bond enthalpy

Question

Given bond enthalpies: C-H = 413 kJ/mol, Cl-Cl = 242 kJ/mol, C-Cl = 328 kJ/mol, H-Cl = 431 kJ/mol. The enthalpy change for CH₄ + Cl₂ → CH₃Cl + HCl is:

RankGuru · Accepted Answer

$\Delta H_{reaction} = \sum 	ext{Bond Enthalpies (Reactants)} - \sum 	ext{Bond Enthalpies (Products)}$
- **Bonds to break (Reactants)**: 1 $C-H$ bond + 1 $Cl-Cl$ bond
$\sum BE_{react} = 413 + 242 = 655 	ext{ kJ/mol}$
- **Bonds to form (Products)**: 1 $C-Cl$ bond + 1 $H-Cl$ bond
$\sum BE_{prod} = 328 + 431 = 759 	ext{ kJ/mol}$
$\Delta H_{rxn} = 655 - 759 = -104 	ext{ kJ/mol}$
- The reaction is exothermic as stronger bonds are formed than broken.

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