If the bond dissociation energies of XY,X2 and Y2 (all diatomic molecules) are in the ratio of 1:1:0.5 and ΔfH for the formation of XY is −200kJmol−1. The bond dissociation energy of X2, will be

A
$300 kJ {mol}^{- 1}$
B
$400 kJ {mol}^{- 1}$
C
$100 kJ {mol}^{- 1}$
D
$800 kJ {mol}^{- 1}$

Solution:

For the formation of XY, we have $\frac{1}{2} X_{2} (g) + \frac{1}{2} Y_{2} (g) ⟶ X Y (g)$

In terms of bond enthalpies, we have

$Δ_{f} H^{\circ} = \frac{1}{2} ε_{X_{2}} + \frac{1}{2} ε_{Y_{2}} - ε_{XY} = \frac{1}{2} ε_{X_{2}} + \frac{1}{2} (\frac{1}{2} ε_{X_{2}}) - ε_{X_{2}} = - \frac{1}{4} ε_{X_{2}}$

Hence $ε_{X_{2}} = - 4 (Δ_{f} H^{\circ}) = - 4 (- 200 kJ {mol}^{- 1}) = 800 kJ {mol}^{- 1}$

If the bond dissociation energies of $XY, X_{2}$ and $Y_{2}$ (all diatomic molecules) are in the ratio of $1 : 1 : 0.5$ and $Δ_{f} H$ for the formation of XY is $- 200 kJ {mol}^{- 1}$ . The bond dissociation energy of $X_{2}$ , will be

Solution:

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