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Equilibrium

40 NEET-UG practice questions on Equilibrium, part of Chemistry. The first 25 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

  1. 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
    Show answer

    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.

  2. 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
    Show answer

    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.

  3. 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
    Show answer

    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.

  4. 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
    Show answer

    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.

  5. 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
    Show answer

    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).

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

  13. Question 14 · Medium

    The solubility product of a sparingly soluble salt MX₂ is 4 × 10⁻¹² at 298 K. Its molar solubility in pure water is

    1. A2 × 10⁻⁶ mol L⁻¹
    2. B1.6 × 10⁻⁴ mol L⁻¹
    3. C2 × 10⁻⁴ mol L⁻¹
    4. D1 × 10⁻⁴ mol L⁻¹

    Answer and explanation in LearnFree account

  14. Question 15 · Easy

    The conjugate base of the hydrogensulphate ion, HSO₄⁻, is

    1. ASO₄²⁻
    2. BHSO₃⁻
    3. CSO₃²⁻
    4. DH₂SO₄

    Answer and explanation in LearnFree account

  15. Question 16 · Medium

    Equal volumes, 50 mL each, of 0.2 M HCl and 0.1 M NaOH are mixed. What is the pH of the resulting solution? (log 5 = 0.7)

    1. A0.7
    2. B1.3
    3. C1.0
    4. D12.7

    Answer and explanation in LearnFree account

  16. Question 17 · Medium

    Given below are two statements: Statement I: For H₂(g) + I₂(g) ⇌ 2HI(g), changing the total pressure at constant temperature does not shift the equilibrium. Statement II: For an exothermic reaction, raising the temperature lowers the value of the equilibrium constant. Choose the correct answer.

    1. AStatement I is correct but Statement II is incorrect
    2. BBoth Statement I and Statement II are correct
    3. CBoth Statement I and Statement II are incorrect
    4. DStatement I is incorrect but Statement II is correct

    Answer and explanation in LearnFree account

  17. Question 18 · Easy

    Assertion A: Adding a catalyst to a reaction mixture at equilibrium does not change the equilibrium constant. Reason R: A catalyst lowers the activation energy of the forward reaction but not of the backward reaction. Choose the correct answer.

    1. ABoth A and R are correct and R is the correct explanation of A
    2. BA is correct but R is not correct
    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

    Answer and explanation in LearnFree account

  18. Question 19 · Hard

    A weak monobasic acid HA (Ka = 1.0 × 10⁻⁶) is present as a 0.01 M solution, which is then diluted 100 times with water. Using 1 − α ≈ 1, what are the pH and the degree of dissociation in the diluted solution?

    1. ApH 4, α = 1%
    2. BpH 6, α = 1%
    3. CpH 5, α = 10%
    4. DpH 5, α = 1%

    Answer and explanation in LearnFree account

  19. Question 20 · Medium

    At a certain temperature, Kc for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is 100. At the same temperature, Kc for SO₃(g) ⇌ SO₂(g) + (1/2)O₂(g) is

    1. A0.1
    2. B10
    3. C0.01
    4. D0.005

    Answer and explanation in LearnFree account

  20. Question 21 · Hard

    In a 2 L closed vessel, 1.0 mol of A(g) and 1.0 mol of B(g) are allowed to reach equilibrium: A(g) + B(g) ⇌ C(g) + D(g); Kc = 4 at this temperature. How many moles of A remain at equilibrium?

    1. A0.17 mol
    2. B0.67 mol
    3. C0.33 mol
    4. D0.20 mol

    Answer and explanation in LearnFree account

  21. Question 22 · Medium

    Arrange the following aqueous solutions at 298 K in increasing order of pH. A. 0.01 M HCl B. 0.001 M NaOH C. 0.001 M H₂SO₄ (assume complete ionisation) D. 0.01 M NaCl Choose the correct answer from the options given below.

    1. AC, A, D, B
    2. BC, A, B, D
    3. CA, C, D, B
    4. DA, C, B, D

    Answer and explanation in LearnFree account

  22. Question 23 · Medium

    Assertion A: At 333 K the pH of pure water is less than 7, yet the water is still neutral. Reason R: The ionic product of water at 298 K is 1.0 × 10⁻¹⁴. Choose the correct answer.

    1. AA is not correct but R is correct
    2. BBoth A and R are correct but R is not the correct explanation of A
    3. CBoth A and R are correct and R is the correct explanation of A
    4. DA is correct but R is not correct

    Answer and explanation in LearnFree account

  23. Question 24 · Medium

    Consider the following statements about buffer solutions. A. A solution containing acetic acid and sodium acetate is an acidic buffer. B. A solution containing ammonium hydroxide and ammonium chloride is a basic buffer. C. A solution of hydrochloric acid and sodium chloride acts as a buffer. D. The pH of an acetic acid–sodium acetate buffer depends only on the pKa and not on the ratio [salt]/[acid]. Choose the correct answer from the options given below.

    1. AA, B and D only
    2. BA, B and C only
    3. CA and B only
    4. DB and D only

    Answer and explanation in LearnFree account

  24. Question 25 · Easy

    Which species can act as a Lewis acid but cannot act as a Brønsted acid?

    1. ANH₄⁺
    2. BH₂O
    3. CHCl
    4. DBF₃

    Answer and explanation in LearnFree account

15 more Equilibrium questions

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