The bond dissociation energies of gaseous $H_{2}, {C l}_{2}$ and $H C l$ are $104, 58$ and $103 k c a l$ respectively. The enthalpy of formation of $H C l$ gas would be

Question

The bond dissociation energies of gaseous - H - 2 - Cl - 2 - and HCl are 104-58 and 103 kcal respectively- The enthalpy of formation of HCl gas would be-44 kcal44 kcal-22 kcal22 kcal

Toppr Admin · Accepted Answer

12H2-12Cl2-x27F6-HCl-x394-H-x2211-BEreactants-x2212-x2211-BEproducts-xA0- -xA0- -xA0- -xA0- -xA0-12BE-H2-12BE-Cl2-x2212-BE-HCl-12-xD7-104-12-xD7-58-x2212-103-xA0- -xA0- -xA0- -xA0- -xA0-xA0- -xA0- -xA0- -xA0- -52-29-x2212-103-x2212-22-xA0-kcal

The bond dissociation energies of gaseous $H_{2}, {C l}_{2}$ and $H C l$ are $104, 58$ and $103 k c a l$ respectively. The enthalpy of formation of $H C l$ gas would be
$- 44 k c a l$
$44 k c a l$
$- 22 k c a l$
$22 k c a l$

$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} ⟶ H C l$

$Δ H = \sum {B E}_{r e a c t a n t s} - \sum {B E}_{p r o d u c t s}$

$= [\frac{1}{2} B E (H_{2}) + \frac{1}{2} B E (C l_{2})] - B E (H C l) = [(\frac{1}{2} \times 104) + (\frac{1}{2} \times 58)] - 103$

$= (52 + 29) - 103 = - 22 k c a l$

The bond dissociation energies of gaseous H2,Cl2 and HCl are 104,58 and 103 kcal respectively. The enthalpy of formation of HCl gas would be−44 kcal44 kcal−22 kcal22 kcal

12H2+12Cl2⟶HClΔH=∑BEreactants−∑BEproducts =[12BE(H2)+12BE(Cl2)]−BE(HCl)=[(12×104)+(12×58)]−103 =(52+29)−103=−22 kcal

The bond dissociation energies of gaseous $H_{2}, {C l}_{2}$ and $H C l$ are $104, 58$ and $103 k c a l$ respectively. The enthalpy of formation of $H C l$ gas would be
$- 44 k c a l$
$44 k c a l$
$- 22 k c a l$
$22 k c a l$

$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} ⟶ H C l$

$Δ H = \sum {B E}_{r e a c t a n t s} - \sum {B E}_{p r o d u c t s}$

$= [\frac{1}{2} B E (H_{2}) + \frac{1}{2} B E (C l_{2})] - B E (H C l) = [(\frac{1}{2} \times 104) + (\frac{1}{2} \times 58)] - 103$

$= (52 + 29) - 103 = - 22 k c a l$