Question

Question 7.18 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)

(i) Write the concentration ratio (reaction quotient), Qc, 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?

Class XI Equilibrium Page 226

Answer 1

CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)

(i) reaction quotient ( Qc) = [CH3COOC2H5][H2O]/[C2H5OH][CH3COOH]

Here, H2O is not excess that's why its concentration isn't constant.

(ii)
$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: CH_3COOH(l)+C_2H_5OH(l) < = > CH_3COOC_2H_5(l)+H_2O(l) \\ at \: t = 0 \: \: \: \: \: \: \: \: 1 \: mol \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0.180mol \\ at \: eqlm \: \: \: (1 - x)mol \: \: \: \: \: \: \: \: \: (0.180 - x)mol \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x$
Given,
[CH3COOC2H5] at equilibrium = 0.171 mol = x
so, x = 0.171 mol

hence, [CH3COOH] = 1 - x = 1 - 0.171 = 0.829
[C2H5OH] = 0.180 - x = 0.180 - 0.171 = 0.009
[CH3COOC2H5] = [H2O] = x = 0.171

now, Kc = [CH3COOC2H5][H2O]/[CH3COOH][C2H5OH]
= 0.171 × 0.171/0.829 × 0.009
= 3.919 ≈ 3.92
hence, Kc = 3.92

(iii) similarly,
at time t ,
[CH3COOH] = 1 - x = 0.5 - 0.214 = 0.786 mol
[C2H5OH] = 0.5 - x = 0.5 - 0.214 = 0.286 mol
[CH3COOC2H5]= x = 0.214 mol
[ H2O] = x = 0.214 mol

reaction quotient ( Qc) = [CH3COOC2H5][H2O]/[CH3COOH][C2H5OH]
= 0.214 × 0.214/0.786 × 0.286
= 0.2037 ≈ 0.204

Here, Qc ≠ Kc hence, equilibrium has not been reached.