The enthalpy of reaction can be found out by finding the difference in the bond energies of reactants and products.
ΔH = Σ Bond Energy of Reactants – Σ Bond energy of Products.
Example:
Calculate DH for the following reaction.
H2 (g) + ½ O2 (g) ---------> H2O
Given
H – H = 433 KJ/mol.
O – O = 499 KJ/mol.
and O – H = 460 KJ/mol.
Solution:
ΔH = Σ B.E of H2 and ½ O2 - Σ B.E of H2O
= (B. E of H2 + ½ B. E. of O2 ) – (B. E. of H2O)
= (B. E. of H – H + ½ B. E. of O – O) – (B. E. of 2 .O – H)
= (433 + ½ x 499) – (2 x 460)
= (433 + 249.5) – (920)
= 682.5 – 920
ΔH = – 237.5 KJ/Mol.
ΔH = Σ Bond Energy of Reactants – Σ Bond energy of Products.
Example:
Calculate DH for the following reaction.
H2 (g) + ½ O2 (g) ---------> H2O
Given
H – H = 433 KJ/mol.
O – O = 499 KJ/mol.
and O – H = 460 KJ/mol.
Solution:
ΔH = Σ B.E of H2 and ½ O2 - Σ B.E of H2O
= (B. E of H2 + ½ B. E. of O2 ) – (B. E. of H2O)
= (B. E. of H – H + ½ B. E. of O – O) – (B. E. of 2 .O – H)
= (433 + ½ x 499) – (2 x 460)
= (433 + 249.5) – (920)
= 682.5 – 920
ΔH = – 237.5 KJ/Mol.
No comments:
Post a Comment