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 _______. - Chemistry

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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

Concept: Enthalpy (H)
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Chapter 4: Chemical Thermodynamics - Exercises [Page 87]

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Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercises | Q 1.1 | Page 87

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