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MP Board Class 11th Maths Solutions Chapter 1 Sets Ex 1.6

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MP Board Class 11th Maths Solutions

MP Board Class 11th Maths Solutions Chapter 1 Sets Ex 1.4

MP Board Class 11th Maths Solutions

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MP Board Class 11th Maths Solutions

MP Board Class 11th Maths Solutions Chapter 1 Sets Ex 1.1

MP Board Class 11th Maths Solutions

MP Board Class 11th Chemistry Important Questions Chapter 7 Equilibrium

Equilibrium Important Questions

Equilibrium Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Which of the following reaction have equal K_c and K_p:
(a) N_2(g) + 3H_2(g) ⇄ 2NH₃
(b) 2H₂S_(g) + 3O_2(g) ⇄ 2SO_2(g) + 2H₂O_(g)
(c) Br₂ + Cl_2(g) ⇄ 2BrCl_(g)
(d) P_4(g) + 6Cl_2(g) ⇄ 4PCl_3(g)
Answer:
(c) Br₂ + Cl_2(g) ⇄ 2BrCl_(g)

Question 2.
For the reaction N_2(g) + 3H_2(g) ⇄ 2NH_3(g), ∆H = 92 kJ, the concentration of NH₃ at the equilibrium increase by temperature:
(a) Increase
(b) No change
(c) Decrease
(d) None of the above
Answer:
(c) Decrease

Question 3.
In one litre container equilibrium mixture of reaction 2H₂S_(g) ⇄ 2H₂S_(g) + S₂S_(g) is filled 6.5 mol H₂S. 0.1 mol H₂ and 0.4 mol S₂ are present in it. Equilibrium constant of this reaction will be:
(a) 0.004 mol litre^-1
(b) 0.080 mol litre^-1
(c) 0.016 mol litre^-1
(d) 0.160 mol litre^-1
Answer:
(c) 0.016 mol litre^-1

Question 4.
Favourable condition for exothermic reaction of ammonia synthesis N₂S_(g) + 3H₂S_(g) ⇄ 2NH_3(g) are:
(a) High temperature and high pressure
(b) High temperature and low pressure
(c) Low temperature and high pressure
(d) Low temperature and pressure
Answer:
(c) Low temperature and high pressure

Question 5.
Oxidation of SO₂ by O₂ in SO₃ is an exothermic reaction manufacture of SO₃ will be maximum if:
(a) Temperature increased and pressure decreased
(b) Temperature decreased and pressure increased
(c) Temperature and pressure both increased
(d) Temperature and pressure both decreased.
Answer:
(b) Temperature decreased and pressure increased

Question 6.
At 440°C HI was heated in a closed vessel till equilibrium is attained. It is dissociated 22%. Equilibrium constant dissociation will be:
(a) 0.282
(b) 0.0796
(c) 6.0199
(d) 1.99
Answer:
(c) 6.0199

Question 7.
K_p and K_c can be expressed as:
(a) K_c = Kp (RT)^∆n
(b) K_p = K_c (RT)Q^∆n
(c) K_p = K_c(RT)^∆n
(d) K_c = K_p (RT)^∆n
Answer:
(c) K_p = K_c(RT)^∆n

How to find kp when initial pressure and total pressure at equilibrium is given.

Question 8.
The equilibrium constant of the reaction H_2(g) + I_2(g) ⇄ 2HI_2(g), is 64. If the volume of the container is reduced to one – fourth of its original volume. The value of the equilibrium constant will be:
(a) 16
(b) 32
(c) 64
(d) 128
Answer:
(c) 64

Question 9.
What would happen to a reversible reaction at equilibrium when an inert gas is added while the presence remain unchanged:
(a) More of the product will be formed
(b) Less of the product will be formed
(c) More of the reactant will be formed
(d) It remains unchanged
Answer:
(d) It remains unchanged.

Question 10.
SO_2(g) + \(\frac{1}{2}\) O_2(g) ⇄ SO_3(g), K₁
2SO_3(g) ⇄ 2SO_2(g) + O_2(g), K₂ which of the following is correct:
(a) K₂ = K₁²
(b) K₂ = K₁^-2
(C) K₂ = K₁
(d) K₂ = K₁^-1
Answer:
(b) K₂ = K₁^-2

Question 11.
Which reaction is not affected by change in pressure:
(a) N_2(g) + O_2(g) ⇄ 2NO_(g)
(b) 2O_3(g) ⇄ 3O_2(g)
(c) 2NO_2(g) ⇄ N₂O_4(g)
(d) 2SO_2(g) + O_2(g) ⇄ 2SO_3(g)
Answer:
(a) N_2(g) + O_2(g) ⇄ 2NO_(g)

Question 12.
For N₂ + 3H₂ ⇄ 2NH₃ + heat:
(a) K_p = K_c
(b) K_p = K_cRT
(c) K_p = K_c (RT)^-2
(d) K_p = K_c(RT)^-1
Answer:
(c) K_p = K_c (RT)^-2

Question 13.
Sodium sulphate dissolve in water with the release of heat. Imagine a saturated solution of sodium sulphate. If temperature is increased then according to Le – Chatelier principle:
(a) Mass of solid will be dissolved
(b) Some solid will be precipitated in solution
(c) Solution will be more saturated
(d) Concentration of solution will be unchangeble.
Answer:
(b) Some solid will be precipitated in solution

Question 14.
For reaction, PCl_3(g) + Cl_2(g) ⇄ PCl_5(g) value of K_c at 250° is 26 value of K_p at this temperature will be:
(a) 0.61
(b) 0.57
(c) 0.83
(d) 0.46
Answer:
(a) 0.61

Question 15.
According to Le – Chatelier principle when heat is given on solid – liquid equilibrium then:
(a) Amount of solid decreases
(b) Amount of liquid decreases
(c) Temperature increases
(d) Temperature decreases
Answer:
(a) Amount of solid decreases

Question 2.
Fill in the blanks:

1. For an endothermic process, PCl₅ ⇄ PCl₃+ Cl₂; ∆H = + Q cal. Hence, in this reaction, temperature should be kept ………………………. and pressure ………………………. (If reaction is to be carried out in forward direction)
2. Ice ⇄ Water – Q cal. High temperature in this reaction favours …………………………. direction while increase in pressure favour reaction in ………………………… direction.
3. According to Ostwald’s dilution law, mathematical relation between degree of dissociation and dissociation constant is expressed by ……………………….. Degree of dissociation of weak electrolyte is inversely proportional to its …………………………..
4. Relation between solubility and solubility product for the reaction AB ⇄ A⁺ + B^– is expressed as ……………………….
5. For the maximum yield of SO₃ in the reaction 2SO₂ + O₂ → 2SO₃, ……………………… temperature and pressure is required.
6. Mixture of acetic acid and sodium acetate is an example of ……………………….. solution
7. Mixture of ammonium hydroxide and ammonium chloride is an example of ……………………… solution
8. Degree of dissociation of a weak electrolyte is inversely proportional to ………………………… of concentration.
9. Ostwald dilution law is not applicable for ……………………..
10. Heneiy’s law is related to solubility of in …………………………. solution
11. Relation between K_p and K_c at constant temperature for reaction is ………………………

Answer:

1. High, low
2. Forward, forward
3. α = \(\sqrt { \frac { K }{ C } } \), square root of concentration
4. K_sp = [A⁺] [B^–]
5. Low temperature, high pressure
6. Acidic buffer
7. Basic buffer
8. Square root
9. Strong electrolyte
10. Gas
11. K_p = K_c × RT^∆n.

Question 3.
Answer in one word/sentence:

1. What is the value of K_p and K_c for the given reaction:
   • PCl₅ ⇄ PCl₃ + Cl₂
   • H₂ + I₂ ⇄ 2HI
2. For the unit of K_c, concentration is expressed in?
3. Manufacture of nitrogen peroxide is exothermic. What should be temperature and pressure for maximum yield?
4. Ammonia gas dissolves in water giving NH₄OH. How is water reacting in it?
5. When NH₄Cl is added to NH₄OH solution, dissociation of NH₄OH decreases, why?
6. pH of water at 298K?
7. What is hydrogen ion concentration in pure water?
8. pH of water at 25°C is 7. If water is heated to 50°C, what change in its pH will occur?
9. Write conjugate base of H₂PO₄^– and HCO₃^–?
10. Name one ion which behaves as both Bronsted acid and base?

Answer:

1. • K_p > K_c
   • K_p = K_c
2. mol/litre
3. Low temperature and high pressure
4. Like acid
5. Common ion effect
6. 7
7. 1 × 10^-7
8. pH value decreases,
9. HPO₄^2- CO₃^2-
10. HCO₃^–

Question 4.
Match the following:
[I]

Answer:

1. (d)
2. (e)
3. (b)
4. (a)
5. (c)
6. (f)

[II]

Answer:

1. (d)
2. (c)
3. (a)
4. (e)
5. (b)

Equilibrium Very Short Answer Type Questions

Question 1.
For unit of Kc the concentration is expressed as?
Answer:
mol/litre.

Question 2.
Ammonia when dissolve in water give NH₄OH, here the water acts as?
Answer:
Like acid.

Question 3.
When NH₄Cl is mixed with NH₄OH solution, then the ionisation of NH₄OH decreases. What is the reason?
Answer:
Due to Common ion effect.

Question 4.
The pH of water at 25°C pH = 7. If water is heated upto 50°C, then what will be the change in pH?
Answer:
The value of pH decreases.

Question 5.
Write the conjugate base of H₂PO₄^– and HC0₃^–?
Answer:
HPO₄^2- and CO₃^2-.

Question 6.
Write the name of an ion which acts as both Bronsted acid and base?
Answer:
HCO₃^–.

Question 7.
Give an example of a salt formed by weak acid and weak base?
Answer:
Ammonium acetate.

Question 8.
What is the pH value of human blood?
Answer:
7.4.

Question 9.
What will happen when HCl gas is passed through NaCl?
Answer:
NaCl will precipitate.

Question 10.
Among conjugate bases CN^– and F^–, which is stronger base?
Answer:
CN^– is stronger base.

Question 11.
What is the change in free energy (∆G) of reversible process at equilibrium?
Answer:
Zero (0).

Question 12.
What is solubility product?
Answer:
The product of the concentration of ions in the saturated solution of a sparingly soluble salt is a constant at a given temperature and is called solubility’ product.

Question 13.
What is buffer solution?
Answer:
The solution in which on adding acid or base in iess quantity, the pH change is negligible, is called buffer solution.

Question 14.
In the exothermic reaction of ammonia formation N_2(g) + 3H_2(g) ⇄ 2NH_3(g), when be the formation of ammonia will be more?
Answer:
At low temperature and high pressure the formation of anunorua is more.

Question 15.
Give an example of acidic buffer?
Answer:
Mixture of acetic acid and sodium acetate.

Question 16.
What is the condition for precipitation?
Answer:
For this ionic product should he more than solubility product.

Question 17.
What is the nature of aqueous solution of KCN?
Answer:
It is of basic nature.

Question 18.
What is the nature of aqueous solution of CH₃COONH₄?
Answer:
It is of neutral nature.

Question 19.
What will be the relation between K_c and K_p for the reaction
PCl₅ ⇄ PCl₃ + Cl₂
Answer:
K_p > K_c.

Question 20.
What is Lewis concept?
Answer:
Acid is an electron pair acceptor and base is an electron pair donor.

Question 21.
What is the definition of acid and base according to Bronsted and Lowry concept?
Answer:
Acid is that which furnishes proton and base accepts the proton.

Question 22.
What is pH?
Answer:
The pH value of a solution is the numerical value of the negative power to which 10 should be raised in order to express the Hydrogen ion concentration of the solution i.e. [H⁺] = 10^-pH or pH = – log [H⁺].

Question 23.
The Ostwald dilution law applied on which electrolyte?
Answer:
On weak electrolytes.

Equilibrium Short Answer Type Questions – I

Question 1.
Explain the effects of catalyst at equilibrium?
Answer:
Effect of catalyst:
Le – Chatelier’s principle ignores the presence of a catalyst since the catalyst cannot displaces the equilibrium and simply reduce the time required for attaining the equilibrium when a catalyst is added to a reversible reaction in equilibrium. (MPBoardSolutions.com) It increases the speed of both forward and backward reaction i.e. r_f and r_b to the same extent. However the addition of a catalyst reduces the time required for a reaction to attain the equilibrium.

Question 2.
Explain Ostwald’s law of dilution?
Answer:
Ostwald’s time dilution law:
Ostwald gave law for weak elecrolytes. By applying law of mass action Ostwald gave a law for expressing the dissociation of weak electrolyte. It states that:
“The degree of dissociation of weak electrolyte is directly proportional to the square root of its dilution.”
α = \(\sqrt { KV } \) = \(\sqrt { \frac { K }{ C } } \)
Where, α = Degree of dissociation, K = Dissociation constant, V = Volume in litre in which one gram equivalent is dissolved, C = No. of gram equivalent in one litre.

Question 3.
What is the effect of pressure on chemical equilibrium?
Answer:
On increasing the pressure on chemical equilibrium, the equilibrium shifts in that direction where the volume decreases i.e., the number of moles decreases.
Example: On combining of SO₂ and O₂, SO₃ is formed and 45.2 kcal energy is liberated.
2SO_2(g) + O_2(g) ⇄ 2SO_3(g); ∆H = -45.2 kcal
In this reaction, 2 moles of SO₂ react with 1 mole of O₂ to form 2 moles of SO₃. So, on increasing the pressure the reaction shifts toward forward direction.

Question 4.
What do you mean by buffer solution?
Answer:
Buffer solution: Solution in which,

1. pH value is definite.
2. pH is not changed on dilution or on keeping for sometime.
3. On adding acid or base in less quantity, pH change is negligible.

Such solutions are called buffer solutions.
Or
Buffer solutions are solutions which retain their pH constant or unaltered.

Question 5.
What are acidic buffer and basic buffer?
Answer:
1. Acidic buffers:
Acidic buffers are formed by mixing an equimolar quantities of weak acid and its salt with a strong base.
Example: Acetic acid and sodium acetate, boric acid and borax, citric acid and so-dium citrate, etc.

2. Basic buffers:
Basic buffers contain equimolar quantities of weak base and its salt with strong acid.
Example: Ammonium hydroxide and ammonium chloride.

Question 6.
On the basis ofthe equation pH = – log[H⁺], the pH of the 10^-8mol dm^-3 HCl solution should be pH = 8. But the observed value is less than 7. Explain the reason?
Answer:
The 10^-8moldm^-3HCl shows that the solution is very dilute. So, we cannot neglect the H₃O⁺ ions formed from water in the solution. So, the total [H₃O⁺ ] = (10 ^-8 + 10^-7) M comes to be near 7 or less than 7 i.e., the pH is less than 7. (As the solution is acidic).

Question 7.
Ammonia is a Lewis base. Why?
Answer:
According to Lewis base, a base, is a substance (molecule or ion) which can donate an electron pair to form a co – ordinate. In the structure of ammonia, nitrogen is present in one lone pair of electron. So, ammonia donate a lone pair electron.
For example:

Question 8.
Write the uses of Buffer solution?
Answer:
Uses of Buffer solution:

1. To keep the pH of the reaction constant during the determination of velocity of chemical reactions.
2. The pH should be between pH – 5 to 6 – 8 in the formation of alcohol by fermentation.
3. Preparation of sugar, paper and electroplating occurs at definite pH.

Question 9.
What is the effect of pressure and temperature on solubility of gases in liquids?
Answer:
1. Effect of Pressure:
On increasing the pressure the solubility of gases increases as the molecules of gases enters in the intermoiecular spaces of solvent.

2. Effect of Temperature:
On increasing the temperature the solubility of gases decreases as the kinetic energy of gas molecules increases.

Question 10.
Write the name of the factors affecting the chemical equilibrium?
Answer:
The factors affecting the chemical equilibrium are:

1. Temperature
2. Pressure
3. Change in concentration
4. Catalyst.

Question 11.
Why the solubility of C0₂ decreases on Increasing the temperature?
Answer:
CO_2(g) + aq ⇄ CO_2(eq)
The solubility of CO₂ in water is an exothermic reaction. So, according to Le – Chatelier’s principle, on increasing temperature the reaction proceeds towards backward direction. That is why on increasing temperature the solubility of CO₂ decreases.

Question 12.
What do you understand by ionization of water?
Answer:
Auto ionisation occurs in water molecules. The ionic equilibrium of water can be shown by following equation:
H₂O + H₂O ⇄ H₃OH⁺ + OH^–
Equilibrium constant K = \(\frac { [H_{ 3 }O^{ + }][OH^{ – }] }{ [H_{ 2 }O]^{ 2 } } \)
⇒ K_H2O = [H₃O⁺][OH^–] [∵ K_H2O = K_w]
K_w = 1 × 10^-14
Here K_w is a constant, called ionic product of water.

