Calculate the enthalpy change for the reaction :-H2(g) + Cl2(g) → 2H

1.	Calculate the enthalpy change for the reaction :-H2(g) + Cl2(g) → 2HCL(g)given that the bond energies H-H,Cl-Cl,and H-Cl bonds are 433, 244 and 431kj/ mol respectively.
Answer» Energy absorbed for dissociation of 1 mole of H-H bonds = 433 kJ/mol Energy absorbed for dissociation of 1 mole of Cl-Cl bonds = 244 kJ/mol Total energy absorbed = 677 kJ/mol Energy released in the formation of 2 moles of H-Cl bond = 431 X 2 = 862 kJ/mol Energy released is greater than Energy absorbed Hence, NET RESULT is the release of energy. Energy released = 862 – 677 = 185 kJ/mol For the given reaction, ∆rH = - 185 kJ/mol

Calculate the enthalpy change for the reaction :-H2(g) + Cl2(g) → 2HCL(g)given that the bond energies H-H,Cl-Cl,and H-Cl bonds are 433, 244 and 431kj/ mol respectively.

Answer»

Energy absorbed for dissociation of 1 mole of H-H bonds = 433 kJ/mol

Energy absorbed for dissociation of 1 mole of Cl-Cl bonds = 244 kJ/mol

Total energy absorbed = 677 kJ/mol

Energy released in the formation of 2 moles of H-Cl bond = 431 X 2

= 862 kJ/mol

Energy released is greater than Energy absorbed

Hence, NET RESULT is the release of energy.

Energy released = 862 – 677

= 185 kJ/mol

For the given reaction, ∆rH = - 185 kJ/mol