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

We are given three reactions with their respective standard Gibbs free energy changes: 1. \(2C_6H_6(l) + 15O_2(g) \rightarrow 12CO_2(g) + 6H_2O(l) \quad \Delta G^{\circ} = -6399 \; kJ\) 2. \(C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta G^{\circ} = -394 \; kJ\) 3. \(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta G^{\circ} = -237 \; kJ\)

Let's consider the given equations and manipulate them so that we can eventually arrive at our desired reaction. Firstly, multiply the first reaction by \(-\frac{1}{2}\), second reaction by -12, and third reaction by -6. This allows us to cancel out the unwanted terms in the reactions: 1. \(-\frac{1}{2} \; \rightarrow C_6H_6(l) - \frac{15}{2}O_2(g) + 6CO_2(g) + 3H_2O(l) \quad \Delta G^{\circ} = \frac{1}{2}(6399 \; kJ)\) 2. \(-12 \; \rightarrow 12C(s) → 12CO_2(g) + 12O_2(g) \quad \Delta G^{\circ} = 12 (394 \; kJ) \) 3. \(-6 \; \rightarrow 3H_2(g) → 6H_2O(l) + 3O_2(g) \quad \Delta G^{\circ} = 6(237 \; kJ)\) Now, sum up the manipulated reactions: \(6C(s) + 3H_2(g) + {\frac{15}{2}}O_2(g) → C_6H_6(l) + 6CO_2(g) + 3H_2O(l) + 12CO_2(g) + 12O_2(g) + 6H_2O(l) + 3O_2(g)\) Now, we can cancel out some terms on the left and right side of the equation: \({\frac{15}{2}}O_2(g) + 12O_2(g) + 3O_2(g) \) will cancel out, and \(6CO_2(g) + 12CO_2(g)\) will cancel out.

Now that we have obtained the desired reaction, we can sum up the ΔG° values of the modified reactions to find the ΔG° for the desired reaction: \(\Delta G^{\circ}_{desired} = \frac{1}{2}(6399 \; kJ) + 12(394 \; kJ) + 6(237 \; kJ) = 3199.5 + 4728 + 1422 = 9349.5 \; kJ\) The standard Gibbs free energy change for the reaction \(6C(s) + 3H_2(g) → C_6H_6(l)\) is \(9349.5 \; kJ\).