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