Gibbs Free Energy and Thermodynamic Favorability
AP Chemistry· AP Chemistry CED — Applications of Thermodynamics· 14 min read
1. Core Concepts: Gibbs Free Energy and Favorability★★☆☆☆⏱ 3 min
Gibbs free energy () is a combined thermodynamic state function that links enthalpy () and entropy () to predict whether a process will be thermodynamically favorable for reactions at constant temperature and pressure — the condition that describes almost all reactions studied in AP Chemistry.
Thermodynamically Favorable Process
A process that will proceed to form products at given conditions without continuous input of external energy, once initiated. This is the AP Chemistry term for what many sources call a "spontaneous" process.
By universal convention: a negative corresponds to a thermodynamically favorable process, while a positive corresponds to an unfavorable process. When , the process is at equilibrium with no net change.
Test your basic understanding:
If a process has kJ/mol, what does this mean?
The process will occur very quickly
The process is thermodynamically favorable
The process is at equilibrium
The process will never occur
Reveal answer
1 —Correct! sign only indicates thermodynamic favorability, not reaction rate.
2. The Fundamental Equation: $\Delta G = \Delta H - T\Delta S$★★★☆☆⏱ 4 min
At constant temperature and pressure, the change in Gibbs free energy for any process is given by the equation:
Where = change in Gibbs free energy (kJ/mol), = enthalpy change (kJ/mol), = absolute temperature in Kelvin, and = entropy change of the system.
If (exothermic) and : at all temperatures (always favorable)
If (endothermic) and : at all temperatures (always unfavorable)
If and have matching signs: temperature determines the sign of and thus favorability
For the evaporation of liquid ethanol to ethanol vapor at 1 atm, kJ/mol and J/mol·K. Is evaporation thermodynamically favorable at (a) 25°C, (b) 100°C?
- 1
Convert temperatures from Celsius to Kelvin:
- 2
Convert to kJ/mol·K to match units of :
- 3
Calculate at 25°C:
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Calculate at 100°C:
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Interpret: At 25°C, so evaporation is unfavorable (liquid ethanol is stable); at 100°C, so evaporation is favorable.
Exam tip:
Always convert from J/mol·K to kJ/mol·K before plugging into the equation. AP exam questions almost always give in joules and in kilojoules, so missing this conversion will give you a wrong sign and incorrect answer.
3. Standard Gibbs Free Energy Calculations★★★☆☆⏱ 3 min
Standard Gibbs free energy change () is the change in Gibbs free energy when reactants in their standard states (1 atm pressure, 1 M concentration, pure solid/liquid, 298 K by default) are converted to products in their standard states. There are two common methods to calculate :
Standard Gibbs Free Energy of Formation
The for formation of 1 mole of a compound from its constituent elements in their standard states. By definition, for any element in its standard state.
When using standard free energies of formation, the reaction is calculated as:
Calculate for the oxidation of iron (rust formation) at 298 K: . Use the values: kJ/mol, kJ/mol, kJ/mol.
- 1
Write the formula for the balanced reaction:
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Substitute the given values:
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Calculate the result:
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Interpret: The negative confirms that rust formation is thermodynamically favorable under standard conditions.
Exam tip:
Remember the minus sign in the formula applies to the entire sum of reactants. If any of reactants is negative, you will subtract a negative which equals adding that value — always write out signs explicitly to avoid arithmetic errors.
4. $\Delta G$, $\Delta G^\circ$, and the Relationship to Equilibrium★★★★☆⏱ 4 min
only describes the Gibbs free energy change when the reaction is at standard state (all reactants and products at 1 M/1 atm, so ). For any non-standard conditions, we calculate using:
When a reaction reaches equilibrium, (no net driving force) and (the equilibrium constant). Substituting these values gives the key relationship connecting thermodynamics and equilibrium:
: Products are favored at equilibrium
: Reactants are favored at equilibrium
: Equal amounts of reactants and products at equilibrium
For the dissolution of calcium hydroxide, , kJ/mol at 25°C. Calculate the solubility product constant for calcium hydroxide at 25°C.
- 1
Convert temperature to Kelvin:
- 2
Convert to J/mol to match units of :
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Rearrange to solve for :
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Substitute values ( J/mol·K):
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Exponentiate to get :
Exam tip:
Always use J/mol·K for Gibbs free energy calculations, not L·atm/mol·K (the gas constant used for ideal gas law problems). Using the wrong R will give you a K that is orders of magnitude off.
