Question 8.3.3: Calculate enthalpy of reaction using bond energies. Calculat......

You are asked to calculate the enthalpy change for a reaction using bond energies.
You are given a chemical equation.
Draw Lewis structures for reactants and products and use them to list all reactant bonds broken and product bonds formed, along with bond energies from Interactive Table 8.3.2.

\overset{Reactants:}{\begin{matrix} \underset{|}{H} \,\,\,\,\,\, \underset{|}{H} \\\quad \quad \quad \enspace \enspace \quad \quad H-C-C-H \quad :\overset{}{\underset{..}{\overset{..}{Cl} }-\overset{..}{\underset{..}{Cl}:} } \\ \overset{|}{H}\,\,\,\,\,\, \overset{|}{H} \end{matrix}}

\overset{Products:}{\begin{matrix} \underset{|}{H} \,\,\,\,\,\, \underset{|}{H}\,\,\,\,\,\, \\\qquad \quad \quad \quad H-C-C-\overset{..}{\underset{..}{Cl}:} \; \quad H-\overset{..}{\underset{..}{Cl:}} \\ \,\,\overset{|}{H}\,\,\,\,\,\,\overset{|}{H} \,\,\,\,\,\, \end{matrix}}

Bonds broken                                              Bonds formed

6 mol C—H    6 × 413 kJ/mol                       5 mol C—H     5 × 413 kJ/mol

1 mol C—C      346 kJ/mol                             1 mol C—C      346 kJ/mol

1 mol Cl—Cl    242 kJ/mol                             1 mol C—Cl    339 kJ/mol

1 mol H—Cl    432 kJ/mol

Use Equation 8.2 to calculate ∆Hº for the reaction.

\Delta H^\circ   = ∑(energies of bonds broken) − ∑(energies of bonds formed)        (8.2)
∆Hº = [6 mol(413 kJ/mol) + 1 mol(346 kJ/mol) + 1 mol(242 kJ/mol)]

− [5 mol(413 kJ/mol) + 1 mol(346 kJ/mol) + 1 mol(339 kJ/mol) + 1 mol(432 kJ/mol)]

= −116 kJ

Notice that five of the C—H bonds and the C—C bond in C_{2} H_{6} are unchanged in the reaction. The enthalpy of reaction can also be calculated using only the energies of the bonds that are broken (1 mol C—H and 1 mol Cl—Cl) and the bonds that are formed (1 mol C—Cl and 1 mol H—Cl).
∆Hº = [1 mol(413 kJ/mol) + 1 mol(242 kJ/mol)] − [1 mol(339 kJ/mol) + 1 mol(432 kJ/mol)]

= −116 kJ