Estimate the ΔH°, in kJ for the following reaction from the bond energies given.

H2 (g) + I2 (g) ---> 2 HI (g)

Bond energy (kJ/mol): H-H = 436 H-I = 297 I-I = 151

A) 884

B) -290

C) +290

D) +7

E) -7 <-----this is the answer that I got

This is how I got it:

ΔH bond breakage:

1 mol H-H bonds = 436 kJ

1 mol I-I bonds = 151 kJ

Total = 436 + 151 = 587

ΔH bond formation:

2 mol H-I bonds = (2 X -297)kJ = -594 kJ

ΔH° rxn = 587 kJ + -594 kJ = -7 kJ

Use the bond energies below to estimate the heat of formation of hydrogen fluoride gas, in kJ/mol.

H-H = 436 kJ/mol H-F = 569 kJ/mol F-F = 159 kJ/mol

1/2 H2 + 1/2 F2 -----> HF

A) +298

B) +271

C) -271<------this is the answer that I got

D) -542

E) -298

This is how I got it:

ΔH bond breakage:

1/2 mol H-H bonds = (1/2 X 436) kJ = 218 kJ

1/2 mol F-F bonds = (1/2 X 159) kJ = 79.5 kJ = 80 kJ

Total = 218 + 80 = 298 kJ

ΔH bond formation:

1 mol H-F bonds = - 569 kJ

ΔH°f = 298 kJ + -569 kJ = -271 kJ

I was just wondering If someone could let me know If I solved both problems right and got the correct answers. Thank You