5. Common Pitfalls
Wrong move:
Forgetting to convert ΔS from J/mol·K to kJ/mol·K in ΔG = ΔH - TΔS, leading to a ΔG with the wrong sign
Why:
AP questions almost always give ΔH in kJ and ΔS in J, so students plug in numbers without checking units
Correct move:
Always check units before plugging in; if they differ, convert ΔS to kJ to match ΔH
Wrong move:
Claiming a reaction with ΔG > 0 will never occur at any observable rate
Why:
Students confuse thermodynamic favorability with kinetic feasibility. ΔG only describes favorability, not how fast the reaction proceeds
Correct move:
State the forward reaction is thermodynamically unfavorable, and note that this tells you nothing about the rate of the reaction
Wrong move:
Claiming ΔG° = 0 at equilibrium
Why:
Students mix up the meaning of ΔG (any conditions) and ΔG° (only standard state)
Correct move:
Remember ΔG is always 0 at equilibrium; ΔG° is only 0 at equilibrium when K = 1
Wrong move:
Treating any process with a positive ΔS of the system as always favorable
Why:
Students forget the second law refers to the entropy of the universe, not just the system
Correct move:
Always use the sign of ΔG, not just ΔS of the system, to determine thermodynamic favorability
Wrong move:
Using R = 0.0821 L·atm/mol·K when calculating K from ΔG°
Why:
Students remember R from gas law problems and use it by mistake
Correct move:
Always reach for R = 8.314 J/mol·K for all Gibbs free energy calculations
Wrong move:
Subtracting a negative ΔG°f value incorrectly, getting a positive ΔG° when it should be negative
Why:
Students forget the formula is products minus reactants, so a negative reactant ΔG°f becomes a positive term
Correct move:
Write all negative signs explicitly before plugging in numbers, e.g., ΔG° = products - (-50) = products + 50
6. Quick Reference Cheatsheet
Category | Formula | Key Notes |
|---|---|---|
Fundamental Gibbs free energy | \Delta G = \Delta H - T\Delta S | Convert ΔS to kJ/mol·K to match ΔH units |
Favorability rule | ΔG < 0 = Favorable; ΔG = 0 = Equilibrium; ΔG > 0 = Unfavorable | Applies to all constant T,P processes |
ΔG° from standard formation | \Delta G^\circ_{\text{rxn}} = \sum n\Delta G^\circ_f(\text{products}) - \sum m\Delta G^\circ_f(\text{reactants}) | ΔG°f = 0 for elements in standard state |
ΔG for non-standard conditions | \Delta G = \Delta G^\circ + RT \ln Q | R = 8.314 J/mol·K, convert ΔG° to J |
ΔG° and equilibrium constant | \Delta G^\circ = -RT \ln K | ΔG° < 0 → K > 1; ΔG° > 0 → K < 1 |
Boundary temperature for favorability | T = \frac{\Delta H}{\Delta S} (at ΔG = 0) | ΔH-,ΔS- → favorable below T; ΔH+,ΔS+ → favorable above T |
Coupled reactions | \Delta G_{\text{total}} = \sum \Delta G_{\text{individual}} | Gibbs free energy adds for sequential reactions |
When this came up on past exams
AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.
- 2023 · FRQ
ΔG calculation from ΔH and ΔS
- 2022 · MCQ
Relate ΔG° to equilibrium K
- 2021 · FRQ
ΔG° from formation values
Going deeper
What's Next
This topic forms the foundational link connecting thermodynamics to two core AP Chemistry topics: chemical equilibrium and electrochemistry, which are heavily tested on both MCQ and FRQ sections. Next, you will apply the relationship between ΔG° and K to predict how equilibrium constants change with temperature, a common multi-part FRQ skill. You will also connect ΔG to cell potential in electrochemistry, using the relation ΔG = -nFE to convert between cell voltage and Gibbs free energy change. Without mastering sign rules and unit conversions for Gibbs free energy, both of these topics will be far more difficult to solve correctly on the exam. This topic also completes the framework of thermodynamics started in Unit 6, giving a complete picture of energy and favorability for chemical processes.
