To find the bond enthalpy of Y-Y in the reaction 1/2 X₂ (g) + 3/2 Y₂ (g) → XY₃ (g), we need to use the concept of bond enthalpy and Hess's Law.

Hess's Law states that the enthalpy change for a reaction is independent of the pathway taken to go from reactants to products. This means that we can calculate the enthalpy change of the reaction by summing up the bond enthalpies of the bonds broken and formed during the reaction.

Let's break down the reaction into individual bond breaking and bond forming steps:

1) Breaking X-X bonds:
In the reactant side, there are 2 X-X bonds, so the energy required to break them is 2 * 380 kJ/mol.

2) Breaking X-Y bonds:
In the reactant side, there are 3/2 * 2 = 3 X-Y bonds, so the energy required to break them is 3 * 150 kJ/mol.

3) Forming XY₃ bonds:
In the product side, there are 2 XY₃ bonds, so the energy released when they are formed is 2 * (bond enthalpy of XY₃).

According to Hess's law, the sum of the bond breaking energies should be equal to the sum of the bond forming energies. Therefore, we can write the equation:

2 * 380 kJ/mol + 3 * 150 kJ/mol = 2 * (bond enthalpy of XY₃)

Simplifying the equation, we have:

760 kJ/mol + 450 kJ/mol = 2 * (bond enthalpy of XY₃)

1210 kJ/mol = 2 * (bond enthalpy of XY₃)

Dividing both sides by 2, we get:

bond enthalpy of XY₃ = 605 kJ/mol

Since the enthalpy change for the reaction is -50 kJ/mol, we can write:

-50 kJ/mol = 605 kJ/mol - bond enthalpy of Y-Y

Rearranging the equation, we find:

bond enthalpy of Y-Y = 605 kJ/mol + 50 kJ/mol

bond enthalpy of Y-Y = 655 kJ/mol

Therefore, the bond enthalpy of Y-Y is 655 kJ/mol.

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