answer:
Specific heat capacity for H₂O = 4.18 JK^-1 g^-1
Molar heat capacity for H₂O = 4.18 × 18 = 75.24 JK^-1 mol^-1.

Question 6.
Why enthalpy of neutralization is less if either the acid or the base or both are weak?
Answer:
A part of the heat is used up for dissociation of the weak acid or weak base or both.

Question 7.
What do you mean by a system?
Answer:
A specified part of the universe that is under thermodynamic observation is called a system.

Question 8.
Define a cyclic process.
Answer:
A process in which a system undergoes a series of changes and ultimately returns to the original state is called a cyclic process. For a cyclic process; ΔU = 0.

Question 9.
Why the entropy of a diamond is less than that of graphite?
Answer:
Diamond is more compact than graphite.

Question 10.
Under what conditions, ΔH of a process is equal to ΔU?
Answer:
At constant temperature and constant volume.

Question 11.
Is the enthalpy of neutralization of HCl is same as that of H2S04? If so, why?
Answer:
Yes. Because both are strong acids and ionized almost completely in aqueous solutions.

Question 12.
Which is a better fuel for the animal body: proteins or carbohydrates?
Answer:
Carbohydrates. They have high calorific value.

Question 13.
Is the experimental determination of enthalpy of formation of CH₄ possible?
Answer:
No.

Question 14.
Can we calculate ΔH of every process from the bond energy data of reactants and products?
Answer:
No.

Question 15.
Define enthalpy of fusion.
Answer:
It is defined as the heat change when one mole of solid changes to its liquid form at its melting point. ‘

Question 16.
For the reaction NaCl (aq) + AgNO₃ (aq) → AgCl(s) + NaNO₃(aq), will ΔH be greater than, equal to or less than ΔE?
Answer:
ΔH will be ΔE.

Question 17.
Write the mathematical relationship between heat, internal energy, and work done on the system.
Answer:
ΔE = q + w.

Question 18.
What is the limitation of the I law of thermodynamics?
Answer:
It cannot tell us about the direction of the process

Question 19.
What is the relationship between ΔH and ΔE?
Answer:
ΔH = ΔE + PΔV = ΔE + Δn_gRT.

Question 20.
State the I law of thermodynamics.
OR
State the Law of conservation of energy.
Answer:
The energy of an isolated system remains conserved.
OR
The energy can neither be created nor destroyed, though it can be converted from one form into another.

Question 21.
Which of the following is a state function?
(i) height of a hill
(ii) distance traveled in climbing the hill
(iii) energy consumed in climbing the hill.
Answer:
Energy consumed in climbing the hill.

Question 22.
What is the internal energy of one mole of a noble gas?
Answer:
U = \(\frac{3}{2}\)RT.

Question 23.
Which of the following are state functions?
(i) q
(ii) heat capacity
(iii) specific heat capacity
(iv) ΔH and ΔU.
Answer:
(iii) specific heat capacity and
(iv) ΔH and ΔU

Question 24.
Name the two most common modes by which a system and surroundings exchange their energy?
Answer:
Heat and work.

Question 25.
What will happen to internal energy if work is done by the system?
Answer:
The internal energy of the system will decrease.

Question 26.
How many times is the molar heat capacity greater than the specific heat capacity of water?
Answer:
18 times.

Question 27.
Neither q nor w is a state function but q + w is a state function. Why?
Answer:
q + w is equal to ΔU, which is a state function.

Question 28.
Name the state function Which remains constant during an isothermal change.
Answer:
Temperature.

Question 29.
What is the value of enthalpy of neutralization of a strong acid and a strong base?
Answer:
– 57.1 kJ/gm equivalent.

Question 30.
Out of 1 mole of H₂O(g) and 1 mole of H₂O(l) which one will have greater entropy?
Answer:
H₂O(g) will have greater entropy.

Question 31.
When a substance is said to be in its standard state?
Answer:
When it is present at 298 K and under one atmospheric pressure.

Question 32.
What is the entropy of the formation of an element in its standard state?
Answer:
By convention, the enthalpies (heats) of the formation of all elements in their most stable form (standard state) is taken as zero.

Question 33.
What is the physical significance of free energy change of a system?
Answer:
The decrease in free energy (- ΔG) is a measure of useful work or network obtainable during the process at constant temp and pressure.

Question 34.
Why is it essential to mention the physical state of reactants and products in thermochemical reactions?
Answer:
It is because the physical state of reactants and products also contributes significantly to the value ΔU or ΔH.

Question 35.
Name two intensive and extensive properties of a system.
Answer:

• Intensive properties: Viscosity, refractive index.
• Extensive properties: Mass, volume, heat capacity, etc.

Question 36.
What is fuel value or calorific value?
Answer:
It is defined as the amount of heat released when 1 gm of fuel or food is burnt completely in air or oxygen.

Question 37.
Is the process of diffusion of gases enthalpy driven or entropy-driven?
Answer:
It is an entropy-driven process.

Question 38.
The value of ΔHsol of NaNO₃ is positive, yet the dissolution is spontaneous, why?
Answer:
There is a large increase in entropy i.e., it is an entropy-driven process.

Question 39.
Write a thermochemical equation of enthalpy of combustion of methanol.
Answer:
2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(1); ΔH = – Q kJ
where Q kJ mol-1 is the enthalpy of combustion.

Question 40.
What is the enthalpy of the formation of Cl₂?
Answer:
Enthalpy of formation of a homonuclear molecule like Cl₂ is zero.

Question 41.
For a reaction also ΔH and ΔS are positive. What is the condition that this reaction occurs spontaneously?
Answer:
To make ΔG negative TΔS > ΔH.

Question 42.
Arrange the following fuels in order of increasing fuel efficiency; kerosene, diesel oil, wax, natural gas.
Answer:
Lower hydrocarbons have higher calorific value and thus are more efficient.
wax < diesel oil < kerosene < natural gas.

Question 43.
What is the sign of ΔS when N₂ and H₂ combine to form NH₃?
Answer:
ΔS is negative as the number of molecules decreases.

Question 44.
How will compare the efficiency of the three given fuels?
Answer:
By comparing their calorific values. Larger the calorific value, the greater the fuel efficiency.

Question 45.
How is a non-spontaneous process made spontaneous?
Answer:
By continuously supplying energy to it from outside.

Question 46.
What is the expression for entropy change for a phase transition? ‘
Answer:
ΔS = \(\frac{q_{\mathrm{rev}}}{\mathrm{T}}\)

Question 47.
Give an example of an isolated system.
Answer:
Thermos.

Question 48.
Is the enthalpy of formation of SnCl₂(s) the same as that of ZnCl₂(s)?
Answer:
No. They are different.

Question 49.
Why we usually study enthalpy change and not internal energy change?
Answer:
Most of the processes including reactions are carried out in open vessels at constant pressure.

Question 50.
What is the enthalpy of one mole of a noble gas?
Answer:
H = U + PV = \(\frac{3}{2}\) RT + RT = \(\frac{5}{2}\) RT.

Question 51.
Which of the thermodynamic properties out of U, S, T, P, V, H, and G are intensive properties and why?
Answer:
T and P, because they depend only upon the nature of the substance.

Question 52.
What is the relationship between q_p and q_v?
Answer:
q_p = q_v + Δng RT, where Δn = n_p– n_r (gaseous).

Question 53.
What is the Gibb’s Helmholtz equation?
Answer:
ΔG = ΔH – TΔS, where ΔG, ΔH, and ΔS are free energy change, enthalpy change, and entropy change respectively.

Question 54.
Comment on the bond energies of four C-H bonds present in CH4?
Answer:
The bond energies of 1st, 2nd, 3rd, and 4th C-H bonds are not equal and so average values are taken.

Question 55.
What is the main limitation of the first law of thermodynamics?
Answer:
It cannot predict the spontaneity of a process.

Question 56.
What is entropy?
Answer:
Entropy is a measure of the randomness/disorder of a system.

Question 57.
A reversible reaction has ΔG° negative for forwarding reaction. What will be the sign of ΔG° for the backward reaction?
Answer:
Negative.