Question 13.
What is concentration Quotient?
Answer:
The ratio of concentration of products and reactants is called concentration quotient. It is denoted by Q. For any reversible reaction the concentration quotient is equal to the equilibrium constant K_c.
K_c = \(\frac { [C]^{ c }[D]^{ d } }{ [A]^{ d }[B]^{ b } } \)
and Q = \(\frac { [C_{ C }]^{ c }[C_{ D }]^{ d } }{ [C_{ A }]^{ d }[C_{ B }]^{ b } } \)
At equilibrium Q = K_c.

Question 14.
What do you understand by Lewis acid and Lewis base? Explain with example?
Answer:
Lewis acid:
Acid is a substance (an atom, molecule or ion) which can accept a pair of electrons to complete its octet.
Example: BF₃, AlCl₃, Br⁺, NO₂⁺ etc.

Lewis base:
Lewis base is a substance (an atom, molecule or ion) which have complete octet of the central metal atom and have a lone pair of electron to donate in a chemical reaction to form co – ordinate bond.
Example:

Question 15.
At 310 K, ionic product of water is 2.7 × 10^-14. at this temperature determine the pH of neutral water?
Solution:
K_w = [H₃O⁺].[OH^–] = 2.7 × 10^-14 (At 310 K)
For reaction H₂O + H₂O ⇄ [H₃O⁺] [OH^–]
[H₃O⁺] = [OH^–]
So, [H₃O⁺] = \(\sqrt { 2.7\times 10^{ -14 } } \) = 1.643 × 10^-7M
∴_pH = -log [H₃O⁺] = -log 1.643 × 10^-7
= 7 + (- 0.2156) = 6.7844.

Question 16.
Determine the concentration quotient for following reactions:

1. CrO₄^-2_(eq) + Pb_(eq) ⇄ PbCrO_4(s)
   
2. CaC0_3(s) ⇄ CaO_(s) + CO_2(g)
   
3. NH_3(eq) + H₂O_(l) ⇄ NH₄⁺_(eq) + OH^–_(eq)
4. H₂O_(l) ⇄ H₂O_(g)

Answer:

Question 17.
What is the importance of solubility product in precipitation of soap?
Answer:
Soap is produced on alkaline hydrolysis of oil and fats. Soaps are actually sodium or potassium salts of higher fatty acids. (MPBoardSolutions.com) Soap is obtained in the form of concentrated liquid. Concentrated salt solution is added for its precipitation. In presence of increased concentration of common sodium ion, ionic product of [Na⁺] [C_n H_2n+1 COO] exceeds its solubility product (K_sp) and it gets precipitated. It is separated by filtration.

Equilibrium Short Answer Type Questions – II

Question 1.
By comparing the values of Kc and how will you find out the following state of the reaction:

1. Resultant reaction proceeds towards forward direction.
2. Resultant reaction proceeds towards backward direction.
3. No change in the reaction.

Answer:

1. If Q_c < K_c; reaction proceeds towards the direction by products (Forward reaction).
2. If Q_c > K_c; reaction proceeds towards reactant side (Backward reaction).
3. If Q_c = K_c; at equilibrium, the reaction mixture remains as before. So no change in the reaction.

Question 2.
The aqueous solution of sodium carbonate is basic in nature, why?
Answer:
Na₂CO₃ ⇄ 2Na⁺ + CO₃^-2
2H – OH ⇄ = 2H⁺ + 2OH^–
Na₂CO₃ + 2H – OH ⇄ 2NaOH + H₂CO₃
Neither solid Na₂CO₃ nor water alone has any action on litmus paper. However aqueous solution of Na₂CO₃ turned red litmus blue. The problem is successfully explained by the Arrhenius theory of ionization in terms of hydrolysis, when a salt is dissolved in water, it undergoes ionization. (MPBoardSolutions.com) The ions of salt interact with opposite ion of water to form acidic basic or neutral solution. This process is called hydrolysis may be defined as the interaction of ion of a salt with oppositely charged ion of water to give acidic or basic solution. Consider the hydrolysis of a general salt (AB).
Dissociation of salt
AB +Water → A⁺ + B^–
Dissociation of water

Question 3.
What is the effect on equilibrium when gases are dissolved in liquids? Explain with example?
Answer:
When the soda water bottle is opened, dissolved CO₂ gas in it comes out rapidly. This is an example of equilibrium of any gas in equilibrium between gas and its liquid. At finite pressure of the gas there exists an equilibrium between soluble and insoluble molecules of the gas.
CO_2(g) ⇄ CO_2(eq)
Henry’s law: The solubility of gas in a given solvent is directly proportional to the pressure to which the gas is subjected, provided the temperature remains the same.
Thus, m ∝ P
or m = KP
Where, K is a constant of proportionality and known as Henry’s constant.
Example: You have seen that when a soda water bottle is opened, the carbon dioxide dissolved in it freezes out rapidly. This can be explained on the basis of Henry’s law. (MPBoardSolutions.com) Carbonated beverages are bottled under pressure to ensure high concentration of carbon dioxide. When the bottle is opened the pressure above the solution falls and the excess carbon dioxide comes out.

Question 4.
What is common ion effect? Explain?
Answer:
Common ion effect:
When, in a solution of weak electrolyte, a solution of strong electrolyte containing the same ion is added, then the ionisation of weak electrolyte decreases. This effect is called common ion effect.
Example: During precipitation of radicals of group II taken care that the radicals of group IV does not get precipitated. (MPBoardSolutions.com) On the basis of this, H₂S gas is passed in presence of HCl in group II. Due to common ion effect dissociation of H₂S is suppressed.

This reduces concentration of sulphide ion and hence only radicals of group II get precipitated as sulphide. For the precipitation of radicals of group IV, H₂S is passed in presence of NH₄OH. H⁺ ion of H₂S combines with OH^– ions of NH₄OH to produce water. This increases dissociation of H₂S. In this way, increase in sulphide ion concentration helps in the precipitation of positive sulphides of group IV.

Question 5.
What do you understand by conjugate acid and conjugate base?
Answer:
The relative strengths of conjugate pairs can be found out, if we know whether forward reaction is favoured or backward reaction is favoured.
For example:

In the above reaction, the reaction proceeds almost to completion. We must therefore conclude that HCl is a stronger acid than H₃O⁺ i.e., HCl has stronger tendency to donate proton than H₃O⁺. Similarly H₂O is stronger base than H₂O has stronger tendency to accept proton than Cl^–. Thus, the strong acid (HCl) has a weak conjugate base Cl^–.
Consider another reaction

For the reaction, backward reaction is strongly favoured. This shows that H₃O⁺ is a stronger acid than CH₃COOH and CH₃COO^– is a stronger base than H₂O. Thus, we again observe that the strong acid (H₃O⁺) has the weak conjugate base (H₂O). Thus, we conclude that, A stronger acid has a weak conjugate base and vice – versa.

Question 6.
Explain the solubility product by giving definition?
Answer:
Solubility product:
The product of the concentration of ions in the saturated solution of a sparingly soluble salt as AgCl is a constant at a given temperature and is called solubility product. (MPBoardSolutions.com) If any sparingly soluble electrolyte at any temperature forms saturated solution, then the equilibrium is

According to law of mass action from (b) and (c),
\(\frac { [A^{ + }][B^{ + }] }{ [AB] } \) = K
∴ [A⁺ ][B⁺ ] = K[AB]
When solution is saturated [AB] will be constant and value of K[AB] will also be a constant and is written as KJ which is the solubility product.

Question 7.
Explain the acid – base concept of Bronsted – Lowry by giving example?
Answer:
Bronsted – Lowry concept of acid and base (1923):
Scientist Bronsted and Lowry gave the theory for acids and bases which is equally applied to aqueous and non – aqueous solutions of acids and bases. According to it, “Acid is that which furnishes proton and base accepts the proton.”

Neutralization reaction:
A reaction in which a proton is transferred from acid to a base or a reaction between H⁺ and OH^– ion to form H₂O molecule.
Relation between acid and base:

That means, every acid has conjugate base and every base has conjugate acid.

According to the definition given by Bronsted and Lowry, acid and base may be in molecular, cationic or anionic states.

Question 8.
Calculate the value of equilibrium constant for following reaction:
PCl₅ ⇄ PCl₃ + Cl₂
Answer:
Let ‘a’ mole of PCl₅, initiate the reaction and x moles dissociate at equilibrium state. If the volume of container is V litre then,

According to law of mass of action
K_c = \(\frac { [PCl_{ 3 }][Cl_{ 2 }] }{ [PCl_{ 5 }] } \)
On putting the values,

Question 9.
With the help of Le – Chatelier’s principle at equilibrium for reaction 2SO₃ + O₂ ⇄ 2SO₃;
∆H = – 188.2 kJ, determine the restrictions for the formation of sulphur trioxide?
Answer:
Le – Chatelier proposed a law. It states “If as equilibrium m £ chemical system is disturbed by changing temperature, pressure or concentration of the system, the equilibrium shift is such a way so that the effect of change gets minimised.”
On the basis of this principle, changes in chemical equilibria and physical equilibria can be explained.
Formation of SO₃ by SO₂ and O₂:
2SO₂ + O₂ ⇄ 2SO₃; ∆H = – 45.2 kcal

1. Effect of concentration:
On increasing concentration of SO₂ and O₂ more SO₃ will be formed.

2. Effect of pressure:
On increasing pressure, the equilibrium shifts on that side in which number of moles are less or towards less volume. In this reaction, 2 moles of SO₂ and 1 mole of O₂ (Total 3 moies) change into 2 moles of SO₃, So, on increasing pressure, more quantity of SO₃ will be formed.

3. Effect of temperature:
Formation of SO₃ is an exothermic reaction, so on increasing temperature, more SO₃ will dissociate and on decreasing temperature, more SO₃ will be formed.

Question 10.
What is the importance of solubility product m the purifkation of salt?
Answer:
Purification of common salt; Common salt contains impurities in it. For removal of these HCl gas is passed through the saturated solution of salt. HCl ionizes tc greater extent being a strong electrolyte.
HCl ⇄ H⁺ + Cl^– (More ionized)
NaCl ⇄ NaCl ⇄ Na⁺ + Cl^–

Equilibrium Long Answer Type Questions – I

Question 1.
Explain the buffering action of alkaline or basic buffer? Give its importance?
Answer:
Bask buffer:
Basic buffer contains equimolar quantities of weak base and sodium citrate etc.
Buffer action of basic buffer: Consider a basic buffer of NH₄OH and NH₄Cl. This buffer solution contains a large amount of NH₄⁺ ions, Cl^– ions and excess of undissociated NH₄OH molecules along with a small amount of OH^– ions.
NH₄Cl → NH₄⁺ + Cl^–
NH₄OH ⇄ NH₄⁺ + OH^–
On adding a drop of HCl, H₃O⁺ ions produced combines with OH^– ions of the buffer to form weakly ionised H₂O molecules. As a result of this ionisation of NH₄OH increases to restore the concentration of OH^– ions. Sc, pH of solution remains unchanged.

When a few drops of NaOH is added, it provides OH^– ions. The additional OH^– ions combines with NH₄⁺ ions to form weakly ionised NH₄OH resulting an increase in ionisation of NH₄OH to restore the concentration of NH₄⁺ ion. Thus, pH remains unchanged.

Importance of Buffer solution:

1. Buffer solutions are very important in laboratories, industries, botanical or zoological, physiology. Human blood is an example of buffer solution which has pH 7.34.
2. In the extraction of phosphate the buffer solution of CH₃COONa and CH₃COOH is used.
3. In industries, for the manufacturing of sugar and paper and electroplating occurs at fixed pH.

Question 2.
What is pH? Explain?
Or, What is meant by pH? What is its relationship with hydrogen ion concentration?
Answer:
To express the acidity or basicity of a solution Sorensen in 1909 established a scale known as the pH scale.
The pH value of a solution is the numerical value of the negative power to which 10 should be raised in order to express the hydrogen ion concentration of the solution, i.e;
[H⁺] = 10^-pH …………… (1)
On taking log of eqn.(1),
or log₁₀[H⁺] = log₁₀ 10^-pH
log₁₀[H⁺] = -pH log₁₀ 10
or pH = -log [H⁺], (∵ log₁₀ 10 = 1)
or pH = log₁₀\(\frac { 1 }{ [H^{ + }] } \)
This is the required relationship between pH and H⁺ ion concentration.
Thus, the pH value of a solution is the logarithm of its hydrogen ion concentration to base 10 with negative sign.
pH scale expresses acidic and basic nature of solution in terms of numbers from 0 to 14. Acidic solutions have pH value less than 7 while basic solutions have pH value greater than 7. pH value of neutral solution is 7.
Uses of pH measurement:

1. In industries:
   • In the manufacturing of alcohol by fermentation, pH is maintained between 5 and 6.8.
   • Manufacturing of paper, sugar and electroplating is being done at fixed pH.
2. In the study of velocities of chemical reactions, for maintaining pH, buffer solutions are used.
3. In qualitative analysis for removal of phosphate to maintain pH, buffer solutions of CH₃COONa and CH₃COOH are used.
4. pH of human blood is 7.34. Medium of stomach is acidic while in the intestine it is basic medium.

Question 3.
Write the characteristics of equilibrium constant?
Answer:
Characteristics of equilibrium constant:
1. The value of equilibrium constant for a reaction is constant at a given temperature and it changes with change in temperature.

2. The value of molar equilibrium constant does not depends upon the initial molar concentration of reactants and products but it depends upon their concentration at equilibrium state.

3. If the reaction is reversed the value of equilibrium constant is inversed. For example,
Then for, H_2(g) + I_2(g) ⇄ 2HI, K_c = 50
2HI_(g) ⇄ H_2(g) + I_2(g)
∴ K’ = \(\frac { 1 }{ K } \) or K’_c = \(\frac{1}{50}\) = 0.02

4. If the equation having equilibrium constant is divided by 2, the equilibrium constant for the new equation is the square root of K (i.e., \(\sqrt { K } \))
For example:

5. If the equation (Having equilibrium constant K) is multiplied by 2, the equilibrium constant for the new equation is square of K (i. e., K²)
K’ = K²

Question 4.
Give only the definition of:

1. Salt hydrolysis
2. Solubility product
3. Common ion effect
4. Buffer solution

Answer:
1. Salt hydrolysis:
The interaction of cation/anion or both with water making the solution acidic or basic is called salt hydrolysis.

2. Solubility product:
At a particular temperature the product of concentration of ions in a saturated solution of a sparingly soluble electrolyte is known as solubility product. (K_sp).

3. Common ion effect:
The dissociation of weak electrolyte (Weak acid or weak base) is suppressed by the addition of a strong electrolyte having a common ion is called common ion effect.

4. Buffer solution:
Buffer solutions are solutions which retain their pH constant or unaltered after addition of less quantity of acid or base.

Question 5.
Before precipitation of IIIrd group radicals as hydroxides on adding NH₄OH, why it is necessary to add NH₄Cl first?
Answer:
Precipitation of hydroxide of group ill:
Radicals of group III i e.. Al³⁺, Fe³⁺, Cr³⁺ gets precipitated as hydroxides. Radicals cf succeeding groups such as Zn²⁺, Ni²⁺, Mg²⁺, etc. are also precipitated as hydroxide. (MPBoardSolutions.com) Solubility product of hydroxide of group III is lesser than that of other as given below:
Table: Solubility Product of Hydroxide at 18°C

Due to the difference in the solubility product of the hydroxides, concentration of OH^– ion is kept low such that only radicals of group III gets precipitated.
Hence, NH₄OH is added before adding NH₄Cl which remains completely ionized. Due to common (NH₄⁺) ion dissociation of weak electrolyte NH₄OH is suppressed.
NH₄Cl ⇄ NH₄⁺ + Cl^– (More ionised)
NH₄OH ⇄ NH₄⁺ + OH^– (Less ionised)
This decreases comcentration of OH^– ion and only hydroxides with low solubility product gets precipated.

Equilibrium Long Answer Type Questions – II

Question 1.
Derive a relation between equilibrium constant K_p and K_c, Or, Prove that K_p = K_cRT^∆n?
Answer:
In year 1867 Guldberg and Waage put forward a relationship between the rate of reaction and the molar concentration of reactants. The reUucrjhip is.0 as law of mass action. (MPBoardSolutions.com) It states that “At constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentration of the reaction each raised to a power equal to its corresponding stoichiometric coefficient that appears in the balanced chemical equation.”
Example: For solids and liquids: Consider a hypothetical reversible reaction in the state of equilibrium.
A + B ⇄ X + Y
Applying law of mass action: Rate of forward reaction
R_f ∝ [A] [B] or R_f = K_f[A][B] ……………… (1)
Where, K_f is rate constant or velocity constant for the forward reaction. Similarly, rate of backward reaction,
R_b ∝ [X] [Y] or R_b = K_b [X][Y] ………………… (2)
Where, K_b rate constant or velocity constant for the backward reaction.
At equlibrium,
Rate of forward reaction = Rate of backward reaction.
K_f[A][B] = K_b[X][Y]

Its value remains constant at particular temperature. The equation (3) is called law of chemical equilibrium. For a general type of the reaction,
aA + bB ⇄ xX + yY

For gaseous reactants and products:
In the expression for K_c, the concentrations of the various species are generally expressed in terms of moles/litre. (MPBoardSolutions.com) However, in case of gaseous reactions the concentrations of gases may also expressed in terms of their partial pressures. Therefore, for a gaseous reaction:
aA + bB ⇄ xX + yY
The law of chemical equilibrium may be expressed as:

Relationship between K_pc:
Let us consider the following general reaction,
aA + bB ⇄ xX + yY
In which all the substances A, B, X and Y are present in gaseous state. For this reaction, K_p and K_c is written as follows:

Since, for an ideal gas,
PV = nRT
∴ P = \(\frac{n}{V}\) RT = CRT ………………… (3)
Where, the term C (equal to \(\frac{n}{V}\)) represents the molar concentration of the gas. On the basis of equation (3), we have
P_A = C_ART = [A] RT
P_B = C_BRT = [B] RT
P_X = C_X RT = [X] RT
P_Y = C_Y RT = [Y] RT
Substituting this value in equation (2), we get

Proved.

