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The bond enthalpies of gas H2 , Cl2 and HCl are 104,58,103 bar3 k-1 mol-1 .calculate the enthalpy of formation of HCL

The bond enthalpies of gas H2 , Cl2 and HCl are 104,58,103 bar3 k-1 mol-1 .calculate the enthalpy of formation of HCL

Grade:11

3 Answers

Arun
25750 Points
6 years ago
1. The chemical equation for the formation of HCl is :

      H2 + Cl2 →2 HCl
2. Bond dissociation energy = Bonds broken - Bonds formed
               = (104 + 58) - 2 (103) = - 44 Kcal.
Rudraksh
13 Points
5 years ago
The chemical equation for the reaction is 
H2 + Cl2 = 2hcl 
Since we have to find out enthalpy of formation we have to deal with 1 mole of product eqn becomes
1/2H2 +1/2 Cl2= HCl
Now bond enthalpy of formation = bond dissociation energy of HCl -1/2 ( B E of H 2 + BE of Cl2)
103-29-52=22 kcal
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

The chemical equation for the reaction is
H2 + Cl2 = 2hcl
Since we have to find out enthalpy of formation we have to deal with 1 mole of product eqn becomes
1/2H2 +1/2 Cl2= HCl
Now bond enthalpy of formation = bond dissociation energy of HCl -1/2 ( B E of H 2 + BE of Cl2)
103-29-52=22 kcal

Thanks and Regards

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