When 1 mole of pure ethyl alcohol `(C_(2)H_(5)OH)` is mixed with `1` mole of acetic acid at `25^(@)C.` the equilibrium mixture contains `2//3` mole each of ester and water
`C_(2)h_(5)OH(l)+CH_(3)COOH(l)hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O(l)`
The `DeltaG^(@)` for the reaction at `298 K` is :

Options

A. `3435` J
B. `4` J
C. `-3435` J
D. zero

Answer

To find the standard free energy change (ΔG°) for the reaction between ethyl alcohol and acetic acid, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction is:
\[ C_{2}H_{5}OH(l) + CH_{3}COOH(l) \rightleftharpoons CH_{3}COOC_{2}H_{5}(l) + H_{2}O(l) \]

### Step 2: Determine initial and equilibrium concentrations
Initially, we have:
- Ethyl alcohol (C₂H₅OH): 1 mole
- Acetic acid (CH₃COOH): 1 mole
- Ester (CH₃COOC₂H₅): 0 moles
- Water (H₂O): 0 moles

At equilibrium, it is given that:
- Ester = \( \frac{2}{3} \) moles
- Water = \( \frac{2}{3} \) moles

Thus, the change in moles for both reactants is:
- Ethyl alcohol remaining = \( 1 - \frac{2}{3} = \frac{1}{3} \) moles
- Acetic acid remaining = \( 1 - \frac{2}{3} = \frac{1}{3} \) moles

### Step 3: Calculate equilibrium concentrations
Assuming the volume of the reaction mixture is \( V \) liters, the concentrations at equilibrium are:
- \[ [C_{2}H_{5}OH] = \frac{1/3}{V} \]
- \[ [CH_{3}COOH] = \frac{1/3}{V} \]
- \[ [CH_{3}COOC_{2}H_{5}] = \frac{2/3}{V} \]
- \[ [H_{2}O] = \frac{2/3}{V} \]

### Step 4: Write the expression for the equilibrium constant (K)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[C_{2}H_{5}OH][CH_{3}COOH]} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{2/3}{V}\right)\left(\frac{2/3}{V}\right)}{\left(\frac{1/3}{V}\right)\left(\frac{1/3}{V}\right)} = \frac{\frac{4}{9}}{\frac{1}{9}} = 4 \]

### Step 5: Calculate ΔG° using the equilibrium constant
The relationship between ΔG° and \( K_c \) is given by:
\[ \Delta G° = -RT \ln K_c \]
Where:
- \( R = 8.314 \, \text{J/mol·K} \)
- \( T = 298 \, \text{K} \)
- \( K_c = 4 \)

Now, calculate \( \ln K_c \):
\[ \ln 4 \approx 1.386 \]

Substituting the values into the equation:
\[ \Delta G° = - (8.314 \, \text{J/mol·K})(298 \, \text{K})(1.386) \]
Calculating this gives:
\[ \Delta G° \approx -3435.02 \, \text{J/mol} \]

### Final Answer:
\[ \Delta G° \approx -3435 \, \text{J/mol} \]