Step-by-step explanation:
To calculate the enthalpy change for the reaction, we need to use the formula:
ΔH = Σ (bond energies of bonds broken) - Σ (bond energies of bonds formed)
First, let's calculate the bond energies of the bonds broken:
2 A-A bonds broken = 2 x 945 kJ/mol = 1890 kJ/mol
6 B-B bonds broken = 6 x 445 kJ/mol = 2670 kJ/mol
The total energy required to break the bonds is:
Σ (bond energies of bonds broken) = 1890 kJ/mol + 2670 kJ/mol = 4560 kJ/mol
Next, let's calculate the bond energies of the bonds formed:
6 AB bonds formed = 6 x 390 kJ/mol = 2340 kJ/mol
3 AB3 bonds formed = 3 x 2 x 390 kJ/mol = 2340 kJ/mol
The total energy released in forming the bonds is:
Σ (bond energies of bonds formed) = 2340 kJ/mol + 2340 kJ/mol = 4680 kJ/mol
Now, we can use the formula to calculate the enthalpy change for the reaction:
ΔH = Σ (bond energies of bonds broken) - Σ (bond energies of bonds formed)
ΔH = 4560 kJ/mol - 4680 kJ/mol
ΔH = -120 kJ/mol
Therefore, the enthalpy change for the reaction A2 + 3B2 → 2AB3 is -120 kJ/mol.