`AB,A_(2)` and `B_(2)` are diatomic molecules. If the bond enthalpies of `A_(2), AB` and `B_(2)` are in the ratio `1:1:0.5` and the enthalpy of formation of `AB` from `A_(2)` and `B_(2)` is `-100kJ mol^(-1)` , what is the bond enthalpy of `A_(2)` ?

Options

A. 400 kJ `mol^(-1)`
B. 200 kJ `mol^(-1)`
C. 300 kJ `mol^(-1)`
D. 100 kJ `mol^(-1)`

Answer

To solve the problem, we need to find the bond enthalpy of \( A_2 \) given the bond enthalpies of \( A_2 \), \( AB \), and \( B_2 \) in the ratio \( 1:1:0.5 \) and the enthalpy of formation of \( AB \) from \( A_2 \) and \( B_2 \) is \( -100 \, \text{kJ mol}^{-1} \).

### Step-by-Step Solution:

1. **Define Variables Based on Ratios**:
- Let the bond enthalpy of \( A_2 \) be \( X \).
- Since the bond enthalpy of \( AB \) is also in the ratio of 1, let the bond enthalpy of \( AB \) be \( X \).
- The bond enthalpy of \( B_2 \) is \( 0.5 \) of \( A_2 \), so it will be \( \frac{X}{2} \).

2. **Write the Reaction**:
- The formation of \( AB \) from \( A_2 \) and \( B_2 \) can be represented as:
\[ A_2 + B_2 \rightarrow 2 AB \]

3. **Use the Enthalpy of Formation**:
- The enthalpy change for the reaction can be expressed as:
\[ \Delta H = \text{(Enthalpy of reactants)} - \text{(Enthalpy of products)} \]
- The enthalpy of the reactants \( A_2 \) and \( B_2 \) is:
\[ \text{Enthalpy of reactants} = X + \frac{X}{2} = \frac{3X}{2} \]
- The enthalpy of the products (since there are 2 moles of \( AB \)) is:
\[ \text{Enthalpy of products} = 2 \times X = 2X \]

4. **Set Up the Equation**:
- Given that the enthalpy of formation \( \Delta H \) is \( -100 \, \text{kJ mol}^{-1} \):
\[ -100 = \frac{3X}{2} - 2X \]

5. **Simplify the Equation**:
- Rearranging the equation gives:
\[ -100 = \frac{3X}{2} - \frac{4X}{2} = \frac{-X}{2} \]
- Multiplying both sides by \( -2 \):
\[ 200 = X \]

6. **Conclusion**:
- The bond enthalpy of \( A_2 \) is \( 400 \, \text{kJ mol}^{-1} \).

### Final Answer:
The bond enthalpy of \( A_2 \) is \( 400 \, \text{kJ mol}^{-1} \).