The ester, ethyl acetate is formed by the reaction between ethanol and acetic acid and their equilibrium is represented as:

$C H_{3} C O O H_{(l)} + C_{2} H_{5} O H_{(l)} ⇌ C H_{3} C O O C_{2} H_{5 (a q)} + H_{2} O_{(l)}$
(a) Write the concentration ratio (reaction quotient), $Q_{e}$ , for this reaction. Note that water is not in excess and is not a solvent in this reaction.

(b) At $293 K$ , if one starts with $1.00$ mole of acetic acid and $0.180$ mole of ethanol, there is $0.171$ mole of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

(c) Starting with $0.500$ mole of ethanol and $1.000$ mole of acetic acid and maintaining it at $293 K$ , $0.214$ mole of ethyl acetate is found after some time. Has equilibrium been reached?

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Solution

(i) The concentration ratio is $Q_{c} = \frac{[C H_{3} C O O C_{2} H_{5}] [H_{2} O]}{[C H_{3} C O O H] [C_{2} H_{5} O H]}$
(ii) The initial concentrations are
$[C H_{3} C O O H] = 1.00$
$[C_{2} H_{5} O H] = 0.18$
$[C H_{3} C O C_{2} H_{4}] = 0$
$[H_{2} O] = 0$

The equilibrium concentrations are:
$[C H_{3} C O O H] = \frac{1 - 0.171}{V} = \frac{0.829}{V}$
$[C_{2} H_{5} O H] = \frac{0.18 - 0.171}{V} = \frac{0.009}{V}$
$[C H_{3} C O C_{2} H_{4}] = \frac{0.171}{V}$
$[H_{2} O] = \frac{0.171}{V}$
The equilibrium constant expression is
$K_{c} = \frac{[C H_{3} C O O C_{2} H_{5}] [H_{2} O]}{[C H_{3} C O O H] [C_{2} H_{5} O H]} = \frac{\frac{0.171}{V} \times \frac{0.171}{V}}{\frac{0.829}{V} \times \frac{0.009}{V}} = 3.92$

(iii) The initial concentrations are:
$[C H_{3} C O O H] = 1.00$
$[C_{2} H_{5} O H] = 0.500$
$[C H_{3} C O C_{2} H_{4}] = 0$
$[H_{2} O] = 0$

The equilibrium concentrations are:
$[C H_{3} C O O H] = \frac{1 - 0.214}{V} = \frac{0.786}{V}$
$[C_{2} H_{5} O H] = \frac{0.500 - 0.214}{V} = \frac{0.286}{V}$
$[C H_{3} C O C_{2} H_{4}] = \frac{0.214}{V}$
$[H_{2} O] = \frac{0.214}{V}$
The expression for the reaction quotient is:
$Q_{c} = \frac{[C H_{3} C O O C_{2} H_{5}] [H_{2} O]}{[C H_{3} C O O H] [C_{2} H_{5} O H]} = \frac{\frac{0.214}{V} \times \frac{0.214}{V}}{\frac{0.786}{V} \times \frac{0.286}{V}} = 0.204$
As the value of the reaction quotient is not equal to the value of the equilibrium constant, the equilibrium is not yet achieved.

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Similar questions

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

(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?

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the

equilibrium is represented as:

(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?

$C H_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l); K_{c} = 4$

In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of little conc. $H_{2} S O_{4}$ . On equilibrium being attained