Consider the following reaction: A2 + B2 → 2AB ΔH = –321 kJ Bond energy (A2)...

Consider the following reaction: A2 + B2 → 2AB ΔH = –321 kJ Bond energy (A2) = 1/2AB Bond energy (B2) =393 kJ/mol. What is the numerical value for Bond energy (A2) ?

Expert Solution

A2 + B2 → 2AB ΔH = –321 kJ

ΔH = ΔH A2 + ΔH B2 - 2* ΔH AB

-321 = ΔH A2 + 393 - 2* 2* ΔH A2 [ (A2) = 1/2AB ]

-321 = ΔH A2 + 393 - 4 ΔH A2

-321 = 393-3 ΔH A2

-3* ΔH A2 = -321-393

-3* ΔH A2 = -714

3* ΔH A2 = 714

ΔH A2 = 714/3 = 238KJ

The bond energy of A2 = 238KJ

queen_honey_blossom answered 2 years ago

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