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The Bond Enthalpy of the Br-Cl Bond Is Equal To \Delta

question 47

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The bond enthalpy of the Br-Cl bond is equal to Δ\Delta E \circ for the reaction BrCl(g) \rarr Br(g) + Cl(g) .Use the following data to find the bond enthalpy of the Br-Cl bond. Br2(l) Br2( g) ΔH=30.91 kJ/molBr2( g) 2Br( g) ΔH=192.9 kJ/molCl2( g) 2Cl( g) ΔH=243.4 kJ/molBr2(l) +Cl2( g) 2BrCl( g) ΔH=29.2 kJ/mol\begin{array}{ll}\mathrm{Br}_{2}(\mathrm{l}) \rightarrow \mathrm{Br}_{2}(\mathrm{~g}) & \Delta \mathrm{H}^{\circ}=30.91 \mathrm{~kJ} / \mathrm{mol} \\\mathrm{Br}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Br}(\mathrm{~g}) & \Delta \mathrm{H}^{\circ}=192.9 \mathrm{~kJ} / \mathrm{mol} \\\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Cl}(\mathrm{~g}) & \Delta \mathrm{H}^{\circ}=243.4 \mathrm{~kJ} / \mathrm{mol} \\\mathrm{Br}_{2}(\mathrm{l}) +\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{BrCl}(\mathrm{~g}) & \Delta \mathrm{H}^{\circ}=29.2 \mathrm{~kJ} / \mathrm{mol}\end{array}

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