Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

19. Chemical Thermodynamics

Gibbs Free Energy Calculations

General Chemistry

19. Chemical Thermodynamics

Gibbs Free Energy Calculations

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Multiple Choice

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° = ?

-1500.0 kJ/mol

-1000.0 kJ/mol

-2000.0 kJ/mol

-1265.4 kJ/mol

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Verified step by step guidance

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).

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.

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° = ?

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.

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).