Bond dissociation enthalpy of $ ext{H}_2$, $ ext{Cl}_2$, and $ ext{HCl}$ are $434$, $242$, and $431 ext{

Question

Bond dissociation enthalpy of $	ext{H}_2$, $	ext{Cl}_2$, and $	ext{HCl}$ are $434$, $242$, and $431 	ext{ kJ mol}^{-1}$ respectively. Enthalpy of formation of $	ext{HCl}$ is:

Accepted Answer

The chemical equation for the formation of $	ext{HCl(g)}$ from its constituent elements in their standard states is:
$\frac{1}{2}	ext{H}_2	ext{(g)} + \frac{1}{2}	ext{Cl}_2	ext{(g)} ightarrow 	ext{HCl(g)}$

The standard enthalpy of formation ($\Delta_f H^{\circ}$) can be calculated from the given bond dissociation enthalpies ($\Delta_{	ext{bond}} H^{\circ}$) using the relation:
$\Delta_f H^{\circ} = \sum \Delta_{	ext{bond}} H^{\circ}(	ext{reactants}) - \sum \Delta_{	ext{bond}} H^{\circ}(	ext{products})$

Substituting the given values into the equation:
$\Delta_f H^{\circ}(	ext{HCl}) = \left[ \frac{1}{2} \Delta_{	ext{bond}} H^{\circ}(	ext{H}_2) + \frac{1}{2} \Delta_{	ext{bond}} H^{\circ}(	ext{Cl}_2) ight] - \left[ \Delta_{	ext{bond}} H^{\circ}(	ext{HCl}) ight]$
$\Delta_f H^{\circ}(	ext{HCl}) = \left[ \frac{1}{2}(434) + \frac{1}{2}(242) ight] - 431$
$\Delta_f H^{\circ}(	ext{HCl}) = (217 + 121) - 431$
$\Delta_f H^{\circ}(	ext{HCl}) = 338 - 431 = -93 	ext{ kJ mol}^{-1}$

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