Question 58.
What is the effect of increasing temperature on the entropy of a substance?
Answer:
It increases.

Question 59.
When is the entropy of a perfectly crystalline solid zero?
Answer:
At absolute zero (O.K).

Question 60.
What is an adiabatic process?
Answer:
In which no eat flow between the system and surroundings.

Thermodynamics Important Extra Questions Short Answer Type

Question 1.
Ice is lighter than water, but the entropy of ice is less than that of water. Explain.
Answer:
Water is the liquid form while ice is its solid form. Molecular motion in ice is restricted than in water, i.e., a disorder in ice is restricted than water, i.e., a disorder in ice is less than in water.

Question 2.
Define spontaneity or-feasibility of a process.
Answer:
Spontaneity or feasibility of a process means its inherent tendency to occur on its own in a particular direction under a given set of conditions.

Question 3.
Enthalpy of neutralization of CH₃COOH and NaOH is 55.9 kJ. What is the value of ΔH for ionization of CH₃COOH?
Answer:
The heat of neutralization of strong acid and strong base + ΔH of ionization of CH₃COOH = Enthalpy of neutralization of CH₃COOH and NaOH
∴ – 57.1 kJ + ΔH of ionisation of CH₃COOH = – 55.9 kJ
∴ ΔH of ionisation of CH₃COOH = (- 55.9 + 57.1) kJ
= 1.2 kJ.

Question 4.
When 1 gm of liquid naphthalene (C10H8) solidifies, 150 J of heat is evolved. What is the enthalpy of fusion of C₁₀H₈?
Answer:
ΔHsolidifcation = – 150 × 128 = – 19200 J = – 19.2 kJ
[∵ M.wt.of C₁₀H₈ = 128]

Question 5.
Why most of the exothermic processes (reactions) are spontaneous?
Answer:
ΔG = ΔH – TΔS; For exothermic reactions,
ΔH is -ve For a spontaneous process ΔG is to be -ve.

Thus decrease in enthalpy (- AH) contributes significantly to the driving force (To make ΔG negative).

Question 6.
What is meant by the term state function? Give examples.
Answer:
A state function is a thermodynamic property that depends upon the state of the system and is independent of the path followed to bring about the change. Internal energy change (ΔU), enthalpy change (ΔH) entropy change (ΔS), and free energy change (ΔG) are examples.

Question 7.
The enthalpy of combustion of sulfur is 297 kJ.
Write the thermochemical equation for the combustion of sulfur. What is the value of Δ_fH of SO₂?
Answer:
S(s) + O₂(g) → SO₂(g); Δ_fH = – 297 kJ
∴ ΔH of SO₂ = – 297 kJ mol^-1

Question 8.
What would be the heat released when 0.35 mol of HC1 in solution is neutralized by 0.25 mol of NaOH solution?
Answer:
HCl and NaOH being strong acid and strong- base is completely ionization in dilute aqueous solutions. The net reaction is H + (0.25 mol) + OH^– (0.25. mol) → H₂O (0.25 mol)

Now the heat of neutralization of 1 mole of a strong acid is – 57.1 kJ
∴ The heat released will be 57.1 × 0.25 kJ = 14.27 kJ.

Question 9.
Predict which of the following entropy increases/ decreases?
(i) A liquid crystallizes into a solid.
Answer:
After crystallization molecules attain an ordered state and therefore entropy decreases.

(ii) Temperature bf a crystalline solid is raised from 0 K to 115 K.
Answer:
When the temperature is raised, a disorder in molecules increases, and therefore entropy increases.

(iii) 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)
Answer:
The reactant is solid and hence has low entropy. Among the products there are two gases and be solid, therefore products represent a condition of higher entropy.

(iv) H₂(g) → 2H(g).
Answer:
Here 2 moles of H atoms have higher entropy than one mole of the hydrogen molecule.

Question 10.
Discuss the effect of temperature on the spontaneity of reactions.
Answer:
Effect of temperature on the spontaneity of reactions:

The terms low temperature and high temperature are relative. For a particular reaction, the high temperature could even mean room temperature.

Question 11.
What is the most important condition for a process to be reversible in thermodynamics?
Answer:
The process should be carried out infinitesimally slowly or the driving force should be infinitesimally greater than the opposing force.

Question 12.
Why heat is not a state function?
Answer:
According to first law of Thermodynamics; ΔU = q + w or q – = ΔU – w. As ΔU is a state function, but w is not a state function, therefore q is also not a state function.

Question 13.
Why the absolute value of enthalpy cannot be determined?
Answer:
As H = U + PV
The absolute value of U – the internal energy cannot be determined as it depends upon various factors whose value, cannot be determined.
∴ The absolute value of H cannot be determined.

Question 14.
Which of the following is/are exothermic and which are endothermic?
(i) Ca(g) → Ca²⁺(g) + 2e^–
Answer:
Endothermic (Ionisation enthalpy is required)

(ii) O(g) + e^– → O^–(g) .
Answer:
Exothermic (first electron affinity-energy is released)

(iii) N^2-(g) + e- → N^3-(g).
Answer:
Endothermic (higher electron affinities are required).

Question 15.
Calculate Δr G^θ for the conversion of oxygen to ozone, \(\frac{3}{2}\)O₂(g) → O₃(g) at 298 K, if K_p for this conversion is 2.47 × 10^-29
Answer:
Δ_r G^θ = – 2.303RT log Kp
and R = 8.314 JK^-1 mol^-1
∴ Δ_r G^θ = – 2.303 × 8.314 × 298 log 2.47 × 10^-29
= 163000 j mol^-1 = 163 kJ mol^-1.

Question 16.
Find the value of the equilibrium constant for the following conversion reaction at 298 K.

Δ_r G^θ = – 13.6 kJ mol^-1.
Answer:
-Δ_r G^θ = 2.303 RT log K

Question 17.
At 60°C, N₂O₄ is 50% dissociated. Calculate Δ_r G^θ this temperature and at one atmospheric pressure.
Answer:
N₂O₄(g) ⇌ 2NO₂(g)
If N₂O₄ is 50% dissociated, the mole fraction of both the substances are given by

Question 18.
Calculate the heat released when 0.5 moles of the nitric acid solution is mixed with 0.2 moles of potassium hydroxide solution.
Answer:
HNO₃ is a strong acid and KOH is a strong base. Therefore, both are completely ionized in water. The net reaction is

∴ Heat evolved = – 57.1 × 0.2 = – 11.42 kJ.

Question 19.
Define Hess’s Law of Constant Heat Summation.
Answer:
The enthalpy change for a reaction remains the same whether it proceeds in one step or in series of steps all’ measurements being done under similar conditions of temperature.

This law is a corollary from I Law of Thermodynamics.

Question 20.
What are the applications of Hess’s Law of constant heat summation?
Answer:

1. It helps to calculate the enthalpies of formation of those compounds which Cannot be determined experimentally.
2. It helps to determine the enthalpy of allotropic transformations like C(graphite) → C (diamond).
3. It helps to calculate the enthalpy of hydration.

Question 21.
Calculate the maximum work obtained when 0.75 mol of an ideal gas expands isothermally and reversible at 27°C from a volume of 15 L to 25 L.
Answer:
For an isothermal reversible expansion of an ideal gas
w = – nRT log \(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\) = – 2.303 nRT log \(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\)

Putting n = 0.75 mol; V₁ = 15 L; V₂ = 25 L, T = 27 + 273 = 300 K R = 8.314 JK^-1 mol^-1.
w = – 2.303 × 0.75 × 8.314 × 300 log \(\frac{25}{15}\)
= – 955.5 J.

Question 22.
What are heat capacities at constant volume and constant pressure? What is the relationship between them?
Answer:
Heat capacity at constant volume (C_v): Heat supplied to a system to raise its temperature through 1°C keeping the volume of the system constant is called its heat capacity at constant volume (C_v).

Heat capacity at constant pressure (C_p): Heat supplied to a system to raise its temperature through 1°C keeping the external pressure constant is called its heat capacity at constant pressure (C_p).

