`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)` ?

Explanation

To solve the problem, we need to find the bond enthalpy of \( A_2 \) given the bond enthalpies of \( A_2 \), \( AB \), and \( B_2 \) are 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 for Bond Enthalpies**:
Let the bond enthalpy of \( A_2 \) be \( x \).
Since the bond enthalpies are in the ratio \( 1:1:0.5 \):
- Bond enthalpy of \( A_2 = x \)
- Bond enthalpy of \( AB = x \)
- Bond enthalpy of \( B_2 = \frac{x}{2} \)

2. **Write the Reaction for Formation of \( AB \)**:
The formation of \( AB \) from \( A_2 \) and \( B_2 \) can be represented as:
\[ \frac{1}{2} A_2 + \frac{1}{2} B_2 \rightarrow AB \]
The enthalpy change for this reaction is given as \( -100 \, \text{kJ mol}^{-1} \).

3. **Set Up the Enthalpy Change Equation**:
The enthalpy change for the reaction can be expressed in terms of bond enthalpies:
\[ \text{Enthalpy change} = \text{Bonds broken} - \text{Bonds formed} \]
For the bonds broken:
- Breaking \( \frac{1}{2} A_2 \) contributes \( \frac{x}{2} \)
- Breaking \( \frac{1}{2} B_2 \) contributes \( \frac{1}{2} \times \frac{x}{2} = \frac{x}{4} \)

For the bonds formed:
- Forming \( AB \) contributes \( x \)

Thus, we can write:
\[ \frac{x}{2} + \frac{x}{4} - x = -100 \]

4. **Combine and Simplify the Equation**:
To combine the terms, we need a common denominator, which is 4:
\[ \frac{2x}{4} + \frac{x}{4} - \frac{4x}{4} = -100 \]
This simplifies to:
\[ \frac{2x + x - 4x}{4} = -100 \]
\[ \frac{-x}{4} = -100 \]

5. **Solve for \( x \)**:
Multiply both sides by -4:
\[ x = 400 \, \text{kJ mol}^{-1} \]

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