Question

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 1

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

Answer 2

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

hope it helps!

Explanation: