To calculate the enthalpy change ($\Delta H$) for the reaction $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ using bond energies, we use the formula: $$ \Delta H = \sum (\text{bond energies of bonds broken}) - \sum (\text{bond energies of bonds formed}) $$ Let's identify the bonds broken in the reactants and bonds formed in the products. Reactants: $\text{C}$ (graphite): The energy given for C(graphite) (717 kJ/mol) represents the energy required to atomize one mole of solid carbon into gaseous carbon atoms. This is an energy input, similar to breaking bonds. $\text{O}_2$: This molecule contains one $\text{O}=\text{O}$ double bond. The energy required to break this bond is 498 kJ/mol. Products: $\text{CO}_2$: This molecule has a linear structure with two $\text{C}=\text{O}$ double bonds. The energy released when these bonds are formed is $2 \times 805 \text{ kJ/mol}$. Step 1: Calculate the total energy required to break bonds (reactants). Energy to atomize C(graphite) = $717 \text{ kJ/mol}$ Energy to break 1 mole of $\text{O}=\text{O}$ bonds = $498 \text{ kJ/mol}$ Total energy for bond breaking = $717 \text{ kJ/mol} + 498 \text{ kJ/mol} = 1215 \text{ kJ/mol}$ Step 2: Calculate the total energy released when bonds are formed (products). Energy released from forming 2 moles of $\text{C}=\text{O}$ bonds = $2 \times 805 \text{ kJ/mol} = 1610 \text{ kJ/mol}$ Step 3: Calculate the enthalpy change ($\Delta H$). $$ \Delta H = (\text{Energy for bond breaking}) - (\text{Energy for bond formation}) $$ $$ \Delta H = 1215 \text{ kJ/mol} - 1610 \text{ kJ/mol} $$ $$ \Delta H = -395 \text{ kJ/mol} $$ Comparing this value to the given options, -394 kJ/mol is the closest. The final answer is $\boxed{\text{A. -394 kJ/mol}}$.