Chemical Equilibrium: NEET MCQs with Answers

15 free questions on Chemical Equilibrium for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Chemical Equilibrium, Law of Mass Action and Equilibrium Constant: Meaning of equilibrium, the concept of dynamic equilibrium. Equilibria involving physical processes: Solid-liquid, liquid - gas and solid-gas equilibria, Henry's law. General characteristics of equilibrium involving physical processes.

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Question 1

Previous-year question
At a given temperature and pressure, the equilibrium constants for the following equilibria are: [Re-NEET 2024]
3A₂ + B₂ ⇌ 2A₃B ; K₁
A₃B ⇌ (3/2)A₂ + ½B₂ ; K₂
Which relation between K₁ and K₂ is correct?
  1. K₁² = 2K₂
  2. K₂ = K₁/2
  3. K₁ = 1/√K₂
  4. K₂ = 1/√K₁
Show answer
Answer: (D) K₂ = 1/√K₁
Explanation: The second equation is the first one reversed and multiplied by ½. Reversing gives 1/K₁, and halving takes the square root, so K₂ = (1/K₁)¹ᐟ² = 1/√K₁. Equivalently K₁ = 1/K₂², so the relation K₁ = 1/√K₂ is incorrect.

Question 2

Previous-year question
For the equilibrium 2NOCl(g) ⇌ 2NO(g) + Cl₂(g), Kc is 3.0 × 10⁻⁶ at 1000 K. Find Kₚ (in bar) for the reaction at this temperature. (Given R = 8.314 J K⁻¹ mol⁻¹) [NEET (Sawai Madhopur) 2024]
  1. 2.494 × 10⁻²
  2. 3.0 × 10⁻⁶
  3. 2.494 × 10⁻⁴
  4. 1.493
Show answer
Answer: (C) 2.494 × 10⁻⁴
Explanation: Δn(gas) = (2 + 1) − 2 = 1, so Kₚ = Kc(RT). The units of R must match the units of K: Kc is in mol L⁻¹ and Kₚ is wanted in bar, so R is written as 0.08314 L bar K⁻¹ mol⁻¹ (the same as 8.314 J K⁻¹ mol⁻¹). Kₚ = 3.0 × 10⁻⁶ × 0.08314 × 1000 ≈ 2.494 × 10⁻⁴ bar. Substituting 8.314 directly with Kc in mol L⁻¹ gives 2.494 × 10⁻², but that number is in kPa, not bar.

Question 3

Previous-year question
For the equilibrium PCl₅ ⇌ PCl₃ + Cl₂ at 500 K, the concentrations are: PCl₅ = 1.40 M, Cl₂ = 1.60 M and PCl₃ = 1.60 M. Calculate Kc. [NEET (Sawai Madhopur) 2024]
  1. 2.6
  2. 1.83
  3. 3.4
  4. 2.00
Show answer
Answer: (B) 1.83
Explanation: For PCl₅ ⇌ PCl₃ + Cl₂, the law of mass action gives Kc = [PCl₃][Cl₂]/[PCl₅]. Substituting the equilibrium concentrations: Kc = (1.60 × 1.60)/1.40 = 2.56/1.40 ≈ 1.83. Note that Kc uses equilibrium (not initial) molar concentrations, and products appear in the numerator raised to their stoichiometric coefficients.

Question 4

Previous-year question
Match the equilibria in List-I with the conditions/parameters in List-II. [NEET (Sawai Madhopur) 2024]
List-I (Equilibria):
A. Liquid ⇌ Vapour
B. Solid ⇌ Liquid
C. Solid ⇌ Vapour
D. Solute (solid) ⇌ Solute (solution)
List-II (Conditions/Parameters):
I. Melting point
II. Boiling point
III. Sublimation point
IV. Saturated solution
V. Unsaturated solution
  1. A-I, B-II, C-III, D-IV
  2. A-III, B-II, C-I, D-IV
  3. A-II, B-I, C-III, D-V
  4. A-II, B-I, C-III, D-IV
Show answer
Answer: (D) A-II, B-I, C-III, D-IV
Explanation: Liquid ⇌ vapour equilibrium exists at the boiling point (A-II); solid ⇌ liquid at the melting point (B-I); solid ⇌ vapour at the sublimation point (C-III). Dissolved solute is in equilibrium with undissolved solid only in a saturated solution (D-IV), not in an unsaturated one.

