For the reaction,
`A hArr B ,deltaH` for the reaction is -33.0 kJ/mol.
Calculate
(i) Equilibrium constant `K_c` for the reaction at 300 K
(ii) If `E_a (f) and E_a(r )` in the ratio 20:31 , calculate `E_a(f)` at 300 K .
Assuming pre-exponential factor same for forward and reverse reaction.

Answer

To solve the problem, we will break it down into two parts as per the question.

### Part (i): Calculate the Equilibrium Constant \( K_c \) for the Reaction at 300 K

1. **Given Data**:
- Reaction: \( A \rightleftharpoons B \)
- \( \Delta H = -33.0 \, \text{kJ/mol} \)
- Temperature \( T = 300 \, \text{K} \)

2. **Convert \( \Delta H \) to J/mol**:
\[ \Delta H = -33.0 \, \text{kJ/mol} = -33000 \, \text{J/mol} \]

3. **Use the Van 't Hoff Equation**:
The relation between the equilibrium constant \( K \) and the change in enthalpy is given by the Van 't Hoff equation:
\[ \ln K = -\frac{\Delta H}{R} \cdot \frac{1}{T} + C \]
For simplicity, we can assume that \( C \) is a constant that will not affect the calculation of \( K \) at a specific temperature.

4. **Using the Ideal Gas Constant \( R \)**:
\[ R = 8.314 \, \text{J/(mol K)} \]

5. **Calculate \( K \)**:
\[ \ln K = -\frac{-33000}{8.314} \cdot \frac{1}{300} \]
\[ \ln K = \frac{33000}{8.314 \times 300} \]
\[ \ln K \approx \frac{33000}{2494.2} \approx 13.24 \]
\[ K = e^{13.24} \approx 5.572 \times 10^5 \]

### Part (ii): Calculate \( E_a(f) \) at 300 K

1. **Given Ratio of Activation Energies**:
- \( E_a(f) : E_a(r) = 20 : 31 \)
- Let \( E_a(f) = 20x \) and \( E_a(r) = 31x \).

2. **Using the Relation Between Activation Energies and Enthalpy**:
The relationship between the activation energies and the enthalpy change is given by:
\[ E_a(r) - E_a(f) = -\Delta H \]
Substituting the values:
\[ 31x - 20x = -(-33000) \]
\[ 11x = 33000 \]
\[ x = 3000 \]

3. **Calculate \( E_a(f) \)**:
\[ E_a(f) = 20x = 20 \times 3000 = 60000 \, \text{J/mol} = 60 \, \text{kJ/mol} \]

### Final Answers:
(i) The equilibrium constant \( K_c \) at 300 K is approximately \( 5.572 \times 10^5 \).

(ii) The activation energy of the forward reaction \( E_a(f) \) at 300 K is \( 60 \, \text{kJ/mol} \).