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* B.E_(C\equiv C))+(2* B.E_(C-H))}+(2* B.E_(H-H)) ]-[(1* B.E_(C-C))+(6* B.E_(C-H))]`

Step-by-step 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* B.E(reactant)]-\sum [n* B.E(product)]`

`\Delta H=[(n_(H-C\equiv C-H)* B.E_(H-C\equiv C-H))+(n_(H_2)* B.E_(H_2)) ]-[(n_(CH_3-CH_3)* B.E_(CH_3-CH_3))]`

`\Delta H=[(1* B.E_(C\equiv C))+(2* B.E_(C-H))}+(2* B.E_(H-H)) ]-[(1* B.E_(C-C))+(6* 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* B.E_(C\equivC))+(2* B.E_(C-H))}+(2* B.E_(H-H)) ]-[(1* B.E_(C-C))+(6* B.E_(C-H))]`