Using the given reactions, calculate the standard free energy of formation, ΔG°f, for phosphoric acid (H3PO4). Given: 1. P4 (s) + 5 O2(g) → P4O10(s) ΔG° = -2697.0 kJ/mol 2. 2 H2(g) + O2(g) → 2 H2O(g) ΔG° = -457.18 kJ/mol 3. 6 H2O(g) + P4O10(s) → 4 H3PO4(l) ΔG° = ?

• A. -1500.0 kJ/mol
• B. -1000.0 kJ/mol
• C. -2000.0 kJ/mol
• D. -1265.4 kJ/mol

Answer: (D)

Verified step by step guidance

1. Identify the target reaction for which you need to find the standard free energy of formation, ΔG°f, for H3PO4. The target reaction is: 4 H3PO4(l) → 6 H2O(g) + P4O10(s).
2. Use Hess's Law, which states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in. This means you can add the given reactions to find the ΔG° for the target reaction.
3. Write the given reactions and their ΔG° values: (1) P4 (s) + 5 O2(g) → P4O10(s) ΔG° = -2697.0 kJ/mol, (2) 2 H2(g) + O2(g) → 2 H2O(g) ΔG° = -457.18 kJ/mol, (3) 6 H2O(g) + P4O10(s) → 4 H3PO4(l) ΔG° = ?
4. Reverse the third reaction to match the target reaction: 4 H3PO4(l) → 6 H2O(g) + P4O10(s). When reversing a reaction, change the sign of ΔG°. Therefore, ΔG° for the reversed reaction is +1265.4 kJ/mol.
5. Add the ΔG° values of the reactions: ΔG° (target) = ΔG° (1) + ΔG° (2) + ΔG° (reversed 3). Substitute the known values and solve for ΔG° (target).