To calculate the enthalpy change for the reaction, we need to use the bond enthalpies of the bonds broken and formed during the reaction. The equation for the enthalpy change is:
ΔH = Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed)
In this reaction, we are breaking two A-B bonds and six B-B bonds, and forming four AB3 bonds. Using the given bond enthalpies, we can calculate the enthalpy change:
ΔH = [2(A-B) + 6(B-B)] - 4(AB3)
ΔH = [2(390) + 6(435)] - 4(0)
ΔH = 780 + 2610
ΔH = 3390 kJ/mol
Therefore, the enthalpy change for the reaction is 3390 kJ/mol.
Calculate the enthalpy change for the reaction. Show your work.
A2 + 3B2 --> 2AB3
bond: Enthalpy:
A-A 945
A-B 390
B-B 435
1 answer
1 answer