From the following thermochemical equations, find out the bond dissociation enthalpy of CH3−H bond.C

Question

From the following thermochemical equations, find out the bond dissociation enthalpy of CH₃−H bond.

CH₃I(g) → CH₃(g) + I(g): ΔH = 54.0 kcal

CH₄(g) + I₂(s) → CH₃I(g) + HI(g): ΔH = 29.0 kcal

HI(g) → H(g) + I(g): ΔH = 79.8 kcal

I₂(s) → 2I(g): ΔH = 51.0 kcal

Accepted Answer

From $(a) + (b) + (c) - (d),$$CH_{3} - H(g) ightarrow CH_{3}(g) + H(g);\Delta H = 54 + 29 + 79.8 - 51 = 111.8	ext{ kcal/mol}$

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