Class 11 Chemistry Chapter 6: Equilibrium worksheet with answer key
A 20-mark worksheet on Equilibrium for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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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.
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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: (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: (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) 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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The two values at the same temperature are equal, because Kc does not depend on the initial concentrations. The value at the higher temperature differs, because an equilibrium constant has one value for a given balanced equation at a given temperature and changes when the temperature changes.
- same temperature: equal, Kc independent of initial concentrations (1)
- higher temperature: different, Kc depends on temperature (1)
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Sodium acetate ionises completely and supplies CH₃COO⁻, an ion already present in CH₃COOH ⇌ H⁺ + CH₃COO⁻. By Le Chatelier's principle the equilibrium shifts towards undissociated acetic acid (the common ion effect), so [H⁺] decreases and the pH rises.
- acetate is a common ion; equilibrium shifts to the left (1)
- [H⁺] falls, so pH rises (1)
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At equilibrium both reactions continue at equal rates, so there is no net change in composition; the equilibrium is dynamic. Evidence: when an H₂–N₂–NH₃ equilibrium mixture is mixed with a D₂–N₂–ND₃ mixture, NH₂D, NHD₂ and HD appear although the amount of ammonia is unchanged, so the reactions have not stopped (the radioactive sugar experiment is also accepted).
- reactions continue at equal rates; dynamic, not static (1)
- names the deuterium (or radioactive sugar) experiment and what it shows (1)
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(a) The levels become constant when, in each cycle, the volume carried from 1 to 2 equals the volume carried from 2 to 1: the two rates become equal. (b) Transfer continues after the levels are constant, just as forward and reverse reactions continue at equilibrium (level stands for concentration). (c) Equilibrium is still reached, but the two constant levels are different, since equal rates now need different levels.
- (a) equal transfer in both directions, i.e. equal rates (1)
- (b) transfer continues with no change in levels (1)
- (c) levels become constant but unequal (1)
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(a) Four moles of gas give two; by Le Chatelier's principle a higher pressure shifts the equilibrium towards NH₃, the side with fewer moles, raising the yield. (b) The reaction is exothermic, so K is larger at low temperature, but at low temperature the reaction is too slow to reach equilibrium in reasonable time; about 500 °C is a compromise between rate and yield. (c) Iron gives a satisfactory rate at a temperature where the equilibrium amount of NH₃ is still reasonably favourable; it does not change K or the equilibrium yield.
- (a) fewer moles of gas on the product side, so high pressure raises yield (1)
- (b) low temperature favours yield but rate is too slow; compromise temperature (1)
- (c) catalyst increases rate without changing equilibrium (1)
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(a) CaF₂(s) ⇌ Ca²⁺ + 2F⁻; in pure water [F⁻] = 2[Ca²⁺] = 2.20 × 10⁻³ mol L⁻¹, so Ksp = [Ca²⁺][F⁻]² = 4S³ = 4 × 1.331 × 10⁻⁹ = 5.32 × 10⁻⁹ ≈ 5.3 × 10⁻⁹. (b) With S = 1.0 × 10⁻⁶, the F⁻ from CaF₂ is negligible, so Ksp = S[F⁻]²: [F⁻]² = 5.3 × 10⁻⁹/1.0 × 10⁻⁶ = 5.3 × 10⁻³, [F⁻] = 7.28 × 10⁻² ≈ 0.073 M. (c) F⁻ is a common ion; by Le Chatelier's principle the added F⁻ pushes CaF₂(s) ⇌ Ca²⁺ + 2F⁻ to the left, so less salt dissolves before Qsp reaches Ksp.
- (a) [F⁻] = 2[Ca²⁺]; Ksp = 4S³ = 5.3 × 10⁻⁹ (1)
- (b) [F⁻] ≈ 0.073 M (1)
- (c) common ion shifts the equilibrium towards the solid (1)