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Step-by-Step Solution
Step 1: Write the formation reaction
The standard enthalpy of formation of HCl is defined for the reaction:
$ \frac{1}{2} \mathrm{H_{2}} + \frac{1}{2} \mathrm{Cl_{2}} \to \mathrm{HCl} $
Given:
• Bond energy of H–H = $430\,\mathrm{kJ\,mol^{-1}}$
• Bond energy of Cl–Cl = $240\,\mathrm{kJ\,mol^{-1}}$
• Enthalpy of formation $\Delta H_f(\mathrm{HCl}) = -90\,\mathrm{kJ\,mol^{-1}}$.
Step 2: Express $\Delta H_f(\mathrm{HCl})$ in terms of bond energies
The standard enthalpy of formation can be written as:
$ \Delta H_f(\mathrm{HCl}) = \Big[\frac{1}{2}D(\mathrm{H-H}) + \frac{1}{2} D(\mathrm{Cl-Cl})\Big] - D(\mathrm{H-Cl}) $
where
$D(\mathrm{H-H}), D(\mathrm{Cl-Cl}),$ and $D(\mathrm{H-Cl})$
are the bond enthalpies (bond energies) for H–H, Cl–Cl, and H–Cl bonds, respectively.
Step 3: Substitute the known values
Substitute the known bond energies and $\Delta H_f$ into the above formula:
$ -90 = \Big[ \frac{1}{2} \times 430 + \frac{1}{2} \times 240 \Big] - D(\mathrm{H-Cl}) $
$ -90 = \big[ 215 + 120 \big] - D(\mathrm{H-Cl}) $
Step 4: Solve for $D(\mathrm{H-Cl})$
$ -90 = 335 - D(\mathrm{H-Cl}) $
$ D(\mathrm{H-Cl}) = 335 + 90 = 425\,\mathrm{kJ\,mol^{-1}} $
Final Answer
The bond enthalpy of HCl is
$\mathbf{425\,kJ\,mol^{-1}}$.