Relationship between C_p and C_v: C_p – C_v = R.

Question 23.
Explain what do you mean by a reversible process.
Answer:
A process or a change is said to be reversible if it can be reversed at any moment by an infinitesimal change. It proceeds infinitesimally slowly by a series of equilibrium states such that the system and surroundings are always in near equilibrium with each other.

Question 24.
Two liters of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its. total volume is 10 liters. How much heat is absorbed and how much work is done in the expansion?
Answer:
We have q = – w: pext(10 – 2) = 0 × 8 = 0
No. work is done; No heat is absorbed.

Question 25.
Define (i) Molar enthalpy of fusion
Answer:
The enthalpy change that accompanies the melting of one mole of a solid substance at its melting point is called the articular enthalpy of fusion.

(ii) Molar enthalpy of vaporisation.
Answer:
The amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1 bar) is called molar enthalpy of vaporization.

Question 26.
How will you arrive at the relationship q_p = q_v + Δn_gRT?
Answer:
Enthalpy change ΔH = q_p; where q_p = heat change at constant pressure,
Internal energy change ΔU = q_v; where q_v = heat change at. constant volume.

Now ΔH = ΔU + PΔV
For ideal gases PV = nRT
∴ ΔH = ΔU + (PV₂ – PV₁)
= ΔU + P(V₂ – V₁) = ΔU + (n₂RT – n₁RT)
= ΔU + RT(n₂ – n₁) = ΔU + Δn_gRT
or
q_p = q_v + Δn_gRT

Question 27.
Define
(i) Specific heat capacity
Answer:
Specific heat capacity: It is the amount of heat required to raise the temperature of 1 gram of the substance through 1°C.

(ii) molar heat capacity.
Answer:
Molar heat capacity: It is the amount of heat required to raise the temperature of one mole of the substance through 1 °C.

Question 28.
Derive the relationship C_p – C_v = R.
Answer:

Question 28.
Derive the expression -ΔG = w non-expansion
Answer:
From I law of thermodynamics

Question 29.
How does the sign of G help in predicting the spontaneity/non-spontaneity of a process?
Answer:

1. IF ΔG is negative, the process is spontaneous.
2. If ΔG = O, the process does not occur and the system is in equilibrium.
3. If ΔG is positive, the process does not proceed in the forward direction.

Question 30.
An exothermic reaction A B is spontaneous in the backward direction. What will be the sign of ΔS for the forward direction?
Answer:
The backward reaction will be endothermic. Thus, the energy factor opposes the backward reaction., As the backward reaction is spontaneous, the randomness factor must favor i.e., ΔS will be positive for the backward reaction or it will be negative for the forward direction.

Thermodynamics Important Extra Questions Long Answer Type

Question 1.
Define
(i) Standard enthalpy of formation.
Answer:
Standard enthalpy of formation: The heat change accompanying the formation of 1 mole df a substance from its elements in their most stable state of aggregation is called its standard enthalpy of formation.
H₂(g) + \(\frac{1}{2}\)O₂(g) H₂O(l); Δ_f He = 285.8 kJ mol^-1

(ii) Standard enthalpy of combustion
Answer:
Standard enthalpy of combustion: It is the heat change accompanying the complete combustion or burning of one mole of a substance in its standard state in excess of air or oxygen.
C₄H₁₀(g) + \(\frac{13}{2}\)O₂(g) → 4CO₂(g) + 5H₂O(1); ΔH^θ = – 2658.0 kJ mol^-1.

(iii) Enthalpy of atomization
Answer:
Enthalpy of atomization: It is defined as the enthalpy change accompanying the breaking of one mole of a substance completely into its atoms in the gas phase.
H₂(g) → 2H(g) Δ_cHe = 435.0 kJ mol^-1

(iv) Enthalpy of solution
Answer:
Enthalpy of solution: It is defined as the heat change when one mole of a substance dissolves in a specified amount of the solvent. The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving 2 moles of the substance in an infinite amount of the solvent.

(v) Lattice enthalpy
Answer:
Lattice Enthalpy: The lattice enthalpy of an ionic compound is the enthalpy change that occurs when one mole of an ionic compound dissociates into its ions in a gaseous state.

(vi) Thermochemical equation.
Answer:
Thermochemical Equation: A balanced chemical equation together with the value of its A^H is called a thermochemical equation.

The above equation describes the combustion of liquid ethanol. The negative sign indicates that tills are an exothermic reaction. We specify the physical state along with the allotropic state of the substance in a thermochemical equation.

Thermodynamics Important Extra Questions Numerical Problems

Question 1.
For the equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) at 298 K, K_c = 1.8 × 10^-7. What is ΔG° for the reaction? (R = 8.314 JK^-1 mol^-1).
Answer:

Question 2.
Calculate the equilibrium constant, K, for the following reaction at 400 K?
2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
Given that Δ_rH° = 80.0 kJ mol^-1 and Δ_rS° = 120 JK^-1 mol^-1.
Answer:

Question 3.
Calculate the standard entropy change for the reaction X ⇌ Y if the value of ΔH° = 28.40 kJ and equilibrium constant is 1.8 × 10^-7 at 298 K and Δ_rG° = 38.484 kJ.
Answer:

= -33.8 JK^-1 mol^-1

Question 4.
Calculate the enthalpy of formation of methane, given that the enthalpies of combustion of methane, graphite, and hydrogen are 890.2 kJ, 393.4 kJ, and 285.7 kJ mol^-1 respectively.
Answer:

Multiply equation (iii) by 2, add it to equation (ii) and subtract equation (i) from their sum
C + 2H₂ → CH₄ .
ΔH = – 393.4 + 2(-285.7) – (-890.2)
= – 74.6 kJ mol^-1
Hence the heat of formation of methane (CH₄) is
Δ_fH = – 74.6 kJ mol^-1.

Question 5.
CO is allowed to expand isothermally and reversibly from 10 m³ to 20 m³ at 300 K and work obtained is 4.75 KJ. Calculate the number of moles of carbon monoxide (CO). R = 8.314 JK^-1 mol^-1.
Answer:

Question 6.
Two moles of an ideal gas initially at 27°C and one atm pressure are compressed isothermally and reversible till the final pressure of the gas is 10 atm. Calculate q, w, and AU for the process.
Answer:
Here n = 2; p₁ = 1 atm; P₂ = 10 atm; T = 300 K
w = 2.303 nRT log \(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\)
= 2.303 × 2 × 8.314 JK mol^-1 × 300 K × log \(\frac{10}{1}\)
= 11488 J .

For isothermal compression of ideal gas
ΔU = 0
Further, ΔU = q + w
∴ q = -w = 11488 J.

Question 7.
The heat of combustion of benzene in a bomb calorimeter (i.e. at constant volume) was found to be 3263.9 kJ mol^-1 at 25°C. Calculate the heat of combustion of benzene at constant pressure.
Answer:
The given reaction is
C₆H₆(l) + \(\frac{15}{2}\) O₂(g) → 6CO₂(g) + 3H₂O(l)
C₆H₆ is a liquid and H20 is a liquid at 25°C.

