Question Detail
Question
Calculate ΔH for CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g) using bond enthalpies: C-H = 413, Cl-Cl = 242, C-Cl = 338, H-Cl = 431 kJ/mol.
Correct Answer
Option A is the correct answer.
Detailed Explanation
Bonds broken: 1 C-H (413) + 1 Cl-Cl (242) = 655 kJ. Bonds formed: 1 C-Cl (338) + 1 H-Cl (431) = 769 kJ. ΔH = 655 - 769 = -114 kJ/mol.
Hint
The notes provide this exact calculation.