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Thermodynamics question

2008 · Q113
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Thermodynamics question

2008 · Q113

NEETChemistryThermodynamicsMCQ+4 / −1
Bond dissociation enthalpy of H2, Cl2 and HCl are 434, 242 and 431 kJ mol−-−1 respectively. Enthalpy of formation of HCl is
  1. A
    −-− 93 kJ mol−-−1
  2. B
    245 kJ mol−-−1
  3. C
    93 kJ mol−-−1
  4. D
    −-− 245 kJ mol−-−1
View written solutionFree

Correct answer: A

12{1 \over 2}21​H2 + 12{1 \over 2}21​Cl2 →\to→ HCl, Δ\Delta ΔHf = ?

Δ\Delta ΔHreaction = [ 12{1 \over 2}21​(B.E)H2 + 12{1 \over 2}21​(B.E)Cl2] - (B.E)HCl

= [217 + 121] – 431

= 338 – 431 = – 93 kJ mol–1

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