Chemical Equilibrium – Factors Affecting Equilibrium and Le-Chatelier's Principles: NEET MCQs with Answers

15 free questions on Chemical Equilibrium – Factors Affecting Equilibrium and Le-Chatelier's Principles 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: Factors Affecting Equilibrium and Le-Chatelier's Principles: Equilibrium involving chemical processes: Law of chemical equilibrium, equilibrium constants (Kp and Kc) and their significance, the significance of G and G in chemical equilibrium, factors affecting equilibrium concentration, pressure, temperature, the effect of catalyst; Le Chatelier’s principle.

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Question 1

Previous-year question
A higher yield of NO in the reaction N₂(g) + O₂(g) ⇌ 2NO(g) [ΔH = +180.7 kJ mol⁻¹] can be obtained at: [NEET 2025]
A. higher temperature
B. lower temperature
C. higher concentration of N₂
D. higher concentration of O₂
Choose the correct answer from the options given below:
  1. A, D only
  2. B, C only
  3. B, C, D only
  4. A, C, D only
Show answer
Answer: (D) A, C, D only
Explanation: The forward reaction is endothermic, so a higher temperature (A) increases the yield of NO, while a lower temperature (B) decreases it. Increasing the concentration of either reactant, N₂ (C) or O₂ (D), shifts the equilibrium forward. Hence A, C and D are correct.

Question 2

Previous-year question
For the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −Q, the forward reaction is favoured by: [Re-NEET 2024]
  1. use of catalyst
  2. decreasing concentration of N₂
  3. low pressure, high temperature and high concentration of ammonia
  4. high pressure, low temperature and higher concentration of H₂
Show answer
Answer: (D) high pressure, low temperature and higher concentration of H₂
Explanation: The forward reaction is exothermic and reduces gaseous moles (4 → 2), so low temperature and high pressure favour it; adding a reactant (H₂) also shifts it forward. A catalyst does not shift equilibrium, and removing N₂, lowering pressure, raising temperature or adding NH₃ all shift it backward.

Question 3

Previous-year question
Hydrolysis of sucrose is given by Sucrose + H₂O ⇌ Glucose + Fructose. If the equilibrium constant Kc is 2 × 10¹³ at 300 K, the value of ΔᵣG° at the same temperature is [NEET 2020]
  1. 8.314 J mol⁻¹K⁻¹ × 300 K × ln(2 × 10¹³)
  2. 8.314 J mol⁻¹K⁻¹ × 300 K × ln(3 × 10¹³)
  3. −8.314 J mol⁻¹K⁻¹ × 300 K × ln(4 × 10¹³)
  4. −8.314 J mol⁻¹K⁻¹ × 300 K × ln(2 × 10¹³)
Show answer
Answer: (D) −8.314 J mol⁻¹K⁻¹ × 300 K × ln(2 × 10¹³)
Explanation: Directly from ΔᵣG° = −RT ln Kc with R = 8.314 J mol⁻¹ K⁻¹, T = 300 K and Kc = 2 × 10¹³. The large positive K makes ΔᵣG° strongly negative, as expected for a reaction that goes essentially to completion.

Question 4

Previous-year question
Which one of the following conditions will favour maximum formation of the product in the reaction A₂(g) + B₂(g) ⇌ X₂(g), ΔᵣH = −x kJ? [NEET 2018]
  1. Low temperature and high pressure
  2. Low temperature and low pressure
  3. High temperature and low pressure
  4. High temperature and high pressure
Show answer
Answer: (A) Low temperature and high pressure
Explanation: The reaction is exothermic, so a low temperature shifts it forward. It also proceeds with Δn = 1 − 2 = −1, so a high pressure shifts it forward as well. Both conditions together maximise X₂.

Question 5

Previous-year question
Which of the following statements is correct for a reversible process in a state of equilibrium? [AIPMT 2015]
  1. ΔG = 2.30 RT log K
  2. ΔG° = −2.30 RT log K
  3. ΔG° = 2.30 RT log K
  4. ΔG = −2.30 RT log K
Show answer
Answer: (B) ΔG° = −2.30 RT log K
Explanation: The thermodynamic relation is ΔG° = −RT ln K = −2.303 RT log K. At equilibrium ΔG itself (not ΔG°) is zero, so the equilibrium constant must be linked to the standard free-energy change.

Question 6

Previous-year question
For a given exothermic reaction, Kₚ and Kₚ′ are the equilibrium constants at temperatures T₁ and T₂ respectively (T₂ > T₁). Assuming that the heat of reaction is constant over this range, it is readily observed that [AIPMT 2014]
  1. Kₚ > Kₚ′
  2. Kₚ < Kₚ′
  3. Kₚ = Kₚ′
  4. Kₚ = 1/Kₚ′
Show answer
Answer: (A) Kₚ > Kₚ′
Explanation: For an exothermic reaction (ΔH negative) the van't Hoff equation ln(K₂/K₁) = −(ΔH/R)(1/T₂ − 1/T₁) gives a decreasing K as T rises. Since T₂ > T₁, K at the lower temperature is larger: Kₚ > Kₚ′.

