Gibbs free energy and thermodynamic favorability — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Definition of Gibbs free energy, standard Gibbs free energy of formation, the Gibbs free energy equation ΔG = ΔH - TΔS, thermodynamic favorability (spontaneity) criteria, relation between ΔG and ΔG°, ΔG = ΔG° + RT lnQ, and the connection between ΔG° and equilibrium constant K.

You should already know: Enthalpy and entropy change calculation from standard values. Basic spontaneity concepts related to entropy. Properties of reaction quotient Q and equilibrium constant K.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Gibbs free energy and thermodynamic favorability?

Gibbs free energy (G, sometimes called Gibbs free enthalpy) is a combined thermodynamic state function that relates enthalpy and entropy to predict whether a process will be thermodynamically favorable (the AP Chemistry term for what many sources call "spontaneous"). Thermodynamic favorability means a process will proceed to form products at given conditions without continuous input of external energy, once initiated. Per the AP Chemistry CED, this topic makes up ~20% of the points for Unit 9 (Applications of Thermodynamics), which itself contributes 7-9% of the total AP exam score. This topic appears in both multiple-choice questions (MCQ) and free-response questions (FRQ), often as multi-part problems connecting thermodynamics to equilibrium, electrochemistry, or biological applications. Unlike entropy alone, which only predicts spontaneity for isolated systems, Gibbs free energy lets us predict favorability for processes at constant temperature and pressure, which describes almost all reactions studied in AP Chemistry. By convention, a negative ΔG corresponds to a thermodynamically favorable process, while a positive ΔG corresponds to an unfavorable process.

2. The Fundamental Gibbs Free Energy Relationship: ΔG = ΔH - TΔS

At constant temperature and pressure (the standard condition for almost all laboratory and natural chemical reactions), the change in Gibbs free energy for any process is given by the fundamental equation: $$\Delta G=\Delta H-T\Delta S$$ where ΔG is the change in Gibbs free energy (units: kJ/mol), ΔH is the enthalpy change (kJ/mol), T is absolute temperature in Kelvin, and ΔS is the entropy change of the system (usually given in J/mol·K, so unit conversion is almost always required here).

The sign of ΔG directly indicates thermodynamic favorability:

• ΔG < 0: Process is thermodynamically favorable (proceeds forward spontaneously)
• ΔG = 0: Process is at equilibrium, no net change occurs
• ΔG > 0: Process is thermodynamically unfavorable (reverse process is favorable)

Because T is multiplied by ΔS, the sign of ΔG (and thus favorability) can change with temperature when ΔH and ΔS have the same sign. If ΔH is negative (exothermic) and ΔS is positive, ΔG is always negative, so the process is favorable at all temperatures. If ΔH is positive (endothermic) and ΔS is negative, ΔG is always positive, so unfavorable at all temperatures. When ΔH and ΔS have matching signs, temperature determines favorability.

Worked Example

For the evaporation of liquid ethanol to ethanol vapor at 1 atm, ΔHvap = 38.6 kJ/mol and ΔSvap = 110 J/mol·K. Is evaporation thermodynamically favorable at (a) 25°C, (b) 100°C?

1. Convert temperatures from Celsius to Kelvin: (a) T = 25 + 273.15 = 298.15 K; (b) T = 100 + 273.15 = 373.15 K.
2. Convert ΔSvap to kJ/mol·K to match ΔHvap units: 110 J/mol·K = 0.110 kJ/mol·K.
3. Calculate ΔG at 25°C: ΔG = 38.6 kJ/mol - (298.15 K)(0.110 kJ/mol·K) = 38.6 - 32.8 = +5.8 kJ/mol.
4. Calculate ΔG at 100°C: ΔG = 38.6 - (373.15)(0.110) = 38.6 - 41.0 = -2.4 kJ/mol.
5. Interpret: At 25°C, ΔG > 0 so evaporation is unfavorable (liquid ethanol is stable); at 100°C, ΔG < 0 so evaporation is favorable.

