Question

In: Chemistry

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) ?

Solutions

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

   

  


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