Question 7

Previous-year question
For the reversible reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat, the equilibrium shifts in the forward direction [AIPMT 2014]
  1. by increasing the concentration of NH₃(g)
  2. by decreasing the pressure
  3. by decreasing the concentrations of N₂(g) and H₂(g)
  4. by increasing the pressure and decreasing the temperature
Show answer
Answer: (D) by increasing the pressure and decreasing the temperature
Explanation: The forward reaction is exothermic and proceeds with a decrease in the number of gaseous moles (4 → 2). High pressure and low temperature therefore both push the equilibrium towards ammonia. The other options each push it backwards.
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Question 8

Previous-year question
[2017] A 20 L container at 400 K contains CO₂(g) at 0.4 atm and an excess of SrO. The volume is decreased by moving a piston. The maximum volume of the container when the pressure of CO₂ attains its maximum value will be (SrCO₃(s) ⇌ SrO(s) + CO₂(g), Kₚ = 1.6 atm):
  1. 10 L
  2. 4 L
  3. 2 L
  4. 5 L
Show answer
Answer: (D) 5 L
Explanation: The CO₂ pressure can rise only until it reaches Kₚ = 1.6 atm; beyond this CO₂ combines with SrO to form SrCO₃. Using P₁V₁ = P₂V₂: 0.4×20 = 1.6×V₂, so V₂ = 5 L.

Question 9

Previous-year question
[2017] Given N₂ + 3H₂ ⇌ 2NH₃ (K₁), N₂ + O₂ ⇌ 2NO (K₂), H₂ + ½O₂ → H₂O (K₃). The equilibrium constant K for 2NH₃ + 5/2 O₂ ⇌ 2NO + 3H₂O is:
  1. K₂K₃³/K₁
  2. K₂K₃/K₁
  3. K₂³K₃/K₁
  4. K₁K₃³/K₂
Show answer
Answer: (A) K₂K₃³/K₁
Explanation: Reversing reaction 1 (1/K₁), adding reaction 2 (K₂) and three times reaction 3 (K₃³) gives the target, so K = K₂K₃³/K₁.

Question 10

Previous-year question
[RE-AIPMT 2015] If the equilibrium constant for N₂(g) + O₂(g) ⇌ 2NO(g) is K, then the equilibrium constant for ½N₂(g) + ½O₂(g) ⇌ NO(g) will be:
  1. K
  2. K²
  3. K^(1/2)
  4. K^(−1/2)
Show answer
Answer: (C) K^(1/2)
Explanation: Halving the balanced equation raises the equilibrium constant to the power ½, so the new constant is K^(1/2).

Question 11

Previous-year question
[AIPMT 2015] If the value of an equilibrium constant for a particular reaction is 1.6×10¹², then at equilibrium the system will contain:
  1. Mostly reactants
  2. Mostly products
  3. Similar amounts of reactants and products
  4. All reactants
Show answer
Answer: (B) Mostly products
Explanation: A very large K (1.6×10¹²) means the equilibrium lies far to the right, so the system contains mostly products.

Question 12

Previous-year question
[AIPMT 2014] For a given exothermic reaction, Kₚ and K′ₚ are the equilibrium constants at temperatures T₁ and T₂ respectively (T₂ > T₁). Assuming the heat of reaction is constant in the range T₁ to T₂, it is readily observed that:
  1. Kₚ > K′ₚ
  2. Kₚ < K′ₚ
  3. Kₚ = K′ₚ
  4. Kₚ = 1/K′ₚ
Show answer
Answer: (A) Kₚ > K′ₚ
Explanation: For an exothermic reaction the equilibrium constant decreases as temperature rises. Since T₂ > T₁, K′ₚ < Kₚ.

Question 13

Previous-year question
[AIPMT 2014] For the reversible reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + Heat, the equilibrium shifts in the forward direction:
  1. By increasing the concentration of NH₃(g)
  2. By decreasing the pressure
  3. By decreasing the concentrations of N₂(g) and H₂(g)
  4. By increasing the pressure
Show answer
Answer: (D) By increasing the pressure
Explanation: The forward reaction reduces gaseous moles (4 → 2), so increasing pressure shifts the equilibrium forward. Adding NH₃, removing N₂/H₂ or lowering pressure all shift it backward.

Question 14

Previous-year question
[AIPMT 2012] For A₂(g) + B₂(g) ⇌ 2AB(g) in a sealed vessel at 527°C, the equilibrium concentrations are [A₂] = 3×10⁻³ M, [B₂] = 4.2×10⁻³ M, [AB] = 2.8×10⁻³ M. The value of Kᶜ is:
  1. 0.62
  2. 4.5
  3. 2.0
  4. 1.9
Show answer
Answer: (A) 0.62
Explanation: Kᶜ = [AB]²/([A₂][B₂]) = (2.8×10⁻³)²/[(3×10⁻³)(4.2×10⁻³)] = 7.84×10⁻⁶/12.6×10⁻⁶ = 0.62.

Question 15

Previous-year question
[AIPMT 2012] Given that the equilibrium constant for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is 278 at a particular temperature, what is the equilibrium constant for SO₃(g) ⇌ SO₂(g) + ½O₂(g)?
  1. 6 × 10⁻²
  2. 1.3 × 10⁻⁵
  3. 1.8 × 10⁻³
  4. 3.6 × 10⁻³
Show answer
Answer: (A) 6 × 10⁻²
Explanation: The required reaction is the reverse of the given one, halved, so K = (1/278)^(1/2) = 1/16.67 ≈ 6 × 10⁻².

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