Question

Given the following reactions and their enthalpies:
ΔH(kJ/mol)−−−−−−−−−−−
H2(g)⟶2H(g)   +436
O2(g)⟶2O(g)   +495
H2+1/2O2(g)⟶H2O(g)   −242

A. Devise a way to calculate ΔH for the reaction

H2O(g)⟶2H(g)+O(g)

B. estimate the H-O bond energy

Answer 1

The reaction is

H2O(g)⟶2H(g)+O(g)

Given:

H2(g)⟶2H(g)   +436 ...................................(1)
O2(g)⟶2O(g)   +495....................................(2)
H2+1/2O2(g)⟶H2O(g)   −242........................(3)

H2O(g)⟶2H(g)+O(g)   ...................................(4)

A) We can derive the given equation from equation 1,2 and 3 as

equation 4 = - equation (3) + equation (1) + 1/2 equation (2)

(Delta H )4 = DeltaH (1) + 1/2 Delta H (2) - Delta H (3)

Delat H rxn = 436 + 495 /2 - (-242) = 925.5 KJ / mole

B) Bond energy and enthalpy of reactin are related as

Delat H rxn = Sum of bond energies of reactants - sum of bond energies of products

For reaction (3)

Delta h rxn = [B.E (H-H)] + 1/2[B.E. O=O] - [2XB.E O-H]

B.E H-H = 436

B.E O=O = 495

-242 = 436 + (495/2) - 2X B.EO-H

Bond energy O-H = 462.75 KJ / mole

+ (