Assertion A: Adding an inert gas at constant volume shifts the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g) to the right. Reason R: At constant volume, adding an inert gas leaves the partial pressures of the reacting gases unchanged. Choose the correct answer.

A is not correct but R is correct

A is correct but R is not correct

Both A and R are correct and R is the correct explanation of A

Both A and R are correct but R is not the correct explanation of A

Equilibrium

13 NEET-UG practice questions on Equilibrium, part of Chemistry. The first 13 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Medium
Assertion A: Adding an inert gas at constant volume shifts the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g) to the right. Reason R: At constant volume, adding an inert gas leaves the partial pressures of the reacting gases unchanged. Choose the correct answer.
1. AA is correct but R is not correct
2. BBoth A and R are correct and R is the correct explanation of A
3. CA is not correct but R is correct
4. DBoth A and R are correct but R is not the correct explanation of A
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Answer: C · A is not correct but R is correct
The number of moles of each reacting gas and the volume stay the same, so their partial pressures and molar concentrations do not change. The reaction quotient still equals K, and the equilibrium does not shift.
- A: A is false, because there is no shift at constant volume.
- B: A is false, so R cannot explain it.
- D: A is false, so this outcome cannot hold.
Remember this
An inert gas added at constant volume does not disturb an equilibrium.
Question 2 · Hard
Take Ksp of Ag₂CrO₄ as 4 × 10⁻¹². By what factor is its solubility in 0.01 M AgNO₃ smaller than its solubility in pure water?
1. A50
2. B250000
3. C2500
4. D10000
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Answer: C · 2500
In water, 4s³ = 4 × 10⁻¹² gives s = 10⁻⁴ mol/L. In 0.01 M AgNO₃, [Ag⁺] ≈ 0.01, so s′ = Ksp/[Ag⁺]² = 4 × 10⁻¹²/10⁻⁴ = 4 × 10⁻⁸ mol/L. The factor = 10⁻⁴/(4 × 10⁻⁸) = 2500.
- A: This finds the water solubility by a square root, as for an AB salt.
- B: This forgets to square [Ag⁺] in the Ksp expression.
- D: This doubles 0.01 M for [Ag⁺], although it is the added salt concentration.
Remember this
With a common ion, [Ag⁺] is fixed by the added salt, but it is still squared in the Ksp expression.
Question 3 · Medium
Match List-I with List-II. List-I (salt in water) A. KCl B. NH₄NO₃ C. Na₂CO₃ D. CH₃COONH₄ List-II (behaviour of the solution) I. Nearly neutral, as both ions hydrolyse to a similar extent II. Basic, as the anion hydrolyses III. Neutral, with no hydrolysis IV. Acidic, as the cation hydrolyses
1. AA-III, B-IV, C-II, D-I
2. BA-III, B-II, C-IV, D-I
3. CA-I, B-IV, C-II, D-III
4. DA-III, B-IV, C-I, D-II
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Answer: A · A-III, B-IV, C-II, D-I
KCl comes from a strong acid and strong base, so neither ion hydrolyses. NH₄⁺ gives an acidic solution and CO₃²⁻ a basic one. In ammonium acetate, Ka and Kb are nearly equal, so the solution is close to neutral.
- B: This swaps the cation and anion hydrolysis of NH₄NO₃ and Na₂CO₃.
- C: This swaps KCl and ammonium acetate, ignoring hydrolysis of both ions in the latter.
- D: This swaps Na₂CO₃ and ammonium acetate, although only carbonate gives a clearly basic solution.
Remember this
Only the ion from the weak parent hydrolyses, and the solution takes the nature of the strong parent.
Question 4 · Medium
In a 1 L flask, 1 mol of H₂ and 1 mol of I₂ are heated and reach equilibrium with 1.6 mol of HI: H₂(g) + I₂(g) ⇌ 2HI(g) Find the value of Kc.
1. A40
2. B8
3. C12.8
4. D64
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Answer: D · 64
Forming 1.6 mol HI uses 0.8 mol each of H₂ and I₂, so 0.2 mol of each remains. Kc = (1.6)²/(0.2 × 0.2) = 2.56/0.04 = 64.
- A: This forgets to square the HI concentration.
- B: This divides 1.6 by 0.2, with no squaring and only one reactant.
- C: This squares HI but divides by only one reactant concentration.
Remember this
Use the equation's coefficients for moles consumed, and raise each concentration to its coefficient in K.
Question 5 · Hard
Find the pH of 0.1 M aqueous CH₃COONa at 25 °C. (pKa of CH₃COOH = 4.74, pKw = 14)
1. A8.87
2. B5.13
3. C9.37
4. D9.87
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Answer: A · 8.87
Kb of acetate = 10⁻¹⁴/10⁻⁴·⁷⁴ = 10⁻⁹·²⁶, so [OH⁻] = √(10⁻⁹·²⁶ × 10⁻¹) = 10⁻⁵·¹³, and pOH = 5.13, giving pH = 8.87. Check: pH = 7 + (1/2)(pKa + log C) = 7 + (1/2)(4.74 − 1) = 8.87.
- B: This is the pOH, not the pH.
- C: This leaves out the log C term.
- D: This takes log C as +1 instead of −1.
Remember this
For a salt of a weak acid and strong base, pH = 7 + (1/2)(pKa + log C).
Question 6 · Medium
A sparingly soluble hydroxide M(OH)₂ has a solubility of 2 × 10⁻⁴ mol/L in water. Find its Ksp.
1. A8 × 10⁻¹²
2. B1.6 × 10⁻¹¹
3. C8 × 10⁻⁸
4. D3.2 × 10⁻¹¹
Answer and explanation in LearnFree account
Question 7 · Hard
Sodium hydroxide (20 mL of 0.1 M) is added to 50 mL of 0.2 M CH₃COOH. Find the pH of the resulting solution. (pKa = 4.74, log 2 = 0.30)
1. A0.94
2. B4.74
3. C5.34
4. D4.14
Answer and explanation in LearnFree account
Question 8 · Medium
Given below are two statements: Statement I: The ionisation constant Ka of a weak acid increases when its solution is diluted. Statement II: The degree of ionisation of a weak acid increases when its solution is diluted. Choose the correct answer.
1. AStatement I is correct but Statement II is incorrect
2. BStatement I is incorrect but Statement II is correct
3. CBoth Statement I and Statement II are correct
4. DBoth Statement I and Statement II are incorrect
Answer and explanation in LearnFree account
Question 9 · Medium
At equilibrium, PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is 50% dissociated, and the total pressure is 3 bar. Find Kp.
1. A0.75 bar
2. B0.25 bar
3. C1 bar
4. D3 bar
Answer and explanation in LearnFree account
Question 10 · Easy
For the gas-phase equilibrium N₂O₄(g) ⇌ 2NO₂(g), how are Kp and Kc related?
1. AKp = Kc(RT)²
2. BKp = Kc/(RT)
3. CKp = Kc(RT)
4. DKp = Kc
Answer and explanation in LearnFree account
Question 11 · Easy
The synthesis N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is exothermic. Which change increases the equilibrium yield of ammonia?
1. ARemoving some H₂ gas
2. BIncreasing the total pressure
3. CAdding an iron catalyst
4. DRaising the temperature
Answer and explanation in LearnFree account
Question 12 · Medium
Arrange the following 0.1 M aqueous solutions at 25 °C in increasing order of pH. A. NH₄Cl B. NaCl C. CH₃COONa D. HCl
1. AD, A, B, C
2. BC, B, A, D
3. CD, B, A, C
4. DA, D, B, C
Answer and explanation in LearnFree account
Question 13 · Easy
Calculate the pH of 0.001 M aqueous NaOH at 25 °C. (Kw = 1 × 10⁻¹⁴)
1. A11
2. B14
3. C3
4. D7
Answer and explanation in LearnFree account

