Nitroglycerine $\left(\mathrm{MW}=227.1\right)$ detonates according to the following equations,

$2{\mathrm{C}}_3{\mathrm{H}}_5{\left({\mathrm{NO}}_3\right)}_3\left(l\right)⟶3 {\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)+6{\mathrm{CO}}_2\left(\mathrm{g}\right)+5{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

The standard molar enthalpies of formation, ${\mathrm{\Delta H}}_{\mathrm{f}}°$ for all the compounds are given below.

${\mathrm{\Delta H}}_{\mathrm{f}}°\left[{\mathrm{C}}_3{\mathrm{H}}_5{\left({\mathrm{NO}}_3\right)}_3\right]=-364 \mathrm{kJ}/\mathrm{mol}$
${\mathrm{\Delta H}}_{\mathrm{f}}°\left[{\mathrm{CO}}_2\left(\mathrm{g}\right)\right]=-393.5 \mathrm{kJ}/\mathrm{mol}$
${\mathrm{\Delta H}}_{\mathrm{f}}°\left[{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right]=-241.8 \mathrm{kJ}/\mathrm{mol}$
${\mathrm{\Delta H}}_{\mathrm{f}}°\left[{\mathrm{N}}_2\left(\mathrm{g}\right)\right]=0 \mathrm{kJ}/\mathrm{mol}$
${\mathrm{\Delta H}}_{\mathrm{f}}°\left[{\mathrm{O}}_2\left(\mathrm{g}\right)\right]=0 \mathrm{kJ}/\mathrm{mol}$

The enthalpy change when $10 \mathrm{g}$ of nitroglycerine is detonated is

• A

  $-100.5 \mathrm{kJ}/\mathrm{mol}$
• Correct answer
  B

  $-62.5 \mathrm{kJ}/\mathrm{mol}$
• C

  $-80.3 \mathrm{kJ}/\mathrm{mol}$
• D

  $-74.9 \mathrm{kJ}/\mathrm{mol}$

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