QJEEM23Ch70d5: JEE Main 2023 Chemistry - Class 11 JEEM23Ch70d5 Solution: and are diatomic molecules. If the bond enthalpies of and are…

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:

$\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}$

The enthalpy change for the reaction, $\Delta H_{f}^{0}$ , is given as:

$\Delta H_{f}^{0} = -200 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The bond enthalpies of $\mathrm{A}_{2}$ , $\mathrm{B}_{2}$ , and $\mathrm{AB}$ are in the ratio of $1: 0.5: 1$ .

Let's denote the bond enthalpies of $\mathrm{A}_{2}$ , $\mathrm{B}_{2}$ , and $\mathrm{AB}$ as $x$ , $0.5x$ , and $x$ , respectively.

The enthalpy change for the reaction can be calculated using the bond enthalpies:

$\Delta H_{f}^{0} = \text{(Sum of bond enthalpies of reactants)} - \text{(Sum of bond enthalpies of products)}$

For the given reaction:

$-200 = (x + 0.5x) - 2x$

Now we can solve for $x$ , which represents the bond enthalpy of $\mathrm{A}_{2}$ :

$-200 = 1.5x - 2x$

$-200 = -0.5x$

$x = \frac{-200}{-0.5}$

$x = 400$

The bond enthalpy of $\mathrm{A}_{2}$ is $400 ~\mathrm{kJ} ~\mathrm{mol}^{-1}$ .