Question 8.
When 0.532 g of benzene (C₆H₆) with boiling point 353 K is burnt with an excess of O₂ in a calorimeter, 22.3 kJ of heat is given out. Calculate ΔH for the combustion process (R = 8.31 JK^-1 mol^-1)
Answer:

Question 9.
Specific heat of an elementary gas is found to be 0.313 Jat constant volume. If the molar mass of the gas is 40 g mol^-1, what is the atomicity of the gas? R = 8.31 JK^-1 mol^-1.
Answer:

Question 10.
Calculate the energy needed to raise the temperature of 10.0 g of iron from 25° C to 500°C if the specific heat of iron is 0.45 J (0°C)^-1g^-1.
Answer:
Energy needed (q) = m × C × ΔT
= 10.0 × 0.45 × (500 – 25) J
= 2137.5 J

Question 11.
Calculate the enthalpy of formation of carbon disulfide given that the enthalpy of combustion of it is 110.2 kJ mol^-1 and those of sulfur and carbon are 297.4 kJ and 394.5 kJ/g atoms respectively.
Answer:
Our aim is C(s) + 2S(s) → CS₂(l); ΔH =?
Given (i) CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g); ΔH = – 110.2 kJ mol^-1
(ii) S(s) + O₂(g) → SO₂(g); ΔH = – 297.4 kj mol^-1
(iii) C(s) + O₂(g) → CO₂(g); ΔH= – 394.5 kj mol^-1

Add (iii) + 2(ii) and subtract (i), it gives, on rearranging
C(s) + 2S(s) → CS₂(l);
ΔH = (- 394.5) +- 2(- 297.4) – (- 110.2)
= -879.1 kj mol^-1

Thus the enthalpy of formation of CS₂ = – 879.1 kJ mol^-1

Question 12.
There are two crystalline forms of PbO; one is yellow and the other is red. The standard enthalpies of formation of these two forms are – 217.3 and – 219.0 kJ mol^-1 respectively. Calculate the enthalpy change for the solid-solid phase transition:
PbO (yellow) → PbO(red)
Answer:
Our aim is PbO (yellow) → PbO (rpd); ΔH =?
Given (i) PbO(s) + \(\frac{1}{2}\)02(g) ) → PbO (yellow); ΔH = – 217.3 kJ mol^-1
(ii) Pb(s) + \(\frac{1}{2}\)O2(g) ) → PbO(red); ΔH = —219.0 kJ mol^-1

Subtracting (1) from (ii), we get
PbO (yellow) → PbO(red); ΔH =- 219.0 – (- 2173) = – 1.7 kJ mol^-1.

Question 13.
The thermite reaction used for welding of metals involves the reaction 2Al (s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)
What is Δ_r, H° at 25°C for this reaction? Given that the standard heats of formation of Al₂O₃ and Fe₂O₃ are – 1675.7 k J and – 828.4 kJ mol^-1 respectively.
Answer:
Our aim is 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s); Δ_rH° =?

Question 14.
Calculate the bond enthalpy of HCI. Given that the bond enthalpies of H₂ and Cl are 430 and 242 kJ mol^-1 respectively and for HCI is -91 kJ mol^-1.
Answer:

Question 15.
CaIculat the enthalpy of hydrogenation of C₂H₂(g) to C₂H₄(g). Given bond energies: C—H = 414.50 kJ mol^-1; C≡C is 827.6 kJ mol^-1, C=C is 606.0 kJ mol^-1; H—H = 430.5 kJ mol^-1.
Answer:

= [827.6 + 2 × 414.0 + 430.5] – [606.0 + 4 × 414.0]
= 175.9 kJ mol^-1.

Question 16.
The entropy change for the vaporization of water is 109 JK^-1 mol^-1. Calculate the enthalpy change for the vaporization of water at 373 K.
Answer:
Δvap H = TΔvap S
= 373 × 10⁹ = 40657 Jmol^-1
= 40.657 kJ mol^-1

Question 17.
Enthalpy and entropy changes of a reaction are 40.63 kJ mol^-1 and 108.8 JK^-1 mol^-1 respectively. Predict the feasibility of the reaction at 27°C.
Answer:
ΔH = 40.63 kJ mol^-1 = 40630 Jmol^-1
ΔS = 108.8 JK^-1 mol^-1
T = 27°C =27 + 273 = 300 K
Now ΔG = ΔH – TΔS
= 40630 – 3 × 108.8 = 7990 J mol^-1

Since AG comes out to be positive (i.e., ΔG > 0), the reaction is not feasible at 27°C in the forward direction.

Question 18.
At 0°C ice and water are in equilibrium and ×H = 6.0 kJ mol^-1 for the process H₂O(s) → H₂O(l). What will be ΔS and
ΔG for the conversion of ice to liquid water.?
Answer:
Since the given process is in equilibrium
ΔG = 0
∴ from the equation ΔG = ΔH – TΔS
ΔH becomes = TΔS

∴ ΔS = \(\frac{6000 \mathrm{~J} \mathrm{~mol}^{-1}}{273 \mathrm{~K}}\) = 21.98 JK^-1 mol^-1.

Question 19.
Calculate the standard- free energy for the reaction
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)
Given that the standard free energies of formation (Δ_fG°) for NH₃(g), NO(g), H₂O(l) are – 16.8, + 86.7, and – 237.2 kJ mol^-1 respectively. Predict the feasibility of the above reaction at the standard state.
Answer:
Given Δ_fG°(NH₃) = – 16.8 kJ mol^-1, Δ_fG°(NO) = + 86.7 kj mol^-1; Δ_fG°(H₂O) = – 237.2 kJ mol^-1

= – [4 × ΔfG°(NH3) + 5 × ΔfG°(O2)]
= [4 × 86.7 + 6(- 237.2)] – [4 × (- 16.8) + 5 × 0]
= – 1009.2 kJ
Since Δ_rG° is negative, the reaction is feasible.

Question 20.
The value of K for the water gas reaction
CO + H₂O ⇌ CO₂ + H₂ is 1.06 × 10⁵ at 25°C. Calculate the standard free energy change for the reaction at 25°C.
Answer:
Δ_rG° = – 2.303 RT log K
= – 2.303 × 8.314 × 298 × log (1.06 × 10⁵)
On solving, Δ_rG° = – 28.38 KJ mol^-1.

Question 21.
Calculate ΔrG° for the conversion of oxygen to ozone. \(\frac{3}{2}\)O2(g) → O3(g) at 298 K. If Kp for the reaction is 2.47 × 10^-29
Answer:
Δ_rG° = – 2.303 RT log K
R = 8.314 JK^-1 mol^-1
∴ Δ_fG° = – 2.303 (8-314 JK^-1 mol^-1) × 298 × (log 2.47 × 10^-29)

On solving
Δ_rG° = 163000 J mol^-1 = 163 kJ mol^-1

Question 22.
Find out the value of the equilibrium constant for the following reaction at 298 K. Standard Gibbs energy change, Δ_rG° at the given temperature is – 13.6 kJ mol^-1.
Answer:
-Δ_rG° = 2.303 RT log K

Question 23.
For oxidation of iron, 4Fe(s) + 3O₂(g) → 2Fe₂O₃ (s) entropy change is – 549.4 JK-1 mol-1 at 298 K. Inspite .of negative entropy change of this reaction, why is the reaction spontaneous? (Δ_rH° = -1648 × 103 J mol^-1).
Answer:
The spontaneity of a reaction is determined from ΔS_total
(which is = Δ_sys + Δ_sutu)

Now ΔS_surr = – \(\frac{\Delta_{r} \mathrm{H}^{0}}{\mathrm{~T}}\) at constant pressure
= \(\frac{-1648 \times 10^{3}}{298 \mathrm{~K}}\) J mol^-1 = 5530 JK^-1 mol^-1

Thus total entropy change for the reaction is
Δ_r S_total = 5530 + (- 549.4) = 4980.6 JK^-1 mol^-1.
Since the total entropy change is positive, therefore, the above reaction is spontaneous.

Question 24.
A swimmer coming from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporization at 100°C. AvapH° for water.
Answer:

No. of moles in 18 g H₂O(l) = \(\frac{18 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 1 mol.
Δ_vap U = Δ_vap H° – pΔV = Δ_vapH – Δn_g RT (assuming steam behaving as an ideal gas)
Δ_vap H° – Δn_g RT = 40.66 kJ mol^-1 – (1)(8.314 JK^-1 mol^-1)(373 K) (10^-3 kJ) ,
∴ Δ_vap U° = 40.66 kJ mol^-1 – 3.10 kJ mol^-1 = 37.56 kJ mol^-1

Question 25.
1 g or graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure according to the reaction
C(graphite) + O₂(g) → CO₂(g). During the reaction, the temperature rises from 298 to 299 K. If the heat, the capacity of the bomb calorimeter is 20.7 kJ K^-1 what is the enthalpy change for the above reaction at 298 K and 1. atm?
Answer:
q = Heat change = C_v × ΔT, where q is the heat absorbed by the calorimeter.
The quantity of heat from the reaction will have the same magnitude, but opposite sign, because heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
∴ q = – C_v × ΔT = – 20.7 kJ K^-1 × (299 – 298) K
= -20.7kJ.

