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NEET Equilibrium MCQs with answers

601 questions on Equilibrium for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

mediumNEET 2022

1. Increasing pressure on N₂+3H₂⇌2NH₃:

  1. Shifts left
  2. Shifts right
  3. No effect
  4. Raises T
Show answer and explanation

Answer: (B) Shifts right

4 moles gas → 2 moles. Higher P → fewer moles side (right) → more NH₃.

Quick tip: High P → fewer moles side. Haber process uses this.

NCERT Class 11, Chapter 7, page 204

easy

2. Given the following two statements regarding a chemical equilibrium: Statement I: A catalyst increases the rate of the forward reaction and the backward reaction by the same factor. Statement II: A catalyst decreases the enthalpy change (ΔH) of the reaction, thereby shifting the equilibrium toward the products. In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

A catalyst provides an alternative pathway with lower activation energy for both forward and reverse reactions equally, leaving the equilibrium position and equilibrium constant unchanged. It does not alter the enthalpy change (ΔH) of the reaction.

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3. Given below are two statements: Statement I: For the equilibrium H₂O(s) ⇌ H₂O(l) at 273 K and 1 atm pressure, the rate of melting of ice equals the rate of freezing of water. Statement II: The temperature at which the liquid phase and solid phase are in dynamic equilibrium at 1 atm is defined as the normal freezing point of the substance. In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Both statements are correct. Dynamic equilibrium between solid and liquid implies opposing rates of melting and freezing are equal. By definition, the temperature at which solid and liquid coexist in dynamic equilibrium at 1 atm is the normal melting or freezing point.

hard

4. A mixture containing 1.0 mol of SO₂ and 1.0 mol of O₂ is placed in a 2.0 L vessel and allowed to reach equilibrium at temperature T according to: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). At equilibrium, 0.60 mol of SO₃ is formed. What is the equilibrium concentration of O₂?

  1. 0.35 mol L⁻¹
  2. 0.70 mol L⁻¹
  3. 0.20 mol L⁻¹
  4. 0.40 mol L⁻¹
Show answer and explanation

Answer: (A) 0.35 mol L⁻¹

From the stoichiometry, formation of 0.60 mol of SO₃ consumes 0.60 mol of SO₂ and (0.60 / 2) = 0.30 mol of O₂. Thus, remaining moles of O₂ at equilibrium = 1.0 - 0.30 = 0.70 mol. Since the vessel volume is 2.0 L, the equilibrium concentration [O₂] = 0.70 mol / 2.0 L = 0.35 mol L⁻¹.

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5. Calculate the pH of a solution prepared by mixing 50 mL of 0.2 M HNO₃ and 50 mL of 0.1 M KOH at 25 °C.

  1. 1.00
  2. 1.30
  3. 2.00
  4. 0.70
Show answer and explanation

Answer: (B) 1.30

Millimoles of H⁺ from HNO₃ = 50 × 0.2 = 10 mmol. Millimoles of OH⁻ from KOH = 50 × 0.1 = 5 mmol. Excess H⁺ = 10 - 5 = 5 mmol in total volume V = 100 mL. Therefore, [H⁺] = 5 mmol / 100 mL = 0.05 M = 5.0 × 10⁻² M. pH = -log(0.05) = -log(5 × 10⁻²) = 2 - log 5 = 2 - 0.699 = 1.30.

easy

6. Given the following two statements regarding a system at chemical equilibrium: Statement I: A catalyst increases the rate of both forward and reverse reactions by the same factor and does not change the position of equilibrium. Statement II: A catalyst alters the value of standard Gibbs free energy change (ΔG°) of the reaction. In light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

A catalyst provides an alternate pathway with a lower activation energy, lowering the energy barrier equally for both forward and reverse reactions; thus, it does not alter the equilibrium composition or the equilibrium constant. Moreover, ΔG° depends strictly on the initial and final states of reactants and products, so a catalyst does not alter ΔG°.

hard

7. For the reversible reaction: 2A(g) + B(s) ⇌ 3C(g); ΔH° = +45 kJ mol⁻¹, the system is at equilibrium. Which of the following simultaneous actions will unequivocally result in an increase in the number of moles of C(g)?

  1. Increasing temperature and decreasing container volume
  2. Increasing temperature and increasing container volume
  3. Decreasing temperature and adding more solid B
  4. Increasing total pressure and removing some solid B
Show answer and explanation

Answer: (B) Increasing temperature and increasing container volume

For the reaction, ΔH° > 0 (endothermic) and Δn_g = 3 - 2 = +1 > 0. Increasing the temperature favors the forward endothermic reaction, producing more C. Increasing the volume lowers the pressure, which favors the side with more gaseous moles (forward reaction, 3 moles of C vs 2 moles of A). Hence, both actions act synergistically to increase the number of moles of C(g).