Question 2.
Derive Ostwald’s dilution law of ionization of weak electrolytes? Write its limitations?
Or, Deduce the relationship between degree of ionization and ionization constant?
Answer:
Ostwald’s dilution law: Weak electrolytes are partially ionised. The ions produced due to ionisation of weak electrolyte exist in dynamic equilibrium with the undissociated molecules.
The fraction of the total number of molecules of electrolyte dissolved, which ionises at equilibrium is called degree of ionisation or degree of dissociation. It is denoted by α.
Consider ‘C’ mol per litre be the initial concentration of weak electrolyte AB dissolved in water. Let α be its degree of ionisation. (MPBoardSolutions.com) Thus, the molar concentration of different species before ionisation and at equilibrium be as given below:

Under normal conditions, value of a is very small for weak electrolyte and it can be neglected in comparison to 1.
∴K = Cα²
or α² = \(\frac{K}{C}\)
∴α = \(\sqrt { \frac { K }{ C } } \)
If V is the volume of the solution in litres containing 1 mole of electolyte, then C = \(\frac{1}{V}\). Thus,
α = \(\sqrt { KV } \)
The above equation helps us to conclude that:
The degree of ionisation is inversely proportional to the square root of the molar concentration or directly propotional to the square root of the volume of the solution containing one mole of electrolyte. This is called Ostwald’s dilution law.

Limitations:
This law is not applicable for strong electrolyte and concentrated solutions because attractive force also acts between the solutions of such electrolytes.

Question 3.
Derive a relationship between pH and pOH value? Or, Prove that pH + pOH = 14.
Answer:
The water ionizes as:
H₂O + H₂O ⇄ H₃O⁺ + OH^–
K = \(\frac { [H_{ 3 }O^{ + }][OH^{ – }] }{ [H_{ 2 }O]^{ 2 } } \)
K [H₂O ]² = [H₃O⁺] [OH^–]
[∵K [H₂O ]² = K_w]
K_w = [H₃O⁺][OH^–] ……………. (1)
K_w is a constant, it is called product solubility.
At 298K, Temperature K_w = 1 × 10^-14.
On putting the value in equation (1)
10^-14 = [H₃O⁺][OH^–]
Taking log on both sides,
-14 log₁₀10 = log₁₀[H₃O⁺] + log₁₀[OH^–]
-14 = log₁₀[H₃O⁺] + log₁₀[OH^–], [∵log₁₀10 = 1]
0r 14 = [-log₁₀[H₃O⁺]] + [-log₁₀[OH^–]]
[∵log₁₀ [H₃O⁺] = pH]
[∵log₁₀ [OH^–] = pOH]
14 = pH + pOH

Question 4.
For the determination of pH value of buffer solution derive Henderson – Hazel equation?
Answer:
Henderson’s equation:
pH of a buffer solution can be calculated with the help of Henderson’s equation. For this consider a buffer of weak acid HA and its salt.
HA ⇄ H⁺ + A^–
K_a = \(\frac { [H^{ + }][A^{ – }] }{ [HA] } \)
K_a is dissociation constant of acid.
or [H⁺] = K_a \(\frac { [HA] }{ [A^{ – }] } \)
Salt is completely ionised while due to presence of excess A^– from the salt, the dissociation of weak acid will be depressed more due to common ion effect.
or [H⁺] = K_a \(\frac { [Acid] }{ [Salt] } \) [A^–] ≈ [Salt]
Taking log value,

This equation is called Henderson equation.
In the same way, for basic buffer solution:

pH of a buffer solution does not change woth dilution because on dilution the ration of conc. of acid or base does not change.

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 6 Thermodynamics

Thermodynamics Important Questions

Thermodynamics Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
In adiabatic expansion of an ideal gas always:
(a) Increase in temperature
(b) ∆H = 0
(c) q = 0
(d) W = 0
Answer:
(c) q = 0

Question 2.
For a reversible process, free energy change at equilibrium:
(a) More than zero
(b) Less than zero
(c) Equal to zero
(d) None of these.
Answer:
(c) Equal to zero

Question 3.
In isothermal expansion of an ideal gas:
(a) 9 = 0
(b) AE = 0
(c) W = 0
(d) dV = Q
Answer:
(b) AE = 0

Question 4.
Hess’s law is an application of the following:
(a) First law of Thermodynamics
(b) Second law of Thermodynamics
(c) Entropy change
(d) Free energy change.
Answer:
(a) First law of Thermodynamics

Question 5.
When the value of heat of neutralization of an acid with a base is 13.7 kcal, then:
(a) Acid and base both are weak
(b) Acid and base both are strong
(c) Acid is strong and base is weak
(d) Acid is weak and base is strong
Answer:
(b) Acid and base both are strong

Question 2.
Fill in the blanks:

1. Enthalpy is an …………………….. property.
2. Nicely closed thermos flask is an example of an ……………………….
3. Value of heat of combustion (∆H) is always ………………………..
4. Extensive property depend on the ………………………. of matter.
5. Value of heat of neutralization is always …………………………. kilocalorie.

Answer:

1. Extensive
2. Isolation
3. Negative
4. Amount
5. – 13.7

Question 3.
Answer in one word/sentence:

1. Tell the value of heat of neutralization of strong acid and strong base?
2. Give two examples of state function?
3. What is entropy?
4. Among NaCl, H₂O_(s) and NH_3(g) value of whose entropy is higher?
5. Expect the system, what is the remaining part of the universe called?
6. System in which changes occur spontaneously and by which entropy of the system increases, what are they called?
7. What is the type of heat of combustion?

Answer:

1. – 57 kJ
2. Enthalpy, Entropy
3. Measure of disorder
4. NH₃
5. Surroundings
6. Spontaneous process
7. Exothermic.

Thermodynamics Very Short Answer Type Questions

Question 1.
Give two examples of state function?
Answer:
Enthalpy and Entropy.

Question 2.
Whose entropy is greater among NaCl, H₂O_(s) and NH_3(g)?
Answer:
NH_3(g).

Question 3.
Who gave the equation ∆G = ∆H – T∆S?
Answer:
Gibbs – Helmholtz.

Question 4.
Write the equation of first law of thermodynamics?
Answer:
∆E = q + W.

Question 5.
What is the value of entropy when ice melts?
Answer:
Entropy increases.

Question 6.
What is Closed system?
Answer:
System which can exchange energy only and not matter with the surroundings.

Question 7.
The value of which enthalpy is always negative?
Answer:
Enthalpy of combustion.

Question 8.
Heat of neutralization of strong acid and strong base is equal to?
Answer:
-13.7 kcal or -57.1 kJ.

Question 9.
What is Adiabatic process?
Answer:
A process in which no exchange of heat between system and surroundings occur is known as adiabatic process.

Question 10.
What is Enthalpy?
Answer:
Heat change at constant pressure is known as enthalpy.

Question 11.
The process in which pressure remains constant is called?
Answer:
Isobaric process.

Question 12.
In Exothermic reaction the value of ∆H is?
Answer:
Negative.

Question 13.
Unit of specific heat capacity is?
Answer:
joule per kelvin per gm.

Question 14.
The relation between ∆G, ∆S and ∆H is given by?
Answer:
∆G = ∆H – T∆S.

Question 15.
Which type of property is heat capacity?
Answer:
Extensive property.

Question 16.
What type of properties are temperature, pressure and surface tension?
Answer:
Intensive property.

Question 17.
What is the unit of molar heat capacity?
Answer:
joule kelvin^-1 mol^-1.

Question 18.
For which process dq = 0?
Answer:
Adiabatic process.

Question 19.
The efficiency of any fuel is measured by which value?
Answer:
Calorific value.

Question 20.
What is entropy?
Answer:
The measurement of degree of disorder or randomness of the molecule of the system.

Question 21.
Write the relation between standard free energy change ∆G° and equilibrium constant (K)?
Answer:
∆G° = -RTlnK.

Question 22.
What is the value of ∆G for Spontaneous process?
Answer:
∆G < 0.

Question 23.
“The electricity obtained from an electrochemical ceil is equivalent to decrease in free energy”. Write expression for this sentence?
Answer:
∆G° = -nE°F.

Thermodynamics Short Answer Type Questions – I

Question 1.
What is System?
Answer:
System:
A specified portion of the universe which is selected for experimental or theoretical investigations is called the system. (MPBoardSolutions.com) In the system, the effects of certain properties such as pressure, temperature, etc. are observed. A system is said to be homogeneous if it consists of only one phase. On the other hand, it is heterogeneous if it consists of more than one phase.

Question 2.
What is process and what are its kinds?
Answer:
The operation which brings about change in the state of a system is called a thermodynamics process.
Thermodynamics process may be further classified as follows:

1. Isothermal process
2. Adiabatic process
3. Isobaric process
4. Isochoric process
5. Reversible process
6. Irreversible process
7. Cyclic process.

Question 3.
Explain Exothermic reaction with example?
Or, Explain why the value of enthalpy change negative for exothermic reaction?
Answer:
The reactions in which heat is emitted are called exothermic reactions. In such reactions total heat content of products is less than that of reactants. Hence, ∆H is negative.
Example: 2NO_(g) → N_2(g) + O_2(g); ∆H = – 180.5 kJ/mol.

Question 4.
Explain extensive and intensive properties?
Answer:
1. Intensive properties:
The properties of the system which are independent of the amount of matter present in it, are called intensive properties.
Example: Temperature, viscosity, surface tension, refractive index, specific heat, density, etc.

2. Extensive properties:
The properties of the system which depend upon the amount of matter present in it, are called extensive properties.
Example: Mass, volume, energy, etc.

Question 5.
State and Explain Zeroth law of Thermodynamics?
Answer:
According to this law, “Two bodies which are separately in thermal equilibrium with a third body are also in thermal equilibrium with each other”.

Question 6.
Explain Enthalpy of neutralization with example?
Answer:
The enthalpy of neutralization is defined as:
“Change in enthalpy when one gram equivalent of an acid is neutralized with one gram equivalent of base in dilute solution at constant temperature.
Example: NaOH_(aq) + HCl_(aq) ⇄  NaCl_(aq) + H₂O_(l); ∆H =- 57.1 kJ

Question 7.
The enthalpy of neutralization of weak acid and weak base is less than enthalpy of neutralisation of strong acid and base. Why?
Answer:
If either acid or base weak then its ionisation in solution remains incomplete. As a result a part of energy liberated during combination of H+ and OH” ion is used up for the ionisation of weak acid and weak base. Therefore, the value of enthalpy of neutralisation of weak acid with strong base or vice – versa is numerically less than – 57.1 kJ.

Question 8.
What is Bond enthalpy and bond dissociation energy?
Answer:
It is a well known fact that during the formation of a chemical bond, energy is required. Therefore the breaking of a bond energy is to be supplied. Thus, the energy required to break a particular bond in a gaseous molecule is called bond dissociation energy.
Example: 2HCl_(g) → H_2(g) + Cl_2(g)

Question 9.
What is the first law of thermodynamics? Write its mathematical form?
Or, Write the first law of thermodynamics and derive the mathematical expression of it?
Answer:
First law of thermodynamics is the law of conservation of energy .The common statement of this law is:
“Energy can neither be created nor be destroyed but it can be converted from one form to another form.” Let internal energy of the system is E₁ and q calorie heat is supplied to the system. E₂ is the energy of the final stage and work done is W. Therefore,
E₂ – E₁ = q + W
or ∆E = q + W.

Question 10.
Define the term Entropy?
Answer:
A change that brings about disorder or randomness is more likely to occur than one that brings about order. To account for the randomness or disorder of a system a state function called entropy was introduced. It is (MPBoardSolutions.com) defined as the measure of degree of disorder or randomness of the molecule of the system. Entropy is represented by symbol S. It is easier to define entropy change than entropy of a system.

Question 11.
Whose entropy is more: Water vapour, water or ice, why?
Answer:
Entropy is the measure of randomness. In solid state the molecules are completely arranged, therefore its entropy is minimum and in gas the molecules move randomly in all directions, so the value of entropy is more. So
S_(ice) < S_(water) < S_{(water vapour)}.

Question 12.
Among NaCl, H₂O and NH₃ whose entropy will be maximum and why?
Answer:
Entropy is the measure of randomness. In solid state the molecules are regularly arranged, so entropy is minimum whereas in gas the molecules move randomly in all directions, so the entropy is maximum. So, in the above example, NaCl is a solid, H₂O is liquid and NH₃ is a gas. So, the entropy of NH₃ is maximum and entropy of NaCl is minimum.

Question 13.
Prove that: P∆V = ∆ntRT?
Answer:
From ideal gas equation
P V = nRT
If the volume of gas at initial state is V₁ and the number of moles of gas is n₁ then,
PV₁ = ∆n₁RT …………….. (1)
If at final state, the volume of gas is V₂ and number of moles of gas is n₂, then,
PV₂ = ∆n₂RT …………. (2)
From eqn. (i) and (2),
P(V₂ – V₁) = (n₂ – n₁)RT
or P∆V = ∆nRT.

Question 14.
What is relation between ∆H and ∆U?
Answer:
If H be the enthalpy of any system and U is the internal energy, then the relation will be
H = U + PV
For enthalpy change, ∆H = ∆U + P∆V
We know that P∆V = ∆nRT
On putting the value,
∆H = ∆U + ∆nRT.

Question 15.
What do you mean by specific heat capacity?
Answer:
It is the heat required to raise the temperature of 1 gram of substance by 1°C. It is denoted by C_s.
C_s = \(\frac{C}{m}\)
Where, C = Heat capacity, C_s = Specific heat capacity, m = Mass of substance.
Its S.I. unit is joule K^-1g^-1.

Question 16.
What do you mean by molar heat capacity?
Answer:
It is the heat required to raise the temperature of 1 mole of substance by 1°C.
Molar heat capacity = \(\frac{C}{n}\) \(\frac{q}{∆T × n}\)
Where C = Heat capacity (Absorbed heat), ∆T = Increase in temperature, n = Number of moles.
Its S.I. unit is joule K^-1 mol^-1.

Question 17.
Write Hess’s law?
Answer:
In 1840, G.H. Hess formulated a law known as Hess’s law. According to law, “The enthalpy change in a physical or chemical change is same whether the process is carried out in one step or in several steps.”

Question 18.
What is Adiabatic process?
Answer:
A process in which no exchange of heat between system and surroundings occur is known as adiabatic process. This process mainly occurs in isolated system. For this type of process dq = 0.

Question 19.
What is standard enthalpy of reaction?
Answer:
The enthalpy change takes place at standard state i.e., at 298 K temperature (25°C), 1 atm pressure (760 mm) condition is called standard enthalpy of reaction. It is denoted by ∆H°_r or ∆_rH°.

Question 20.
What is enthalpy of solution? Explain with example?
Answer:
Enthalpy change taking place during the dissolution of one mole of a substance in excess of a solvent such that further addition of solvent does not produce any heat change is known as enthalpy of solution.
Example: KCl_(s) + aq → KCl_(aq)

Question 21.
What is enthalpy of hydration? Explain with example?
Answer:
Enthalpy change taking place when one mole of anhydrous salt combines with the required number of moles of water to form hydrated salt is called enthalpy of hydration.
Example: CuSO_4(s) + 5H₂O_(l) → CuS0₄.5H₂O ∆H = – 78.2kJ.

Thermodynamics Short Answer Type Questions – II

Question 1.
What is law of energy conservation? Write its mathematical expression. Or, What is the first law of thermodynamics? Deduce its mathematical expression?
Answer:
This law was first expressed by Meyer and Helmholt. According to this law, “Energy cannot be created or destroyed although one form and vice – versa”.
Or
Whenever a certain quantity of energy in one form disappear, an equivalent amount of energy in another form reappear.

