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NTA Abhyas NEET2020ChemistryChemical EquilibriumPractice

In one litre container, ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as CH 3 COOH l ⁡ + C 2 H 5 OH l ⁡ ⇌ C H 3 COOC 2 H 5 l ⁡ + H 2 O l ⁡ At 293 K, if one starts with 1.00 mole of acetic acid and 0.18 mole of ethanol, there is 0.171 mole of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

Options

  1. AK c = 3.92
  2. BK c = 2.56
  3. CK c = 4.89
  4. DK c = 6.23

Correct answer

A. K c = 3.92

Step-by-step solution

CH 3 COOH l ⁡ + C 2 H 5 OH l ⁡ ⇌ C H 3 COOC 2 H 5 l ⁡ + H 2 O l ⁡ Initial conc. 1.00 mol 0.180 mol 0 0 Equili. conc. (1 - x) mol (0.180 - x) mol x mol x mol Given, [ CH 3 COOC 2 H 5 ] equilibrium = 0 . 171 mol = x K c = CH 3 COOC 2 H 5 H 2 O CH 3 COOH C 2 H 5 OH K c = 0.171 × 0.171 1 - 0.171 × 0.180 - 0.171 K c = 0.171 × 0.171 0.829 × 0.009 = 3.919 ≈ 3.92

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