It is interesting to compare the "fuel value" of a hydrocarbon in a hypothetical world where oxygen is not the combustion agent. The enthalpy of formation of \(\mathrm{CF}_{4}(g)\) is \(-679.9 \mathrm{~kJ} / \mathrm{mol}\). Which of the following two reactions is the more exothermic? $$ \begin{aligned} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{CH}_{4}(g)+4 \mathrm{~F}_{2}(g) & \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) \end{aligned} $$

We are given the enthalpy of formation of \(\mathrm{CF}_{4}(g)\) as -679.9 kJ/mol. We need to find the enthalpy change of both reactions. To do this, we need standard enthalpy of formation values for the other substances involved in each reaction. Remember that the standard enthalpy of formation for an element in its natural state is zero. We should be able to find the standard enthalpy of formation values for the other substances in a textbook or online. Here are the \(\Delta H_{\text{f}}^\circ\) for the required substances: \(\Delta H_{\text{f}}^\circ(\mathrm{CH}_4) = -74.9 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{f}}^\circ(\mathrm{O}_2) = 0 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{f}}^\circ(\mathrm{CO}_2) = -393.5 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{f}}^\circ(\mathrm{H}_2\mathrm{O}) = -285.8 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{f}}^\circ(\mathrm{F}_2) = 0 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{f}}^\circ(\mathrm{HF}) = -273.3 \, \mathrm{kJ/mol}\)

We can calculate the enthalpy change of the first reaction using the formula: \(\Delta H_{\text{rxn}} = \Sigma n \Delta H_{\text{f}}^{\circ}(\text{products}) - \Sigma m \Delta H_{\text{f}}^\circ(\text{reactants})\) For the first reaction: \(\Delta H_{\text{rxn}1} = [1 \times \Delta H_{\text{f}}^\circ(\mathrm{CO}_2) + 2 \times \Delta H_{\text{f}}^\circ(\mathrm{H}_2\mathrm{O})] - [1 \times \Delta H_{\text{f}}^\circ(\mathrm{CH}_4) + 2 \times \Delta H_{\text{f}}^\circ(\mathrm{O}_2)]\) By substituting the values obtained in Step 1, we get: \(\Delta H_{\text{rxn}1} = [1 \times (-393.5) + 2 \times (-285.8)] - [1 \times (-74.9) + 2 \times (0)] = -802.3 \, \mathrm{kJ/mol}\)

Now, let's calculate the enthalpy change of the second reaction using the same formula used above. For the second reaction: \(\Delta H_{\text{rxn}2} = [1 \times \Delta H_{\text{f}}^\circ(\mathrm{CF}_4) + 4 \times \Delta H_{\text{f}}^\circ(\mathrm{HF})] - [1 \times \Delta H_{\text{f}}^\circ(\mathrm{CH}_4) + 4 \times \Delta H_{\text{f}}^\circ(\mathrm{F}_2)]\) By substituting the values, we get: \(\Delta H_{\text{rxn}2} = [1 \times (-679.9) + 4 \times (-273.3)] - [1 \times (-74.9) + 4 \times (0)] = -1824.3 \, \mathrm{kJ/mol}\)

Now that we have both enthalpy changes, let's compare them to find which reaction is more exothermic. \(\Delta H_{\text{rxn}1} = -802.3 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{rxn}2} = -1824.3 \, \mathrm{kJ/mol}\) Comparing them, it is clear that \(\Delta H_{\text{rxn}2} < \Delta H_{\text{rxn}1}\), which means that the second reaction is more exothermic \((\)gaining more heat\()\) than the first one.