The homogeneous reversible reaction, `C_(2)H_(5)OH+COOHhArrCH_(3)COOC_(2)H_(5)+H_(2)O` is studied at various initial concentrations of the reactants at constant temperature. Calculate `K` in each case.
`{:(,"Moles of acid", "Moles of alcohol","Moles of ester"),(,"Per litre (initial)","Per litre(initial)","Per litre at equilibrium"),((i),1,1,0.637),((ii),1,4,0.93):}`

Answer

To calculate the equilibrium constant \( K \) for the given reversible reaction \( C_2H_5OH + COOH \rightleftharpoons CH_3COOC_2H_5 + H_2O \) at two different sets of initial concentrations, we will follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K \).
The equilibrium constant \( K \) for the reaction is given by the formula:
\[ K = \frac{[CH_3COOC_2H_5][H_2O]}{[C_2H_5OH][COOH]} \]
Where:
- \([CH_3COOC_2H_5]\) is the concentration of the ester at equilibrium.
- \([H_2O]\) is the concentration of water at equilibrium.
- \([C_2H_5OH]\) is the concentration of ethanol at equilibrium.
- \([COOH]\) is the concentration of acetic acid at equilibrium.

### Step 2: Analyze the first case.
For the first case:
- Initial moles of acid (COOH) = 1 M
- Initial moles of alcohol (C2H5OH) = 1 M
- Moles of ester at equilibrium = 0.637 M

At equilibrium:
- Moles of COOH at equilibrium = \( 1 - 0.637 = 0.363 \) M
- Moles of C2H5OH at equilibrium = \( 1 - 0.637 = 0.363 \) M
- Moles of H2O at equilibrium = 0.637 M (since it is produced in a 1:1 ratio with the ester)

Substituting these values into the equilibrium expression:
\[ K = \frac{(0.637)(0.637)}{(0.363)(0.363)} \]

### Step 3: Calculate \( K \) for the first case.
Calculating the values:
\[ K = \frac{0.637^2}{0.363^2} = \frac{0.405769}{0.131769} \approx 4.83 \]

### Step 4: Analyze the second case.
For the second case:
- Initial moles of acid (COOH) = 1 M
- Initial moles of alcohol (C2H5OH) = 4 M
- Moles of ester at equilibrium = 0.93 M

At equilibrium:
- Moles of COOH at equilibrium = \( 1 - 0.93 = 0.07 \) M
- Moles of C2H5OH at equilibrium = \( 4 - 0.93 = 3.07 \) M
- Moles of H2O at equilibrium = 0.93 M (since it is produced in a 1:1 ratio with the ester)

Substituting these values into the equilibrium expression:
\[ K = \frac{(0.93)(0.93)}{(3.07)(0.07)} \]

### Step 5: Calculate \( K \) for the second case.
Calculating the values:
\[ K = \frac{0.93^2}{3.07 \times 0.07} = \frac{0.8649}{0.2149} \approx 4.32 \]

### Final Results:
- For the first case, \( K \approx 4.83 \)
- For the second case, \( K \approx 4.32 \)