The bond enthalpy of the Br-Cl bond is equal to $\Delta$H° for the reaction
BrCl(g)$\rarr$ Br(g)+ Cl(g).
Use the following data to find the bond enthalpy of the Br-Cl bond.

A)14.6 kJ/mol
B)203.5 kJ/mol
C)219.0 kJ/mol
D)438.0 kJ/mol
E)407.0 kJ/mol

Question

The bond enthalpy of the Br-Cl bond is equal to $\Delta$H° for the reaction 
BrCl(g)$\rarr$ Br(g)+ Cl(g).
Use the following data to find the bond enthalpy of the Br-Cl bond.

A)14.6 kJ/mol 
B)203.5 kJ/mol 
C)219.0 kJ/mol 
D)438.0 kJ/mol 
E)407.0 kJ/mol

Quizplus · Accepted Answer

Using Hess's law, we can write the given reaction as:
Br(g) + Cl(g) → BrCl(g) (reverse of the given reaction)  ΔH° = -F1F1F1S1ΔH°
Adding this reaction to the given reaction:
BrCl(g) + Br(g) + Cl(g) → Br(g) + Cl(g) + BrCl(g)  ΔH° = 0
Now, we can use bond enthalpies:
ΔH° = bond enthalpy of Br-Cl bond + bond enthalpy of Br-Br bond + bond enthalpy of Cl-Cl bond - bond enthalpy of Br-Cl bond
0 = (413 + 193 + 242 - x) kJ/mol
x = 848 kJ/mol
Now, we can use the given equation:
F1F1F1S1ΔH° = bond enthalpy of Br-Cl bond
F1F1F1S1(-848 kJ/mol) = -219.0 kJ/mol
Bond enthalpy of the Br-Cl bond is 219.0 kJ/mol.

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