Here, negative sign indicates the exothermic nature of the reaction.
Thus, ΔU for the combustion of 1 g of graphite = – 20.7 kJ K^-1 For combustion of 1 mol of graphite
= – \(\frac{12.0 \times(-20.7)}{1}\) = – 2.48 × 10² kJ mol^-1

Since Δng_g = 0
∴ ΔH = ΔU = – 2.48 × 10² kJ mol^-1.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 5 States of Matter. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 5 Important Extra Questions States of Matter

States of Matter Important Extra Questions Very Short Answer Type

Question 1.
What is the pressure of a gas? What is its S.I. unit?
Answer:
The force exerted by the gas molecules per unit area on the walls of the container is equal to its pressure SI. unit of pressure is the pascal (Pa).
1 Pa = Nm^2- =1 kg m^-1s^-2.

Question 2.
What does the temperature of gas represent?
Answer:
The temperature of the gas represents the average kinetic energy of the gas molecules as we know K.E. ∝ \(\sqrt{T}\).

Question 3.
Why is it not possible to cool gas to o K?
Answer:
This is because all gases condense to liquids or solids before this temperature is reached.

Question 4.
What property of molecules of real gases is indicated by van der Waal’s constant ‘a’?
Answer:
Intermolecular attraction.

Question 5.
What do you understand by standard temperature?
Answer:
The standard temperature is 0°C or 273 K.

Question 6.
What is the effect of temperature on the vapour pressure of a liquid?
Answer:
Vapour pressure increases with rising in temperature.

Question 7.
What is an ideal gas equation?
Answer:
For given ideal gas PV = nRT
where P = gas pressure;
V = volume,
R = gas constant
T = Temp. (Kelvin scale);
n – no. of moles of gas.

Question 8.
Which state of matter has a definite volume, but no definite shape?
Answer:
Liquid.

Question 9.
What is the value of the gas constant in S.I. units?
Answer:
8.314 JK^-1 mol^-1.

Question 10.
What are the S.I. units of surface tension?
Answer:
Nm^-1

Question 11.
What is the compressibility factor?
Answer:
Z = \(\frac{PV}{nRT}\).

Question 12.
What is the equation of state for real gases?
Answer:
van der Waal’s equation(P + \(\frac{a}{\mathrm{~V}^{2}}\))(v – b) = RT for 1 mole.

Question 13.
How is the pressure of a gas related to its density at a particular temperature?
Answer:
d = \(\frac{MP}{RT}\)

Question 14.
How is the partial pressure of a gas in a mixture related to the total pressure of the gaseous mixture?
Answer:
The partial pressure of a gas = Mole fraction of that gas × total pressure.

Question 15.
What is the relationship between average kinetic energy , and the temperature of a gas?
Answer:
K.E. = \(\frac{3}{2}\)kT
where k is Boltzmann constant = \(\frac{\mathrm{R}}{\mathrm{N}_{0}}\).

Question 16.
What is the significance of van der Waal’s constant ‘a’ and ‘b’?
Answer:
‘a’ is a measure of the magnitude of the intermolecular forces of attraction while ‘b’ is a measure of the effective size of the as molecules.

Question 17.
Arrange the solid, liquid and gas in order of energy-giving reasons.
Answer:
Solids < liquid < gas. This is because a solid absorbs energy to change into a liquid which further absorbs energy to change into a gas.

Question 18.
What is the effect of pressure on the boiling point of a liquid?
Answer:
The boiling point increases as the prevailing pressure increases.

Question 19.
Why are vegetables cooked with difficulty at a hill station?
Answer:
The atmospheric pressure decreases as we go up. Therefore at the hills due to the lowering of atmospheric pressure, boiling points lowered.

Question 20.
The size of the weather balloon keeps on becoming larger as it rises to high altitude. Explain why?
Answer:
At higher altitudes, the external pressure on the balloon decreases and therefore, its size increases.

Question 21.
How is the pressure of a given sample of a gas related to the temperature at a constant volume?
Answer:
P ∝ T, i.e., Pressure varies as absolute temperature.

Question 22.
How is the pressure of a gas related to the number of molecules of a gas at constant volume and temperature?
Answer:
P ∝ N; N = No. of molecules of a gas.

Question 23.
What type of graph do we get when we plot a graph PV against P? What is shown by this graph?
Answer:
It is a straight line parallel to the r-axis. It shows that PV is constant at different pressures. It shows Boyle’s Law.

Question 24.
At what temperature will oxygen molecules have the same kinetic energy as ozone molecules at 30°C?
Answer:
AT 30°C. Kinetic energy depends only on the absolute temperature and not on the identity of the gas.

Question 25.
At a particular temperature why vapours pressure of acetone is less than that of ether?
Answer:
This is because intermolecular forces in acetone are more than those present in ether.

Question 26.
Out of NH₃ and N₂ which will have
(i) a larger value of V and
Answer:
NH₃ will have a large value of ‘a’ because of hydrogen bonding.

(ii) larger value of ‘b’?
Answer:
N2 should have a large value of ‘b’ because of its larger molecular size.

Question 27.
Why are the gases helium and hydrogen not liquefied at room temperature by applying very high pressure?
Answer:
Because their critical temperature is lower than room temperature. Gases cannot be liquefied above the critical temperature by applying even very high pressure.

Question 28.
What will boil at a higher temperature at sea level or at the top of mountains?
Answer:
Water will boil at a higher temperature at sea level.

Question 29.
Under what conditions do real gases tend to show ideal gas behaviour?
Answer:
When the pressure of the gas is very low and the temperature is high.

Question 30.
Both propane (C₃Hg) and carbon dioxide (CO₂) diffuse at the same rate under identical conditions of temperature and pressure. Why?
Answer:
Both propane (C₃Hg) and carbon dioxide (CO₂) have the same molar mass (44 gm).

Question 31.
If the number of moles of a gas is doubled by keeping the temperature and pressure constant, what will happen to the volume?
Answer:
The volume will also double as V ∝ n according to Avogadro’s Law.

Question 32.
What is a Triple point?
Answer:
The temperature at which solid, liquid and vapour; i.e., all the three states of the substance exist together is called the triple point.

Question 33.
Is Dalton’s law of partial pressures valid for a mixture of SO₂ and O₂?
Answer:
No, the law holds good only for those gases which do not react with each other.

Question 34.
Under what conditions a gas deviates from ideal gas behaviour?
Answer:
It deviates at low temperature and high pressure.

Question 35.
Molecule A is twice as heavy as molecule B. Which of these has higher kinetic energy at any temperature?
Answer:
K.E. of a molecule is directly proportional to temperature and is independent of its mass. So both the molecule A and B at any temperature will have equal kinetic energy.

Question 36.
How will you define pascal?
Answer:
It is defined as the pressure exerted by a force of one newton on an area of one meter2.
Pa = 1 Nm^-2.

Question 37.
How will you define London or dispersion forces?
Answer:
The forces of attraction between the induced momentary dipoles are called London or dispersion forces.

Question 38.
What is Absolute Zero?
Answer:
The lowest possible hypothetical temperature of – 273°C at which gas is supposed to have zero volume is called Absolute zero.

Question 39.
What is the nature of the gas constant R?
Answer:

= work done per degree per mole.

Question 40.
What is the value of R in SI units?
Answer:
R = 8.314 JK^-1 mol^-1 = 8.314 k Pa dm³ K^-1 mol^-1

Question 41.
What is meant from Boyle point or Boyle temperature?
Answer:
The temperature at which a real gas behaves like an ideal gas over an appreciable pressure range is called Boyle point or Boyle temperature.

Question 42.
How will you define viscosity?
Answer:
The internal resistance to the flow of a fluid is called its viscosity.

Question 43.
What is the effect of temperature on the viscosity of a liquid?
Answer:
The viscosity of a liquid decreases with an increase in temperature.

