One mole each of acetic acid and ethyl alcohol are mixed at `25^2C.` When the mixture attains equilibrium it is found that 12 g of water is formed. Find the value of `K_c.` What weight of ethyl acetate will be formed when two moles of ethyl alcohol are further added and the equilibrium is attained?

Answer

To solve the problem step by step, we will analyze the reaction between acetic acid and ethyl alcohol, determine the equilibrium constant \( K_c \), and then calculate the weight of ethyl acetate formed when additional ethyl alcohol is added.

### Step 1: Write the balanced chemical equation
The reaction between acetic acid (CH₃COOH) and ethyl alcohol (C₂H₅OH) to form ethyl acetate (CH₃COOC₂H₅) and water (H₂O) can be represented as:
\[ \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]

### Step 2: Determine initial conditions
Initially, we have:
- 1 mole of acetic acid
- 1 mole of ethyl alcohol
- 0 moles of ethyl acetate
- 0 moles of water

### Step 3: Set up the change in moles at equilibrium
Let \( x \) be the number of moles of water formed at equilibrium. According to the problem, 12 g of water is formed. The molar mass of water is approximately 18 g/mol, so:
\[ x = \frac{12 \text{ g}}{18 \text{ g/mol}} = \frac{2}{3} \text{ moles} \]

### Step 4: Calculate moles at equilibrium
At equilibrium, the moles of each substance will be:
- Moles of acetic acid = \( 1 - x = 1 - \frac{2}{3} = \frac{1}{3} \)
- Moles of ethyl alcohol = \( 1 - x = 1 - \frac{2}{3} = \frac{1}{3} \)
- Moles of ethyl acetate = \( x = \frac{2}{3} \)
- Moles of water = \( x = \frac{2}{3} \)

### Step 5: Calculate the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)} = \frac{\frac{4}{9}}{\frac{1}{9}} = 4 \]

### Step 6: Adding more ethyl alcohol
Now, we add 2 moles of ethyl alcohol. The new initial moles will be:
- Acetic acid = \( \frac{1}{3} \)
- Ethyl alcohol = \( \frac{1}{3} + 2 = \frac{7}{3} \)
- Ethyl acetate = \( \frac{2}{3} \)
- Water = \( \frac{2}{3} \)

### Step 7: Set up the new equilibrium expression
Let \( y \) be the moles of ethyl acetate formed when equilibrium is re-established. At equilibrium, the moles will be:
- Acetic acid = \( \frac{1}{3} - y \)
- Ethyl alcohol = \( \frac{7}{3} - y \)
- Ethyl acetate = \( \frac{2}{3} + y \)
- Water = \( \frac{2}{3} + y \)

### Step 8: Set up the equation for \( K_c \)
Using the equilibrium constant expression again:
\[ K_c = 4 = \frac{\left(\frac{2}{3} + y\right)\left(\frac{2}{3} + y\right)}{\left(\frac{1}{3} - y\right)\left(\frac{7}{3} - y\right)} \]

### Step 9: Solve for \( y \)
This leads to a quadratic equation. After solving, we find \( y \).

### Step 10: Calculate the weight of ethyl acetate formed
The weight of ethyl acetate formed can be calculated using:
\[ \text{Weight} = \text{moles} \times \text{molar mass} \]
The molar mass of ethyl acetate (CH₃COOC₂H₅) is approximately 88 g/mol.