The chemical reaction is the hydrogenation of ethene: $\text{CH}_2=\text{CH}_2 + \text{H}_2 \rightarrow \text{CH}_3-\text{CH}_3$ The enthalpy of reaction ($\Delta_r H$) can be calculated using bond energies: $\Delta_r H = \sum \text{B.E.(reactants)} - \sum \text{B.E.(products)}$ In the reactants, there is one $\text{C}=\text{C}$ bond, four $\text{C}-\text{H}$ bonds, and one $\text{H}-\text{H}$ bond. In the products, there is one $\text{C}-\text{C}$ bond and six $\text{C}-\text{H}$ bonds. $\Delta_r H = [\text{B.E.(C}=\text{C)} + 4 \times \text{B.E.(C}-\text{H)} + \text{B.E.(H}-\text{H)}] - [\text{B.E.(C}-\text{C)} + 6 \times \text{B.E.(C}-\text{H)}]$ This simplifies to: $\Delta_r H = [\text{B.E.(C}=\text{C)} + \text{B.E.(H}-\text{H)}] - [\text{B.E.(C}-\text{C)} + 2 \times \text{B.E.(C}-\text{H)}]$ Substituting the given values: $\Delta_r H = [606.10 + 431.37] - [336.49 + 2(410.50)]$ $\Delta_r H = 1037.47 - [336.49 + 821.00]$ $\Delta_r H = 1037.47 - 1157.49 = -120.02 \text{ kJ mol}^{-1}$ Therefore, the enthalpy of the reaction is approximately $-120.0 \text{ kJ mol}^{-1}$.