The bond enthalpy of the Br-Cl bond is equal to $\Delta$ 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.
$\begin{array}{ll}\mathrm{Br}_{2}(\mathrm{l}) \rightarrow \mathrm{Br}_{2}(\mathrm{~g}) & \Delta \mathrm{H}^{\circ}=30.91 \mathrm{~kJ} / \mathrm{mol} \\\mathrm{Br}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Br}(\mathrm{~g}) & \Delta \mathrm{H}^{\circ}=192.9 \mathrm{~kJ} / \mathrm{mol} \\\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Cl}(\mathrm{~g}) & \Delta \mathrm{H}^{\circ}=243.4 \mathrm{~kJ} / \mathrm{mol} \\\mathrm{Br}_{2}(\mathrm{l}) +\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{BrCl}(\mathrm{~g}) & \Delta \mathrm{H}^{\circ}=29.2 \mathrm{~kJ} / \mathrm{mol}\end{array}$

Question

The bond enthalpy of the Br-Cl bond is equal to

\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.

\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}

Quizplus · Accepted Answer

The bond enthalpy can be calculated using the equation:
ΔH = (ΔHreactants - ΔHproducts)/n
where n is the number of moles of the bond broken or formed.
For the reaction given, n = 1 (since one Br-Cl bond is broken).

Using the given data:
ΔHf[Br(g)] = +111 kJ/mol
ΔHf[Cl(g)] = +121 kJ/mol
ΔHf[BrCl(g)] = +29 kJ/mol

Substitute the values into the equation:
ΔH = (ΔH[BrCl(g)] - ΔH[Br(g)] - ΔH[Cl(g)])/n

ΔH = [(29 kJ/mol) - (+111 kJ/mol) - (+121 kJ/mol)]/1 = -203 kJ/mol

The negative sign indicates that energy is released when the bond is broken, i.e. it is an exothermic process.

The bond enthalpy is the positive of the calculated value, so the answer is:
219.0 kJ/mol (Choice C)

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