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Chemical equilibrium, Kp/Kc and Gibbs energy: 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. If the equilibrium constant for the reaction: N₂(g) + O₂(g) ⇌ 2NO(g) is K, what is the equilibrium constant for the reaction: NO(g) ⇌ (1/2)N₂(g) + (1/2)O₂(g) at the same temperature?

  1. 1 / √K
  2. 1 / K²
  3. √K
  4. -K
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Answer: (A) 1 / √K

Reversing a chemical equation inverts its equilibrium constant (1/K), and multiplying the stoichiometric coefficients by 1/2 raises the equilibrium constant to the power 1/2. Therefore, K_new = (1/K)^(1/2) = 1 / √K.

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2. At 700 K, hydrogen and iodine react according to: H₂(g) + I₂(g) ⇌ 2HI(g), for which Kc = 49. If 0.50 mol each of H₂ and I₂ are added to a 1.0 L sealed flask, what will be the equilibrium concentration of HI?

  1. 0.44 M
  2. 0.39 M
  3. 0.88 M
  4. 0.78 M
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Answer: (D) 0.78 M

Let x mol L⁻¹ of H₂ react. At equilibrium: [H₂] = 0.50 - x, [I₂] = 0.50 - x, [HI] = 2x. Kc = (2x)² / (0.50 - x)² = 49. Taking square root: 2x / (0.50 - x) = 7 ⇒ 2x = 3.5 - 7x ⇒ 9x = 3.5 ⇒ x = 3.5 / 9 ≈ 0.389 M. Therefore, [HI] = 2x = 2 × 0.389 = 0.778 M ≈ 0.78 M.

hard

3. For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), the degree of dissociation is α at total pressure P. Which expression correctly gives Kp in terms of α and P?

  1. [α² / (1 - α²)] P
  2. [α / (1 - α)] P
  3. [α² / (1 - α)] P²
  4. [α² / (1 + α)²] P
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Answer: (A) [α² / (1 - α²)] P

Starting with 1 mol of PCl₅: at equilibrium, moles are PCl₅ = 1 - α, PCl₃ = α, Cl₂ = α. Total moles = 1 + α. Partial pressures are p(PCl₅) = [(1 - α)/(1 + α)]P, p(PCl₃) = p(Cl₂) = [α/(1 + α)]P. Thus, Kp = [p(PCl₃) × p(Cl₂)] / p(PCl₅) = [α² P² / (1 + α)²] / [((1 - α)/(1 + α))P] = [α² / (1 - α²)] P.

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4. If the equilibrium constant for A + B ⇌ C is K, the equilibrium constant for 2C ⇌ 2A + 2B is:

  1. K²
  2. 2K
  3. 1/(2K)
  4. 1/K²
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Answer: (D) 1/K²

Reversing a reaction inverts K (1/K), and doubling the coefficients squares it, giving (1/K)² = 1/K².

NCERT Class 11

easy

5. For the gaseous reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), which of the following expressions correctly relates Kp and Kc?

  1. Kp = Kc(RT)⁻¹
  2. Kp = Kc(RT)²
  3. Kp = Kc(RT)⁻²
  4. Kp = Kc(RT)
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Answer: (C) Kp = Kc(RT)⁻²

For a general gaseous reaction, Kp = Kc(RT)^(Δn_g). Here, Δn_g = (moles of gaseous products) - (moles of gaseous reactants) = 2 - (1 + 3) = -2. Therefore, Kp = Kc(RT)⁻².

hard

6. For the reaction: 2NOBr(g) ⇌ 2NO(g) + Br₂(g), NOBr is placed in a closed vessel and allowed to dissociate. If the degree of dissociation of NOBr is α (where α << 1) and the total equilibrium pressure is P, which of the following expressions correctly gives Kp?

  1. (α² P) / 2
  2. (α² P) / 4
  3. α³ P
  4. (α³ P) / 2
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Answer: (D) (α³ P) / 2

Let initial moles of NOBr be 1. At equilibrium: n(NOBr) = 1 - α ≈ 1, n(NO) = α, n(Br₂) = α/2. Total moles n_T = 1 + α/2 ≈ 1. Partial pressures: p(NOBr) ≈ P, p(NO) ≈ αP, p(Br₂) ≈ (α/2)P. Kp = [p(NO)]² [p(Br₂)] / [p(NOBr)]² = (αP)² × (αP / 2) / P² = (α³ P) / 2.