Question 44.
How will you convert pressure in atmospheres into SI units?
Answer:
1 atm = 1,01,325 Pa or Nm^-2 or 1 bar = 10⁵ Pa.

Question 45.
What do you understand from surface tension?
Answer:
The surface tension of a liquid is defined as the force acting at right angles to the surface along a one-centimetre length of the. liquid.

Question 46.
What is the effect of the increase in temperature on the surface tension of a liquid?
Answer:
Surface tension decreases in increasing the temperature.

Question 47.
What is the difference between normal boiling point and standard boiling point?
Answer:
When the external pressure is equal to one atmospheric pressure the boiling point is referred to as a normal boiling point, when it is one bar, the boiling point is called its standard boiling point.

Question 48.
What is the difference between vapour and gas?
Answer:
When gas is below its critical temperature, it is called vapour.

Question 49.
What happens if a liquid is heated to the critical temperature of its vapours?
Answer:
The meniscus between the liquid and the vapour disappears, the surface tension of the liquid becomes zero.

Question 50.
Why falling liquid drops are spherical?
Answer:
The surface area of a sphere is minimum. In order to have minimum surface area drops of the liquid become spherical.

States of Matter Important Extra Questions Short Answer Type

Question 1.
Ammonia and Sulphur dioxide gases are prepared in two comers of a laboratory. Which gas will be detected first by a student working in the middle of the laboratory and why?
Answer:
Molecular mass of NH₃ = 17 Molecular mass of SO₂ = 64
NH₃ is a lighter gas and diffuses at a faster speed than SO₂
∴ NH₃ gas will be detected first.
[∵ Acc. to Graham’s law of diffusion \(\frac{r_{1}}{r_{2}}=\sqrt{\frac{d_{2}}{d_{1}}}\)

Question 2.
What is the effect of hydrogen bonding on the viscosity of a liquid?
Answer:
Hydrogen bonding leads to an increase in the effective size of the moving unit in the liquid. Due to an increase in the size and mass of the molecule, there is greater interval resistance of the flow of the liquid. As a result, the viscosity of the liquid rises.

Question 3.
Which are the two faulty assumptions in the kinetic theory of gases.
Answer:

1. There is no force of attraction between the molecules of the gas.
2. The volume of the molecules of the gas is negligibly small as compared to the total space (volume) occupied by the gas.

Question 4.
What is the relationship between the density and molar mass of a gaseous substance? Derive it.
Answer:
Ideal gas equation is PV = nRT
or \(\frac{n}{V}=\frac{p}{R T}\)
Replacing n by \(\frac{m}{M}\), we get

Question 5.
What is meant by the term: Non-ideal or real gas?
Answer:
The gas which does not obey the Cas law:
Boyle’s law, Charles’ law, Avogadro^ law at all temperatures and pressures is a called-non-ideal or real gas. Most of the real gases show ideal behaviour at low pressure and high temperature.

Question 6.
Derive the ideal gas equation PV = nRT.
Answer:
According to Boyle’s law V ∝ \(\frac{1}{P}\) if n and T are constant.
According to Charles’ law V ∝ T at constant P.and n
According to Avogadro’s law V ∝ n at constant T and P
Combining the three laws
In
V ∝ \(\frac{Tn}{P}\)
or
PV ∝ nT
or
PV = nRT where R is a constant of proportionality called universal gas constant.

Question 7.
Why liquids have a definite volume, but no definite shape?
Answer:
It is due to the fact that in liquids intermolecular forces are strong enough to hold the molecules together, but these are strong enough to hold the molecules together, but these forces are not strong enough to fix them into definite or concrete positions as in solids. Hence they possess fluidity but no definite shape.

Question 8.
How do the real gases deviate from ideality above and below the Boyle point?
Answer:
Above their Boyle point, real gases show positive deviations from ideality and the values of Z are greater than one. The forces of attraction between the molecules are very feeble. Below the Boyle point, real gases first show a decrease, in Z value with increasing pressure, the value Of Z increases continuously.

Question 9.
Write down the van der Waals, equation for n moles of a real gas. What do the constants ‘a’ and ‘b’ stand for?
Answer:
(P + \(\frac{a n^{2}}{V^{2}}\)) (V – nb) = nRT
where p, Vindicate gas pressure and its volume. T is the Kelvin temperature of the gas. R is gas constant. Value of ‘a’ is a measure of the magnitude of intermolecular attractive forces within the molecule and is independent of temperature and pressure, ‘rib’ is approximately the total volume occupied by the molecules themselves. ‘a’ and ‘b’ are called van der Waals constants and their value depends upon the nature of the gas.

Question 10.
How is compressibility factor Z related to the real and ideal volume of the gas?
Answer:
Z = \(\frac{P \mathrm{~V}_{\text {real }}}{n \mathrm{RT}}\) …(1)

If the gas shows ideal behaviour, then
Videal = \(\frac{nRT}{P}\) ….(2)

On putting the titis value of nRT in equation (1), we get
Z = \(\frac{\mathrm{V}_{\text {real }}}{\mathrm{V}_{\text {ideal }}}\)

Thus the compressibility factor Z is the ratio of the actual molar volume of a gas to the molar volume of it. if it were an ideal gas at that temperature and pressure.

Question 11.
What do the critical temperature, critical pressure, and critical volume for a gas stand for?
Answer:
The critical temperature is the temperature above which a gas cannot be liquified however large may be the pressure applied on it. The pressure sufficient to liquiíy a gas at its critical temperature is called its critical pressure. The volume of the gas at its crìtical temperature is called its critical volume’.

Question 12.
What do the absolute zero and absolute scale of temperature stand for?
Answer:
The lowest possible hypothetical temperature of – 273°C at which gas is supposed to have zero volume is called Absolute! zero.

Lord Kelvin suggested a new scale of temperature starting with – 273°C and it’s zero. This scale of temperature is called the absolute scale of temperature.

Absolute or Kelvin temperature is given by T, where
T = t°C + 273
t°C stands for the centigrade scale of temperature.

Question 13.
Give one application of Dalton’s law of partial pressures.
Answer:
One application of Dalton’s law of partial pressures is in determining the/pressure of dry gas. Gases are generally collected over water and, so they contain water vapours. The pressure exerted by the water Vapours at-a particular temperature is called the Aqueous Tension. By subtracting the aqueous tension from the ‘ vapour pressure of the moist gas, the pressure of the dry gas can be calculated.
P_{dry gas}= P_{moist gas} – Aqueous Tension (at t°C).

Question 14.
How will you calculate the partial pressure of gas using Dalton’s Law of partial pressures?
Answer:
In a mixture of non-reacting gases A, B, C etc., if each gas is considered to be an ideal gas, then
P_A = n_A \(\frac{RT}{V}\)
P_B = n_B \(\frac{RT}{V}\)
P_C = n_C \(\frac{RT}{V}\)
where n_A, n_B n_C stand for their respective moles in the same vessel .
[V = constant, keeping T constant]
Then according to Dalton’s law ‘

Thus, particle pressure of a. gas – Mole fraction of A × Total pressure.

Question 15.
Is there any effect of the nature of a liquid and temperature on the surface tension of a liquid?
Answer:

1. Effect of nature of the liquid: Surface tension is a property that arises due to the intermolecular forces of attraction among the molecules of the liquid.
2. The surface tension of the liquids generally decreases with an increase in temperature and becomes zero at the critical temperature.

Question 16.
Leaving electrostatic forces that exist between oppositely charged ions, which other intermolecular attractive forces exist between atoms or ions. Mention them.
Answer:

1. van der Waals forces
2. London Forces or Dispersion Forces
3. Dipole-dipole forces
4. Dipole-induced dipole forces
5. Hydrogen bond.

Question 17.
What is the relationship between intermolecular forces and thermal interactions?
Answer:
Intermolecular forces tend to keep the molecules together but the thermal energy of the molecules tends to keep them apart. Three states of matter are the result of a balance between intermolecular forces and the thermal energy of the molecules.
Gas → Liquid → Solid
intermolecular forces increases →
Gas ← Liquid ← Solid
increase in Thermal energy ←

Question 18.
What are the main characteristics associated with gases?
Answer:

1. Gases are highly compressible.
2. Gases exert pressure equally in all directions.
3. Gases have a much lower density than solids or liquids.
4. Gases do not have definite shape and volume.
5. Gases intermix freely and completely in all proportions.