Mathematical form:
Let us suppose that system has internal energy equal to U₁. If it absorbs heat energy 17 from the surroundings then internal energy will increase and becomes U₁ + q. If the work is done on the system then its final internal energy will become
U₂ = U₁ + q + w
U₂ – U₁ = q + w,
∆U – q + w, (∴ U₂ – U₁ = ∆U
If the work done on the system is w, then
∆U = q + w
If work done by the system
∆U = q – w

Question 2.
What is Enthalpy of fusion? Explain with example?
Answer:
Enthalpy change taking place during the conversion of 1 mole of a solid substance into liquid at its melting point is known as enthalpy of fusion.
Example: Enthalpy of fusion of ice at 273 K is 6.0 kJ.
H₂O_(s) → H₂O_(l);
i.e 6.0 kJ of energy is absorbed for the conversion of 1 mole of ice into water. ∆H = +6.0 KJ

Question 3.
For an isolated system if ∆U = 0, what will be ∆S?
Answer:
For an isolated system, AU = 0 and for a spontaneous process, the change in entropy should be positive. For example: For a closed container, which is an isolated system, two gases A and B are diffused (MPBoardSolutions.com) Both gases A and B are separated by a kinetic separator. When the separator is removed, the gases starts fusing with each other and randomness increases in the system. For this process, ∆S > 0 and ∆U = 0.
Again ∆S = \(\frac { q_{ rev } }{ T } \) = \(\frac { \Delta H }{ T } \) = \(\frac { \Delta U+P\Delta V }{ T } \) = \(\frac { P\Delta V }{ T } \) [∴∆U = 0]
So, T∆S or ∆S > 0.

Question 4.
On the basis of the following equations, write a note on thermodynamic stability of NO_(g):
\(\frac{1}{2}\) N_2(g) + \(\frac{1}{2}\) O_2(g) → NO_(g); ∆_rH° = 90kJ mol^-1
NO_(g) + \(\frac{1}{2}\) O_2(g) → NO_2(g); ∆_rH° = 94kJ mol^-1
Answer:
NO_(g) is unstable, as the preparation of NO is an endothermic reaction. But preparation of NO_2(g) occurs because it is an exothermic reaction (energy evolved). So unstable NO_(g) is converted into NO_2(g).

Question 5.
At equilibrium state, whose value will be zero ArG or ArG°?
Answer:
∆_rG = ∆_rG° + RTlnK
At equilibrium, 0 = ∆_rG° + RTlnK
or ∆_rG° = – RTlnK
∆_rG° = 0 (When K = 1)
For other values K, ∆_rG° will be zero.

Question 6.
What is the meaning of calorific value of any fuel? Explain with example?
Answer:
The heat or energy evolved in joule or calorie by the (combustion) burning of 1 gram food or fuel is known as calorific value of fuel.
C₂H₁₂O_6(s) + 6O_2(g) → 6CO_2(g) + 6H_2(g); ∆H = -2840 kJ
In this process from 1 mole glucose or 180 gm = 2480 kJ energy obtained
So, energy obtained from 1 gm glucose = \(\frac{2480}{180}\) = 15.78 kJ/gm.
So, calorific value of glucose is 15.78 kJ/gm.

Question 7.
Explain heat of vapourization and heat of reaction?
Answer:
Heat or Enthalpy of vapourization : Enthalpy change taking place during the conversion of 1 mole of liquid into vapours at its boiling point and 1 atm pressure is called enthalpy of vapourization.
Example:
H₂O ⇄ H₂O_(g); ∆H = +40.7 kJ

Heat of reaction:
Enthalpy change taking place when number of moles of reactants as represented by the chemical equation have completely reacted is known as enthalpy of reaction. It is denoted by ∆H_f.
Example: C_(s) + O_2(g) → CO_2(g); ∆H_f = – 393.5 kJ/mol

Question 8.
Explain enthalpy of fusion and enthalpy of sublimation with example?
Answer:
Enthalpy of fusion:
Enthalpy change taking place during the conversion of I mole of a solid substance into liquid at its melting point is known as enthalpy of fusion.
H₂O_(s) ⇄ H₂O_(l); ∆H = +6.01 kJ
Enthalpy of Sublimation:
Enthalpy change taking place when one mole of solid changes into vapours without passing into intermediate liquid state at a temperature below its melting point is known as enthalpy of sublimation.
I_2(s) ⇄  I_2(g). ∆H = +62.4 kJ

Question 9.
How the change in entropy takes place in vapourisation process?
Answer:
Entropy of vapourisation:
Change in entropy when one mole of a liquid at its boiling point changes to the vapour state at the same temperature.
∆S_vapour = S_(vapour) – ∆S_(liquid) = \(\frac { \Delta H_{ vap } }{ T_{ b } } \)
Here, ∆_vap H is the latent heat of vapourisation (enthalpy of vapourisation) and T_b is the boiling point when liquid is converted into vapour, entropy of the system increases. Thus, ∆S_vap is +ve.

Question 10.
What is Internal Energy? Is its absolute value can be determined?
Answer:
Physical and chemical process occurs by some energy change. This energy may appear in the form of light, heat and work. (MPBoardSolutions.com) This evolution and absorption of energy clearly shows that every’ substance or system is associated with some definite amount of inherent energy’. The actual value of this inherent energy depends upon:

1. Chemical nature of substance
2. Conditions, like temperature, pressure and volume, and
3. Composition of the substance.

Thus, “The energy stored within a substance is called internal energy or intrinsic energy”.
Actually, internal energy is the sum of various forms of energy such as; electronic energy E_e, nuclear energy E_n, chemical bond energy E_c, potential energy E_p and kinetic energy E_k . Kinetic energy is the sum of translational energy (E_t), vibrational energy (E_v) and rotational energy (E_r).
It is represented by the symbol ‘U’,
U = E_e + E_n + E_c + E_p + E_k
It may be noted that, absolute value of internal energy cannot be determined, because it is not possible to determine the exact values for the constituent energies, such as: translational, vibrational, rotational energies etc.

Question 11.
Expansion of any gas in vacuum is called free expansion. 1 L of an ideal gas expands isothermally upto 5 L, then determine the change in internal energy and work done?
Solution:
Work done, W = – P_(external) (V₂ – V₁)
When P_(external) = 0,
So, W = – 0 (5 – 1) = 0
For isothermal expansion,
∆U = 0
So, ∆T = 0.

Question 12.
An ideal gas filled in a cylinder (according to figure.) is compressed in a single step with external pressure P_(external) Then what will be the work done on the gas? Explain with graph?
Solution:
Let the initial volume of the gas is V_i and pressure of cylinder is P. On compressing the gas by pressure P the final volume of gas is V_f.
So, change in volume ∆V = (V_f – V_i)
If W is the work done by the piston on the system
W = P_(external) (- ∆V)
W = P_(external) (V_f – V_i)
This can be shown in the figure by (P – V) graph, The work done is equal to ABV_fV_i. The positive sign shows that the work is done on the system.

Thermodynamics Long Answer Type Questions – I

Question 1.
What is heat capacity? Deduce the expression C_p – C_v = R?
Answer:
Heat Capacity: It is equal to the amount of heat required to raise the temperature of the system through 1°C. Its unit is JK^-1.
Relationship between C^p and C^v:
At constant volume: q_v = C_v∆T = ∆U
At constant pressure: q_p = C_p∆T = ∆H
∆H and ∆U are related to each other as
∆H = ∆U + ∆n_gRT
or ∆H = ∆U + ∆n_g(PV)
For 1 mole of ideal gas PV = RT
∴ ∆H = ∆U + ∆(RT)
or ∆H = ∆U + R∆T
on putting the value of ∆H and ∆U
C_p∆T = C_v∆T = C_v∆T + R∆T
Dividing whole equation by ∆T,
C_p = C_v + R
C_p – C_v = R
Thus, value of C_p is always more than C_v and the difference between them is about 2 calories or 8.314 joule.
This relationship is known as Meyer’s relationship.
The ratio C_p/C_v:
The ratio of molar heat capacities at constant pressure (C_p) to that at constant volume (C_v) is represented by γ. Value of γ gives information about the atomicity of the gas. Thus,
For monoatomic gases γ = 1.67
For diatomic gases γ = 1.40
For triatomic gases γ = 1.30

Question 2.
Explain Enthalpy of combustion? Write its uses also?
Answer:
Enthalpy of combustion:
The enthalpy change taking place when one mole of substance is completely oxidised or burnt in presence of excess of oxygen is known as enthalpy of combustion. For example, the enthalpy of combustion of methane is 890.4 kJ.
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(g) ∆Hc = – 890.4 kJ

Carbon on the other hand is oxidised to carbon monoxide and carbon dioxide.
C_(s) + \(\frac{1}{2}\) O_2(g) → CO_(g); ∆H = – 110.5kJ
and C_(s) + O_2(g) → CO_2(g); ∆H_c = – 393.5 kJ

In this case enthalpy of combustion of carbon is – 393.5 kJ and not – 110.5 kJ as formation of CO is a result of incomplete combustion of carbon.
Uses of Enthalpy of combustion:

1. To determine the calorific value of fuel.
2. To determine the enthalpy of reaction of compounds.
3. Determination of structure of compounds.
4. To calculate the calorific value of food.

This enthalpy calculator calculates the enthalpy (the measure of heat content) of a substance.

Question 3.
Prove that qr = ∆H_p?
Or, Prove that at constant pressure and constant temperature the heat of reaction is equal to the change in enthalpy of the system?
Answer:
Suppose enthalpy, internal energy and volume of a system in initial state are H₁, U₁ and V₁ respectively and after gaining heat these values becomes H₂, U₂ and V₂ respectively, then according to definition.
H₁ = U₁ + PV₁, (in initial state) ……………… (1)
H₂ = U₂ + PV₂, (in final state) …………….. (2)
Subtracting eqn. (1) from eqn eqn. (2),
H₂ – H₁ = U₂ – U₁ + P(V₂ – V₁)
or ∆H = ∆U + P∆V
Where ∆H is enthalpy change, ∆U is change in internal energy and ∆V is change in volume. Therefore at constant pressure enthalpy change is equal to sum of internal energy change and expansion type of mechanical work.
According to first law of thermodynamics,
∆U = q – P∆V
q = ∆U + P∆V
= (U₂ – U₁) + P(V₂ – V₁)
= (U₂ + PV₂) – (U₁ + PV₁)
= H₂ – H₁
= ∆H
∴ ∆H = q_p
Thus, enthalpy change represents the heat change occurring at constant temperature and pressure. It is noteworthy that though q is path dependent, q_p is not because ∆H is a state function.

Question 4.
Differentiate between Reversible and Irreversible processes?
Answer:
Differences between Reversible and Irreversible process:
Reversible process:

1. It is carried out infinitesimally slowly i.e., the difference between driving force and the opposing force is very very small.
2. It is an ideal process requiring infinite time for completion.
3. Equilibrium is not disturbed at any stage during the process.
4. Work obtained is maximum.
5. It is an imaginary process and cannot be realised in actual practice.

Irreversible process:

1. This process is carried out rapidly /.e.,the difference between driving force and the opposing force and the opposing force is quite large.
2. It is a spontaneous process requiring finite time for completion.
3. Equilibrium may occur only after the completion of the process.
4. Work obtained is not maximum.
5. It is a natural process which occurs in
6. particular direction under given set of conditions.

Question 5.
What are the factors affecting enthalpy of reaction?
Answer:
Factors on which enthalpy of a reaction (∆H) depends: Enthalpy of a reaction i.e., ∆H depends upon the following factors:
1. Physical state of reactants and products:
Enthalpy of reaction is affected by the physical state of reactants and products because latent heat of substance is also involved. For example, value of enthalpy of reaction for the formation of liquid water and water vapour is different.
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 286 kJ
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 249 kJ

2. Quantities of the reactants involved:
Enthalpy of reaction is affected by the physical state of reactants and products because latent heat of substance is also involved. For example, value of enthalpy of reaction for the formation of liquid water and water vapour is different.
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 286 kJ
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 249 kJ

3. Allotropic modifications:
Allotropes of an element may have different enthalpies. For example, enthalpy change during combustion of graphite and diamond is – 393.5 kJ/mol^-1 and – 395.4 kJ/mol^-1 respectively.
C_(graphite) + O_2(g) → CO_2(g); ∆H = – 393.5 kJ
C_(diamond) + O_2(g) → CO_2(g); ∆H = – 395.4 kJ

4. Temperature:
Value of enthalpy of reaction is dependent on the temperature at which the reaction is carried out. For example, at 25°C enthalpy of formation of HCl_(g) is 184.6 kJ while at 75°C it is 184.4 kJ.
H_2(g) + Cl_2(g) → 2Hl_(g); ∆H = 184.6 kJ at 25°C
H_2(g) + Cl_2(g) → 2HCl_(g); ∆H = 184.4 kJ at 75°C

Question 6.
Prove that at constant volume q_v = ∆U?
Answer:
When a reaction occurs at constant volume, then no work is done by the system. So W = 0.
So, ∆U = q + W
On putting the value, ∆U = q
So, at constant volume the energy is absorbed which increases the internal energy of the system. So,
∆U = q + W = q + p∆V, (W = P∆V)
Since, the reaction occurs at constant volume. So, ∆V = 0.
On putting vallue, ∆U = q_v

Question 7.
From following data determine the heat of reaction of CH₄ or enthalpy, ∆H:
C_(s) + O_2(g) → CO₂; ∆H = – 97k cal ………… (1)
2H_2(g) + O_2(g) → 2H₂O_(g) ∆H = – 136k cal …………. (2)
CH₄ + 2O_2(g) → CO_2(g) + 2H₂O_(g); ∆H = – 212k cal ………………… (3)
Solution:
To determine
C_(s) + 2H_2(g) → cH_4(g); ∆H = ?
On adding eqn.(1) and (2),

Thermodynamics Long Answer Type Questions – II

Question 1.
What is Hess’s Law of constant heat summation? Explain with an example?
Answer:
In 1840, G.H. Hess gave an important law of constant heat summation according to which, “The enthalpy change in a particular reaction is always constant and does not depend on the path in which reaction takes place”.
Or
“The enthalpy change in a physical or chemical process is the same whether the process is carried out in one or in several steps.”
This law is based on the law of conservation of energy. Suppose that the conversion of substance A to (MPBoardSolutions.com) substance Z takes place in a single step by first method and through several steps in second method. In single step by first method and through several steps in second method. In single step:
A → Z + Q₁

Where, Q₁ is the energy in several steps:
A → B + q₁
B → C + q₂
C → Z + q₃
Total energy evolved in several steps = q₁ + q₂ + q₃
= Q₂ calories (suppose)
Acoording to Hess’s law,
Q₁ = Q₂
Suppose Hess’s law is incorrect and Q₂ > Q₁. In this stage if we convert A to Z by several steps and then Z directly to A, then heat equal to (Q₂ – Q₁) is produced. By repeating this cyclic process several times, an (MPBoardSolutions.com) unlimited amount of heat (energy) may be produced in an isolated system. But this is against the law of conservation of energy.
Practically also, Hess’s law is proved to be true.
Example: Carbon can be directly burnt to produce C0₂ or in the second method it is first converted to carbon monoxide and then oxidized to carbon dioxide.

Energy evolved by both the methods is nearly same. Different of 0.3 kcal is due to experimental error.
∴ ∆H = ∆H₁ + ∆H₂

Question 2.
Prove that ∆H = ∆U + P∆V?
Or, Explain the relationship between ∆H and ∆U?
Answer:
Relation between AH and AU: In case of solids and liquids, the difference between ∆H and ∆U is not significant but in gases it is significant. Let us consider a reaction involving gases. Let the process be isothermal and carried out at constant pressure (P). If V_A is the total volume of the gaseous reactants and V_B be the total volume of gaseous products, also n_A be the number of moles of gaseous reactants and n_B be the number of moles of gaseous products. Then
PV_A = n_ART ……………………. (1)
and PV_B = n_BRT ………………….. (2)
Substarcting eqn. (1) from eqn. (2) we get
PV_B – PV_A = (n_B – n_A)RT
or P(V_B – V_A) = (n_B – n_A)RT
P∆V = ∆nRT
For gaseous reactants PV_R = n_RRT …………………… (3)
and for gaseous products PV_p = n_pRT …………………… (4)
Substracting eqn. (3) from eqn. (4), we get
P(V_p – V_R) = (n_p – n_R) RT
or P∆V = ∆n_gRT ……………………. (5)
But enthalpy change ∆H = ∆U + P∆V …………………….. (6)
Substracting the value of P∆V from eqn. (5) into eqn. (6), we get
∆H = ∆U + ∆n_gRT ………………. (7)
Thus, using above eqn. (7) ∆H can be converted into ∆U or vice – versa.eqn. (7) can be wriien as
q_p = q_v + ∆n_gRT ………………….. (8)
Because ∆H = q_p and ∆U = q_v.
Conditions:

1. If the number of moles of products is greater than that of reactants than ∆n will be +ve and ∆H is greater than ∆U. So,
∆H = ∆U + ∆nRT

2. If the number of moles of reactants is greater than the number of moles of products then ∆n will be – ve and the value of ∆H is less than ∆U.
∆H = ∆U – ∆nRT

3. If the No. of moles of reactants is equal to No. of moles of products then ∆n = 0, then in this condition ∆H = ∆U.

Question 3.
Deduce an expression for PV work done?
Answer:
Let us, consider a cylinder, fitted with a weightless, frictionless piston having a cross – sectional area A, filled with 1 mole of an ideal gas. The total volume of gas is V_i and pressure inside the cylinder is P_in.
Suppose, external pressure on the gas is P_ex which is slightly greater than the internal pressure of the gas. (MPBoardSolutions.com) Due to this difference in pressure the gas is compressed till the pressure inside becomes equal to P_ex Suppose, the change is achieved in one single step and the final volume of the gas is V_f The gas is compressed and suppose the piston moves a distance l. Work done during compression is
W = Force × Displacement = F × l

or F = P × ABV
∴W = P × A × large
or W = – P∆V, [∴ A × l = ∆V]

The negative sign in the expression is required to obtain conventional sign W. In compression, V_f < V_i and therefore, (V_f – V_i) or ∆V is – ve.
Hence, W will come out to be +ve from the above expression. In expansion type work V_f > V_i and value of ∆V is positive, therefore, work done will be negative.