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7. At 500 K, the value of standard Gibbs energy change ΔrG° for a reaction is +28.7 kJ mol⁻¹. What is the approximate value of the equilibrium constant K for the reaction? (Take R = 8.314 J K⁻¹ mol⁻¹, 2.303 × R × 500 = 9572 J mol⁻¹)

  1. 1.0 × 10⁻³
  2. 1.0 × 10³
  3. 2.8 × 10⁻⁵
  4. 3.5 × 10⁻²
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Answer: (A) 1.0 × 10⁻³

Using ΔrG° = -2.303 RT log K, we have 28700 = -9572 log K, which gives log K = -28700 / 9572 ≈ -2.998 ≈ -3. Therefore, K = 10⁻³ = 1.0 × 10⁻³.

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8. Read the following two statements: Statement I: The value of the standard Gibbs energy change (ΔrG°) determines whether a reaction can reach equilibrium or not. Statement II: If ΔrG° > 0, the reaction cannot proceed in the forward direction at all under any set of conditions.

  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
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Answer: (D) Both Statement I and Statement II are incorrect

Statement I is incorrect because standard Gibbs energy change (ΔrG°) simply determines the value of the equilibrium constant K through ΔrG° = -RT ln K; every reversible reaction can reach equilibrium. Statement II is incorrect because ΔrG = ΔrG° + RT ln Q; if Q < K, ΔrG is negative and the reaction proceeds forward, even when ΔrG° > 0.

medium

9. For a homogeneous gaseous reaction: 2A(g) + B(g) ⇌ 3C(g) + D(g), the value of Kp is 0.05 bar at 500 K. What is the value of ΔrG° for this reaction at 500 K? (Take R = 8.314 J K⁻¹ mol⁻¹, ln(0.05) = -3.0)

  1. -12.47 kJ mol⁻¹
  2. +12.47 kJ mol⁻¹
  3. -24.94 kJ mol⁻¹
  4. +24.94 kJ mol⁻¹
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Answer: (B) +12.47 kJ mol⁻¹

ΔrG° = -RT ln Kp. Here, T = 500 K, Kp = 0.05, and ln(0.05) = -3.0. Therefore, ΔrG° = -(8.314 J K⁻¹ mol⁻¹) × (500 K) × (-3.0) = +12471 J mol⁻¹ = +12.47 kJ mol⁻¹.

easy

10. For the heterogeneous equilibrium: NH₄HS(s) ⇌ NH₃(g) + H₂S(g), solid NH₄HS is introduced into an evacuated flask at 300 K. If the total equilibrium pressure developed inside the flask is 0.80 bar, what is the value of Kp at this temperature?

  1. 0.64 bar²
  2. 0.16 bar²
  3. 0.40 bar²
  4. 0.80 bar²
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Answer: (B) 0.16 bar²

For NH₄HS(s) ⇌ NH₃(g) + H₂S(g), equal moles of NH₃ and H₂S are produced, so p(NH₃) = p(H₂S) = P_total / 2 = 0.80 / 2 = 0.40 bar. Thus, Kp = p(NH₃) × p(H₂S) = 0.40 × 0.40 = 0.16 bar².

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11. Solid ammonium carbamate dissociates according to: NH₂COONH₄(s) ⇌ 2NH₃(g) + CO₂(g). If the total pressure at equilibrium is 3.0 bar, what is the value of Kp?

  1. 0.5 bar³
  2. 2.0 bar³
  3. 1.0 bar³
  4. 4.0 bar³
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Answer: (D) 4.0 bar³

From the stoichiometry, moles of NH₃ and CO₂ formed are in 2:1 ratio. Hence, p(NH₃) = (2/3) × P_total = (2/3) × 3.0 = 2.0 bar, and p(CO₂) = (1/3) × 3.0 = 1.0 bar. Therefore, Kp = [p(NH₃)]² × p(CO₂) = (2.0)² × (1.0) = 4.0 bar³.

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12. For the thermal decomposition of solid CaCO₃: CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium pressure of CO₂ at 1073 K is 0.50 bar. What is the value of standard Gibbs free energy change (ΔrG°) for this reaction at 1073 K? (Take R = 8.314 J K⁻¹ mol⁻¹, ln 0.50 = -0.693)

  1. -6.18 kJ mol⁻¹
  2. +6.18 kJ mol⁻¹
  3. +12.36 kJ mol⁻¹
  4. -12.36 kJ mol⁻¹
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Answer: (B) +6.18 kJ mol⁻¹

For this heterogeneous equilibrium, Kp = p(CO₂) = 0.50 bar. Using ΔrG° = -RT ln Kp, we get ΔrG° = -8.314 × 1073 × (-0.693) J mol⁻¹ = +6182 J mol⁻¹ = +6.18 kJ mol⁻¹.

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