Question

Bond dissociation enthalpy of $H_2,C{l}_2$ and $HCl$ are $434,242$ and $431{\text{kJ mol}}^{-1}$ respectively.
Enthalpy of formation of $HCl$ in ${\text{kJ mol}}^{-1}$ is $X$, then value of $X$ is:

Answer 1

$H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)\Delta H_{reaction}=\sum (B.E{)}_{reactant}-\sum (B.E{)}_{product}=\left[\right(B.E{)}_{H-H}+\left(B.E{)}_{Cl-Cl}\right]-[2\times (B.E{)}_{H-Cl}]=434+242-(431\times 2)\Delta H_{reaction}=-186{\text{kJ mol}}^{-1}$
Heat of formation is the amount of heat eveolved or absorbedwhen one mole of substance is directly obtained from its constituent element.
Hence enthalpy of formation of $HCl=-\frac{186}{2}{\text{kJ mol}}^{-1}=-93{\text{kJ mol}}^{-1}$