Question 19.
What do you understand by the term? Isotherms and Isobars?
Answer:
Isotherms are the curves plotted by varying pressure against volume keeping temperature constant.

Each line obtained by plotting volume against temperature keeping, pressure constant is called an Isobar.

Question 20.
What are the forces which are responsible for the viscosity of a liquid? Why is glycerol more viscous than water?
Answer:
The forces responsible for the viscosity of the liquids are

• Hydrogen bonding
• van der Waals torches.

Glycerol possesses greater viscosity than water because glycerol has extensive hydrogen bonding in its molecules due to the pressure of three-Oil groups in it as compared to the hydrogen bonding in water molecules.

States of Matter Important Extra Questions Long Answer Type

Question 1.
Derive the Ideal gas equation PV = nRT.
Answer:
Let a certain mass of a gas at pressure P₁ and temperature
T1 occupy a volume V₁. On changing the pressure and temperature respectively to P₂ and T₂, let the volume of the gas change to V₂.
(i) Let us first change the pressure P₁ to P₂ at constant temperature T₁. Then, according to Boyle’s law, volume V is given by,
P₂V = P₁V₁
or
V = \(\frac{P_{1} V_{1}}{P_{2}}\) …(1)

(ii) Now keeping the pressure constant at P₂, heat the gas from temperature T₁ to T₂, the volume changes from V to V₂.
By Charle’s Law, we must have,

The numerical value of k depends upon the amount of gas and the units in which P and V are expressed. By Avogadro’s law, for one mole of any gas, the value of \(\frac{PV}{T}\) will be the same and is represented
\(\frac{PV}{T}\) = R
or
PV = RT.
for n moles of an ideal gas
PV = nRT

Question 2.
State the fundamental assumptions of the kinetic theory of gases
Or
What are the postulates of the kinetic theory of gases?
Or
Write a short note on “Microscopic Model of a gas”.
Answer:
To explain the general properties of gases to provide some theoretical explanation for various gas laws (stated on basis of experimental studies), various, scientists Bernoulli, Clausius, Maxwells, Boltzmann and others, gave ‘Kinetic Theory of Gases’ which explains qualitatively as well as quantitatively the various aspects of gas behaviour.

The important postulates of the theory are:

1. All gases are made up of a very large number of minutes particles called molecules.
2. The molecules are separated from one another by large distances. The empty spaces among the molecules are so large that the actual volume of the molecules is negligible as compared to the total volume of the gas.
3. The molecules are in a state of constant rapid motion in all directions. During their motion, they keep on colliding with one another and. also with the walls of the container and, thus, change their directions.
4. Molecular collisions are perfectly elastic, no loss of energy occurs when the molecules collide with one another or with the walls of the container. However, there may be. redistribution of energy during the collisions.
5. There are no forces of interaction (attractive or repulsive) between molecules. They move completely independent of one another.
6. The pressure exerted by the gas is due to the bombardment of its molecules on the walls of the container per unit area.
7. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature.

Question 3.
Explain the following observations.
(a) Aerated water bottles are kept underwater during summer.
Answer:
Aerated water contains CO₂ gas dissolved in an aqueous solution under pressure and the bottles are well stoppered. As in summer temperature increases and we know that the solubility of the gases decreases with increase in temperature and as a result move of gas is expected to be generated may be large in quantity and hence pressure exerted by the gas may be very high and the bottle may explode. So, to decrease the temperature and hence to avoid explosion of the bottles.

(b) Liquid ammonia bottle is cooled before opening the seal.
Answer:
Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in the volume of the gas. As a result, the gas will come out of the bottle with force. This will lead to the breakage of the bottle. Cooling under tap water will result in a decrease in volume. It reduces the chances of an accident.

(c) The tyre of an automobile is inflated at lesser pressure in summer than in winter.
Answer:
The pressure of the air is directly proportional to the temperature. During summer due to high temperature, the pressure in the tyre will be high as compared to that in water. The tube may burst under high pressure in summer. Therefore, it is advisable to inflate the tyre to lesser pressure in summer than in winter.

(d) The size of the weather balloon becomes larger and larger as it ascends up to higher altitudes.
Answer:
Answered somewhere else in the book.

States of Matter Important Extra Questions Numerical Problems

Question 1.
In a hospital, an oxygen cylinder holds 10 L of oxygen at 200 atm pressure. If a patient breathes in 0.50 ml of oxygen at 1.0 atm with each breath, for how many breaths the cylinder will be sufficient. Assume that all the data is at 37°C.
Answer:
10 L at 200 atm =? L at 1 atm. [Temp, is constant at 37°C]
∴ Boyle’s law P₁V₁ = P₂V₂ can be applied
200 × 10 = 1 × V₂
or
V₂ = 2000 L

Number of breaths = \(\frac{\text { Total volume }}{\text { Volume inhaled per breath }}\)
= \(\frac{2000 \mathrm{~L}}{0.5 \times 10^{-3} \mathrm{~L}}\)
= 4 × 10⁶

Question 2.
Calculate the pressure of 1 × 10²² molecules of oxygen gas when enclosed ¡n a vessel of 5-litre capacity at a temperature of 27°C.
Answer:
Number f moles of O₂ = \(\frac{1 \times 10^{22}}{6.02 \times 10^{23}}\)

= 0.166 × 10^-1 mol
= 1.66 × 10² mol
The pressure exerted by O₂ molecules can be calculated by the equation
PV = nRT [Ideal gas equation]

Question 3.
When a ship is sailing in the Pacific Ocean where the temperature is 23.4°C, a balloon is filled with 2 L air. What will be the volume of the balloon when the ship reaches the Indian Ocean, where the temperature is 26.1°C.
Answer:
V₁ = 2 L
T₁ = 23.4 + 273 = 296.4K
T₂ = 26.1 + 273 = 299.1 K
From Charle’s Law

Question 4.
At what temperature centigrade, will the volume of a gas at 0°C double itself, pressure remaining constant?
Answer:
Let the volume of the gas at 0°C = V₁ mL
Final Volume = V₂ = 2V₁ mL [Given]
T₁ = OC + 273 = 273 K, T₂ =?

Since Pressure remains constant
∴ Charles law can be applied

Question 5.
A bulb ‘B’ of the unknown volume containing gas at one atmospheric pressure is connected to an evacuated bulb of 0.5-litre capacity through a stop-cock. On opening the stop-cock, the final pressure was found to be 570 mm at the same temperature. ‘What is the volume of bulb ‘B’?
Answer:
Let the volume of the bulb = V₁; P, = 1 atm
Evacuated Bulb has volume = 0.5 L;
Total volume of both bulbs after the opening of stop-cock = (V₁ + 0.5) L i.e. V₂ = (V₁ + 0.5):
P₂ = 570 mm = \(\frac{570}{760}\) atm

Apply Boyle’s law P₁V₁ = P₂V₂
1 × V₁ = \(\frac{570}{760}\) × (V₁ + 0.5)

On solving the above equation
V₁ = 1.5 L
∴ Volume of the unknown bulb ‘B’ = 1.5 L

Question 6.
1 mole of SO2 occupies 1.5 L at 25°C. Calculate the pressure exerted by gas assuming that gas does not obey the ideal gas equation. (Given a = 3.6 atm L²mol^-2, b = 0.04 L mol^-1) where a, b are van der Waal’s constants.
Answer:

Question 7.
A gas occupies a volume of 2.5 L at 9 × 10⁵ Nm^-2. Calculate the additional pressure required to decrease the volume of the gas to 1.5 L, keeping the temperature constant.
Answer:
V₁ = 2.5 L; P₁ = 9 × 10⁵ Nm^-2
V₂ = 1.5 L; P₂ =?

Since temperature is constant, Boyle’s law is applied.