Equilibrium

Assertion A: Adding an inert gas at constant volume shifts the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g) to the right. Reason R: At constant volume, adding an inert gas leaves the partial pressures of the reacting gases unchanged. Choose the correct answer.

Take Ksp of Ag₂CrO₄ as 4 × 10⁻¹². By what factor is its solubility in 0.01 M AgNO₃ smaller than its solubility in pure water?

In a 1 L flask, 1 mol of H₂ and 1 mol of I₂ are heated and reach equilibrium with 1.6 mol of HI: H₂(g) + I₂(g) ⇌ 2HI(g) Find the value of Kc.

Find the pH of 0.1 M aqueous CH₃COONa at 25 °C. (pKa of CH₃COOH = 4.74, pKw = 14)

A sparingly soluble hydroxide M(OH)₂ has a solubility of 2 × 10⁻⁴ mol/L in water. Find its Ksp.

Sodium hydroxide (20 mL of 0.1 M) is added to 50 mL of 0.2 M CH₃COOH. Find the pH of the resulting solution. (pKa = 4.74, log 2 = 0.30)

Given below are two statements: Statement I: The ionisation constant Ka of a weak acid increases when its solution is diluted. Statement II: The degree of ionisation of a weak acid increases when its solution is diluted. Choose the correct answer.

At equilibrium, PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is 50% dissociated, and the total pressure is 3 bar. Find Kp.

For the gas-phase equilibrium N₂O₄(g) ⇌ 2NO₂(g), how are Kp and Kc related?

The synthesis N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is exothermic. Which change increases the equilibrium yield of ammonia?

Arrange the following 0.1 M aqueous solutions at 25 °C in increasing order of pH. A. NH₄Cl B. NaCl C. CH₃COONa D. HCl

Calculate the pH of 0.001 M aqueous NaOH at 25 °C. (Kw = 1 × 10⁻¹⁴)