$$\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB} . \Delta H_{f}^{0}=-200 \mathrm{~kJ} \mathrm{~mol}^{-1}$$

$\mathrm{AB}, \mathrm{A}_{2}$ and $\mathrm{B}_{2}$ are diatomic molecules. If the bond enthalpies of $\mathrm{A}_{2}, \mathrm{~B}_{2}$ and $\mathrm{AB}$ are in the ratio $1: 0.5: 1$, then the bond enthalpy of $\mathrm{A}_{2}$ is ____________ $\mathrm{kJ} ~\mathrm{mol}^{-1}$ (Nearest integer)

To find the bond enthalpy of $\mathrm{A}_{2}$, we can use the given information about the reaction and the bond enthalpies' ratio. The reaction is: <br/><br/> $\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}$ <br/><br/> The enthalpy change for the reaction, $\Delta H_{f}^{0}$, is given as: <br/><br/> $\Delta H_{f}^{0} = -200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ <br/><br/> The bond enthalpies of $\mathrm{A}_{2}$, $\mathrm{B}_{2}$, and $\mathrm{AB}$ are in the ratio of $1: 0.5: 1$. <br/><br/> Let's denote the bond enthalpies of $\mathrm{A}_{2}$, $\mathrm{B}_{2}$, and $\mathrm{AB}$ as $x$, $0.5x$, and $x$, respectively. <br/><br/> The enthalpy change for the reaction can be calculated using the bond enthalpies: <br/><br/> $$\Delta H_{f}^{0} = \text{(Sum of bond enthalpies of reactants)} - \text{(Sum of bond enthalpies of products)}$$ <br/><br/> For the given reaction: <br/><br/> $-200 = (x + 0.5x) - 2x$ <br/><br/> Now we can solve for $x$, which represents the bond enthalpy of $\mathrm{A}_{2}$: <br/><br/> $-200 = 1.5x - 2x$ <br/><br/> $-200 = -0.5x$ <br/><br/> $x = \frac{-200}{-0.5}$ <br/><br/> $x = 400$ <br/><br/> The bond enthalpy of $\mathrm{A}_{2}$ is $400 ~\mathrm{kJ} ~\mathrm{mol}^{-1}$.