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Factors affecting equilibrium and Le Chatelier principle: NEET MCQs with answers

90 questions on this topic of Equilibrium (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. For the system: C(s) + CO₂(g) ⇌ 2CO(g), ΔH > 0, an inert gas (Argon) is introduced at constant volume. The equilibrium will:

  1. Remain unaffected
  2. Shift in the backward direction
  3. Shift in the forward direction
  4. Shift initially forward and then backward
Show answer and explanation

Answer: (A) Remain unaffected

Addition of an inert gas at constant volume does not change the partial pressures or molar concentrations of the reacting gaseous species. Hence, the reaction quotient Q remains equal to K, and the equilibrium remains unaffected.

medium

2. In the manufacture of hydrogen gas by the water-gas shift reaction: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g); ΔH < 0, which method will increase the equilibrium yield of H₂(g) without changing the value of the equilibrium constant?

  1. Continuous removal of CO₂(g) as it forms
  2. Increasing the temperature of the reactor
  3. Increasing the total pressure on the system
  4. Adding an efficient iron-based catalyst
Show answer and explanation

Answer: (A) Continuous removal of CO₂(g) as it forms

Continuous removal of a product (CO₂) causes the system to shift forward to produce more products by Le Chatelier's principle, thus increasing the yield of H₂. Since temperature is held constant, the equilibrium constant K does not change.

hard

3. Consider the solid-gas heterogeneous equilibrium: 2NaHCO₃(s) ⇌ Na₂CO₃(s) + CO₂(g) + H₂O(g); ΔH° > 0. At equilibrium, the total pressure in the vessel is P. If the volume of the container is suddenly halved at the same temperature, what will be the final total pressure when equilibrium is re-established (assuming both solids remain present)?

  1. 2P
  2. P / 2
  3. P
  4. 4P
Show answer and explanation

Answer: (C) P

For this heterogeneous equilibrium, Kp = p(CO₂) × p(H₂O) = (P/2) × (P/2) = P²/4. Since Kp depends only on temperature and both solids are still present, Kp must remain constant. Therefore, when the system re-establishes equilibrium at the original temperature, the total pressure must return to its original value P.

medium

4. Given the following two statements regarding chemical equilibrium: Statement I: A positive value of Δn(g) means addition of an inert gas at constant pressure shifts the equilibrium in the forward direction. Statement II: Addition of an inert gas at constant volume to any gaseous equilibrium increases the total pressure and shifts the equilibrium towards fewer moles of gas. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (C) Statement I is true but Statement II is false

At constant pressure, adding an inert gas increases the total volume, causing the system to shift towards the side with more moles of gas (forward if Δn(g) > 0). At constant volume, the partial pressures of reactants and products remain unchanged, so the equilibrium position does not shift at all.

easy

5. For the reaction: 2A(g) ⇌ B(g) + C(g), the value of Kp is 4 at 300 K and 9 at 400 K. What can be concluded about the enthalpy change (ΔH) of the forward reaction?

  1. ΔH = 0
  2. ΔH < 0
  3. ΔH > 0
  4. ΔH cannot be predicted from this data
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Answer: (C) ΔH > 0

According to the van 't Hoff equation, when the equilibrium constant K increases with an increase in temperature (from 4 at 300 K to 9 at 400 K), the forward reaction must be endothermic, meaning ΔH > 0.

hard

6. For the gaseous equilibrium: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), which of the following expressions correctly gives the effect of increasing the total pressure P on the equilibrium position in terms of the mole fractions x (where Kx is the equilibrium constant in terms of mole fractions)?

  1. Kx = Kp × P⁻²
  2. Kx = Kp × P⁻¹
  3. Kx = Kp × P²
  4. Kx = Kp × P
Show answer and explanation

Answer: (D) Kx = Kp × P

For the given reaction, Δn = 2 - (2 + 1) = -1. Since Kp = Kx × P^(Δn), we have Kp = Kx × P⁻¹, which rearranges to Kx = Kp × P. Therefore, increasing the total pressure P increases Kx, which shifts the equilibrium forward to yield more SO₃.

easy

7. For the system: CO(g) + 2H₂(g) ⇌ CH₃OH(g); ΔH = -90.7 kJ mol⁻¹. Under which set of conditions will the equilibrium yield of methanol (CH₃OH) be maximized?

