MCQ
Calculate the enthalpy change for the reaction

$H_2 + F_2 \longrightarrow 2HF$

given that
Bond energy of $H-H$ bond $= 434\, kJ/mol$
Bond energy of $F-F$ bond $= 158\, kJ/mol$
Bond energy of $H-F$ bond $= 565\, kJ/mol$

.....$ kJ$

  • A
    $538$
  • $-538$
  • C
    $27$
  • D
    $-27$

Answer

Correct option: B.
$-538$
b
$\Delta H = \sum {{{\left( {{\Delta _{B.E}}H} \right)}_{{\mathop{\rm Re}\nolimits} act}} - } \sum {{{\left( {{\Delta _{B.E}}H} \right)}_{\Pr oduct}}} $

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