Question

4. Set-up ONLY the equation for determining the bond dissociation energy ΔH° for the hydrogenation of acetylene, C2H2 in the reaction below by showing what bonds are being broken (yes, draw Lewis Structures!) and what bonds are being formed:   H-C=C-H(g) + 2 H2(g) → CH3-CH3(g)

Answer 1

Answer: $\Delta H=[(1\times B.E_{C\equiv C})+(2\times B.E_{C-H})}+(2\times B.E_{H-H}) ]-[(1\times B.E_{C-C})+(6\times B.E_{C-H})]$

Explanation:

The balanced chemical reaction is:

$H-C\equiv C-H(g)+2H_2(g)\rightarrow CH_3-CH_3(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]$

$\Delta H=[(n_{H-C\equiv C-H}\times B.E_{H-C\equiv C-H})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{CH_3-CH_3}\times B.E_{CH_3-CH_3})]$

$\Delta H=[(1\times B.E_{C\equiv C})+(2\times B.E_{C-H})}+(2\times B.E_{H-H}) ]-[(1\times B.E_{C-C})+(6\times B.E_{C-H})]$

where,

n = number of moles

Thus the equation for determining the bond dissociation energy ΔH° for the hydrogenation of acetylene is $\Delta H=[(1\times B.E_{C\equivC})+(2\times B.E_{C-H})}+(2\times B.E_{H-H}) ]-[(1\times B.E_{C-C})+(6\times B.E_{C-H})]$