Question 8.
What volume of air will be expelled from a vessel containing 400 cm³ at 7° C when it is heated to 27° C at the same pressure?
Answer:
As the pressure is constant, Charles law is applied

428.6 cm³ is the volume after expansion
∴ Volume expelled = (428.6 – 400) cm³ = 28.6 cm³.

Question 9.
A 10.0 L container is filled with a gas to a pressure of 2.0 atm at 0°C. At what temperature will the pressure of the gas inside the container be 2.50 atm.
Answer:
As the volume of the container remains constant, Gay- Lussac’s Law/Amonton’s Law is applicable
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
\(\frac{2}{273}=\frac{2.50}{\mathrm{~T}_{2}}\)
T₂ = 341 K
or
t₂ = (341 – 273)°C = 68°C.

Question 10.
One litre flask containing vapours of methyl alcohol (Mol. mass = 32) at a pressure of 1 atm and 25°C was evacuated till the final pressure was 10^-3 mm. How many molecules of methyl alcohol were left in the flask?
Answer:
P₁ = 10^-3 mm; V₁ = 1000 cm³; T₁ = 298 K
Converting this volume to volume, at STP, where T₂ = 2/3 K and P₂ = 760 mm

Now, 22400 cm³ at STP contains 6.02 × 10²³ molecules

∴ 1.205 × 10^-3 cm³ at STP contains
= \(\frac{6.02 \times 10^{23}}{22400}\) × 1.205 × 10^-3 molecules
= 3.24 × 10¹⁶ molecules.

Question 11.
A sealed tube that can withstand a pressure of 3 atm is filled with air at 27°C and 760 mm pressure. Find the temperature above which it will burst?
Answer:
Applying gas equation \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\) as volume of the sealed tube remains constant .
\(\frac{1}{300}=\frac{3}{\mathrm{~T}_{2}}\)
T₂ = 900 K
or
t₂ = 900 – 273 = 627°C.

Question 12.
The temperature at the foot of a mountain is 30°C and pressure is 760 mm. Whereas at the top of the mountain these are 0°C and 710 mm. Compare the densities of the air at the foot and at the top.
Answer:

∴ the ratio of densities of air at the foot and the top of the mountain = 0.964: 1.

Question 13.
10.0 g of O₂ were introduced into an evacuated vessel of 5-litre capacity maintained at 27°C. Calculate the pressure of the gas in the atmosphere in the container.
Answer:
Volume of the gas = V = 5 L;
Wt. of O₂ = 10.0 g
Molar mass of O₂ = 32.0
∴ No. of moles = \(\frac{10}{32}\)

T = 27 + 273 = 300 K; R = 0.0821 L atm K^-1 mol^-1
From ideal gas equation,
PV = nRT, we get
P = \(\frac{10}{32}\) × 0.0821 × 300 = 1.54 atm.

Question 14.
8.0 g of methane is placed in a 5 L container at 27°C. Find Boyle constant.
Answer:
Pressure × Volume is called Boyle’s constant
PV = nRT = \(\frac{W}{M}\) RT ,
= \(\frac{8}{16}\) × 0.0821 × 300
= 12.315 L atm

Question 15.
The density of A gas is 3.80 g L^-1 at STP. Calculate its density at 27°C and 700 torr pressure
Answer:
d = \(\frac{PM}{RT}\)
For the same gas at two different pressures and temperatures,

Question 16.
If the density of a gas at sea level and 0°C is 1.29 kg m^-3, what will be its molar mass (Assume that pressure is equal to 1 bar).
Answer:

Question 17.
A gaseous mixture contains 56 g N₂, 44 g CO₂ and 16 g CH₄ The total pressure of the mixture is 720 mm Hg. What is the partial pressure of CH₄?
Answer:

Question 18.
A 2 L flask contains 1.6 g of methane and 0.5 g of hydrogen at 27°C. Calculate the partial pressure of each gas in the mixture and hence calculate the total pressure.
Answer:
1.6 g CH₄ = \(\frac{1.6}{16}\) = 0.1 Mole
16
Partial pressure of CH4 (pCH4) = CH4 × \(\frac{RT}{P}\)

Question 19.
One mole of S02 gas occupied a volume of 350 mL at 27°C and 50 atm pressure. Calculate the compressibility factor of the gas. Comment on the type of deviation shown by the gas from ideal behaviour.
Answer:
Compressibility factor, Z =
Here n = 1, P = 50 atm, V = 350 × 10^-3 L = 0.35 L
R = 0.0821 L atm K^-1 mol^-1; T = 27 + 273 = 300 K
∴ Z = \(\frac{50 \times 0.35}{1 \times 0.0821 \times 300}\) = 0.711
For an ideal gas Z = 1
As for the given gas Z < 1, it shows a negative deviation, i.e., it is more compressible than expected from ideal behaviour.

Question 20.
A mixture of CO and CO₂ is found to have a density of 1.5 g L^-1 at 20°C and 740 mm pressure. Calculate the composition of the mixture.
Answer:
1. Calculation of average molecular mass of the mixture
M = \(\frac{d R T}{P}=\frac{1.50 \times 0.0821 \times 293}{\frac{740}{760}}\) = 37.06

2. Calculation of percentage composition
Let mol% of CO in the mixture = x
Then mol% of CO? in the mixture = 100 – x

Average molecular mass

Question 21.
A 5-L vessel contains 14 g of nitrogen. When heated to 1800 K, 30% of molecules are dissociated into atoms. Calculate the pressure of the gas at 1800 K.
Answer:
N₂ ⇌ 2N

Initial moles = \(\frac{1.4}{28}\) = 0.05
Moles left = 0.05 – \(\frac{30}{100}\) × 0.05 = 2 × 0.015 = 0.03

∴ Total no. of moles = 0.035 + 0.030 = 0.065
i.e. n = 0.065 mol, V = 5 L, T = 1800 K; P =?
P = \(\frac{n \mathrm{RT}}{\mathrm{V}}=\frac{0.065 \times 0.0821 \times 1800}{5}\)
= 1.92 atm.

Question 22.
Calculate the temperature of 2 moles of S02 gas contained in a 5 L vessel at 10 bar pressure. Given that for SO₂ gas van der Waal’s constant is a – 6.7 bar L² mol^-2 and b = 0.0564 L mol^-1.
Answer:
According to van der Waal’s equation

Question 23.
A given mass of a gas occupies 919.0 mL in a dry state at STP. The same mass when collected over water at 15°C and 750 mm pressure occupies on litre volume. Calculate the vapour pressure of water at 15°C.
Answer:
If p is the vapour pressure of water at 15°C, then P₂ = 750 – p
From the gas equation \(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\), we get
\(\frac{760 \times 919}{273}=\frac{(750-p) \times 1000}{288}\)
or
p = 13.3 mm

∴ Vapour pr. of water = 13.3 mm.

Question 24.
A steel tank containing air at 15 atm pressure at 15°C ¡s provided with a safety valve that will yield at a pressure of 30 atm. To what minimum temperature must the air be heated to below the safety valve?
Answer:
\(\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}\)
i.e., \(\frac{15}{30}=\frac{288}{\mathrm{~T}_{2}}\)
or
T₂ = 576 K
or
t₂°C = 576 – 273 = 303°C

Question 25.
Calculate the pressure exerted by 110 g of CO₂ in a vessel of 2 L capacity at 37°C. Given that the van der Waals constants are a = 3.59 L² atm mol^-2 and b = 0.0427 L mol^-1. Compare the value with the calculated value if the gas were considered ideal.
Answer:
According to van der Waals equation

110
Here, n = \(\frac{110}{44}\) = 2.5 moles. Putting the given values, we get
P = \(\frac{2.5 \times 0.0821 \times 310}{(2-2.5 \times 0.0427)}-\frac{3.59 \times 2.5}{2}\)
= 33.61 atm – 5.61 atm = 28.0 atm

If the gas were considered an ideal gas. applying ideal gas equation
PV = nRT
or
P = \(\frac{n \mathrm{RT}}{\mathrm{V}}\)
∴ P = \(\frac{2.5 \times 0.0821 \times 310}{2}\) = 31.8 atm