One mole each of `CH_3CO_2H and CH_3CH_2OH` are heated in presence of little conc. `H_2SO_4` so as to establish the following equilibrium
`CH_3CO_2H+ CH_3CH_2OH hArr CH_3CO_2CH_2CH_3+H_2O` `K_c = 4`
The moles of `CH_3 CO_2CH_2CH_3` formed at equilibrium is

Options

A. 1 mol
B. 2 mol
C. `1/2` mol
D. `2/3` mol

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation and identify the initial moles.
The reaction given is:
\[ \text{CH}_3\text{CO}_2\text{H} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{CO}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O} \]

Initially, we have:
- Moles of CH₃CO₂H = 1 mole
- Moles of CH₃CH₂OH = 1 mole
- Moles of CH₃CO₂CH₂CH₃ = 0 moles
- Moles of H₂O = 0 moles

### Step 2: Define the change in moles at equilibrium.
Let \( \alpha \) be the amount of CH₃CO₂H and CH₃CH₂OH that reacts at equilibrium. Therefore, at equilibrium, we have:
- Moles of CH₃CO₂H = \( 1 - \alpha \)
- Moles of CH₃CH₂OH = \( 1 - \alpha \)
- Moles of CH₃CO₂CH₂CH₃ = \( \alpha \)
- Moles of H₂O = \( \alpha \)

### Step 3: Write the expression for the equilibrium constant \( K_c \).
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{CH}_3\text{CO}_2\text{CH}_2\text{CH}_3][\text{H}_2\text{O}]}{[\text{CH}_3\text{CO}_2\text{H}][\text{CH}_3\text{CH}_2\text{OH}]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{\alpha}{V}\right)\left(\frac{\alpha}{V}\right)}{\left(\frac{1 - \alpha}{V}\right)\left(\frac{1 - \alpha}{V}\right)} = \frac{\alpha^2}{(1 - \alpha)^2} \]
Given \( K_c = 4 \):
\[ 4 = \frac{\alpha^2}{(1 - \alpha)^2} \]

### Step 4: Solve for \( \alpha \).
Taking the square root of both sides:
\[ 2 = \frac{\alpha}{1 - \alpha} \]
Cross-multiplying gives:
\[ 2(1 - \alpha) = \alpha \]
Expanding and rearranging:
\[ 2 - 2\alpha = \alpha \implies 2 = 3\alpha \implies \alpha = \frac{2}{3} \]

### Step 5: Calculate the moles of CH₃CO₂CH₂CH₃ formed at equilibrium.
From our definition of \( \alpha \):
- Moles of CH₃CO₂CH₂CH₃ formed = \( \alpha = \frac{2}{3} \)

### Final Answer:
The moles of CH₃CO₂CH₂CH₃ formed at equilibrium is \( \frac{2}{3} \) moles.

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