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Question: Given that 2H<sub>2</sub>(g) + O<sub>2</sub>(g) → H<sub>2</sub>O(g) , ∆H = –115.4 kcal the bond en...

Given that

2H2(g) + O2(g) → H2O(g) , ∆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

A

110.6 kcal / mol

B

–110.6 kcal

C

105 kcal / mol

D

None

Answer

110.6 kcal / mol

Explanation

Solution

We know that heat of reaction

∆H = ΣB.E. (reactant ) – ΣB.E (product)

For the reaction,

2H–H(g) + O = O (g) → 2H – O–H(g)

∆H = –115.4 kcal, B.E. of H–H = 104 kcal

B.E. of O=O = 119 kcal

Since one H2O 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 = (2×104)+119+115.44\frac{(2 \times 104) + 119 + 115.4}{4}

= 110.6 kcal mol–1