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Given: $Δ_{f} H^{\circ} (UF, g) = 92.0 kJ {mol}^{- 1}$ and $Δ_{f} H^{\circ} (U, g) = 536 kJ {mol}^{- 1}$ . If bond energy of $F - F$ is $155 kJ {mol}^{- 1}$ , then the bond dissociation energy of $U - F$ will be

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420 kJ {mol}^{- 1}

521.5 kJ {mol}^{- 1}

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detailed solution

Correct option is B

Given information are as follows

$(i) U (s) + \frac{1}{2} F_{2} (g) \to UF (g) Δ_{f} H^{\circ} = 92.0 kJ {mol}^{- 1}$

$(ii) U (s) \to U (g)$ $Δ_{f} H^{\circ} = 536 kJ {mol}^{- 1}$

$(i i i) \frac{1}{2} F_{2} (g) \to F (g)$ $Δ H^{\circ} = (1 / 2) (155 kJ {mol}^{- 1})$

For the bond dissociation energy, we have to calculate $Δ H$ for the reaction

$UF (g) \to U (g) + F (g)$

This reaction can be obtained by the following manipulations

$- Eq \cdot (i) + Eq \cdot (ii) + Eq \cdot (iii)$

Hence,

$ε (U - F) = [- 92 + 536 + (1 / 2) (155)] kJ {mol}^{- 1} = 521.5 kJ {mol}^{- 1}$

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Given: ΔfH∘(UF,g)=92.0 kJ mol−1 and ΔfH∘(U,g)=536 kJ mol−1. If bond energy of F-F is 155 kJ mol−1, then the bond dissociation energy of U−F will be