Some bacteria use light energy to convert carbon dioxide and water to glucose and oxygen: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \quad \Delta G^{\circ}=2870 \mathrm{~kJ}\) at \(25^{\circ} \mathrm{C}\) Other bacteria, those that do not have light available to them, couple the reaction $$ \mathrm{H}_{2} \mathrm{~S}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{S}(s) $$ to the glucose synthesis above. Coupling the two reactions, the overall reaction is \(24 \mathrm{H}_{2} \mathrm{~S}(g)+6 \mathrm{CO}_{2}(g)+6 \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+18 \mathrm{H}_{2} \mathrm{O}(l)+24 \mathrm{~S}(s)\) Show that the reaction is spontaneous at \(25^{\circ} \mathrm{C}\).

To combine the two reactions, we will first multiply the second reaction by some factor so that the number of water molecules in both reactions is equal. Then we'll add the first reaction to the adjusted second reaction to obtain the overall reaction. Looking at the given reactions: Reaction 1: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g)\) Reaction 2: \(\mathrm{H}_{2} \mathrm{S}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{S}(s)\) We can adjust the stoichiometry of Reaction 2: \(24 \times\) (Reaction 2): \(24 \mathrm{H}_{2} \mathrm{S}(g)+12 \mathrm{O}_{2}(g) \longrightarrow 24 \mathrm{H}_{2} \mathrm{O}(l)+24 \mathrm{S}(s)\) Now, we can add Reaction 1 to the adjusted Reaction 2: Reaction 1 \(+\) \(24 \times\) (Reaction 2): \(24 \mathrm{H}_{2} \mathrm{S}(g)+6 \mathrm{CO}_{2}(g)+6 \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+18 \mathrm{H}_{2} \mathrm{O}(l)+24 \mathrm{S}(s)\) This is the overall reaction.

We are given the ΔG value for Reaction 1 at 25ºC: ΔG¹ = 2870 kJ To find the ΔG for \(24 \times\) (Reaction 2), we can use tabulated standard Gibbs free energy values of formation for each compound: ΔG = Σ[nΔG(products)] - Σ[nΔG(reactants)] ΔG² = 24*(ΔG(1 mol of H₂O) + ΔG(1 mol of S)) - 24*(ΔG(1 mol of H₂S) + 0.5*ΔG(1 mol of O₂)) Looking up the values in a table, we have: - ΔG(1 mol of H₂O) = -237.2 kJ/mol - ΔG(1 mol of S) = 0 kJ/mol (since it's a standard state) - ΔG(1 mol of H₂S) = -33.6 kJ/mol - ΔG(1 mol of O₂) = 0 kJ/mol (since it's a standard state) Using these values: ΔG² = 24*(-237.2 + 0 - (-33.6) - 0.5*0) = -4891.2 kJ Now, we can add the ΔG values of reactions to calculate the ΔG for the overall reaction: ΔG_total = ΔG¹ + ΔG² = 2870 - 4891.2 = -2021.2 kJ

Since ΔG_total < 0, the reaction is spontaneous at 25ºC.