This expression is useful for all types of PVwork and for irreversible flow.
Now, we will calculate, the work done during expansion of ideal gas in a reversible manner and in isothermal condition.

Question 4.
What is free energy? Derive its mathematical form and write Gibbs – Helmholtz equation?
Answer:
The free energy of a system is defined as the maximum amount of energy of the system which can be converted into useful work.
Free energy is related to enthalpy (H), entropy (S) and absolute temperature (T) as,
G = H – TS
Since, we know H = E + PV
∴ G = E + PV – TS
Change in free energy may be given as,
∆G = ∆E + ∆(PV) – ∆(TS)
If the process is carried out at constant temperature and pressure, then
∆(TS) = T∆S and ∆(PV) = P∆V
∆G_Tp = ∆E + P∆V – T∆S or
∆G = ∆H – T∆S
The above equation is called Gibbs – Helmholtz equation and it helps to predict the spontaneity of a process.
Free energy change and spontaneity:
For a system which is not isolated with surroundings
∆S_total = ∆S_system + ∆S_surrounding …………………… (1)
when reaction takes place at constant temperature and constant pressure, heat is supplied to surrounding.
∆S_surrounding = \(\frac { -q_{ p } }{ T } \) = \(\frac { -\Delta H }{ T } \), (q_p = ∆H at constant pressure) ………………………. (2)
From eqns. (1) and (2),
∆S_total = ∆S_system – \(\frac { -\Delta H }{ T } \) …………….. (3)
Since, all the quantities on the right – hand side are system properties, the subscript ‘system’ is not used in equations.
Multiplying both sides by T, we get
T∆S_total = T∆S – ∆H, (Where, ∆S = ∆S_system)
or -T∆Sc = ∆H – T∆S …………………… (4)
For Gibbs free energy (G),
G = H – TS
∆G = ∆H – T∆S – S∆T ………………… (5)
or ∆G = ∆H – T∆S_total
For the process taking place at constant temperature and constant pressure, eqn. (5) will be
So that, ∆G = ∆H – T∆S ……………………… (6)
Comparing eqns. (4) and (6),
∆G = – T∆S_total ………………….. (7)
So that, ∆G = – ve(for sontaneous chnages)
We know that, for spontaneous chemical change ∆S_total is positive. Eqn. (7) shows that the spontaneity of a change can be predicted on the basis of the value of ∆G.
Three special cases may be considered according to eqn. (7):

1. If AG is negative, the change is spontaneous.
2. If AG is zero, the system is in equilibrium.
3. If AG is positive, the change is non – spontaneous.

Conditions for spontaneity of a process (conditions for ∆G to be – ve):

1. If AH is negative and AS is positive, AG would certainly be negative and the process will be spontaneous.
2. If AH is negative and AS is also negative, then AG would be negative if AH is greater than TAS in magnitude.

Question 5.
Explain the determination of internal energy ∆U by Bomb calorimeter under the following heads:

1. Labelled diagram of the apparatus,
2. Explanation of the process
3. Calculations.

Answer:
Experimental determination of change in internal energy:
The change in internal energy in a chemical reaction is determined with the help of an apparatus called bomb calorimeter. (MPBoardSolutions.com) It is made up of steel so that it can bear high pressure developed during the chemical reaction taking place in the calorimeter. The inner side of the steel vessel is coated with some non – oxidizable metal like Pt or Au. It is also fitted with a pressure tight screw – cap. The two electrodes are connected to each other through a platinum wire dipped in a platinum cup.

A small known mass of the substance under investigation is taken in the platinum cup. The bomb is filled with excess of oxygen under a pressure of 20 – 25 atm and sealed. Now it is kept in an insulated water – bath which contains a known amount of water. The water – bath is also provided with a thermometer and mechanical stirrer.

The initial temperature of water is noted and the reaction (i.e., combustion of the sample) is started by passing an electric current through the Pt wire. (MPBoardSolutions.com) The heat evolved during the chemical reaction raises the temperature of water which is recorded by the thermometer. When rise in temperature and the heat capacity of the calorimeter are known, the amount of heat evolved in the chemical reaction can be calculated. This will be equal to the change in internal energy (∆E) of the reaction.

Calculation: Let W = Mass of calorimeter in gm, w = Water equivalent of calorimeter, bomb stirrer, etc. t°C = Rise in temperature, x = Mass of compound ignited in gm and m = Molecular mass of the compound.
Heat produced by x gm compound = (W + w) t calories
∴ Heat of combustion of the compound at constant volume,
∆U or ∆E = \(-\frac { m }{ x } \) (W + w) t calorie/mol.
Heat is evolved so negative sign is used.
Using the equation ∆H = ∆U + ∆ng RT heat of combustion ∆H at constant pressure can be caluculated.

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure Important Questions

Chemical Bonding And Molecular Structure Very Short Answer Type Questions

Question 1.
What type of bond is present generally in same atoms?
Answer:
Covalent bond.

Question 2.
Which hybridization is present in ammonia (NH₃)?
Answer:
sp³.

Question 3.
What is the reason for high boiling point of water?
Answer:
Presence of H – bond between molecules of water.

Question 4.
What is the dipole moment of C0₂?
Answer:
Zero.

Question 5.
What type of bonds are directional?
Answer:
Covalent bonds.

Question 6.
What is the bond angle in water?
Answer:
104°5′.

Question 7.
What is the structure of [Ni(CN)₄]^2-?
Answer:
Square planner.

Question 8.
Which bond is present in s – s overlapping?
Answer:
cr (Sigma) bond.

Question 9.
What is the structure of diamond?
Answer:
Crystal lattice (Tetrahedral).

Question 10.
Which bond is present in sidewise overlapping of p – p orbitals?
Answer:
n – bond.

Question 11.
Which type of hybridization found in PCl₅?
Answer:
sjy’d hybridization.

Question 12.
What is full form of LCAO?
Answer:
Linear Combination of Atomic Orbitals.

Question 13.
What is the structure of NH₃?
Answer:
Trigonal bipyramidal.

Question 14.
What is dipole moment (µ) of a linear covalent molecule?
Answer:
Zero.

Question 15.
What is the unit of electron gain enthalpy?
Answer:
eV per atom or kJ per mole.

Question 16.
What is the formula of bond order?
Answer:
Bond order = \(\frac{1}{2}\) (N_b – N_a).

Question 17.
What is the symbol of superoxide and peroxide?
Answer:
Superoxide – O^–₂ and peroxide – O^2-₂.

Question 18.
What do you mean by bond order?
Answer:
The number of electrons present between two atoms of molecules or ions.

Question 19.
More polarizing power and more polarisability increases which property of molecule?
Answer:
Covalent property.

Question 20.
What is determined by Born – Haber cycle?
Answer:
Lattice energy.

Question 21.
What is full form of VSEPR?
Answer:
Valence Shell Electron Pair Repulsion.

Chemical Bonding And Molecular Structure Short Answer Type Questions – I

Question 1.
What is the electronic theory of covalency? Write its main postulates?
Answer:
The electronic theory of valency and its main postulates are as follows:

1. The covalency of any element depends upon the no.of electrons present in its valence shell.
2. All the elements have the tendency to acquire the nobel gas configuration.
3. The electrons present in valence shell are called valence electrons and when the electron comes out than the vacant space is called kernel.
4. If an element is unstable, than it works for its stability for this it gives and takes the electron. On this basis the bonds are of three types:
   1. Ionic bond
   2. Covalent bond and
   3. Co – ordinate bond.

Question 2.
What do you mean by lone pair of electron?
Answer:
The electron pair which present in the valence orbital of the element and does not take part bond formation is called lone pair electron.
Example: In the H₂O atom the lone pairs of electrons on an atom:

Bond pairs = 2, Lone pair electrons = 2.

Question 3.
What do you mean by dipole moment?
Answer:
Dipole moment is defined as, the product ofthe magnitude of charge on any one of the atoms and distance between them. It is represented by Greek letter µ(mu).
Mathematically, dipole moment is expressed as µ = e × d
Where, e is charge on any one of the atoms and d is distance between the atoms.
As e is of the order of 10^-10 esu while d is of the order of 10^-8 cm µ is of the order 10¹⁸ esu cm and this unit of p is known as Debye (D). Thus,
1D = 1 × 10^-18esu cm

Question 4.
Is He₂ molecule is possible? Clearify it.
Answer:
He₂ molecule is not possible.
2He → 1s²
Two electrons are present in Is orbital of He atom. It is complete and stable orbital and so it cannot accept an extra electron. The bond order is zero. So the formation of He₂ molecule is impossible.

Question 5.
What is the total number of σ (sigma) and π (pi) bond are present in following molecules:

1. C₂H₂
2. C₂H₄

Answer:

1. C₂H₂
2. C₂H₄

Question 6.
What do you mean by hydrogen bond?
Answer:
Hydrogen bond is defined as the electrostatic force of attraction which exists between the covalently bonded hydrogen atom of one molecule and the electronegative atom of the other molecule. Hydrogen bond is of two types:

1. Intermolecular hydrogen bonding
2. Intra – molecular hydrogen bonding.

Question 7.
At normal temperature H20 is in liquid state but H2S is in gaseous state. Why?
Answer:
Hydrogen bonding affects the physical state of the molecule. For example, H₂0 and H₂S are the hydrides of group 16 elements, but H₂O is liquid whereas H₂S is gas at room temperature. Actually, H₂O fulfils the (MPBoardSolutions.com) condition of hydrogen bonding therefore, in H₂O hydrogen bonding is found and they get associated, grow in molecular size and exist in liquid state. The H₂S molecule, the magnitude of hydrogen bonding is negligible, therefore, molecules remains separated from each other and acquires gaseous state at room temperature.

Question 8.
Why, viscosity of glycerol is more than ethanol?
Answer:
In ethanol molecule, one hydroxyl gap is present whereas in every glycerol, three OH – groups are present which form hydrogen bond. Therefore, in glycerol hydrogen bond formation is more in comparison to ethanol. That is why, the viscosity of glycerol is more than ethanol.

Question 9.
Why σ – bond is stronger than π – bond?
Answer:
The strength of any bond depends upon the overlapping limit, σ – bonds are formed by axial overlapping whereas σ – bonds are formed by sidewise overlapping. So, the extent of overlapping in σ – bond is more than π – bond. So σ – bond is more stronger than π – bond.

Question 10.
Why HF molecule is more polar than HI?
Answer:
The electronegativity of F is more than I. So, the displacement of electron in covalency in HF is more than HI, resultant the deviation of charges in HF is greater than HI. So, HF is more polar than HI.

Question 11.
C – Cl bond is polar but CCl₄ is non – polar, why? Give the reason?
Answer:
In C – Cl bond the electronegativity of Cl is more than C, due to this the electrons in covalent bond shifts towards Cl, due to which partial +ve charge appears on C and partial -ve charge develops once and this bond become polar in nature. Whereas the structure of CCl₄ is symmetric, due to which the dipole moment of C – Cl bond cancelled each other. So, CCl₄ molecule is non – polar.

Question 12.
What do you mean by resonance?
Answer:
Some compounds cannot be represented by a single definite structure, rather more than one structure is required by none of them is able to explain all the known properties of the compound alone. Thus, the various structures written for a compound to explain the known properties of it completely are called resonating structure. This phenomenon is called resonance.

Question 13.
What are the conditions for resonance?
Answer:
The conditions are as follows:

1. The heat of formation of each resonating structure should be same.
2. The arrangement of atoms in each formula should be same.
3. The number of unpaired electron in each structures should be same.

Question 14.
What is resonance energy ?
Answer:
The difference between the actual energy of the resonance hybrid and the most stable one of the resonating structures is called resonance energy.

Question 15.
Determine the bond order in N₂?
Answer:
N₂ (14 electrons): The electronic configuration
σ1s², σ^*1s², σ2s², σ^*2s², (π2p_x² = π2p_y²), σ2p^z2
Bond order = \(\frac{1}{2}\) (N_b – N_a) = \(\frac{1}{2}\) × (10 – 4) = 3.

Question 16.
On the basis of molecular orbital theory explain that the Be₂ atom is not formed?
Answer:
₄Be: Electronic configuration = 1s² 2s²
Be₂ (molecule) (4 + 4 = 8e^–)
Electronic configuration = σ1s², σ^*1s², σ2s², σ^*2s²
Bond order = \(\frac{1}{2}\) (N_b – N_a) = \(\frac{1}{2}\) × (4 – 4) = 0.
So, the Be₂ molecule does not formed.

Question 17.
Write the definition of hybridization?
Answer:
The process of intermixing of atomic orbitals of nearly equal energy and proper symmetry giving rise to equal number of new orbitals of same energy is called hybridization and the orbitals so formed hybridized orbital.

Question 18.
What do you understand by lattice energy and solvation energy?
Answer:
Lattice energy:
Once the gaseous ions are formed, the ions of opposite charges come close together and pack up three – dimensionally in a definite geometric pattern to form ionic crystal (Crystal lattice). Since, the packing of ions of opposite charges takes place as a result of attractive force between them, the process is accompanied (MPBoardSolutions.com) with the release of energy referred to as lattice enthalpy. Lattice enthalpy may be defined as; the amount of energy released when one mole of ionic solid is formed by the close packing of its constituents. It is denoted by ∆_LH and negative in nature.

Solvation energy:
The energy released when an ion get soluble in water is called solvation energy.

Question 19.
Whose boiling points are high electrovalent compounds or covalent compounds, why?
Answer:
High boiling points: The boiling points of ionic solids are very high. This is due to strong electrostatic force of attraction between the oppositely charged ions. To change the physical state of ionic compounds, high temperature is required.

Question 20.
Why BaS0₄ is insoluble in water?
Answer:
The solubility of any ionic compound depends upon the lattice energy and solvation energy. If the lattice energy of any compound is more than solvation energy, than ionic compound is insoluble in water. The solvation energy of BaS04 is less than lattice energy. So it is insoluble in water.

Question 21.
If Be – H bond is polar, the dipole moment of Be – H₂ is zero. Why?
Answer:
BeH₂ is linear. The bond moment present in opposite direction cancelled each other. That is why Be the dipole moment of BeH₂ is zero.

Chemical Bonding And Molecular Structure Short Answer Type Questions – II

Question 1.
Why ice is lighter than water? Explain?
Or, Density of ice is less than water. Why?
Answer:
Density of ice is less than water:
In ice each oxygen atom is tetrahedrally surrounded by four hydrogen atoms in which two hydrogen atoms are linked to oxgyen atom by covalent bond and other two hydrogen atoms are linked by hydrogen bond.

The molecules of H₂O are not packed closely. This H gives rise to open cage like structure for ice having a larger volume for the given mass of water. Thus, density of ice is less than water. Ice is actually hydrogen bonded crystal. (MPBoardSolutions.com) Three dimensional structure of protein and nucleic acids like biologically important substances is due to hydrogen bond. Energy of hydrogen bond is between 3.5 kJmol^-1 and 8 kJmol^-1. Thus, hydrogen bond is stronger than van der Waals force and weaker than covalent bond.

Question 2.
Write the rules of hybridization?
Answer:
Conditions for hybridization: Following are the conditions for hybridization:

1. The orbitals of one and same atom participate in hybridization. Only the orbitals and not electrons get hybridized.
2. The energy difference between the hybridizing orbitals should be small.
3. Promotion of electron is not essential prior to hybridization.
4. It is not necessary that only the half-filled orbitals may participate in hybridiza¬tion. In some cases, even the filled orbitals may participate in hybridization.

