Question
Bond dissociation enthalpy of $${H_2},C{l_2}$$ and $$HCl$$ are 434, 242 and 431 $$kJ\,mo{l^{ - 1}}$$ respectively. Enthalpy of formation of $$HCl$$ is
A.
$$93\,kJ\,mo{l^{ - 1}}$$
B.
$$ - 245\,kJ\,mo{l^{ - 1}}$$
C.
$$ - 93\,kJ\,mo{l^{ - 1}}$$
D.
$$245\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 93\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{Given,}}\,\,\Delta {H_{H - H}} = 434\,kJ/mol \cr
& \Delta {H_{Cl - Cl}} = 242\,kJ/mol \cr
& \Delta {H_{H - Cl}} = 431\,kJ/mol \cr
& \frac{1}{2}{H_2} + \frac{1}{2}C{l_2} \to HCl,\Delta {H_r} = ? \cr
& \Delta {H_r} = \frac{1}{2} \times \Delta {H_{H - H}} + \frac{1}{2} \times \Delta {H_{Cl - Cl}} - \Delta {H_{H - Cl}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \times 434 + \frac{1}{2} \times 242 - 431 \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = 217 + 121 - 431 \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = - 93\,kJ/mol \cr} $$