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Asked by g_archanasharma | 08 Feb, 2019, 17:49: PM
let the bond dissociation energy of,
X2= x kJ/mol
XY=x kJ/mol
Y2=0.5x kJ/mol
We have,
1/2 X2 + 1/2Y2 → XY ΔHf = -200 kJ/mol
We know that,
ΔHreaction = (bond dissociation energy of reactants) - ( bond dissociation energy of products)
= ![x over 2 plus fraction numerator 0 times 5 x over denominator 2 end fraction minus x equals negative 200
fraction numerator 1.5 straight x over denominator 2 end fraction minus x equals space minus space 200
fraction numerator 0.5 x over denominator 2 end fraction space equals space 200
0.25 x space equals space 200 space space
straight x space space equals space 800 space kJ divided by mol](https://images.topperlearning.com/topper/tinymce/cache/10938a055ab55919786b9d1a9fe6ff7f.png)
![x over 2 plus fraction numerator 0 times 5 x over denominator 2 end fraction minus x equals negative 200
fraction numerator 1.5 straight x over denominator 2 end fraction minus x equals space minus space 200
fraction numerator 0.5 x over denominator 2 end fraction space equals space 200
0.25 x space equals space 200 space space
straight x space space equals space 800 space kJ divided by mol](https://images.topperlearning.com/topper/tinymce/cache/10938a055ab55919786b9d1a9fe6ff7f.png)
Bond dissociation energy of X2 is 800 kJ/mol.
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