Given the following average bond dissociation enthalpies (in $ ext{kJ mol}^{-1}$): $ ext{H-H} = 43

Question

Given the following average bond dissociation enthalpies (in $	ext{kJ mol}^{-1}$): $	ext{H-H} = 436$, $	ext{C}\equiv	ext{C} = 812$, $	ext{C-C} = 348$, $	ext{C-H} = 412$. Calculate the standard enthalpy of reaction ($\Delta H_{rxn}$) for the complete hydrogenation of one mole of ethyne ($	ext{HC}\equiv	ext{CH}$) to ethane ($	ext{H}_3	ext{C-CH}_3$).

Professor PC Thomas Classes · Answer

$	ext{HC}\equiv	ext{CH} + 2	ext{H}_2 ightarrow 	ext{H}_3	ext{C-CH}_3$

$\Delta H_{rxn} = \sum 	ext{Bonds Broken} - \sum 	ext{Bonds Formed}$

Bonds Broken (Reactants): $1(	ext{C}\equiv	ext{C}) + 2(	ext{C-H}) + 2(	ext{H-H})$

$812 + 2(412) + 2(436) = \mathbf{2508 	ext{ kJ}}$
Bonds Formed (Products):$1(	ext{C-C}) + 6(	ext{C-H})$
$348 + 6(412) = \mathbf{2820 	ext{ kJ}}$
$\Delta H_{rxn}$ $= 2508 - 2820 = \mathbf{-312 	ext{ kJ mol}^{-1}}$