Practice question
Question
Using bond enthalpies (C-H = 413 kJ/mol, Cl-Cl = 243 kJ/mol, C-Cl = 328 kJ/mol, H-Cl = 431 kJ/mol), calculate Δ H for CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g).
Explanation
Bonds broken: C-H (413) + Cl-Cl (243) = 656 kJ. Bonds formed: C-Cl (328) + H-Cl (431) = 759 kJ. Δ H = 656 - 759 = -103 kJ/mol.
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