Equilibrium — Class 11 MCQs with Answers

Equilibrium — important questions & MCQs with answers (Class 11 Chemistry)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   Chemical equilibrium is:

   A.Static state with no reactions

   B.All product

   C.Dynamic — forward and reverse rates equal✓ Correct

   D.All reactant

   Answer: C. Dynamic — forward and reverse rates equal

   Explanation: Equilibrium is dynamic, not static: forward and reverse reactions both keep running, but at equal rates so concentrations stay constant. The trap is calling it ‘no reaction’ — nothing stops, it only stops changing.

2. Q2Easy

   pH of pure water at 25°C:

   A.1

   B.0

   C.14

   D.7✓ Correct

   Answer: D. 7

   Explanation: Pure water self-ionises: Kw = [H⁺][OH⁻] = 1×10⁻¹⁴, and since [H⁺] = [OH⁻] here, each equals 1×10⁻⁷ M, giving pH = −log(10⁻⁷) = 7. Picking 14 confuses pH with pOH’s scale limit.

3. Q3Easy

   The relation between Kp and Kc is:

   A.Kp = Kc(RT)^Δn✓ Correct

   B.Kp = Kc/RT

   C.Kp = Kc + RT

   D.Kp = Kc

   Answer: A. Kp = Kc(RT)^Δn

   Explanation: Kp and Kc connect through the ideal-gas pressure–concentration link: Kp = Kc(RT)^Δn, where Δn = moles gaseous products − reactants. Writing Kp = Kc/RT wrongly fixes Δn at −1 instead of computing it.

4. Q4Easy

   Ionic product of water Kw at 25°C equals:

   A.1 × 10⁻⁴

   B.1 × 10⁻⁷

   C.1 × 10⁻¹²

   D.1 × 10⁻¹⁴✓ Correct

   Answer: D. 1 × 10⁻¹⁴

   Explanation: Water’s self-ionisation constant Kw = [H⁺][OH⁻] = 1×10⁻¹⁴ at 25°C, fixed by measuring pure water’s tiny ionisation. Confusing it with 10⁻⁷ mixes up [H⁺] itself with the product [H⁺][OH⁻].

5. Q5Easy

   A catalyst at equilibrium:

   A.Shifts equilibrium toward products

   B.Increases K

   C.Does not change the position of equilibrium✓ Correct

   D.Decreases K

   Answer: C. Does not change the position of equilibrium

   Explanation: A catalyst speeds up both forward and reverse reactions by the same factor, so the equilibrium position and K stay unchanged — it only gets the system to equilibrium faster, never shifting yield toward products.

6. Q6Easy

   Conjugate base of HCl is:

   A.H₃O⁺

   B.Cl⁻✓ Correct

   C.OH⁻

   D.H₂

   Answer: B. Cl⁻

   Explanation: A conjugate base is what remains after an acid donates its proton: HCl → H⁺ + Cl⁻, so Cl⁻ is the conjugate base. H₃O⁺ is instead the conjugate acid formed when water accepts that proton.

7. Q7Medium

   Le Chatelier's principle states equilibrium shifts to:

   A.Stop

   B.Side with more atoms

   C.Random side

   D.Counteract the disturbance✓ Correct

   Answer: D. Counteract the disturbance

   Explanation: Le Chatelier’s principle: when a system at equilibrium is disturbed by concentration, pressure, or temperature change, it shifts in the direction that partially counteracts that disturbance — not one that stops or randomises the reaction.

8. Q8Medium

   An acid by Brønsted definition:

   A.Donates a proton✓ Correct

   B.Accepts a proton

   C.Donates electron pair

   D.Accepts electron pair

   Answer: A. Donates a proton

   Explanation: Brønsted–Lowry defines an acid as a proton (H⁺) donor. Accepting a proton instead describes a Brønsted base, while donating or accepting an electron pair are the separate Lewis acid/base definitions.

9. Q9Medium

   For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kp/Kc =

   A.1

   B.1/RT

   C.(RT)²

   D.RT✓ Correct

   Answer: D. RT

   Explanation: PCl₅⇌PCl₃+Cl₂ has Δn = moles gas products − reactants = 2 − 1 = 1, so Kp = Kc(RT)^Δn = Kc(RT)¹, meaning Kp/Kc = RT. Picking (RT)² wrongly doubles Δn instead of using the real mole difference.

10. Q10Medium

    pH of 10⁻⁸ M HCl at 25°C is approximately:

    A.8

    B.6.96✓ Correct

    C.7

    D.6

    Answer: B. 6.96

    Explanation: Below 10⁻⁶ M, water’s own H⁺ can’t be ignored: solving [H⁺]²−C[H⁺]−Kw=0 with C=10⁻⁸ gives [H⁺]≈1.05×10⁻⁷, pH≈6.98 — within 0.3% of the keyed 6.96. Ignoring water gives pH 8 instead.

11. Q11Medium

    If Q > Kc, the reaction:

    A.Stops

    B.Proceeds forward

    C.Is at equilibrium

    D.Proceeds in reverse direction✓ Correct

    Answer: D. Proceeds in reverse direction

    Explanation: When Q > K, there’s more product than the equilibrium ratio allows, so the system reacts in reverse, consuming excess product and rebuilding reactants until Q falls back to K. Only Q < K drives it forward.

12. Q12Hard

    For aA + bB ⇌ cC + dD, Kc =

    A.[A][B][C][D]

    B.[A]^a[B]^b / [C]^c[D]^d

    C.[A]+[B] / [C]+[D]

    D.[C]^c[D]^d / [A]^a[B]^b✓ Correct

    Answer: D. [C]^c[D]^d / [A]^a[B]^b

    Explanation: The law of mass action puts products over reactants, each raised to its own stoichiometric coefficient: Kc = [C]^c[D]^d/[A]^a[B]^b. Writing reactants on top instead inverts the whole expression.

13. Q13Hard

    A buffer solution resists:

    A.Changes in pH on addition of small amount of acid/base✓ Correct

    B.Heat

    C.Light

    D.Pressure

    Answer: A. Changes in pH on addition of small amount of acid/base

    Explanation: A buffer pairs a weak acid/base with its conjugate, so added H⁺ or OH⁻ is consumed by that reserve instead of changing free [H⁺] much — it resists pH shifts from small acid or base addition, not heat, light, or pressure.

14. Q14Hard

    For PCl₅ ⇌ PCl₃+Cl₂ at total P, in terms of α: Kp =

    A.α²/(1+α²)

    B.αP/(1−α)

    C.α/(1−α²)

    D.α²P/(1−α²)✓ Correct

    Answer: D. α²P/(1−α²)

    Explanation: With total moles (1+α) and mole fractions x(PCl₃)=x(Cl₂)=α/(1+α), x(PCl₅)=(1−α)/(1+α), partial pressures give Kp = [αP/(1+α)]²/[(1−α)P/(1+α)] = α²P/(1−α²). Missing the (1+α) factor is the usual slip.

Equilibrium — FAQs

More Class 11 Chemistry chapters

• 01Some Basic Concepts of Chemistry
• 02Structure of Atom
• 03Classification of Elements and Periodicity
• 04Chemical Bonding and Molecular Structure
• 05Thermodynamics
• 07Redox Reactions

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