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 need to establish the equilibrium for the given reaction and calculate the moles of `CH3CO2CH2CH3` formed at equilibrium.

### Step-by-Step Solution:

1. **Write the Balanced Reaction:**
The reaction is:
\[ CH_3CO_2H + CH_3CH_2OH \rightleftharpoons CH_3CO_2CH_2CH_3 + H_2O \]

2. **Initial Moles:**
Initially, we have:
- Moles of `CH3CO2H` = 1
- Moles of `CH3CH2OH` = 1
- Moles of `CH3CO2CH2CH3` = 0
- Moles of `H2O` = 0

3. **Change in Moles at Equilibrium:**
Let \( x \) be the number of moles of `CH3CO2CH2CH3` formed at equilibrium. Then, the changes in moles will be:
- Moles of `CH3CO2H` = \( 1 - x \)
- Moles of `CH3CH2OH` = \( 1 - x \)
- Moles of `CH3CO2CH2CH3` = \( x \)
- Moles of `H2O` = \( x \)

4. **Equilibrium Expression:**
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[CH_3CO_2CH_2CH_3][H_2O]}{[CH_3CO_2H][CH_3CH_2OH]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{x \cdot x}{(1 - x)(1 - x)} = \frac{x^2}{(1 - x)^2} \]

5. **Given Value of \( K_c \):**
We know that \( K_c = 4 \). Therefore, we can set up the equation:
\[ \frac{x^2}{(1 - x)^2} = 4 \]

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

7. **Conclusion:**
The moles of `CH3CO2CH2CH3` formed at equilibrium is \( \frac{2}{3} \).

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

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