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Given that D(H-H) and D(F-F) in H2 and F2 are 436 and 158kJ mol-1, estimate the bond dissociation enthalpy of H-F using a simple additivity rule. Compare the answer with the experimental value of 570kJ mol-1

Respuesta :

Explanation:

Equation of the reaction:

H2(g) + F2(g) --> 2HF(aq)

1/2H2(g) + 1/2F2(g) --> HF(aq)

D(H-H) in H2 = 436 kJ/mol

D(F-F) in F2 = 158kJ/mol

ΔH bond breakage (dissociation):

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

= 218 kJ

1/2 mol F-F bonds = (1/2 X 158) kJ = = 80 kJ

Total = 218 + 80 = 298 kJ

ΔH bond formation:

1 mol H-F bonds = - 570 kJ

= DHreactant - DHproduct

ΔH°f = 298 kJ + -570 kJ = -272 kJ

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