Question

Calculate ΔH_f of HCl if bond energy of H - H bond is 437 kJ Cl - Cl bond is 244, and H - C is 433 kJ mol^-1.

Answer 1

1/2H₂(g) + 1/2Cl₂(g) > HCl(g) 

ΔH = 1/2 B_{H – H} + B_{O – O} = 1/2 × 437 + 1/2 × 244 -433 

= 218.5 kJ + 122 kJ - 433 kJ = -92.5 kJ mol^-1