Given that 2H 2 (g) + O 2 (g) ⎯ ⎯ → → H 2 O(g) , Δ Δ H = –115.4 kcal the bond energy of H–H and O = O bond respectively is 104 kcal and 119 kcal, then the O–H bond energy in water vapour is
Text Solution
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We know that heat of reaction
Δ Δ H = Σ Σ B.E. (reactant ) – Σ Σ B.E (product)
For the reaction,
2H–H(g) + O = O (g) ⎯ ⎯ → → 2H – O–H(g)
Δ Δ H = –115.4 kcal, B.E. of H–H = 104 kcal
B.E. of O=O = 119 kcal
Since one H 2 O molecule contains two O–H bonds
–115.4 = (2 × × 104) + 119 – 4 (O–H) bond energy
∴ ∴ 4 (O–H) bond energy = ( 2 × × 104) +119+115.4
i.e., O–H bond energy = 
= 110.6 kcal mol –1
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