Given the following data: $$\begin{array}{lr}2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta G^{\circ}=-6399 \mathrm{~kJ} \\\\\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{array}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$

Start by looking at the three given reactions. The target reaction is: $$6 C(s) + 3 H_2(g) \longrightarrow C_6H_6(l)$$ We need to manipulate the given reactions so that we can add them together to get the target reaction. This may involve reversing and/or scaling the reactions.

Reverse the first reaction in order to obtain \(\Delta G^{\circ}\) for: $$12 CO_2(g) + 6 H_2O(l) \longrightarrow 2 C_6H_6(l) + 15 O_2(g)$$ The new free energy change associated with this reaction is: \(\Delta G^{\circ}_{1\_reversed} = 6399 \ \text{kJ}\) Now, divide both sides of the reversed reaction by 2: $$6 CO_2(g) + 3 H_2O(l) \longrightarrow C_6H_6(l) + 7.5 O_2(g)$$ The new free energy change associated with this reaction is: \(\Delta G^{\circ}_{1\_scaled} = \frac{1}{2} * 6399 \ \text{kJ} = 3199.5 \ \text{kJ}\)

Multiply the second reaction by 6 to have the same number of carbon atoms as in the target reaction: $$6 C(s) + 6 O_2(g) \longrightarrow 6 CO_2(g)$$ The new free energy change associated with this reaction is: \(\Delta G^{\circ}_{2\_scaled} = 6 * (-394) \ \text{kJ} = -2364 \ \text{kJ}\)

Multiply the third reaction by 3 to have the same number of hydrogen atoms as in the target reaction: $$3 H_2(g) + 1.5 O_2(g) \longrightarrow 3 H_2O(l)$$ The new free energy change associated with this reaction is: \(\Delta G^{\circ}_{3\_scaled} = 3 * (-237) \ \text{kJ} = -711 \ \text{kJ}\)

Now, add the reactions from Steps 2, 3, and 4. This will cancel out unwanted terms and result in the target reaction. \( (6 CO_2(g) + 3 H_2O(l) \longrightarrow C_6H_6(l) + 7.5 O_2(g)) \newline(+ 6 C(s) + 6 O_2(g) \longrightarrow 6 CO_2(g)) \newline(+ 3 H_2(g) + 1.5 O_2(g) \longrightarrow 3 H_2O(l))\) The sum of these reactions gives us the target reaction: $$6 C(s) + 3 H_2(g) \longrightarrow C_6H_6(l)$$ Now, sum up the associated free energy changes for each reaction: \(\Delta G^{\circ}_{target} = \Delta G^{\circ}_{1\_scaled} + \Delta G^{\circ}_{2\_scaled} + \Delta G^{\circ}_{3\_scaled} = 3199.5 + (-2364) + (-711) = 124.5 \ \text{kJ}\) Thus, the standard Gibbs free energy change for the formation of benzene, \(C_6H_6(l)\), from its elements is \(\Delta G^{\circ}_{target} = 124.5 \ \text{kJ}\).