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Question: Answered & Verified by Expert
Bond enthalpies of $\mathrm{H}_{2}, \mathrm{X}_{2}$ and $\mathrm{HX}$ are in the ratio 2: 1: 2 . If enthalpy of formation of $\mathrm{HX}$ is $-50 \mathrm{~kJ}$ $\mathrm{mol}^{-1},$ the bond enthalpy of $\mathrm{X}_{2}$ is
ChemistryChemical Bonding and Molecular StructureJEE Main
Options:
  • A $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • B $300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • C $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • D $400 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution:
1910 Upvotes Verified Answer
The correct answer is: $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\frac{1}{2} \mathrm{H}_{2}+\frac{1}{2} \mathrm{X}_{2} \longrightarrow \mathrm{HX}$
Let the bond enthalpy of $\mathrm{X}-\mathrm{X}$ bond be $x$.
$\Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HX})=-50$
$=\frac{1}{2} \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{X}-\mathrm{X}}-\Delta \mathrm{H}_{\mathrm{H}-\mathrm{X}}$
$=\frac{1}{2} 2 x+\frac{1}{2} x-2 x=\frac{-x}{2}$
$\therefore x=50 \times 2=100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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