Q. The bond dissociation energies of gaseous $H_{2},\ Cl_{2}$ and HClare 104, 58 and 103 kcal respectively. The enthalpy of formation of HClgas would be
A) $- 44$ kcal
B) $44$ kcal
C) $- 22$ kcal
D) 22 kcal
Explanation
(c)
Required equation is $\frac{1}{2}H_{2} + \frac{1}{2}Cl_{2} \rightarrow HCl$
$\mathrm{\Delta}H = \sum_{}^{}{BE_{(products)}} - \sum_{}^{}{BE_{(reactants)}}$
$= BE(HCl) - \left\lbrack \frac{1}{2}BE\left( H_{2} \right) + \frac{1}{2}BE\left( Cl_{2} \right) \right\rbrack$
$= - 103 - \left\lbrack \frac{1}{2}( - 104) + \frac{1}{2}( - 58) \right\rbrack$
$= - 103 - ( - 52 - 29) = - 22$ kcal
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.