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At standard conditions, if the change in the enthalpy for the following reaction is $-109\ kJ\,mol^{-1}$: $H_2(g)+Br_2(g)\rightarrow2HBr(g)$. Given that bond energy of $H_2$ and $Br_2$ is $435\ kJ\,mol^{-1}$ and $192\ kJ\,mol^{-1}$, respectively, what is the bond energy (in kJ mol$^{-1}$) of HBr?
A
259
B
368
C
736
D
518
Detailed Solution
$\Delta H=\text{(bonds broken)}-\text{(bonds formed)}=(435+192)-2\times BE(HBr)$. $-109=627-2\,BE(HBr)\Rightarrow2\,BE(HBr)=736\Rightarrow BE(HBr)=368\ kJ\,mol^{-1}$.