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

Explanation

To solve the problem, we need to determine the bond enthalpy of the diatomic molecule \( A_2 \) 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 the Bond Enthalpies**:
Let the bond enthalpy of \( A_2 \) be \( E_{AA} \), the bond enthalpy of \( AB \) be \( E_{AB} \), and the bond enthalpy of \( B_2 \) be \( E_{BB} \).
From the problem, we have the ratio:
\[ E_{AA} : E_{AB} : E_{BB} = 1 : 1 : 0.5 \]
This implies:
\[ E_{AB} = x, \quad E_{AA} = x, \quad E_{BB} = 0.5x \]

2. **Write the Enthalpy of Formation Equation**:
The enthalpy of formation of \( AB \) from \( A_2 \) and \( B_2 \) can be expressed as:
\[ \Delta H_f (AB) = E_{AB} - \left(\frac{1}{2} E_{AA} + \frac{1}{2} E_{BB}\right) \]
Given that \( \Delta H_f (AB) = -100 \, \text{kJ mol}^{-1} \), we can substitute the bond enthalpy values:
\[ -100 = x - \left(\frac{1}{2} x + \frac{1}{2} (0.5x)\right) \]

3. **Simplify the Equation**:
Now simplify the right-hand side:
\[ -100 = x - \left(\frac{1}{2} x + \frac{0.25x}{2}\right) \]
\[ -100 = x - \left(\frac{1}{2} x + 0.25x\right) \]
\[ -100 = x - \frac{3}{4} x \]
\[ -100 = \frac{1}{4} x \]

4. **Solve for \( x \)**:
Rearranging gives:
\[ x = -100 \times 4 = 400 \, \text{kJ mol}^{-1} \]

5. **Determine the Bond Enthalpy of \( A_2 \)**:
Since \( E_{AA} = x \), we find:
\[ E_{AA} = 400 \, \text{kJ mol}^{-1} \]

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

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