Question 3.
Explain covalent bond with example?
Answer:
Lewis – Langmuir theory:
Lewis and Langmuir suggested that, atoms may combine by sharing of electrons in their outermost shell to complete their respective octet. The shared electrons becomes the property of both the atoms. This types of linkage is known as covalent linkage or covalent bond. Thus,

1. The force which binds atoms of same or different elements by mutual sharing of electrons is called a covalent bond.

2. This type of bond is formed between two similar non-metalic elements (A, A) or (B, B) or dissimilar atoms (A and B).
Example:
Chlorine molecule:
Both the chlorine atoms (Z = 17) contain 7 electrons in their valence shells and short in one electron each. They share one electron pair in which an electron is contributed by both as shown below:

Question 4.
Differentiate between atom and ion?
Answer:
Differences between Atom and Ion:
Atom:

1. Atoms are electroneutral.
2. Not present in free state.
3. Takes part in chemical reaction.
4. No. of e^– and proton is same.

Ion:

1. Ions are charged.
2. Ions present in free state.
3. Not take part in chemical reaction.
4. No. of e^– is more than proton.

Question 5.
Differentiate between Sigma (σ) and Pi (π) bond?
Answer:
Differences between Sigma (σ) and Pi (π) bond:
Sigma (σ) bond:

1. This bond is formed by end to end or head on overlapping of orbitals along the inter nuclear axis.
2. This is formed by overlapping of s – s, s – p or p – p orbitals.
3. Overlapping is large, hence it is strong bond.
4. Free rotation about σ – bond is possible.
5. Electron cloud is symmetrical about inter nuclear axis.

Pi (π) bond:

1. This bond is formed by the sidewise overlapping of orbitals.
2. This is formed by overlapping of p – p orbitals only.
3. Overlapping is small, hence it is weak bond.
4. Free rotation about a π – bond is not possible.
5. Electron cloud of π – bond is unsymmetrical.

Question 6.
Write difference between s – and p – orbitals?
Answer:
Differences between s – and p – orbitals:
s – orbital:

1. They are oval – symmetrical.
2. Non – directional
3. 1 = 0 and m = 0

p – orbitals:

1. They are dumbelled shape and lines symmetry of axis.
2. Directional
3. l = 1 and m = -1, 0, +1

Question 7.
Explain inter molecular and intramolecular hydrogen bonds with example?
Answer:
Types of hydrogen bond:
1. Intermolecular hydrogen bonding:
When these atoms (hydrogen and electronegative atom) are of different molecules, it is called intermolecular hydrogen bonding as in H₂O, HF, C₂H₅OH etc.

Hydrogen bond is represented by dotted lines. Many molecules of HF associates and form (HF)_n. On the same way molecules of water and alcohols are linked with hydrogen bonds.

2. Intramolecular hydrogen bonding:
If these atoms (hydrogen and electronegative atoms) are present in same molecule, this type of hydrogen bonding is called intramolecular hydrogen bonding.
e.g., o – nitrophenol

Question 8.
Among NH₃ and NF₃ whose dipole moment is high. Why?
Answer:
Both these molecules have pyramidal shape with one lone pair of electron on nitrogen atom. As fluorine is more electronegative than hydrogen therefore N – F bond should be more polar than N – H bonds. Consequently, the resultant dipole moment of NF₃ should be much larger than that of NH₃. However, the dipole moment of NH₃ (µ = 1.47D) is larger than that of NF₃ (µ = 0.24D).

The anomalous behaviour can be explained due to the presence of lone pair on nitrogen. In case of NH₃, the orbital dipole due to lone pair of electron and the bond moments of three N – H bonds are in same direction. Therefore, it adds on the resultant dipole moment of the N – H bonds. On the other hand in case of NF₃, the orbital dipole moment is in the opposite direction to resultant dipole moment of three N – F bonds. Thus, the lone pair moment cancels the resultant N – F bond moments as shown in figure. Consequently, the dipole moment of NF₃ is low.

Question 9.
Is according to following equation, is the hybridization changes in B and N:
BF₃ + NH₃ → F₃B.NH₃.
Answer:
In BF₃ three bonded pair and zero lone pair electrons are present. Due to this B is sp² hybridized and in NH₃ three bonded pair and one lone pair of electron is present. So, N is sp₃ hybridized. After reaction the hybridization of B becomes sp₃ but the hybridization of N remains same as N gives its lone pair to B atom.

Question 10.
Explain the change in hybridization in A1 atom in following reaction:
AlCl₃ + cl^– → AlCl^–₄
Answer:
The electronic configuration of Al is:
At ground state = ₁₃Al = ls² 2s² 2p⁶ 3s² 3p_x¹
At excited state = 1s², 2s², 2p⁶ 3s² 3p_x¹ _y¹
In the formation of AlCl₃, Al is sp² hybridized and its geometry is trigonal bipyramidal. Whereas in the formation of AlCl₄^–, due to inclusion of 3p_z orbital. Al is sp³ hybridized and its geometry is tetrahedral.

Question 11.
What are the postulates of orbital overlap concept of covalent bond?
Answer:
According to this concept:
1. The covalent bond is formed due to partial overlap of the two half-filled atomic orbitals of the valence shells of the combining atoms. Partial overlap means that a part of the electron cloud of each of the two half – filled domic orbitals becomes common. As a result the probability of finding electrons in the region of overlap is much more at the other places. This reduces the intemuclear repulsion and hence decreases the energy.

2. The orbitals undergoing overlap must have electrons with opposite spins.

3. Greater the extent of overlapping, stronger is the bond formed.

4. Larger the size of the orbitals, less effective is the overlapping and thus weaker is the bond formed.

Question 12.
Explain that the geometry of PCl₅ is trigonal bipyramidal and that of IF₅ is pyramidal?
Answer:
PCl₅: P is central metal atom.

As P in PCl₅ in sp³d hybridized so its geometry is square pyramidal.
IF₅: Central Metal atom is I (Z = 53)

In IF₇ , I is sp³d²

Question 13.
What are σ bond and π bond? Explain with example?
Answer:
Hybridization is defined as, “The process of intermixing of atomic orbitals of nearly equal energy and proper symmetry giving rise to equal number of new orbitals of same energy is called hybridization and the orbitals so formed hybridized orbitals.”

Sigma bond:
The bond formed by overlapping of two orbitals along their axis is called a sigma (σ) bond. The line joining the two nuclei of the combining atoms is called the intemuclear axis or bond axis.
Example: This type of overlapping takes place between s – s orbital, s – p orbital and pz – pz orbitals.

Pi – bond:
The bond formed by the lateral overlapping of two p – orbitals (p_x – p_x) (p_y – p_y) is called π – bond.

It is important to note that overlapping at both the lobes of the p – orbital occurs in Pi – bond whereas in case of sigma bond the overlap occurs in a single region.

Question 14.
Give the reason for the difference in properties of two allotropes diamond and graphite of carbon?
Answer:
Diamond and graphite are two allotropes of carbon. But due to difference in C arrangement their properties are different. In diamond the C atom is sp3 hybridized. Every C atom attached with four cations form tetrahedral geometry. So it forms a lattice structure and so hard and have high melting point.

In graphite every C atom is sp² hybridized, i.e. each C is surrounded by three cations and fourth valency of C is unstable. In graphite different layers are present which are joined together with weak vander Waals’ forces. That is why graphite is soft and due to presence of free electron it conducts electricity.

Question 15.
Explain the hybridized structure of acetylene by diagram?
Answer:
Formation of ethyne or acetylene (HC = CH): In the formation of acetylene molecule, each carbon atom undergoes ip – hybridization leaving two 2p – orbitals in the original unhybridized state. The two sp – hybrid orbitals of carbon atom are linear and are directed at an angle of 180°. Whereas the two unhybridized p – orbitals remain perpendicular to ip – hybrid orbital and also perpendicular to each other.

In the formation of acetylene, ip – hybrid orbital of One C – atom overlap with ip – hybrid orbital of another C – atom along the intemuclear axis forming a σ – bond. The second sp – hybrid orbital of each C – atom overlaps with the half – filled 1s – orbital of H – atom again along intemuclear axis thus forming a-bonds. (MPBoardSolutions.com) Each of the two unhybridized orbitals of both the carbon atoms overlap. Sidewise to form two π – bonds. Thus, all the carbon and hydrogen atoms are linear and there is electron cloud above and below, in the front and at the back of the C – C axis. In other words, there is electron cloud all around the intemuclear axis thus giving a cylindrical shape as represented in fig.

Question 16.
With tetrahedral geometry CH₄ molecule have a possible geometry of square plannar. In which the H atoms are present at the four corners of the square. Explain that the CH₄ molecule is not have square plannar geometry?
Answer:
The electronic configuration of C is:
In ground state ₆C = 1s² 2s² 2p_x¹2p_y¹
In excited state ₆C = 1s² 2s¹2p_x¹ 2p_y¹2p_z¹
sp³ hybridisation.
In CH₄ molecule carbon is sp³ hybridized. So its geometry is tetrahedral. For square plannar geometry dsp² hybridization is necessary. But due to absence of d – orbital in C atom. This geometry is impossible. With this according to VSEPR concept the bonded electrons in C atom is present at four comers of tetrahedron. The bond angle in tetrahedron is 109°28′ and in square plannar 90°. So in case of tetrahedral geometry the repulsion of electrons is less than in square plannar geometry.

Question 17.
On the basis of hybridization explain that the structure of BeCl₂ is linear?
Answer:
Formation of BeCl₂:
In the compound (BeF₂, BeCl₂, etc.), beryllium shows a covalency of two. In order to explain the formation of two equivalent bonds with beryllium its 25 – electron from the ground state (₄Be, Is² 2s²) is excited to 2p – orbital (1s²2s¹ 2p¹]) 2s and 2p – orbitals get mixed up to two equivalent sp – hybrid orbitals which make an angle of 180° with each other and oriented linearly. Each sp – orbitals overlap with half – filled p – orbital of chlorine (1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹) atoms to form two sigma bonds. Thus, the shape of BeCl₂ is linear.

Question 18.
Explain the formation of N₂ molecule on the basis of orbital theory as overlapping?
Answer:
π – bond formation of N₂ molecules:
Nitrogen molecule has a triple – bond consiting of one σ and two π – bonds (\(N\overset { \pi }{ \underset { \pi }{ \equiv N } } \)). Nitorgem atom has three half – filled p – orbitals.
₇N : 1s² 2s² 2p_x¹2p_y¹ 2p_z¹

When 2p_x orbital of each nitrogen atom overlaps co – axially, a σ – bond is formed. The 2p_y and 2p_z orbitals of one N atom overlap N atom to overlap laterally to form two π – bonds.

Chemical Bonding And Molecular Structure Long Answer Type Questions – I

Question 1.
Write the postulates of Valence bond theory? Write its limitation also?
Answer:
Valence bond theory was given by Heitler and London which is modified by Pauling and Slater. The postulates of the theory are:

1. The covalent bonds are formed by the partial overlapping of atomic orbitals (half filled).
2. In the orbitals taking part in overlapping electrons with opposite spin are present.
3. Strength of bond depends upon the extent of overlapping.
4. Strong directional bonds are formed between the orbitals of same stability, same energy and same symmetry.

Limitations of Valence Bond Theory:

1. According to this theory, no unpaired e^– is present in O² molecules. So the nature of O² is paramagnetic but O² is diamagnetic.
2. Not explain about the formation of coordinate bonds.
3. This theory is failed to explain the formation of H₂⁺ molecules.
4. Doesn’t give any information about resonance.

Question 2.
Write the condition for the formation of molecular orbitals by linear combination of atomic orbitals?
Answer:
Conditions for the combination of Atomic Orbitals:
Molecular orbital is formed by the linear combination of atomic orbitals. There are certain conditions for the effective linear combination of atomic orbitals. These conditions are:

1. The combining atomic orbitals should have same or nearly same energy: This means that in the formation of a homonuclear diatomic molecule Is atomic orbital of one atom will undergo linear combination with 1s atomic orbital of the other atom, but not with the 2s atomic orbital because the energy of the 2s orbital is appreciably higher than that of 1s atomic orbital. Similarly, because of the energy difference between 2s and 2p atomic orbitals, they will also not combine to form molecular orbitals.

2. There should be maximum overlap of atomic orbitals:
Greater the overlap, greater will be the charge density between the nuclei of a molecular orbital. This condition is often referred to as the principle of maximum overlap.

3. The atomic orbitals should have the same symmetry about the molecular axis:
This condition is known as symmetry condition for the combination of atomic orbitals. Taking the Z – axis as the molecular axis, the following pairs of atomic orbitals will not combine to form any molecular orbital, because of their different symmetries.

1. s – p_x pair
2. s – p_y pair
3. p_x – p_y pair and
4. ↔p_y – p_z pair.

This means that, s – s, px- px, py -py and Pz~Pz combinations are allowed because combining atomic orbitals have the same symmetry. A ,pz orbital, however, is able to combine with an s-orbital since, they have the same symmetry.

Question 3.
Differentiate between Ionic compounds and Covalent compounds?
Answer:
Differences between Ionic and Covalent compounds:

Question 4.
On what factors the formation of ionic bond depends?
Answer:
Factors influencing Ionic Bond formation:
The formation of ionic bond (Electrostatic force of attraction) depends upon the following factors:
1. Ionisation enthalpy:
One of the combining atoms (metal) must have low ionisa-tion enthalpy. So that, cation formation becomes easy. For example, alkali and alkaline earth metals of periodic table has tendency to form positive ion because they have comparatively low ionisation energy.

2. Electron gain enthalpy:
The electrons released in the formation of cation are to be accepted by the other atom taking part in the ionic bond formation. The electron accepting tendencies of an atom depends upon the electron gain enthalpy. (MPBoardSolutions.com) It may be defined as: Energy released when an isolated gaseous atom takes up an electron to form an anion. Greater the negative electron gain enthalpy, easier will be the formation of anion or negative ion. The halogen present in group 17 have the maximum tendency to form anions as they have very high negative electron gain enthalpy.

3. Lattice energy:
The amount of energy released when one mole of ionic solid is formed by the close packing of its constituents. It is denoted by ∆_LH and negative in nature.
A_(g)⁺ + B_(g)^– → A⁺B_(s)^– + Lattice enthalpy (∆_LH)
Thus, greater the magnitude of -ve lattice energy, more will be the stability of the ionic bond or ionic compound.

Question 5.
Explain the main points of Molecular orbital theory?
Answer:
The main points of Molecular orbital theory are:
1. In a molecule, electrons are present in new orbitals called molecular orbitals. These molecular orbitals are characterised by a set of quantum numbers just like atomic orbitals.

2. Molecular orbitals are formed by combination of atomic orbitals of equal energies (in case of homonuclear molecules) or of comparable energies (in case of heteronuclear molecules).

3. The number of molecular orbitals formed is equal to the number of atomic orbitals undergoing combination.

4. Two molecular orbitals are formed by combination of two atomic orbitals one of these two molecular orbitals has a lower energy and the other has a higher energy than either of the combining atomic orbitals. The molecular orbital with lower energy is called bonding molecular orbital and the other is called antibonding molecular orbital.

Question 6.
What is resonance? Explain with example?
Answer:
When properties of a molecule are not explained by one structure and two or more than two structures are assigned to express its characteristics, it is said that molecule is resonance hybrid of these structures and this property is known as resonance. Different resonating structures are exhibited by using sign (↔) in between these structures.
Example: Carbon dioxide (CO₂) is represented by following three structures:

Chemical Bonding And Molecular Structure Long Answer Type Questions – II

Question 1.
Show the molecular orbital energy levels of N₂ by diagram?
Answer:
N₂ molecule: Each nitrogen atom contains seven electron. Thus total 14 electrons are filled in seven orbitals of increasing energy.

Molecular orbital structure of N₂ molecule will be as follows:

Where KK represents closed shell structure (σ _1s)²(\(({ \sigma _{ 1s } })^{ 2 }\))(\(({ \sigma ^{ * }_{ 1s } })^{ 2 }\)
This structure shows that it contains 10 bonding and 4 antibonding electrons.
Bond order = \(\frac{1}{2}\) [N_b – N_a] = \(\frac{1}{2}\) [10 – 4] = 3.
Value of bond order is more. Hence value of bond energy should also be high. Experimental value of bond energy is 945 kJ mol^-1 which proves the presence of paired electrons in nitrogen molecule. Thus, it is a diamagnetic molecule.

How to Calculate Bond Order. Calculations Step by Step.

Question 2.
Give the important applications of dipole moment?
Answer:
Applications of Dipole moment:
1. Comparison of relative polarity:
It is possible by comparing the value of dipole moment e.g., HF (1.98 D) is more polar than HCl (1.03D).

2. Predicting the nature of molecules:
Molecules with specific dipole moments are polar in nature while those with zero value are non – polar. Thus, BeF₂ (µ – 0D) is non – polar while H₂O (µ = 1.84D) is polar.

