ThermochemistryHard
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.
Options
A.−33.5 kcal/mol
B.33.5 kcal/mol
C.18 kcal/mol
D.−9 kcal/mol
Solution
$CH_{4}(g) \rightarrow CH_{3}(g) + H(g)$
$103 = \left\lbrack \Delta_{f}H_{CH_{3}(g)} + \frac{103}{2} \right\rbrack - ( - 18) $$$\Rightarrow \Delta_{f}H_{CH_{3}(g)} = 33.5\text{ kcal/mol}$$
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