31. XeF (g) + H2(g) 2HF(g) + Xe(g); AH = – 430 kJusing the following

1.	31. XeF (g) + H2(g) 2HF(g) + Xe(g); AH = – 430 kJusing the following bond energies : H-H = 435 kJ/mol, H-F = 565 kJ/molCalculate the average bond energy of Xe-F in XeF2
Answer» The average BOND energy of Xe-F in $XeF_2$ is 132.5kJ Explanation: The BALANCED chemical reaction is, $XeF_2(g)+H_2(g)\rightarrow 2HF(g)+Xe(g)$ $\Delta H=-430kJ$ The EXPRESSION for enthalpy change is, $\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]$ $\Delta H=[(n_{XeF_2}\times B.E_{XeF})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{HF}\times B.E_{HF})+(n_{Xe}\times B.E_{Xe})]$ $\Delta H=[(2\times B.E_{Xe-F}+(1\times B.E_{H-H}) ]-[(2\times B.E_{H-F})+(1\times 0)]$ where, n = number of moles Now put all the given VALUES in this expression, we get $-430kJ=[(2\times B.E_{Xe-F})+(1\times 435)]-[(2\times 565)]$ $B.E_{Xe-F}=132.5kJ$ $XeF_2$ Therefore, the average bond energy of Xe-F in is 132.5kJ Learn More about average bond energy https://brainly.com/question/14286998 https://brainly.com/question/11655289