Bond Enthalpy Question - 5

QuestionEstimate the enthalpy change for the reaction of methane with chlorine in UV light:CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)Mean bond enthalpies / kJ mol^-1: C-H 413, Cl-Cl 243, C-Cl 346, H-Cl 432Step 1: compare the molecules. CH₄ has 4 C-H bonds. CH₃Cl has 3 C-H bonds and 1 C-Cl bond. So overall one C-H bond is broken and three C-H bonds are unchanged.Step 2: bonds broken.
1 C-H + 1 Cl-Cl = 413 + 243 = 656 kJStep 3: bonds made.
1 C-Cl + 1 H-Cl = 346 + 432 = 778 kJStep 4: subtract.ΔH = 656 - 778 = -122 kJ mol^-1Full-count checkBroken: 4(413) + 243 = 1895 kJ. Made: 3(413) + 346 + 432 = 2017 kJ. ΔH = 1895 - 2017 = -122 kJ mol^-1. The three unchanged C-H bonds cancel, as expected.ExtensionIf chlorine is in excess, substitution continues. Each further step, such as CH₃Cl + Cl₂ → CH₂Cl₂ + HCl, breaks the same types of bond (one C-H and one Cl-Cl) and makes the same types (one C-Cl and one H-Cl). Using mean bond enthalpies, every step therefore has the same estimate of -122 kJ mol^-1, even though the real values differ slightly. This shows a limitation of mean values: they cannot tell apart bonds in different environments.