3. Calculation of percentage ionic character:
% ionic character = \(\frac{Observed dipole moment}{Caluculated dipole moment}\) × 100 (100% ionic character)
or % I.C. = \(\frac { \mu _{ obs } }{ \mu _{ cal } } \) × 100
For example, the observed dipole moment of HCl molecule is 1.03D. For 100% ionic character i.e., complete transfer of electron charge on H⁺ and Cl^– ions would be equal to one unit (4.8 × 10^-10e.s.u.)× (1.275 × 10^-8 each. The bond length of H – Cl bond is 1.275 × 10^-8cm. Therefore, dipole moment for complete electron transfer
µ = q × d = (4.8 × 10^-10 e.s.u) × (1.275 × 10^-8cm)
= 6.12 × 10^-18 e.s.u cm = 6.12D
Observed dipole moment, µ(obs) = 1.03D
% ionic character = \(\frac { 1.03 }{ 6.12 } \) × 100 = 16.83 %

4. Dipole moment of symmetric molecules is zero, but they have two or more than two polar bonds. It is applied for the measurement of symmetry.

5. Distinction between ortho, meta and para isomers of aromatic compounds:
In general, the dipole moments follow the order: ortho > meta > para e.g., In dichlorobenzene, the dipole moments of o, m and p isomers are: 2.54 D, 1.48 D and 0 respectively.

Question 3.
What is valence shell electron pair repulsion theory? Write its limitations? Or, Explain valence shell electron pair repulsion theory with example?
Answer:
VSEPR theory:
On the basis of this theory, “When central atom is surrounded by only bonded electron pairs, in that case the geometry of the molecule will be ordinary, but if the central atom is surrounded by bonded electron pairs as well as lone pairs or non – bonded pairs, then the geometry of die molecule will become abnormal.” This (MPBoardSolutions.com) means that the repulsion between the non – bonded or lone pair of electrons and bonded electron pairs become greater than that of repulsion between only bonded electron pairs. Repulsion

Between the electron pairs is in the following order:
lone pair – lone pair > lone pair – bonded pair > bonded pair – bonded pair

Shapes of some molecules accroding to VSEPR theory:
1. Shape of CH₄:
In methane, central atom carbon is surrounded by four electron pairs and four C – H bonds. This molecule has tetrahedral geometry. Shared electrons are at the comers of tetrahedron for maximum separation. Bond angle is 109°28′.

2. Shape of H_2O:
In water molecule central oxygen atom is surrounded by four electron pairs, two of which are lone pair of electrons. Thus, Ip – lp and lp – bp repulsion exist. Due to this, bond angle reduces to 104.5° in place of 109°28′ and shape becomes V shaped.

3. Shape of NH₃:
In ammonia, central N atom is surrounded by four electron pairs, so shape of molecule is tetrahedral. According to VSEPR theory, the four groups around the central atom of ammonia should be tetrahedrally arranged at bond angle of 109°28’.

But, the measured bond angle is 107°. This is explained on the basis of repulsive effect of the lone pair of electrons on bonding electrons. In ammonia molecule there is a lone pair of electrons on the N atom. Thus, the shape of NH₃ molecule is distorted and it looks like pyramidal and it is polar in nature.

Example:

Limitations of VSEPR theory:
VSEPR theory no doubt, theoretically gives the shapes of simple molecules but could not explain them and also has limited application. To overcome these limitations, two important theories based on quantum mechanical principles are commonly used. These are:

1. Valence bond theory (VBT) and
2. Molecular orbital theory (MOt).

Question 4.
Structure of two molecules are given:

1. Among these which contain intermolecular hydrogen bond and intramolecular hydrogen bond?
2. The melting point of any compound depends upon hydrogen bonding also, on the basis of this explain which one have high M.P.
3. The solubility of any compound depends upon its tendency to formed H – bonding with water among these who will form H – bond with water easily.

Answer:
1. Compound (a) forms intramolecular H – bond. When H – bond is present between the atoms of same molecule than it is called intramolecular H – bond.
In ortho nitrophenol [(a)] H – bond is present between two O – atoms.

In compound (b) inter molecular H – bond forms.
In p – nitrophenol [(b)] between – N0₂ and – OH vacant space is present. So among H – atom of one molecule and O – atom of another molecule, H – bond is formed.

2. The melting point of compound (b) is high because many molecules are forming H – bond.

3. Due to intramolecular H – bonding, compound (a) will not form H – bond with water so it is less soluble in water. Whereas compound (b) forms H – bond easily with water so soluble in water.

Question 5.
Draw molecular orbital diagram for 0₂ molecule?
Answer:
Oxygen molecule, (0₂): Each oxygen atom has eight electrons. When two oxygen atom combine, molecular orbitals are formed. These molecular orbitals have following configuration:

Its molecular orbital diagram is:

From the above configuration we have,
N_b = 10, N_a = 6
∴ Bond order = \(\frac{1}{2}\) [N_b – N_a] = \(\frac{1}{2}\) [10 – 6] = 2
Hence, there is a double bond in oxygen molecule. Due to the presence of two unpaired electrons it is paramagnetic.

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 5 States of Matter

States of Matter Important Questions

States of Matter Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
At the time of opening the bottle of ammonia, it can be recognized from a distance because:
(a) It is very reactive
(b) It diffuses very fast
(c) It possess a pungent smell
(d) It is lighter than air
Answer:
(b) It diffuses very fast

Charles’ Law Calculator uses formula that relates the initial and final volume and temperature values of an ideal gas at constant pressure.

Question 2.
Who established the relationship between density and rate of diffusion of a gas:
(a) Boyle
(b) Charles
(c) Graham
(d) Avogadro
Answer:
(c) Graham

Question 3.
The value of in Calorie (approx):
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 4.
Value of gas constant R is:
(a) 8314 × 10⁷ ergs/degree/mol
(b) 83.14 × 10⁶ ergs/degree/mol
(c) 83.14 × 10⁵ ergs/degree/mol
(d) 8.314 × 10⁷ ergs/degree/mol
Answer:
(d) 8.314 × 10⁷ ergs/degree/mol

Question 5.
Absolute temperature is:
(a) 0° C
(b) – 100° C
(c) – 273° C
(d) – 373°C
Answer:
(c) – 273° C

Question 6.
To get general gas equation which two laws are combined:
(a) Charle’s law and Dalton’s law
(b) Graham’s law and Dalton’s law
(c) Boyle’s law and Charle’s law
(d) Avogadro’s law and Dalton’s law
Answer:
(c) Boyle’s law and Charle’s law

Question 7.
Which is correct in the following:
(a) r.m.s. velocity = 0.9212 × average velocity
(b) average velocity = 0.9212 × r.m.s. velocity
(c) r.m.s. velocity = 0.9013 × average velocity
(d) average velocity = 0.9013 × r.m.s. velocity
Answer:
(b) average velocity = 0.9212 × r.m.s. velocity

Question 8.
At constant volume the pressure of monoatomic gas depends on:
(a) Thickness of the wall of the vessel
(b) Absolute temperature
(c) Atomic number of the element
(d) Number of valence electron.
Answer:
(b) Absolute temperature

Question 9.
Behaviour of real gases near to that of ideal gases if:
(a) Temperature is low
(b) Pressure is high
(c) Low pressure and high temperature
(d) Gas is monoatomic
Answer:
(c) Low pressure and high temperature

Question 10.
At mountains of high altitude, water boils at lower temperature because of:
(a) Low atmospheric pressure
(b) High atmospheric pressure
(c) Hydrogen bonding in water is more strong at height
(d) Water vapour is lighter than liquid
Answer:
(a) Low atmospheric pressure

Question 11.
If molecular masses of two gases A and B are 16 and 64 respectively ratio of rates of diffusion A and B will be:
(a) 1:4
(b) 4:1
(c) 2:1
(d) 1:2
Answer:
(c) 2:1

Question 12.
Gases deviate from ideal behavior at high pressure because:
(a) At pressure number of colloision of molecule increases
(b) Attraction between molecules increases at high pressure
(c) Size of molecule decreases at high pressure.
(d) Molecule become steady at high pressure.
Answer:
(b) Attraction between molecules increases at high pressure

Question 13.
Distance on which magnitude of energy is minimum is called:
(a) Atomic radius
(b) Lattice radius
(c) Critical distance
(d) Molecular distance
Answer:
(a) Atomic radius

Question 14.
Diffusion rate of methane is twice than that of gas x, molecular mass of gas x will be:
(a) 64
(b) 32
(c) 40
(d) 8.0
Answer:
(a) 64

Question 15.
Part of van der Waals’ equation which illustrate the inter molecule force of real gases:
(a) (V – b)
(b) RT
(c) [P+\(\frac{a}{VL}\)]
(d) (RT)^-1
Answer:
(c) [P+\(\frac{a}{VL}\)]

Question 16.
Temperature which is same in both Celsius scale and Fahrenheit scale:
(a) 0°C
(b) 32°F
(c) – 40°C
(d) 40°C
Answer:
(c) – 40°C

Question 2.
Fill in the blanks:

1. With increase in temperature viscocity ………………………..
2. S.I. unit of surface tension is ………………………….
3. ………………………… is more heavy among dry air and moist air
4. AT N.T.P., volume of gas equal to Avogrado’s numbers ………………………….
5. Unit of vander Waals’ constant ‘b’ is ………………………..
6. …………………………. has minimum value among average velocity, root mean square velocity and most probable velocity
7. Process of diffusion of air from the puncture of an autommobile tube is known as ……………………….
8. …………………………… is obtained if a graph is plotted between volume and absolute temperature
9. Apparatus used for measuring gas pressure is …………………………
10. S.I. unit of viscosity is ………………………….
11. Absolute temperature of ideal gas is ………………………… to the average kinetic theory of the molecule.
12. A liquid which is permanently super cooled is called ………………………..
13. According to kinetic theory, the average kinetic energy of a gas is directly proportional to its …………………………. temperature.
14. The kinetic energy of one mole of a gas is equal to …………………………..
15. Root mean square velocity is …………………………..

Answer:

1. Decreases
2. Nm^-1
3. Dry air
4. 22.41
5. Litre/mol
6. Most probable velocity
7. Effiission
8. Straight line
9. Monometer
10. Nm^-2S
11. Proportional
12. Glass
13. Absolute
14. \(\frac{3}{2}\) RT
15. \(\sqrt { \frac { 3PV }{ M } } \) or \(\sqrt { \frac { 3RT }{ M } } \)

Question 3.
Answer in one word/sentence:

1. Vaporization of ethanol is faster as compared to water?
2. The resistance produced in the flow of a liquid is called?
3. Unit of surface tension is?
4. Molecular mass of two gases A and B are 36 and 64 respectively. What will be the ratio of diffusion of the two gases?
5. Write S.I. unit of pressure?
6. If rate of diffusion of oxygen is r, then tell the rate of diffusion of hydrogen?
7. Compressibility factor of ideal gases is?
8. Vander Waals’ equation is?
9. Adiabatic expression of ideal gas is?
10. Number of electron in outer most orbit of crypton is?

Answer:

1. Value of molar vaporization enthalpy of water is high
2. Viscosity
3. Dyne per cm
4. 4:3
5. pascal
6. r₁ = 4r [\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { m_{ 2 } }{ m_{ 1 } } } \)]
7. 1.0
8. [P + \(\frac { an^{ 2 } }{ V^{ 2 } } \)] [V – nb] = nRT
9. q = 0
10. 8 electron

Question 4.
Match the following:
[I]

Answer:

1. (f)
2. (a)
3. (b)
4. (a)
5. (d)
6. (c)

[II]

Answer:

1. (d)
2. (c)
3. (e)
4. (a)
5. (b)

States of Matter Very Short Answer Type Questions

Question 1.
The resistance produced in the flow of a liquid is called?
Answer:
Viscosity.

Question 2.
Unit of surface tension is?
Answer:
Dyne per cm.

Question 3.
Write S.I. unit of pressure?
Answer:
Pascal.

Question 4.
Name the scientist who developed relation between density and rate of diffusion of gas?
Answer:
Graham’s.

Question 5.
What is the Value of gas constant in S.I. unit?
Answer:
8.314 JK^-1 mol^-1

Question 6.
1 Pascal is equal to?
Answer:
1 Nm².

Question 7.
Who established the relationship between density and rate of diffusion of gas?
Answer:
Graham’s.

Question 8.
Absolute temperature is?
Answer:
~273°C.

Question 9.
The value of R in calorie (approx)?
Answer:
2.

Question 10.
Behaviour of real gases is near to that of ideal gases of?
Answer:
Low pressure and high temperature.

Question 11.
At mountains or high altitude, water boils at lower temperature because of?
Answer:
Low atmospheric pressure.

Question 12.
To get general gas equation. Which two laws are combined?
Answer:
Boyle’s law and Charle’s law.

Question 13.
At constant volume the pressure of gas depends upon?
Answer:
On absolute temperature.

Question 14.
Part of van der Waals’ equation which illustrate the inter – molecular forces of real gases?
Answer:
P + \(\frac { a }{ V^{ 2 } } \).

Question 15.
Temperature which is same in both Celsius scale and fahrenheit scale?
Answer:
– 40°C.

Question 16.
What is the formula for kinetic theory of gases?
Answer:
PV = \(\frac{1}{3}\) mnv².

Question 17.
The average kinetic energy of gases is proportional to?
Answer:
Absolute temperature.

Question 18.
What is the kinetic energy of 1 mole of gas?
Answer:
\(\frac{3}{2}\) RT.

Question 19.
What is the formula of Root mean square velocity?
Answer:
\(\sqrt { \frac { 3PV }{ M } } \) or \(\sqrt { \frac { 3RT }{ M } } \).

Question 20.
Which type of Crystals are diamond and ice?
Answer:
Diamond – Covalent crystal
Ice – Ionic crystal.

Question 21.
What is Critical temperature?
Answer:
The temperature above which the gas cannot be liquefied.

Question 22.
Poise is the unit of which basic property?
Answer:
Viscosity (1 poise = dynes/cm²s).

Question 23.
What is the unit of a (volume correction), b (pressure correction) in van der Waals’ equation?
Answer:
a (Volume correction) = atm L² mol^-2
b (Pressure correction) = L mol^-1

States of Matter Short Answer Type Question – I

Question 1.
Explain Anisotropic and Isotropic?
Answer:
Anisotropic:
The crystalline solid exhibits different physical properties in the , three direction. In this way, crystalline solids are called anisotropic.

Isotropic:
The amorphous solid exhibits same physical properties in all directions. Due to this property they are called isotropic.

Question 2.
Define the term absolute zero? Write its value in centigrade scale?
Answer:
The lowest possible temperature at which all the gases are supposed to occupy zero volume is called absolute zero. The actual value of absolute zero is – 273.15°C. It is , related to temperature in centigrade scale by this relation.
t°C = t + 273 K.

Question 3.
What is unit cell?
Answer:
Smallest unit is a crystal which is formed by systematic arrangements of constituent particles as atom, molecule or ions, is called unit cell. The unit cell generates the whole lattice translation.

Question 4.
What is crystal lattice?
Answer:
Geometry or shape of any crystal in which unit cells are arranged systematically and three dimensionally is called crystal lattice.

Question 5.
How does volume of balloon used for weather study, change with height?
Answer:
At height, atmospheric pressure decreases. The volume of gas inside the balloon increases with decrease in pressure. A stage comes when due to decrease in atmospheric pressure in larger extent, volume increases and balloon bursts.

Question 6.
In winter season, a layer of ice is formed in the lake but the fishes and other organisms present in the lake remain alive. Why?
Answer:
The maximum density of water is at 4°C but below 4°C temperature the density decreases. When the temperature of lake decreases then the water present on the surface become denser and goes downward. This occurs upto the level when the temperature rises to t 4°C. (MPBoardSolutions.com) If the temperature of the upper layer is less than 4°C, the water remains on the upper surface and converts into ice slowly and the water below the surface remains as such due to high density and remains as liquid. That is why, the fishes and micro – organisms remains alive.

Question 7.
Why is the density of hot gas is less in comparison to cold gas?
Answer:
According to Charle’s law, volume of any gas of definite mass is directly proportional to absolute temperature. On increasing the temperature volume of gas also increases, but increase in volume results decrease in density.

Question 8.
Why one feel sluggish, breathlessness and headache at high altitude?
Answer:
At high altitude, the pressure is less and the corresponding volume of air is more, Thus, air becomes less dense at high altitude and the oxygen in air becomes insufficient for normal breathing. This causes what is known as altitude sickness.

Question 9.
What is Critical temperature?
Answer:
The temperature to which gas must be cooled before it can be liquefied by com-pression is known as critical temperature and is represented by Tc.
Example: Critical temperature of CO₂ gas is 31.1°C or 304.1K.

Question 10.
What is critical pressure and critical volume?
Answer:
Critical pressure:
The minimum pressure required to liquefy the gas at its critical temperature is known as critical pressure and denoted by P_c.
Example: Critical pressure of CO₂ is 72.8 atm.
Critical Volume: The volume occupied by 1 mole of gas at the critical temperature and critical pressure is known as critical volume and denoted by V_c.
Example: Critical volume of C0₂ gas is 94 cm³/mol.

