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If the bond dissociation energies of the diatomic molecules $XY, X_2$ and $Y_2$ are in the ratio of 1 : 1 : 0.5 and $\Delta_fH$ for the formation of XY is -200 kJ $mol^{-1}$. What is the bond dissociation energy of $X_2$ will be


(a) 100 kJ $mol^{-1}$
(b) 200 kJ $mol^{-1}$
(c) 800 kJ $mol^{-1}$
(d) 400 kJ $mol^{-1}$
Can you answer this question?
 
 

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Answer: 800 kJ $mol^{-1}$
$X-X + Y-Y \rightarrow 2 X-Y$ ; $\Delta H = -200 \times 2 $ kJ
Let, B.E. of X-X =a
B.E. of X-Y = a
and B.E. of Y-Y = a/2
$\Delta H = -400$ kJ
$\Delta H = B.E. (X-X) + B.E. (Y-Y) - 2 \times B.E. (X-Y)$
or, -400 kJ = a + 0.5 a - 2a
or, 0.5 a = 400 kJ
or, a = 800 kJ
B.E. of $X_2$ = a = 800 kJ $mol^{-1}$
answered Feb 26, 2014 by mosymeow_1
 

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