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8. The molar solubility of AgCl in water is 1 × 10⁻⁴ mol/L. Its solubility product is:

  1. 1 × 10⁻⁴
  2. 2 × 10⁻⁸
  3. 1 × 10⁻⁸
  4. 4 × 10⁻⁸
Show answer and explanation

Answer: (C) 1 × 10⁻⁸

For a 1:1 salt, Ksp = s² = (1 × 10⁻⁴)² = 1 × 10⁻⁸.

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9. Henry's law constant for gas 'A' in water at 293 K is 1.0 × 10⁵ bar. If the partial pressure of gas 'A' over water is 0.5 bar, what is the number of millimoles of 'A' dissolved in 10 mol of water?

  1. 0.005 mmol
  2. 0.5 mmol
  3. 5.0 mmol
  4. 0.05 mmol
Show answer and explanation

Answer: (D) 0.05 mmol

According to Henry's law, p = K_H · x_A. Thus, x_A = p / K_H = 0.5 / (1.0 × 10⁵) = 5.0 × 10⁻⁶. Since x_A ≈ n_A / n_water, n_A = 5.0 × 10⁻⁶ × 10 mol = 5.0 × 10⁻⁵ mol = 0.05 mmol.

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10. For a reversible reaction A + B ⇌ C + D at dynamic equilibrium, addition of a catalyst:

  1. Increases the rates of both forward and backward reactions by the same factor.
  2. Increases the rate of the forward reaction more than the backward reaction.
  3. Shifts the equilibrium toward products by lowering ΔG° of the reaction.
  4. Decreases the time required to reach equilibrium while altering the equilibrium constant K_c.
Show answer and explanation

Answer: (A) Increases the rates of both forward and backward reactions by the same factor.

A catalyst provides an alternative reaction pathway with a lower activation energy for both the forward and backward directions equally. Thus, it increases both the forward and reverse rates by the exact same factor without altering the equilibrium position or K_c.

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11. Under what condition is the standard Gibbs free energy change (ΔrG°) of a reversible reaction equal to zero?

  1. When the equilibrium constant K = 1
  2. When the reaction quotient Q = 0
  3. When the reaction quotient Q = 1
  4. When the equilibrium constant K = 0
Show answer and explanation

Answer: (A) When the equilibrium constant K = 1

The relationship between standard Gibbs energy and equilibrium constant is ΔrG° = -RT ln K. For ΔrG° = 0, ln K = 0, which means K = 1.

easy

12. At a given temperature, a gas has a solubility of 0.050 g L⁻¹ in water when its partial pressure is 760 mm Hg. What will be the mass of the gas dissolved per litre of water if the partial pressure of the gas is increased to 1140 mm Hg at the same temperature?

  1. 0.075 g L⁻¹
  2. 0.025 g L⁻¹
  3. 0.033 g L⁻¹
  4. 0.100 g L⁻¹
Show answer and explanation

Answer: (A) 0.075 g L⁻¹

According to Henry's law, mass of dissolved gas m = k · p. Therefore, m₂/m₁ = p₂/p₁. Here, m₂ = 0.050 × (1140 / 760) = 0.050 × 1.5 = 0.075 g L⁻¹.

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13. According to the Lewis concept of acids and bases, which of the following species behaves as a Lewis acid?

  1. CH₄
  2. NH₃
  3. OH⁻
  4. CO₂
Show answer and explanation

Answer: (D) CO₂

In carbon dioxide (CO₂), the central carbon atom is bonded to two highly electronegative oxygen atoms via double bonds. Nucleophiles can attack the electrophilic carbon atom causing one pair of π-electrons to shift to oxygen, making CO₂ behave as a Lewis acid.

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14. The conjugate base of H₂O is:

  1. O²⁻
  2. H₂
  3. OH⁻
  4. H₃O⁺
Show answer and explanation

Answer: (C) OH⁻

When water donates a proton it forms OH⁻, its conjugate base. When it accepts a proton it forms H₃O⁺, its conjugate acid.

NCERT Class 11

easy

15. For the equilibrium: CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔrH° = -90.7 kJ mol⁻¹. Which of the following changes will decrease the equilibrium yield of methanol (CH₃OH)?

  1. Removing CH₃OH(g) continuously from the system
  2. Increasing the total pressure by reducing the volume
  3. Adding more H₂(g) to the container
  4. Increasing the temperature at constant volume
Show answer and explanation

Answer: (D) Increasing the temperature at constant volume

The forward reaction is exothermic (ΔrH° < 0). According to Le Chatelier's principle, increasing the temperature favours the endothermic reverse reaction, thereby shifting the equilibrium to the left and decreasing the yield of methanol.

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