Question

Bond energies of H-H and Cl-Cl are $430kJmo{l^{ - 1}}$ and $242kJmo{l^{ - 1}}$ respectively. $\vartriangle {H_{{f_{}}}}$ for HCl is $91kJmo{l^{ - 1}}$. What will be the bond energy of H-Cl bond (per mole value)?
A.$672kJ$
B.$182kJ_{}^{}$
C.$245kJ$
D.$88kJ$

Answer 1

Bond energy is the amount of energy required to break one mole of a molecule to its constituent atoms.bond energy, also called the mean bond enthalpy or average bond enthalpy is the measure of bond strength in a chemical bond.

Complete step by step answer:
Bond energy of H-H bond can be represented as,
${H_2} \to 2H$ $:\vartriangle {H_H} = 430kJmo{l^{ - 1}}$
i.e. bond energy of H-H is the energy required to break the H-H bond in ${H_2}$ to two hydrogen atoms.
Bond energy of Cl-Cl bond can be represented as,
$C{l_2} \to 2Cl$ $:\vartriangle {H_{Cl}} = 242kJmo{l^{ - 1}}$
i.e. bond energy of Cl-Cl is the energy required to break the Cl-Cl bond in $C{l_2}$ to two chlorine atoms.
In the same way, bond energy of H-Cl can be represented as,
$HCl \to H + Cl$ $:\vartriangle {H_{HCl}} = ?$
i.e. bond energy of H-Cl is the energy required to break the H-Cl bond in HCl to hydrogen and chlorine atoms. We need to calculate this value.
Given that the enthalpy of formation of HCl, $\vartriangle {H_{{f_{}}}}(HCl)$ is $91kJmo{l^{ - 1}}$.We can represent this as,
$\dfrac{1}{2}{H_2} + \dfrac{1}{2}C{l_2} \to HCl$ $:\vartriangle {H_f}(HCl) = 91kJmo{l^{ - 1}}$
$\vartriangle {H_{{f_{}}}}$ for HCl can also be written as,
$\vartriangle {H_f}(HCl)$ = $\dfrac{{Bond{\text{ }}energy{\text{ }}of{\text{ }}H - H}}{2}{\text{ }} + {\text{ }}\dfrac{{Bond{\text{ }}energy{\text{ }}of{\text{ }}Cl - Cl{\text{ }}}}{2}-{\text{ B}}ond{\text{ }}energy{\text{ }}of{\text{ }}H - Cl$
From this equation we can write,
Bond energy of H-Cl bond = $\dfrac{{Bond{\text{ }}energy{\text{ }}of{\text{ }}H - H}}{2}{\text{ }} + {\text{ }}\dfrac{{Bond{\text{ }}energy{\text{ }}of{\text{ }}Cl - Cl{\text{ }}}}{2}-{\text{ }}\vartriangle {{\text{H}}_f}(HCl)$
Substituting the values,
Bond energy of H-Cl bond = $\dfrac{{430}}{2} + \dfrac{{242}}{2} - 91 = 215 + 121 - 91 = 245kJmo{l^{ - 1}}$
i.e. Bond energy of H-Cl bond is $245kJmo{l^{ - 1}}$.
Hence the correct option is C.

Additional information-
Enthalpy of formation is the enthalpy change when one mole of the compound is formed from the constituent elements. Hence the enthalpy of formation of H-Cl is the enthalpy change when one mole of the compound is formed from hydrogen and chlorine.

Note:
While calculating enthalpy of formation/bond energy we should be aware of the sign. The enthalpy change will be negative if the heat is released and positive if heat is absorbed.