NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | CHEMISTRY IMPORTANT QUESTIONS PART-1 | CHAPTER -7 | EQUILIBRIUM | EDUGROWN |

In This Post we are  providing Chapter-7 EQULIBRIUM NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON EQUILIBRIUM

Question 1.
Justify the statement that water behaves like acid as well as a base on the basis of the protonic concept.

Answer:
Water ionizes as H₂O + H₂O ⇌ H₃O⁺ + OH^–
With strong acids, water behaves as a base by accepting a proton from an acid.
HCl + H₂O ⇌ H₃O⁺ (aq) + Cl^– (aq)
While with bases, water behaves as an acid by liberating a proton
NH₃ + H₂O ⇌ NH₄⁺ (aq) + OH (aq).

Question 2.
What is p^OH? What is its value for pure water at 298 K?

Answer:
p^OH = – log [OH^–]
p^H + p^OH = 14 for pure water at 298 K
p^H = 7
or
p^OH of water at 298 = 7.

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

Answer:

Question 4.
An aqueous solution of CuSO₄ is acidic while that of Na₂SO₄ is neutral. Explain.

Answer:
CuSO₄ + 2H₂O ⇌ Cu(OH)₂ + H₂SO₄ (weak base strong acid)
CuS04 is the salt of weak base Cu(OH)₂ and a strong acid H₂SO₄.
Thus the solution will have free H⁺ ions and will, therefore, be acidic.

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

Question 5.
The dissociation constants of HCN, CH₃COOH, and HF are 7.2 × 10^-10, 1.8 × 10^-5, and 6.7 × 10^-4 respectively. Arrange them in increasing order of acid strength.

Answer:
More the value of Ka, the stronger the acid
Their Ka1S are 6.7 × 10^-4 > 1.8 × 10^-5 > 7.2 × 10^-10
∴ HCN < CH₃COOH < HF.

Question 6.
The dissociation of PCl₅ decreases in presence of Cl₂. Why?

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

Question 7.
The dissociation of HI is independent of pressure while dissociation of PCl₅ depends upon the pressure applied. Why?•

Answer:
For 2HI ⇌ H₂ + I₂
K_c = x24(1−x)2
where x = degree of dissociation

For PCl₅ ⇌ PCl₃ + Cl₂
K = x2V(1−x)2; V = Volume of container.

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

Question 8.
The reaction between ethyl acetate and water attains a state of equilibrium in an open vessel, but not the decomposition of CaCO3. Explain.

Answer:
CH₂ COOC₂H₅ (l) + H₂O (l) ⇌ CH₃COOH (l) + C₂H₅OH (l)
Here both reactants and products are liquids and they will not escape from the vessel even if it is open. Therefore equilibrium is attained.
CaSO₃ (s) ⇌ CaO (s) + CO₂ (g)
Here CO₂ is a gas. It will escape from the vessel if it is open and so backward reaction cannot take place. Therefore equilibrium is attained.

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

Answer:

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

Question 10.
Why NH₄Cl is added in precipitating III group hydroxides before the addition of NH4OH?

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

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

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

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

Answer:

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

Question 14.
The following reaction has attained equilibrium
CO (g) + 2H₂ (g) ⇌ CH₃OH (g); ΔH° = – 92.0 kJ mol^-1.
What will happen if

(i) the volume of the vessel is suddenly reduced to half?
Answer:
K_c = [CH₃OH]/[CO][H₂]², K_p = PCH₃OH/PCO × PH₂
When the volume of the vessel is reduced to half, the concentration of each reactant or product becomes double. Thus
Q_c = 2[CH₃OH]/2[CO] × {2[H₂]}² = i K..
As Q_c < K_p, equilibrium will shift in the forward direction.

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

Question 15.
How does the degree of ionization of a weak electrolyte vary with concentration? Give exact relationship. What is this law called?

Answer:
α = Ki/c−−−−√ . It is called Ostwald’s dilution law.
(K_i is ionization constant and c is the molar concentration).