Question

calculate ΔHr of the following reaction
H2(g) + I2 (g) -> 2HI (g)
given that bond energies of H-H , I-I AND H-I are 433,151 and 299 kj/mol respectively

Answer 1

ΔH bond breakage: 
1 mol H-H bonds = 433 kJ
1 mol I-I bonds = 151 kJ 
Total = 433 + 151 = 584

ΔH bond formation:
2 mol H-I bonds = (2 X -299)kJ = -598 kJ

ΔH° rxn = 584 kJ + -598 kJ = -14 kJ

Answer 2

Given:

1 mol H-H bonds = 433 kJ/mol

1 mol I-I bonds = 151 kJ /mol

1 mol H-I bonds = 299 kJ/mol

Reaction = H₂ + I₂ → 2HI

To Find:

ΔHr of the given reaction

Solution:

ΔHr of the given reaction is -14 kJ.

The enthalpy change of a reaction represented as ΔHr, is the net enthalpy change when the reactants of the given reactions are converted into products.

ΔHr = ΔHf of products - ΔHf of recatants

= Bond Energies of Reactants - Bond energies of Products

BE of products = 2 X ΔBE (HI)

= 2 X 299 kJ

= 598kJ

BE of reactants = ΔBE (H₂) + ΔBE (I₂)

= 433 + 151

= 584kJ

Substituting,

ΔH° rxn = 584 kJ - 598 kJ

= -14 kJ