Question 5

Previous-year question
For the reaction A + B ⇌ C + D at 300 K, the equilibrium concentrations of A, B, C and D are 2, 3, 10 and 6 mol L⁻¹ respectively. What is ΔG° for the reaction? (R = 2 cal mol⁻¹ K⁻¹) [NEET 2023]
  1. −137.26 cal
  2. −1381.80 cal
  3. −13.73 cal
  4. 1372.60 cal
Show answer
Answer: (B) −1381.80 cal
Explanation: K = [C][D]/[A][B] = (10 × 6)/(2 × 3) = 10. ΔG° = −2.303RT log K = −2.303 × 2 × 300 × log 10 = −2.303 × 600 × 1 = −1381.8 cal. A negative ΔG° is consistent with K > 1.

Question 6

Previous-year question
The reaction A + B ⇌ C + D takes place in a 1 L container at 300 K. Initially one mole each of A and B is present, and at equilibrium 0.7 mol of C has been formed. The equilibrium constant (Kc) for the reaction is: [NEET (Overseas) 2022]
  1. 1.2
  2. 6.2
  3. 5.4
  4. 9.7
Show answer
Answer: (C) 5.4
Explanation: At equilibrium: C = D = 0.7 mol L⁻¹ and A = B = 1 − 0.7 = 0.3 mol L⁻¹. Kc = [C][D]/[A][B] = (0.7 × 0.7)/(0.3 × 0.3) = 0.49/0.09 ≈ 5.4.

Question 7

Previous-year question
Equivalent amounts of H₂ and I₂ are heated in a closed vessel until equilibrium is reached. If 80% of the hydrogen is converted into HI, what is Kc at this temperature? [NEET 2021]
  1. 64
  2. 16
  3. 0.25
  4. 4
Show answer
Answer: (A) 64
Explanation: H₂ + I₂ ⇌ 2HI. Starting with 1 mol each, 0.8 mol H₂ reacts, so at equilibrium H₂ = 0.2, I₂ = 0.2 and HI = 1.6 mol (Δn = 0, so volume cancels). Kc = (1.6)²/(0.2 × 0.2) = 2.56/0.04 = 64.
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Question 8

Previous-year question
A 20 litre container at 400 K holds CO₂(g) at 0.4 atm together with an excess of solid SrO. The volume is now decreased by moving a piston. Given SrCO₃(s) ⇌ SrO(s) + CO₂(g) with Kₚ = 1.6 atm, the maximum volume of the container when the pressure of CO₂ attains its maximum value will be [NEET 2017]
  1. 10 litre
  2. 4 litre
  3. 2 litre
  4. 5 litre
Show answer
Answer: (D) 5 litre
Explanation: CO₂ cannot exceed the pressure fixed by Kₚ, which is 1.6 atm; beyond that it is removed as SrCO₃. Before any reaction occurs, Boyle's law gives P₁V₁ = P₂V₂, so 0.4 × 20 = 1.6 × V₂ and V₂ = 5 litre. This is the largest volume at which CO₂ reaches its maximum pressure.

Question 9

Previous-year question
Given N₂ + 3H₂ ⇌ 2NH₃ (K₁), N₂ + O₂ ⇌ 2NO (K₂) and H₂ + ½O₂ ⇌ H₂O (K₃), the equilibrium constant K for 2NH₃ + (5/2)O₂ ⇌ 2NO + 3H₂O is [NEET 2017]
  1. K₂K₃³/K₁
  2. K₂K₃/K₁
  3. K₂³K₃/K₁
  4. K₁K₃³/K₂
Show answer
Answer: (A) K₂K₃³/K₁
Explanation: The target equation is obtained as (reverse of reaction 1) + (reaction 2) + 3 × (reaction 3). Reversing inverts the constant, and multiplying an equation by n raises its constant to the n-th power, so K = (1/K₁) × K₂ × K₃³ = K₂K₃³/K₁.

Question 10

Previous-year question
For the liquid ⇌ vapour equilibrium, the correct relation describing the variation of the vapour pressure with temperature is [NEET 2016]
  1. d(ln P)/dT = ΔHᵥ/RT²
  2. d(ln P)/dT = ΔHᵥ/RT
  3. d(ln P)/dT = −ΔHᵥ/T²
  4. d(ln P)/dT² = −ΔHᵥ/RT²
Show answer
Answer: (A) d(ln P)/dT = ΔHᵥ/RT²
Explanation: This is the Clausius–Clapeyron equation in differential form: d(ln P)/dT = ΔHᵥ/RT², where ΔHᵥ is the enthalpy of vaporisation. Because ΔHᵥ is positive, vapour pressure always rises as temperature rises. Written against 1/T it becomes d(ln P)/d(1/T) = −ΔHᵥ/R, which is the familiar straight-line form.

