Given the bond enthalpies H-O (464),Cl-Cl (243),Cl-O (205),and H-Cl (431)in kJ/mol,compute ΔH° in kJ/mol for: H₂O(g)+ Cl₂(g)→ HOCl(g)+ HCl(g).

Question

Quizplus · Accepted Answer

The reaction involves breaking the H-O and Cl-Cl bonds and forming the Cl-O and H-Cl bonds. The ΔH° for the reaction can be calculated using the bond enthalpies given: ΔH° = Bonds broken - Bonds formed = (464 + 243) - (205 + 431) = 707 - 636 = +71 kJ/mol.

Given the Bond Enthalpies H-O (464),Cl-Cl (243),Cl-O (205),and H-Cl (431)in