A$$_2$$ + B$$_2$$ $$\to$$ 2AB. $$\Delta H_f = -200$$ kJ mol$$^{-1}$$
AB, A$$_2$$ and B$$_2$$ are diatomic molecules. If the bond enthalpies of A$$_2$$, B$$_2$$ and AB are in the ratio 1 : 0.5 : 1, then the bond enthalpy of A$$_2$$ is _____ kJ mol$$^{-1}$$ (Nearest integer)

Given: $$\text{A}_2 + \text{B}_2 \to 2\text{AB}$$, $$\Delta H_f = -200$$ kJ mol$$^{-1}$$.

Bond enthalpies of A$$_2$$, B$$_2$$, and AB are in the ratio $$1 : 0.5 : 1$$.

Let the bond enthalpy of A$$_2$$ = $$x$$ kJ/mol.

Then bond enthalpy of B$$_2$$ = $$0.5x$$ kJ/mol, and bond enthalpy of AB = $$x$$ kJ/mol.

$$\Delta H_f = -200$$ kJ/mol is the enthalpy of formation per mole of AB.

For the reaction $$\text{A}_2 + \text{B}_2 \to 2\text{AB}$$, the total enthalpy change is:

$$\Delta H_{\text{rxn}} = 2 \times (-200) = -400 \text{ kJ}$$

Using bond enthalpies (bonds broken $$-$$ bonds formed):

$$\Delta H_{\text{rxn}} = [\text{BE}(\text{A}_2) + \text{BE}(\text{B}_2)] - [2 \times \text{BE}(\text{AB})]$$

$$-400 = [x + 0.5x] - [2x]$$

$$-400 = 1.5x - 2x = -0.5x$$

$$x = \frac{400}{0.5} = 800 \text{ kJ mol}^{-1}$$

The bond enthalpy of A$$_2$$ is $$\mathbf{800}$$ kJ mol$$^{-1}$$.