Given the following data: $$ \begin{aligned} 2 \mathrm{H}_{2}(g)+\mathrm{C}(s) & \longrightarrow \mathrm{CH}_{4}(g) & & \Delta G^{\circ}=-51 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & \Delta G^{\circ}=-474 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g) & & \Delta G^{\circ}=-394 \mathrm{~kJ} \end{aligned} $$ Calculate \(\Delta G^{\circ}\) for \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(l)\).

Focus on each species in the final reaction and identify how the given reactions need to be manipulated to obtain them. For CH4(g), this can be done by using the first provided reaction as is. For 2 O2(g), use the second provided reaction and divide it by 2. For CO2(g), utilize the third provided reaction as it is. For 2 H2O(l), refer to the second provided reaction and divide it by 2.

Now that we have manipulated the given reactions to feature the desired species and stoichiometry, write them down with their \(\Delta G^{\circ}\) values: 1. \(2 \mathrm{H}_{2}(g) + \mathrm{C}(s) \rightarrow \mathrm{CH}_{4}(g)\ \ \ \Delta G^{\circ}_{1} = -51\ \mathrm{kJ}\) 2. \(\mathrm{H}_{2}(g) + \dfrac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2}\mathrm{O}(l)\ \ \ \Delta G^{\circ}_{2} = -237\ \mathrm{kJ}\) 3. \(\mathrm{C}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)\ \ \ \Delta G^{\circ}_{3} = -394\ \mathrm{kJ}\) We divided the second provided reaction by 2 to get the value for the required one mole of H2O(l). In doing so, we should also divide its original \(\Delta G^{\circ}\) value by 2.

Combine the manipulated reactions from Step 2 to generate the desired equation: (1) + (2) + (2) + (3): \(\mathrm{CH}_{4}(g) + 2\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\)

By summing up the respective \(\Delta G^{\circ}\) values for the manipulated reactions, we obtain the \(\Delta G^{\circ}\) value for the desired reaction: \(\Delta G^{\circ}_{final} = \Delta G^{\circ}_{1} + 2 \Delta G^{\circ}_{2} + \Delta G^{\circ}_{3} \) \(\Delta G^{\circ}_{final} = (-51\ \mathrm{kJ}) + 2(-237\ \mathrm{kJ}) + (-394\ \mathrm{kJ})\) \(\Delta G^{\circ}_{final} = -51\ \mathrm{kJ} - 474\ \mathrm{kJ} - 394\ \mathrm{kJ}\) \(\Delta G^{\circ}_{final} = -919\ \mathrm{kJ}\) Therefore, the standard Gibbs free energy change for the reaction \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(l)\) is \(\Delta G^{\circ} = -919\ \mathrm{kJ}\).