Question

In the esterification ${C_2}{H_5}OH\left( l \right) + C{H_3}COOH\left( l \right) \rightleftharpoons C{H_3}COO{C_2}{H_5}\left( l \right) + {H_{2}}O\left( l \right)\;$ an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction $= {\text{ }}0.333$ Calculate the equilibrium constant:
A) $K = 2$
B) $K = 4$
C) $K = 8$
D) None of these.

Answer 1

In a chemical reaction, when both the reactants and the products are in a concentration which does not change with time any more, it is said to be in a state of chemical equilibrium. Here to calculate the equilibrium constant $\left( {{K_c}} \right)$ using the equation of equilibrium concentration.

Complete answer:
When the chemical is in equilibrium, the ratio of the products to the reactants is called equilibrium constant. For a general reaction, $aA\; + \;bB\; \rightleftarrows cC\; + \;dD$
If, The reaction is in equilibrium $\left( {{K_c}} \right)$ = equilibrium constant
$\left( {{K_c}} \right)$ = ${\text{ }}\dfrac{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}{{\;{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}$
We will begin this problem by calculating the changes in concentration as the system goes to equilibrium. Then we determine the equilibrium concentrations and, finally, the equilibrium constant.
Let initially, $1{\text{ }}mole$ of ethanol and $1{\text{ }}mole$ of acetic acid be present.
Let $x{\text{ }}moles$ of ethanol react with $x{\text{ }}moles$ of acetic acid to reach equilibrium.
Now, In the esterification process
${C_2}{H_5}OH\left( l \right) + C{H_3}COOH\left( l \right) \rightleftharpoons C{H_3}COO{C_2}{H_5}\left( l \right) + {H_{2}}O\left( l \right)\;$

| ${C_2}{H_5}OH$ | CH3COOH| CH3COOC2H5| H2O
---|---|---|---|---
Initial Moles| $1$ | $1$| $0$ | 0
Moles at equilibrium| $(1 - x)$ | $(1 - x)$| $x$ | $x$

The total number of moles at equilibrium = $\left( {1 - x + \;1 - x{\text{ }} + {\text{ }}x{\text{ }} + {\text{ }}x{\text{ }}} \right){\text{ }} = {\text{ }}2$
Given, the mole fraction of water is $= {\text{ }}0.333$
The mole fraction of water at equilibrium is =${\text{ }}\dfrac{x}{2}$
So mole fraction of water, $\dfrac{x}{2} = 0.333$
$\Rightarrow x = 2{\text{ }} \times 0.333 = 0.666$
Hence, $x\; = 0.666\;$ and $\left( {1 - x} \right) = 0.334$
The equilibrium constant expression is
$\Rightarrow Kc\; = \;\dfrac{{\left[ {C{H_3}COO{C_2}{H_5}} \right]{\text{ }}\left[ {{H_2}O} \right]}}{{\left[ {{C_2}{H_5}OH} \right]{\text{ }}\left[ {C{H_3}COOH} \right]}}\;$= $\dfrac{{\left[ {0.666} \right]{\text{ }}\left[ {0.666} \right]}}{{\left[ {0.334} \right]{\text{ }}\left[ {0.334} \right]}}\;$ = $4$

**So the correct option is option B. $K = 4$

Note:**
It is reaction specific and at a constant temperature, it is fixed. A catalyst changes the rate of forward and backward reactions equally not to affect the value of the equilibrium constant. Factors that affect equilibrium constant are: Change in concentration of any product or reactant, Change in temperature of the system.