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Question

Bond dissociation enthalpies of H2(g),Cl2 and HCl(g) are 104, 58 and 103 kcal respectively. The enthalpy of formation of HCl is

A
- 44 kcal
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B
- 88 kcal
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C
- 22 kcal
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D
- 11 kcal
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Solution

The correct option is A - 44 kcal
H2(g)+Cl2(g)2HCl(g)
(BE of reactants)(BE of products)=ΔHr
=104+582×103

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