ThermochemistryHard

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

Options

A.125.2 kcal/mol
B.91.7 kcal/mol
C.101.9 kcal/mol
D.111.8 kcal/mol

Solution

From $(a) + (b) + (c) - (d),$

$CH_{3} - H(g) \rightarrow CH_{3}(g) + H(g);\Delta H = 54 + 29 + 79.8 - 51 = 111.8\text{ kcal/mol}$

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