The energies of activation for forward and reverse reaction for `A_(2)+B_(2) hArr 2AB` are `180 kJ "mol"^(-1)` and `200 kJ "mol"^(-1)` respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by `100 kJ "mol"^(-1)`. The enthalpy change of the reaction `(A_(2) + B_(2) rarr 2 AB)` in the presence of catalyst will be (in `kJ "mol"^(-1)`):

Options

A. 300
B. 120
C. 280
D. `-20`

Explanation

To solve the problem, we need to determine the enthalpy change (ΔH) of the reaction \( A_2 + B_2 \rightarrow 2AB \) in the presence of a catalyst, given the activation energies for the forward and reverse reactions and how they change with the catalyst.

### Step-by-Step Solution:

1. **Identify the given activation energies:**
- Activation energy for the forward reaction (\( E_{a,f} \)) = 180 kJ/mol
- Activation energy for the reverse reaction (\( E_{a,r} \)) = 200 kJ/mol

2. **Determine the new activation energies in the presence of a catalyst:**
- The catalyst lowers the activation energy for both reactions by 100 kJ/mol.
- New activation energy for the forward reaction (\( E_{a,f}' \)):
\[ E_{a,f}' = E_{a,f} - 100 = 180 - 100 = 80 \text{ kJ/mol} \]
- New activation energy for the reverse reaction (\( E_{a,r}' \)):
\[ E_{a,r}' = E_{a,r} - 100 = 200 - 100 = 100 \text{ kJ/mol} \]

3. **Calculate the enthalpy change (ΔH) using the formula:**
- The enthalpy change for the reaction is given by the difference between the activation energies of the forward and reverse reactions:
\[ \Delta H = E_{a,f}' - E_{a,r}' \]
- Substitute the values:
\[ \Delta H = 80 \text{ kJ/mol} - 100 \text{ kJ/mol} = -20 \text{ kJ/mol} \]

4. **Conclusion:**
- The enthalpy change of the reaction \( A_2 + B_2 \rightarrow 2AB \) in the presence of a catalyst is \( -20 \text{ kJ/mol} \).

### Final Answer:
\[ \Delta H = -20 \text{ kJ/mol} \]