Class 11 Chemistry Chapter 6: Equilibrium MCQs with answers
10 multiple-choice questions on Equilibrium, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.
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Answer: (a) the beaker is an open system, so condensation cannot keep pace with evaporation and equilibrium is never reached
In the open beaker the vapour spreads into the room, so the rate of condensation stays far below the rate of evaporation and equilibrium cannot be reached. In the closed flask evaporation goes on, but condensation rises to match it, so the level stays constant.
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Answer: (c) below 100 °C, because the pressure over it is less than 1.013 bar
Water boils when its vapour pressure equals the pressure above it. The normal boiling point, 100 °C, is for 1.013 bar; at 0.70 bar the vapour pressure reaches the outside pressure at a lower temperature, so the boiling point falls with altitude.
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Answer: (d) NH₃ < H₂O < HF, and HF < HCl < HI
Across a period the polarity of the H–A bond decides: electronegativity of A rises from N to O to F, so NH₃ < H₂O < HF. Down a group the H–A bond strength decides: the bond weakens as A grows, so HF < HCl < HI and H₂O < H₂S. Only (d) has both trends the right way.
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Answer: (b) raising the pressure of CO₂ above the liquid
By Henry's law the mass of gas dissolved in a given mass of solvent at a given temperature is proportional to the pressure of the gas above it. Warming lowers the solubility; shaking only helps equilibrium to be reached sooner.
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Answer: (a) 4.0 × 10⁻¹²
[X⁻] = 2S, so S = [M²⁺] = 1.0 × 10⁻⁴ mol L⁻¹. Ksp = [M²⁺][X⁻]² = (1.0 × 10⁻⁴)(2.0 × 10⁻⁴)² = 4.0 × 10⁻¹². Taking S = 2.0 × 10⁻⁴ and using 4S³ gives 3.2 × 10⁻¹¹; forgetting the square gives 2.0 × 10⁻⁸.
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Answer: (c) L² mol⁻²
Kc = [NH₃]²/[N₂][H₂]³, so its unit is (mol L⁻¹)²/(mol L⁻¹)⁴ = (mol L⁻¹)⁻² = L² mol⁻².
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Answer: (d) 1/8
Reversing the equation inverts K (1/64); halving the coefficients takes the square root: (1/64)^½ = 1/8.
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Answer: (b) N₂(g) + O₂(g) ⇌ 2NO(g)
Kp = Kc(RT)^Δn, so Kp = Kc only when Δn = 0. For N₂ + O₂ ⇌ 2NO, Δn = 2 − 2 = 0; the others have Δn = +1, −1 and +1.
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Answer: (c) C(s) + H₂O(g) ⇌ CO(g) + H₂(g)
RT = 0.0831 × 500 = 41.6, so (RT)^Δn = RT and Δn = +1. Counting gases only (pure solids are left out): (a) Δn = −1; (b) 1 − 1 = 0; (c) 2 − 1 = +1; (d) 0. Only (c) fits.
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Answer: (a) HCl < NH₄Cl < NaCl < CH₃COONa
HCl is a strong acid (pH 1). NH₄⁺ hydrolyses to give H⁺, so NH₄Cl is mildly acidic. NaCl, from a strong acid and a strong base, is neutral (pH 7). Acetate hydrolyses to give OH⁻, so CH₃COONa is basic.