For the equilibrium,
`A(g) rarr B(g) , DeltaH` is -40 KJ/mol.
If the ratio of the activation energies of the forward `(E_(f))` and reverse `(E_(b))` reactions is `(2)/(3)` then:

Options

A. `E_(f) =60" kJ/mol, "E_(b) =100" kJ/mol"`
B. `E_(f)=30" kJ/mol, "E_(b) =70" kJ/mol"`
C. `E_(f)=80" kJ/mol, "E_(b) =120" kJ/mol"`
D. `E_(f)=70" kJ/mol", E_(b) =30" kJ/mol"`

Explanation

To solve the problem, we need to find the activation energies \( E_f \) (forward reaction) and \( E_b \) (reverse reaction) given the enthalpy change \( \Delta H \) and the ratio of the activation energies.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- The enthalpy change \( \Delta H \) for the reaction \( A(g) \rightarrow B(g) \) is given as \( -40 \, \text{kJ/mol} \).
- The ratio of the activation energies is given as \( \frac{E_f}{E_b} = \frac{2}{3} \).

2. **Set Up the Variables:**
- Let \( E_f = X \) (activation energy of the forward reaction).
- Then, according to the ratio, \( E_b = \frac{3}{2}X \) (activation energy of the reverse reaction).

3. **Use the Relationship Between Enthalpy and Activation Energies:**
- The relationship between the enthalpy change and the activation energies is given by:
\[ \Delta H = E_f - E_b \]
- Substituting the values we have:
\[ -40 = X - \frac{3}{2}X \]

4. **Simplify the Equation:**
- Combine the terms on the right side:
\[ -40 = X - 1.5X = -0.5X \]

5. **Solve for \( X \):**
- Rearranging gives:
\[ -40 = -0.5X \implies X = 80 \, \text{kJ/mol} \]
- Thus, \( E_f = 80 \, \text{kJ/mol} \).

6. **Calculate \( E_b \):**
- Now, substitute \( X \) back to find \( E_b \):
\[ E_b = \frac{3}{2} \times 80 = 120 \, \text{kJ/mol} \]

### Final Answers:
- \( E_f = 80 \, \text{kJ/mol} \)
- \( E_b = 120 \, \text{kJ/mol} \)