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

Hint: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.