Question 11.
Why are tyres of automobile inflated to lesser pressure in summer than in winter?
Answer:
As the automobiles move the temp, of tyre increases due to friction against the road. Consequently the air inside the tyre expand thereby the pressure exerted by air against the wall of tyre also increases. (MPBoardSolutions.com) In summer, there is increase in temperature hence increase in pressure is much more. This may leads bursting at tyre. In order to check bursting of tyre, the tyre are inflated with looser amount of air than in winter.

Question 12.
What would be the S.I. unit for the quantity PV²T²/n?
Answer:
\(\frac { PV^{ 2 }T^{ 2 } }{ n } \) = \(\frac { (Nm^{ -2 })(m^{ 3 })^{ 2 }(K)^{ 2 } }{ mol } \) = Nm⁴K²mol^-1.

Question 13.
Explain on the basis of Charle’s law that minimum possible temperature is – 273°C
Answer:
According to Charle’s law:
V_t = V₀ [1 + \(\frac { t }{ 273 } \)]
At t = – 273°C V_t = V₀ [ 1 – \(\frac { 273}{ 273 } \)] = 0
Therefore at – 273°C, the volume of gas becomes 0 and below this temperature the volume becomes – ve which is meaningless.

Question 14.
Why are liquid drops spherical?
Answer:
Small drops are spherical in shape:
Surface tension tries to decrease the surface area of a given liquid for a given volume. Therefore, drops of liquid are spherical because for a given volume sphere has minimum volume.

Question 15.
What is Root Mean Square velocity and Average velocity?
Answer:
1. Root Mean Square velocity:
It is defined as root of, mean of, square of, velocity of large no. of molecules of same gas. It is denoted by V.
V = \(\sqrt { \frac { v_{ 1 }^{ 2 }+v_{ 2 }^{ 2 }+v_{ 3 }^{ 2 }…..v_{ n }^{ 2 } }{ n } } \)
2. Average velocity:
It is defined as average of, velocity of all the molecules present in gas. It is denoted by V_a.
V_a = \(\frac { v_{ 1 }+v_{ 2 }+v_{ 3 }…..v_{ n } }{ n } \)

Question 16.
Explain the difference between Evaporation and Boiling?
Answer:
Differences between Evaporation and Boiling:
Evaporation:

1. Evaporation decreases spontaneously and occur at all temperatures.
2. Evaporation is a process of the liquid surface.
3. Evaporation is a slow process.

Boiling:

1. Boling takes place only when the vapour pressure of this liquid becomes equal to atmospheric pressure.
2. Boling is a process of the entire liquid and occurs in the form of bubbles inside the liquid.
3. Boling is a fast process.

Question 17.
What do you mean by compressibility factor of gases?
Answer:
The ratio of observed volume and caluculated volume of a gas at a given temperature and pressure is known as compressibility factor. It is denoted by Z.
Thus,

or Z = \(\frac { PV }{ nRT } \)
For ideal gases, PV = nRrt
∴For ideal gas, Z = 1.

Question 18.
What is the effect of pressure on melting of ice?
Answer:
By increasing pressure there occur tremendous increase in kinetic energy of molecules, due to this at low temperature, the molecules move freely, hence on increasing pressure the ice below its melting point converted into liquid.

Question 19.
Mountaineers carry oxygen cylinders with them at the lance of climbing mountains. Why?
Answer:
Atmospheric pressure is relatively low at heights. Quantity of oxygen is low in mountain and climbers feel difficulties in breathing. Therefore, they carry oxygen cylinders along with them.

Question 20.
Define viscosity of liquid. Explain the effect of temperature on viscosity?
Answer:
Resistance in flow of any liquid is called viscosity. Such resistance is produced due to internal friction of different layers of liquid. (MPBoardSolutions.com) When temperature is increased, the cohesive force, which opposes liquid flow, decreases and molecular velocity increases. Due to this, viscosity decreases.

Question 21.
On same temperature when ether and water pour on different hands, then ether seems to be more colder than water. Why?
Answer:
In ether the intermolecular attractive forces between the molecules is less in comparison to water, so ether evaporates more quickly than water and the energy required for evaporation is absorbed from hand, that is why ether seems to be colder.

Question 22.
What is surface tension? Write its S.L unit?
Answer:
It is an important property of a liquid which is related with interatomic attraction force. The molecules present inside the liquid is attracted equally by molecules present in all direction. (MPBoardSolutions.com)
But molecule present on the surface of liquid is attracted by molecules at bottom and in sides, as a result the molecules at surface are pulled downward and nature of surface is to lessen the area. Due to compactness, the surface of liquid behaves as a stretched membrane. This effect is called surface tension.

“Surface tension is a measure of work which is necessary to increase the unit cross-section of liquid.”
Its S.I. unit is Joule/metre² or Newton metre.

Question 23.
The compressibility factor Z of a gas is as follows:

1. What will be the value of Z for an ideal gas?
2. What will be the effect on Z above Boyle’s temperature for real gas?

Answer:

1. For ideal gas, compressibility factor Z = 1.
2. Above Boyle’s temperature, real gases show positive deviation. So, Z > 1.

Question 24.
What is Ideal gas? Write its characteristics?
Answer:
Ideal Gas:
The gases which obey Ideal gas equation under all conditions of temperature and pressure is called ideal gas.
Characteristics:

1. At constant temperature, product of pressure and volume of ideal gas are always constant. Therefore, horizontal line should be obtained in a graph. If graph is plotted between PV and P.
2. If an ideal gas is called at constant pressure, then its volume requestly decreased and become at – 273°C.
3. There is no force of attraction between gas molecules.
4. The compressibility factor of ideal gas is equal to one.

Question 25.
What is Real gas? What are its properties?
Answer:
Gas which does not obey Boyle’s law, Charle’s law and Ideal gas equation strictly is called Real gas.
The gases which does not follow ideal gas equation behaviour under all condtions of temperature and pressure called real gas.
Properties:

1. They do not follow gas law at low temperature and high pressure.
2. At – 273°C their volume is not zero because most of the gases converted into liquid on cooling.
3. The attractive force between gas molecule is negligible.
4. The compressibility factor of real gas is not equal to zero.

States of Matter Short Answer Type Questions – II

Question 1.
State and explain Boyle’s Law?
Answer:
According to this law: “At constant temperature, the volume of a known amount of gas is inversely proportional to the pressure.”
P ∝ \(\frac{1}{V}\) (at constant temperature)
⇒P = Constant × \(\frac{1}{V}\)
⇒ PV = Constant.
Thus, “at constant temperature, product of volume of a given mass of gas and pressure remain constant.”
At initial condition, P₁V₁ = K …………….. (1)
At final condition, P₂V₂ = K ………………. (2)
From eqn. (1) and eqn. (2).
P₁V₁ = P₂V₂

Question 2.
What is the nature of gas constant R?
Answer:
We know that,
PV = nRT
R = \(\frac{PV}{nT}\)

Question 3.
Explain concept of absolute zero from Charle’s law?
Answer:
Charle’s law: According to this law “At a constant pressure the volume of certain mass of a gas increases or decreases by of its previous volume for every 1°C changes in temperature (increases or decreases).”
Suppose V₀ is the volume of certain mass of gas at 0°C then,
Volume of the gas at 1°C temperature = V₀ [1 + \(\frac{1}{273}\)]
Volume of the gas at t°C temperature = V₀ [1 + \(\frac{t}{273}\)]
Volume of the gas at – 1°C temperature = V₀ [1 – \(\frac{t}{273}\)]
Volume of gas at – 273°C = V₀ [1 – \(\frac{273}{273}\)] = 0
Thus decrease of temperature results in the decrease in volume of the gas and ultimately the volume should become zero at – 273°C. (MPBoardSolutions.com) This lowest possible temperature at which all the gases are suppossed to occupy zero volume is called absolute zero.

Question 4.
What is Gay Lussac’s law?
Answer:
Gay Lussac law:
According to this law, “At constant volume the pressure of a given mass of gas is directly proportional to its absolute temperature.”
P ∝ T (Mass and volume are constant)
⇒ P = K × T
⇒ \(\frac{P}{T}\) = K
Suppose at initial condition,
\(\frac { P_{ 1 } }{ T_{ 1 } } \) = K
At final condition,
\(\frac { P_{ 2 } }{ T_{ 2 } } \) = K
From eqn. (1) and (2),
\(\frac { P_{ 1 } }{ T_{ 1 } } \) = \(\frac { P_{ 2 } }{ T_{ 2 } } \)

Question 5.
State and explain Avogadro’s law?
Answer:
According to this law, “Equal volume of all gases under identical conditions of temperature and pressure contain equal number of molecules”.
V ∝N (at constant temperature and pressure) …………. (1)
At constant temperature and pressure number of moles of a gas n is directly proportional to number of molecules N.
Hence, N ∝n
⇒\(\frac{V}{n}\) = Constant
Suppose at initial condition volume of gas is V₁ and no. of mole of gas is n₁ hence
\(\frac { V_{ 1 } }{ n_{ 1 } } \) = Constant …………… (2)
Similarly at final condition no. of moles and volume of gas is n₂ and V₂, hence
\(\frac { V_{ 2 } }{ n_{ 2 } } \) = Constant …………… (3)
From eqns. (2) and (3)
\(\frac { V_{ 1 } }{ n_{ 1 } } \) = \(\frac { V_{ 2 } }{ n_{ 2 } } \)

Question 6.
Write the applications of Graham’s law of diffusion?
Answer:
1. To determine the density and molecular weight of a gas:
If the time of diffusion and density of a gas is known and the time of diffusion of other gas is known, then the density and molecular weight of other gas can be calculated.

2. Marsh gas indicator:
The persons working in the mines get aware by the leakage of the poisonous gases by this indicator.

3. In separation of gases:
The gases can be separated easily due to difference in the rate of diffusion of gases.

4. Smell:
Bad smell and poisonous gases get separated due to diffusion in air.

Question 7.
Derive Charle’s law on the basis of Kinetic gas theory?
Answer:
According to Kinetic gas theory, kinetic energy of gases is directly proportional to absolute temperature.
K.E ∝T
\(\frac{1}{2}\) mnv² ∝ T
⇒\(\frac{1}{2}\) mnv² = KT
⇒\(\frac{3}{2}\) × \(\frac{1}{3}\) mnv² = KT
⇒\(\frac{1}{3}\) mnv² = \(\frac{2}{3}\) KT
⇒PV = \(\frac{2}{3}\) KT,
⇒V = \(\frac{2}{3}\) \(\frac{K}{P}\).T
At constant pressure \(\frac{2}{3}\) \(\frac{K}{P}\) = constant
V = constant ∝ T
V ∝ T

Question 8.
How are rates of diffusion of different gases compared?
Answer:
Let two gases are A and B, equal volume V of both gases diffuse in times t_A and t_B respectively.
Rate of diffusion of gas A,
r _A = \(\frac { V }{ t_{ A } } \)
Rate of diffusion of gas B,
r _B = \(\frac { V }{ t_{ B } } \)
∴\(\frac { r_{ A } }{ r_{ B } } \) = \(\sqrt { \frac { d_{ B } }{ d_{ A } } } \)
From equation (3) and equation (4),
\(\frac { t_{ B } }{ t_{ A } } \) = \(\sqrt { \frac { d_{ B } }{ d_{ A } } } \).

Question 9.
The ratio between the rate of diffusion of an unknown gas (x) s&d CO₂ Is 40:45 Find out the molecular mass of unknown gas (x).
Solution:
According to Graham’s law of diffusion,
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { M_{ 2 } }{ M_{ 1 } } } \)
Given, r₁:r₂ = 40:45 M₂(CO₂) = 44, M₁ = ?
⇒\(\frac{40}{45}\) = \(\sqrt { \frac { 44 }{ M_{ 1 } } } \)
⇒ \(\frac { (40)^{ 2 } }{ (45)^{ 2 } } \) = \(\sqrt { \frac { 44 }{ M_{ 1 } } } \)
M₁ = \(\frac { 44\times 45\times 45 }{ 40\times 40 } \) = 55.68.

Question 10.
If relative density of chlorine is 36, diffusion of 25 volume of hydrogen takes 40 sec. under the condition how much time will be taken for the diffusion of 30 volume of chlorine?
Solution:
Hydrogen d₁ = 1, r₁ = \(\frac{25}{40}\), Chlorine d₂ = 36, r₂ = \(\frac{30}{t}\)

States of Matter Long Answer Type Questions

Question 1.
State and explain Graham’s law of diffusion?
Answer:
Graham’s Law of Diffusion:
The rate of diffusion of gas under similar condition of temperature and pressure is inversely proportional to the square roots of their density.”
Thus, Rate of diffusion ∝\(\frac { 1 }{ \sqrt { density } } \)
⇒r ∝\(\frac { 1 }{ \sqrt { d } } \)
If the rate of diffusion of two gases are r₁ and r₂ and their density are d₁ and d₂ respectively.
r₁ = K \(\frac { 1 }{ \sqrt { d_{ 1 } } } \)
r₁ = K \(\frac { 1 }{ \sqrt { d_{ 2 } } } \)
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { d_{ 2 } }{ d_{ 1 } } } \)
∵ M. Mass = 2 × Vapour density = \(\frac { Molecular\quad mass }{ 2 } \) = Vapour density
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { d_{ 2 } }{ d_{ 1 } } } \) = \(\sqrt { \frac { M_{ 2 } }{ M_{ 1 } } } \).

Question 2.
Write difference between Real and Ideal gas?
Answer:
Differences between Ideal gas and Real gas:
Ideal gas:

1. Ideal gas obeys die equation, PV = RT at all temperature and pressure.
2. There is no ideal gas, they are hypothetical.
3. Total volume of gas molecules is supposed to be negligible in comparison to total volume.
4. There is no attraction force between gas molecules.

Real gas:

1. Real gas obeys PV = RT only at low pressure and high temperature.
2. All existing gases are real gases and show deviation from ideal gas behaviour, positive or negative.
3. Volume of gas molecules are not negligible in comparison to total volume.
4. Gas molecules attract each other.Therefore, total pressure is less than ideal gas.

Question 3.
What are the main differences between Crystalline and Amorphous solids?
Answer:
Differences between Crystalline and Amorphous solids:
Crystalline:

1. In it the constituent particles are arranged in regular manner.
2. Melting point of crystalline solid is fixed.
3. They are an anisotropic i.e. some of their physical properties are different in different directions.
4. They are rigid and their shape is not distorted by mild distorting tone.
5. They have a definite heat of fusion.
6. They are true solid in real meaning.

Amorphous solids:

1. In it the constituent particles are arranged in irregular manner.
2. Melting point of amorphous solid is not fixed.
3. They are Isotropic i.e. their physical properties are same in all directions.
4. They are not very rigid, they can be distorted easily.
5. They do not have definite heat of fusion.
6. They are super cooled liquid.

Question 4.
Using state equation clarify that the density of a gas at given temperature is proportional to pressure of gas?
Answer:
PV = nRT
PV = \(\frac{m}{M}\) RT,
Or P = \(\frac{mRT}{VM}\),

Or P = \(\frac{dRT}{M}\)
Or d = \(\frac{PM}{RT}\)
if T = known constant
∴d ∝P

Question 5.
Derive Ideal gas equation on the basis of kinetic gas equation?
Answer:
According to Kinetic gas theory, “ The average kinetic energy of gas molecules is directly proportional to absolute temperature.”
Average Kinetic energy = \(\frac{1}{2}\) mnv²
\(\frac{1}{2}\) mnv² ∝ T
⇒\(\frac{1}{2}\) mnv² = KT
⇒\(\frac{3}{2}\) × \(\frac{1}{3}\) mnv² = KT
⇒\(\frac{1}{3}\) mnv² = \(\frac{2}{3}\) KT,
⇒\(\frac{PV}{T}\) = R
⇒ PV = RT.

Question 6.
Explanation of Dalton’s law on the basis of Kinetic gas theory?
Answer:
If the volume of container V litre and no. of moles of gas A is n₁ ar.d mass of each particles in m₁ R.M.S. velocity is V₁ then,
P_A = \(\frac{1}{3}\) \(\frac { m_{ 1 }n_{ 1 }v_{ 1 }^{ 2 } }{ V } \)
Kinetic gas equation PV = \(\frac{1}{3}\) mnv²
For B gas number of moles = n₂
Mass = m₂
R.M.S. Velocity = v₂
P_B = \(\frac{1}{3}\) \(\frac { m_{ 2 }n_{ 2 }v_{ 2 }^{ 2 } }{ V } \)
If both the gases are kept in same container total pressure of mixture
P = \(\frac{1}{3}\) \(\frac { m_{ 1 }n_{ 1 }v_{ 1 }^{ 2 } }{ V } \) + \(\frac{1}{3}\) \(\frac { m_{ 2 }n_{ 2 }v_{ 2 }^{ 2 } }{ V } \)
⇒ P = P_A + P_B
so in general, for more than two gases
P = P_A + P_B + P_C + …………………..
It is Daltons law of partial pressure.

MP Board Class 11 Chemistry Important Questions