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Question
Select the most appropriate option.
Bond enthalpies of H–H, Cl–Cl, and H–Cl bonds are 434 kJ mol–1, 242 kJ mol–1, and 431 kJ mol–1, respectively. Enthalpy of formation of HCl is _______.
Options
245 kJ mol–1
–93 kJ mol–1
–245 kJ mol–1
93 kJ mol–1
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Solution
Bond enthalpies of H–H, Cl–Cl, and H–Cl bonds are 434 kJ mol–1, 242 kJ mol–1, and 431 kJ mol–1, respectively. Enthalpy of formation of HCl is `bb(underline(–93 kJ mol^(–1))`.
Explanation:
ΔrH° = ∑ ΔH° (reactant bonds) – ∑ΔH° (products bonds)
\[\ce{H2_{(g)} + Cl2_{(g)} → 2HCl_{(g)}}\]
∴ ΔrH° = [1 mol × 434 kJ mol–1 + 1 mol × 242 kJ mol–1] – [2 mol × 431 kJ mol–1]
= –186 kJ
∴ \[\ce{H2_{(g)} + Cl2_{(g)} → 2HCl_{(g)}, ΔrH° = -186 kJ}\]
For enthalpy of formation of HCl, the reaction is
\[\ce{1/2 H2_{(g)} + 1/2 Cl2_{(g)} -> HCl_{(g)}}\],
ΔrH° =`(- 186 "kJ")/(2 "mol")`= –93 kJ mol–1
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