Question 11

Previous-year question
If the equilibrium constant for N₂(g) + O₂(g) ⇌ 2NO(g) is K, the equilibrium constant for ½N₂(g) + ½O₂(g) ⇌ NO(g) will be [AIPMT 2015]
  1. K¹ᐟ²
  2. ½K
  3. K
  4. K²
Show answer
Answer: (A) K¹ᐟ²
Explanation: Every coefficient has been halved, so the equilibrium constant is raised to the power ½, giving K¹ᐟ² = √K. Note that halving the coefficients is not the same as halving K.

Question 12

Previous-year question
If the equilibrium constant of a particular reaction is 1.6 × 10¹², the equilibrium mixture will contain [AIPMT 2015]
  1. mostly reactants
  2. mostly products
  3. comparable amounts of reactants and products
  4. only reactants
Show answer
Answer: (B) mostly products
Explanation: A very large K (typically K > 10³) means the numerator, which carries the product concentrations, greatly outweighs the denominator. The reaction therefore proceeds nearly to completion and the mixture is overwhelmingly product.

Question 13

Which of the following statements about equilibria in physical processes is incorrect?
  1. For H₂O(s) ⇌ H₂O(l) at constant pressure, the melting point changes with the amount of ice present
  2. For sugar(s) ⇌ sugar(solution), the concentration of sugar in the saturated solution is constant at a given temperature
  3. For H₂O(l) ⇌ H₂O(g) in a closed vessel, the pressure of water vapour is constant at a given temperature
  4. For CO₂(g) ⇌ CO₂(aq), the ratio [CO₂(aq)]/[CO₂(g)] is constant at a given temperature
Show answer
Answer: (A) For H₂O(s) ⇌ H₂O(l) at constant pressure, the melting point changes with the amount of ice present
Explanation: At constant pressure, a solid and its liquid coexist at one fixed melting point, which does not depend on how much ice or water is present. The other three statements give the quantity that stays constant for each process at a given temperature. These are the vapour pressure, the solubility of the solute, and the ratio of dissolved gas to gas above the liquid.

Question 14

Water at the same temperature is kept in an open beaker and in a sealed flask standing in a thermostat bath. Which conclusion about liquid–vapour equilibrium is correct?
  1. Equilibrium is set up in both vessels, because the temperature is the same in each
  2. Equilibrium is set up only in the sealed flask, because the vapour cannot escape; the exchange of heat with the bath does not prevent it
  3. Equilibrium is set up only in the open beaker, because the vapour can move freely above the liquid
  4. Equilibrium is set up in neither vessel, because it can be reached only in an isolated system
Show answer
Answer: (B) Equilibrium is set up only in the sealed flask, because the vapour cannot escape; the exchange of heat with the bath does not prevent it
Explanation: Liquid–vapour equilibrium needs a closed system, so that molecules leaving the liquid stay above it and the rate of condensation can rise to match the rate of evaporation. In the open beaker vapour escapes, and the water simply keeps evaporating. An isolated system is not required, because a closed vessel that exchanges heat with a thermostat can still reach equilibrium at constant temperature.

Question 15

Water boils below 100 °C at a hill station, while food cooks faster in a pressure cooker. Which single idea about the liquid–vapour equilibrium explains both observations?
  1. Boiling point depends only on the amount of water, which is less in a cooker than in an open vessel
  2. A liquid boils when its vapour pressure equals the external pressure, so boiling point rises and falls with the external pressure
  3. Rate of evaporation at a fixed temperature depends only on the external pressure, so boiling point stays fixed
  4. The vapour pressure of water at any temperature falls as the external pressure rises, so boiling is delayed
Show answer
Answer: (B) A liquid boils when its vapour pressure equals the external pressure, so boiling point rises and falls with the external pressure
Explanation: Vapour pressure of water depends on temperature, and boiling occurs when it equals the pressure above the liquid. At a hill station the atmospheric pressure is lower, so this condition is met below 100 °C. In a sealed cooker the steam raises the pressure above 1 atm, so water stays liquid at a higher temperature and food cooks faster.

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