One mole of ethanol is treated with one mole of ethanoic acid at `25^@C`. One -fourth of the acid changes into ester at equilibrium . The equilibrium constant for the reaction will be

Options

A. `1//9`
B. `4//9`
C. 9
D. `9//4`

Answer

To find the equilibrium constant for the reaction between ethanol and ethanoic acid, we can follow these steps:

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

### Step 2: Identify the initial concentrations
Initially, we have:
- 1 mole of ethanol
- 1 mole of ethanoic acid

### Step 3: Determine the change in concentration
According to the problem, one-fourth of the acid changes into ester at equilibrium. This means:
- Change in ethanoic acid = -0.25 moles
- Change in ethanol = -0.25 moles
- Change in ester = +0.25 moles

### Step 4: Calculate the equilibrium concentrations
At equilibrium, the concentrations will be:
- Ethanol: \( 1 - 0.25 = 0.75 \) moles
- Ethanoic acid: \( 1 - 0.25 = 0.75 \) moles
- Ester: \( 0 + 0.25 = 0.25 \) moles

### Step 5: Write the expression for the equilibrium constant (K)
The equilibrium constant \( K \) for the reaction is given by:
\[ K = \frac{[\text{Ester}]}{[\text{Ethanol}][\text{Ethanoic Acid}]} \]

### Step 6: Substitute the equilibrium concentrations into the expression
Substituting the values we found:
\[ K = \frac{0.25}{(0.75)(0.75)} \]

### Step 7: Calculate the equilibrium constant
Calculating the denominator:
\[ (0.75)(0.75) = 0.5625 \]
Now substituting back:
\[ K = \frac{0.25}{0.5625} \]
\[ K = \frac{0.25 \times 16}{9} = \frac{4}{9} \]

### Final Answer
The equilibrium constant \( K \) for the reaction is:
\[ K = \frac{4}{9} \]

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