Question

The ester, ethyl acetate is formed by the reaction between ethanol and acetic acid and equilibrium is represented as:
CH$_3$COOH$_{(l)}$ + C$_2$H$_5$OH$_{(l)}$ $\rightleftharpoons$ CH$_3$COOC$_2$H$_5$ $_{(aq)}$ + H$_2$O$_{(l)}$
a.) Write the concentration ratio (reaction quotient), Q$_e$, or 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 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?

Answer 1

Hint: Solve the questions partwise. First we have to write a reaction quotient, and the chemical reaction is given. In the second part, equilibrium constant can be found by the initial concentration, and the concentration at the equilibrium. In the third part calculate the reaction quotient, and compare it with the equilibrium constant.

Complete step-by-step answer:
First, we will write down the reaction quotient for the general reaction. If we consider a general reaction as:

aA + bB $\rightleftharpoons$ cC + dD
Now, we have the reaction for the following reaction, i.e.
Q$_e$ = $\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$
So, we can say that it is the ratio of products to that of reactants.
Now, we will answer each part step by step.
The (a) part, we have to write the reaction quotient. The chemical reaction is given in the question i.e.
CH$_3$COOH$_{(l)}$ + C$_2$H$_5$OH$_{(l)}$ $\rightleftharpoons$ CH$_3$COOC$_2$H$_5$ $_{(aq)}$ + H$_2$O$_{(l)}$
For the reversible reaction, the reaction quotient can be expressed as:
Q$_e$ = $\dfrac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$
We can say that the reaction quotient represents the products in the numerator, and the reactants in the denominator. Here, water is considered as a product as mentioned it is not solvent.
Now, the part (b), let us calculate the equilibrium constant.
CH$_3$COOH$_{(l)}$ + C$_2$H$_5$OH$_{(l)}$ $\rightleftharpoons$ CH$_3$COOC$_2$H$_5$ $_{(aq)}$ + H$_2$O$_{(l)}$

Initial conc.	1/V M	0.18/V M	0	0
Conc. At the Equilibrium	(1-0.171)/V	(0.18-0.171)/V	0.171/ V M	0.171/V M
= 0.829/V M	= 0.009/V M

From the table we can say that the volume of mixture is considered as V, and the water is also considered as mentioned.
Now, the equilibrium constant can be expressed as:
K$_c$ = $\dfrac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$
Now, we will substitute the values for the reactants, and the products as mentioned in the table, then it will be written as:
K$_c$ = $\dfrac{\dfrac{0.171}{V}\times\dfrac{0.171}{V}}{\dfrac{0.829}{V}\times\dfrac{0.829}{V}}$
K$_c$ = 3.919 approximate value.
Now, the part (c), it is similar to the part (b); but the values are different. Here, we will calculate the reaction quotient. So, let us make a table first.
CH$_3$COOH$_{(l)}$ + C$_2$H$_5$OH$_{(l)}$ $\rightleftharpoons$ CH$_3$COOC$_2$H$_5$ $_{(aq)}$ + H$_2$O$_{(l)}$

Initial conc.	1.0/V M	0.5/V M	0	0
Conc. At the Equilibrium	(10-0.214)/V	(0.5-0.214)/V	0.214/ V M	0.214/V M
= 0.786/V M	= 0.286/V M

Now, the reaction quotient can be expressed as:
Q$_e$ = $\dfrac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$
We have calculated in the part (a), so if we substitute the values from the table then;
Q$_e$ = $\dfrac{\dfrac{0.214}{V}\times\dfrac{0.214}{V}}{\dfrac{0.786}{V}\times\dfrac{0.286}{V}}$
Q$_e$ = 0.2038 approximate value.
So, we can say that Q$_e$ < K$_c$, it means that the equilibrium is not attained.
In the last, we can conclude that the value of equilibrium constant is 3.919, and the reaction quotient is 0.2038.

Note:
There is no need of confusion while calculating the equilibrium constant, and the reaction quotient. The both factors represent the reactants in the denominator, and the products in the numerator as mentioned in the chemical reaction. The most important water will be considered while calculation.