Question

From the following bond energies:

H – H bond energy: 431.37 kJ mol⁻¹

C = C bond energy: 606.10 kJ mol⁻¹

C – C bond energy: 336.49 kJ mol⁻¹

C – H bond energy: 410.50 kJ mol⁻¹

Enthalpy for the given reaction will be:

\[\begin{array}{cc}
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{...................}\ce{H}\phantom{...}\ce{H}\phantom{....}\\
\phantom{.}|\phantom{....}|\phantom{....................}|\phantom{....}|\phantom{.....}\\
\ce{C = C + H - H -> H - C - C - H}\\
\phantom{.}|\phantom{....}|\phantom{....................}|\phantom{....}|\phantom{.....}\\
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{...................}\ce{H}\phantom{...}\ce{H}\phantom{....}
\end{array}\]

Options

• 553.0 kJ mol⁻¹

• −120.0 kJ mol⁻¹

• −243.6 kJ mol⁻¹

• 1523.6 kJ mol⁻¹

Answer 1

−120.0 kJ mol⁻¹

Explanation:

Enthalpy of reaction = \[\ce{{B.E.}_{(Reactant)} - {B.E.}_{(Product)}}\]

= \[\ce{[{B.E.}_{(C = C)} + 4{B.E.}_{(C - H)} + {B.E.}_{(H - H)}] - [{B.E.}_{(C - C)} + 6{B.E.}_{(C - H)}]}\]

= [606.1 + ((4 × 410.5) + 431.37)] − [336.49 + (6 × 410.5)]

= −120.0 kJ mol⁻¹