Question 18 of 69advanced🔧 ApplyNumerical5 marks

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

CH3COOH(l)+C2H5OH(l)⇌CH3COOC2H5(l)+H2O(l)CH_3COOH (l) + C_2H_5OH (l) \rightleftharpoons CH_3COOC_2H_5(l) + H_2O (l)

(i) Write the concentration ratio (reaction quotient), QcQ_c, for this reaction (note: water is not in excess and is not a solvent in this reaction) (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

(i)
Answer

Qc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]Q_c = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}

(ii)
Answer

Kc=3.92K_c = 3.92

(iii)
Answer

No, equilibrium has not been reached because Qc<KcQ_c < K_c.

Explanation

For part (i), since water is not in excess and is not a solvent, its concentration is included in the reaction quotient. For part (ii), an ICE table is used to find the equilibrium moles of all reactants and products. The volume term cancels out in the equilibrium expression, allowing the use of mole values directly to calculate KcK_c. For part (iii), the reaction quotient QcQ_c is calculated using the given mole values. Comparing QcQ_c with the previously calculated KcK_c determines if the system is at equilibrium.

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