Question

Calculate the bond enthalpy of $H - Cl$ given that the bond enthalpies of ${H_2}$ and $Cl{}_2$ are 435.4 and $242.8KJmo{l^{ - 1}}$ respectively and the enthalpy of formation of $HCl$ is $- 92.2kJmol$.

Answer 1

Bond enthalpy or bond energy is defined as the amount of energy required to break one mole of the stated bond. It is basically the quantity that offers insight into the strength of the chemical bond.
Formula used: $\Delta {H_{rxn}} =$ $\dfrac{{sum\,of\,bond\,enthalpies\, - enthalpy\,of\,formation}}{2}$

Complete step by step answer:
Bond enthalpy and enthalpy of a reaction helps to understand how a chemical system uses energy during reactions. It is basically the energy needed to break or to form a bond.
Further, depending on whether the enthalpy of reaction is positive or negative, we can determine whether the reaction is endothermic or exothermic.
Now, we have to calculate the bond enthalpy of $H - Cl$ in the given question.
So, $\Delta {H_{HCl}} = - 92.2kJmo{l^{ - 1}}$ (given)……………………………………………………$(1)$
Further, ${H_2} + C{l_2} \to 2HCl$ so,$\Delta H = 2 \times - 92.2kJmo{l^{ - 1}}$
$= 184.4kJmo{l^{ - 1}}$
We have been given the bond enthalpies of ${H_2}$and $C{l_2}$ i.e.
${H_2} + 2H \to 430kJmo{l^{ - 1}}$ ……………………………………………………………………..$(2)$
$C{l_2} + 2Cl \to 242kJmo{l^{ - 1}}$ ………………………………………………………………………$(3)$
Therefore, according to the formula
$\Delta {H_{HCl}} =$ $\dfrac{{sum\,of\,bond\,enthalpies\, - enthalpy\,of\,formation}}{2}$
$= \dfrac{{(435.4 + 242.8) - ( - 184.4)kJmo{l^{ - 1}}}}{2}$
$= \dfrac{{862.6}}{2}$
$= 431.3kJmo{l^{ - 1}}$

Hence, bond enthalpy of $H - Cl$is$431.3kJmo{l^{ - 1}}$.

Note:
Bond enthalpy can be calculated directly if everything is in gaseous state but if it is in liquid state then extra energy is needed to convert it from liquid state to gaseous state. Moreover, if a molecule has several bonds, then bond enthalpy is calculated for each bond and then the average value is considered.