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Question
Calculate enthalpy of formation of HCl if bond enthalpies of H2, Cl2 and HCl are 434 kJ mol-1, 242 kJ mol–1 and 431 kJ mol–1 respectively.
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Solution
∆rH° = Σ∆H°(reactant bonds) − Σ∆H°(product bonds)
\[\ce{H_{2(g)} + Cl_{2(g)} -> 2HCl_{(g)}}\]
∴ ∆rH° = [1 mol × 434 kJ mol−1 + 1 mol × 242 kJ mol−1 - [2 mol × 431 kJ mol−1]
= - 186 kJ
∴ \[\ce{H_{2(g)} + Cl_{2(g)} -> 2HCl_{(g)}}\], ∆rH° = −186 kJ
For enthalpy of formation of HCl, the reaction is
\[\ce{\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{2(g)} -> HCl_{(g)}}\]
∆rH° = `(- 186 "kJ")/(2 "mol")` = - 93 kJ mol–1
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