Consider the gas-phase reaction: with an overall . Using the given bond dissociation energies (, ), — Thermodynamics and Thermochemistry Chemistry Question
Question
Consider the gas-phase reaction: $XeF_2(g) + H_2(g) \rightarrow 2HF(g) + Xe(g)$ with an overall $\Delta H^\circ = -430 \text{ kJ}$. Using the given bond dissociation energies ($BE_{H-H} = 435 \text{ kJ mol}^{-1}$, $BE_{H-F} = 565 \text{ kJ mol}^{-1}$), calculate the average bond energy of the $Xe-F$ bond.
Answer: A
💡 Solution & Explanation
The reaction enthalpy can be estimated via $\Delta H_{rxn} = \sum BE(reactants) - \sum BE(products)$. Thus, $-430 = [2 \times BE(Xe-F) + BE(H-H)] - [2 \times BE(H-F)]$. Substitute the numerical values: $-430 = 2 \times BE(Xe-F) + 435 - 2(565) \Rightarrow -430 = 2 \times BE(Xe-F) - 695$. Solving gives $2 \times BE(Xe-F) = 265 \Rightarrow BE(Xe-F) = 132.5 \text{ kJ mol}^{-1}$.
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