Given the bond dissociation enthalpy of CH3−H bond as 103 kcal/mol and the enthalpy of formation of

Question

Given the bond dissociation enthalpy of CH3−H bond as 103 kcal/mol and the enthalpy of formation of CH4(g) as −18 kcal/mol, find the enthalpy of formation of methyl radical. The dissociation energy of H2(g) into H (atoms) is 103 kcal/mol.

Accepted Answer

$CH_{4}(g) ightarrow CH_{3}(g) + H(g)$$103 = \left\lbrack \Delta_{f}H_{CH_{3}(g)} + \frac{103}{2} ightbrack - ( - 18) $$$\Rightarrow \Delta_{f}H_{CH_{3}(g)} = 33.5	ext{ kcal/mol}$$

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