Q7 / 125 · BITSAT 2012 · Chemistry

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

  • Correct answer
    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

$\frac{1}{2} \mathrm{H}_{2}+\frac{1}{2} \mathrm{X}_{2} \longrightarrow \mathrm{HX}$ Let the bond enthalpy of $\mathrm{X}-\mathrm{X}$ bond be $x$. $\Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HX})=-50$ $=\frac{1}{2} \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{X}-\mathrm{X}}-\Delta \mathrm{H}_{\mathrm{H}-\mathrm{X}}$ $=\frac{1}{2} 2 x+\frac{1}{2} x-2 x=\frac{-x}{2}$ $\therefore x=50 \times 2=100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Hence, the correct answer is option: 'A'.

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