Question

In the preparation of ethyl acetate, it is observed that 1 mole of acetic acid and ethyl alcohol each are taken initially $\dfrac{2}{3}$ mole of ethyl acetate and water each is formed at equilibrium. Calculate the equilibrium constant of the reaction.

Answer 1

Hint: In a chemical reaction, at equilibrium, the equilibrium constant of concentration is the ratio of product concentration to the concentration of reactant raised to their stoichiometric coefficient. The equilibrium constant of concentrated is represented by ${K_c}$.

Complete step by step answer:
In a chemical reaction, the equilibrium constant gives a relationship among the product species and the reactant species after the reaction reaches its equilibrium state. The equilibrium constant is given by K.
The preparation of ethyl acetate is shown below.
${C_2}{H_5}OH + C{H_3}COOH \to C{H_3}COO{C_2}{H_5} + {H_2}O$
In this reaction, ethanol reacts with acetic acid to form ethyl acetate.
Given,
1 mol of acetic acid$(C{H_3}COOH)$ and ethyl alcohol$({C_2}{H_5}OH)$ are present initially.
 $\dfrac{2}{3}$ mol of acetic acid $(C{H_3}COOH)$and ethyl alcohol $({C_2}{H_5}OH)$are consumed to give $\dfrac{2}{3}$ mol ethyl acetate $(C{H_3}COO{C_2}{H_5})$ and water$({H_2}O)$.
Number of moles of acetic acid left$= 1 - \dfrac{2}{3}$
Number of moles of ethyl alcohol left$= 1 - \dfrac{2}{3}$
Therefore, the number of moles of water $({H_2}O)$and ethyl acetate $(C{H_3}COO{C_2}{H_5})$ are $\dfrac{2}{3}$ each.
In writing the equilibrium constant for the reaction, the products are written in the numerator and the reaction is written in the denominator.
The equilibrium constant is given as shown below.
${K_c} = \dfrac{{[C][D]}}{{[A][B]}}$

For the given reaction the equilibrium constant is given as shown below.
${K_c} = \dfrac{{[C{H_3}COO{C_2}{H_5}][{H_2}O]}}{{[C{H_3}COOH][{C_2}{H_5}OH]}}$
${K_c} = \dfrac{{\dfrac{2}{3} \times \dfrac{2}{3}}}{{\dfrac{1}{3} \times \dfrac{1}{3}}}$
$\Rightarrow {K_c} = \dfrac{4}{9} \times \dfrac{9}{1}$
$\Rightarrow {K_c} = 4$

Thus, the equilibrium constant of the reaction is 4

Note: Here, the ${K_c}$ denotes that the equilibrium constant is measured in terms of moles per liter or molarity. When the equilibrium constant is calculated in terms of partial pressure it is denoted by ${K_P}$. The high value of ${K_c}$ and ${K_P}$shows that a large amount of conversion is done and the product is formed.