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

The target reaction for which we have to find the standard Gibbs free energy change is: \( \mathrm{CH}_{4}(g)+2\ \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2\ \mathrm{H}_{2}O(l) \)

We need to find a way to combine the given reactions in a way that will result in the target reaction. After examination, we can see that adding the 1st and 2nd reactions, and then the 3rd reaction in reverse, will resemble our target reaction: (1) \( 2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \longrightarrow \mathrm{CH}_{4}(g) \) (2) \( 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2}O(l) \) (3, reverse) \( \mathrm{CO}_{2}(g) \longrightarrow \mathrm{C}(s)+\mathrm{O}_{2}(g) \) >equals> \( \mathrm{CH}_{4}(g) + 2\ \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + 2\ \mathrm{H}_{2}O(l) \)

To find the standard Gibbs free energy change for the target reaction, we will sum the respective \(\Delta G^{\circ}\) values of the reactions we used in Step 2, considering the appropriate signs: \(\Delta G^{\circ}_{\text{target}} = \Delta G^{\circ}_{1} + \Delta G^{\circ}_{2} - \Delta G^{\circ}_{3}\) (Note that we subtracted \(\Delta G^{\circ}_{3}\) since we used the reverse form of reaction 3) Now, plug in the given values: \(\Delta G^{\circ}_{\text{target}} = (-51\ \mathrm{kJ}) + (-474\ \mathrm{kJ}) - (-394\ \mathrm{kJ})\) \(\Delta G^{\circ}_{\text{target}} = -51\ \mathrm{kJ} - 474\ \mathrm{kJ} + 394\ \mathrm{kJ}\) \(\Delta G^{\circ}_{\text{target}} = -131\ \mathrm{kJ}\) Thus, the standard Gibbs free energy change for the target reaction is \(\Delta G^{\circ} = -131\ \mathrm{kJ}\).