Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
$C H_{3} C O O H (l) + C_{2} H_{5} O H (l) \leftrightarrow C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)$

(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 some time. Has equilibrium been reached?

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Solution

(i) $R e a c t i o n q u o t i e n t, Q_{c} = \frac{[C H_{3} C O O C_{2} H_{5}] [H_{2} O]}{[C H_{3} C O O H] [C_{2} H_{5} O H]}$

(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.
The given reaction is:
$\begin{matrix} C H_{3} C O O H_{(l)} & + & C_{2} H_{5} O H_{(l)} & \leftrightarrow & C H_{3} C O O C_{2} H_{5 (l)} & + & H_{2} O_{(l)} I n i t i a l c o n c . & \frac{1}{V} M & \frac{0.18}{V} M & 0 & 0 A t e q u i l i b r i u m & \frac{1 - 0.171}{V} & \frac{0.08 - 0.171}{V} & \frac{0.171}{V} M & \frac{0.171}{V} M = \frac{0.829}{V} M & = \frac{0.009}{V} M \end{matrix}$
Therefore, equilibrium constant for the given reaction is:
$K_{c} = \frac{[C H_{3} C O O C_{2} H_{5}] [H_{2} O]}{[C H_{3} C O O H] [C_{2} H_{5} O H]} = \frac{\frac{0.071}{V} \times \frac{0.171}{V}}{\frac{0.829}{V} \times \frac{0.009}{V}} = 3.919$
=3.92 (approximately)

(iii) Let the volume of the reaction mixture be V.
$\begin{matrix} C H_{3} C O O H_{(l)} & + & C_{2} H_{5} O H_{(l)} & \leftrightarrow & C H_{3} C O O C_{2} H_{5 (l)} & + & H_{2} O_{(l)} I n i t i a l c o n c . & \frac{1.0}{V} M & \frac{0.5}{V} M & 0 & 0 A f t e r s o m e t i m e & \frac{10 - 0.214}{V} & \frac{0.5 - 0.214}{V} & \frac{0.214}{V} M & \frac{0.214}{V} M = \frac{0.786}{V} M & = \frac{0.286}{V} M \end{matrix}$
Therefore, the reaction quotient is,
$Q_{c} = \frac{[C H_{3} C O O C_{2} H_{5}] [H_{2} O]}{[C H_{3} C O O H] [C_{2} H_{5} O H]} = \frac{\frac{0.214}{V} \times \frac{0.214}{V}}{\frac{0.786}{V} \times \frac{0.286}{V}} = 0.2037 = 0204 (a p p r o x i m a t e l y)$
Since $Q_{C} < K_{c}$ , equilibrium has not been reached.

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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?

Q. 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.

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?

Q. What is true for the equilibrium reaction :
Acetic acid+ ethanol

^{+}

Ethyl acetate +

H_{2} O

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

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