  1. Low temperature and high pressure
  2. High temperature and low pressure
  3. High temperature and high pressure
  4. Low temperature and low pressure
Show answer and explanation

Answer: (A) Low temperature and high pressure

The forward reaction is exothermic (ΔH < 0), so lower temperature favors the forward reaction according to Le Chatelier's principle. Additionally, the forward reaction proceeds with a decrease in gaseous moles (3 moles → 1 mole), so higher pressure favors the forward reaction. Therefore, low temperature and high pressure maximize the yield.

hard

8. Consider the homogeneous gaseous reaction: 2A(g) + B(g) ⇌ 2C(g); ΔH° = -60 kJ mol⁻¹. At equilibrium, the volume of the reaction container is suddenly compressed to half of its original volume at constant temperature. Once the new equilibrium is established, how does the final concentration of C compare with its initial equilibrium concentration [C]₀?

  1. It is less than [C]₀
  2. It is equal to [C]₀
  3. It is greater than [C]₀ but less than 2[C]₀
  4. It is greater than 2[C]₀
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Answer: (D) It is greater than 2[C]₀

When the volume is halved, instantaneous concentrations of all species double: [C] becomes 2[C]₀ and Q_c = (2[C]₀)² / ((2[A]₀)²(2[B]₀)) = Q_c,initial / 2 = K_c / 2. Because Q_c < K_c, the system shifts further in the forward direction (fewer moles) to restore equilibrium, increasing the concentration of C beyond 2[C]₀.

medium

9. For H₂(g) + I₂(g) ⇌ 2HI(g), increasing the total pressure:

  1. Has no effect on the equilibrium position
  2. Shifts it to the right
  3. Shifts it to the left
  4. Doubles K
Show answer and explanation

Answer: (A) Has no effect on the equilibrium position

The number of moles of gas is the same on both sides (Δn = 0), so a change in pressure does not shift the equilibrium.

NCERT Class 11

easy

10. Given the following two statements regarding a gaseous equilibrium: Statement I: For an endothermic reaction, the value of equilibrium constant increases as temperature is raised. Statement II: According to Le Chatelier's principle, the absorption of heat shifts the equilibrium in the endothermic direction. Choose the correct option:

  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

By the van 't Hoff equation (d ln K / dT = ΔH° / RT²), K increases with T when ΔH° > 0. Le Chatelier's principle explains this physically: when heat is supplied (raising T), the system absorbs heat by shifting in the endothermic direction, resulting in higher product concentrations and a larger K. Both statements are correct.

medium

11. Consider the equilibrium: Fe³⁺(aq) (pale yellow) + SCN⁻(aq) (colourless) ⇌ [Fe(SCN)]²⁺(aq) (deep red). Which of the following reagents, when added to the equilibrium mixture, will decrease the red colour intensity?

  1. FeCl₃(aq)
  2. KSCN(aq)
  3. H₂C₂O₄(aq)
  4. HCl(aq)
Show answer and explanation

Answer: (C) H₂C₂O₄(aq)

Addition of oxalic acid (H₂C₂O₄) forms a stable complex [Fe(C₂O₄)₃]³⁻ with Fe³⁺ ions. This removes free Fe³⁺ from the solution, shifting the equilibrium in the backward direction and reducing the intensity of the deep red [Fe(SCN)]²⁺ complex.

hard

12. For the dissociation reaction: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), argon gas is added at constant total pressure P. How does this addition affect the degree of dissociation (α) of PCl₅ and the partial pressure of Cl₂?

  1. α increases, partial pressure of Cl₂ decreases
  2. α decreases, partial pressure of Cl₂ increases
  3. Both α and partial pressure of Cl₂ increase
  4. α remains unchanged, partial pressure of Cl₂ decreases
Show answer and explanation

Answer: (A) α increases, partial pressure of Cl₂ decreases

At constant total pressure, adding an inert gas increases the total volume, decreasing the partial pressures of all existing gases. The equilibrium shifts toward the side with more moles of gas (forward direction), so α increases. However, the dilution effect outweighs the shift, resulting in a net decrease in the partial pressure of Cl₂.

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