Exam tip: Always convert ΔS from J/mol·K to kJ/mol·K before plugging into ΔG = ΔH - TΔS. AP exam questions almost always give ΔS in joules and ΔH in kilojoules, so missing this conversion will give you a wrong sign and incorrect answer.

3. Standard Gibbs Free Energy Calculations

Standard Gibbs free energy change (ΔG°) is the change in Gibbs free energy when reactants in their standard states (1 atm pressure, 1 M concentration, pure solid/liquid, 25°C = 298 K by default) are converted to products in their standard states. There are two common methods to calculate ΔG° for a reaction:

1. From standard Gibbs free energy of formation ($\Delta G_f^{\circ }$): $\Delta G_f^{\circ }$ is the ΔG for formation of 1 mole of a compound from its constituent elements in their standard states. By definition, $\Delta G_f^{\circ }=0$ for any element in its standard state. The formula for reaction ΔG° is: $$\Delta G_{\text{rxn}}^{\circ }=\sum n\Delta G_f^{\circ }(\text{products})-\sum m\Delta G_f^{\circ }(\text{reactants})$$ where n and m are stoichiometric coefficients of products and reactants, respectively.

2. From standard enthalpy and entropy of reaction: We can also use the fundamental Gibbs equation for standard state conditions: $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$, where T is 298 K unless another temperature is specified.

Worked Example

Calculate ΔG° for the oxidation of iron (rust formation) at 298 K: $4\text{Fe}(s)+3{\text{O}}_2(g)\to 2{\text{Fe}}_2{\text{O}}_3(s)$. Use the values $\Delta G_f^{\circ }({\text{Fe}}_2{\text{O}}_3(s))=-742.2$ kJ/mol, $\Delta G_f^{\circ }(\text{Fe}(s))=0$ kJ/mol, $\Delta G_f^{\circ }({\text{O}}_2(g))=0$ kJ/mol.

1. Write the ΔG°rxn formula for the balanced reaction: $\Delta G_{\text{rxn}}^{\circ }=2\Delta G_f^{\circ }({\text{Fe}}_2{\text{O}}_3)-[4\Delta G_f^{\circ }(\text{Fe})+3\Delta G_f^{\circ }({\text{O}}_2)]$.
2. Plug in the given values: $\Delta G_{\text{rxn}}^{\circ }=2(-742.2)-[4(0)+3(0)]$.
3. Calculate the result: $\Delta G_{\text{rxn}}^{\circ }=-1484.4$ kJ/mol.
4. Interpret: The negative ΔG° confirms that rust formation is thermodynamically favorable under standard conditions.

Exam tip: Remember the minus sign in the ΔG°rxn formula applies to the entire sum of reactants. If any ΔG°f of reactants is negative, you will subtract a negative, which equals adding that value — always write out signs explicitly to avoid arithmetic errors.

4. ΔG, ΔG°, and the Relationship to Equilibrium

ΔG° describes the Gibbs free energy change only when the reaction is at standard state (all reactants and products are at 1 M/1 atm, so Q = 1). For any non-standard conditions (any concentration or pressure other than standard), we calculate ΔG using: $$\Delta G=\Delta G^{\circ }+RT\ln Q$$ where R = 8.314 J/mol·K (the gas constant in energy units), T is absolute temperature, and Q is the reaction quotient for the current reaction conditions.

When a reaction reaches equilibrium, ΔG = 0 (no net driving force) and Q = K (the equilibrium constant). Substituting these into the equation above gives the key thermodynamic relationship connecting favorability and equilibrium: $$\Delta G^{\circ }=-RT\ln K$$ This relation tells us:

• ΔG° < 0 → ln K > 0 → K > 1: Products are favored at equilibrium
• ΔG° > 0 → ln K < 0 → K < 1: Reactants are favored at equilibrium
• ΔG° = 0 → ln K = 0 → K = 1: Equal amounts of reactants and products at equilibrium

Worked Example

For the dissolution of calcium hydroxide, ${\text{Ca(OH)}}_2(s)\rightleftharpoons {\text{Ca}}^{2+}(aq)+2{\text{OH}}^{-}(aq)$, ΔG° = +39.0 kJ/mol at 25°C. Calculate the solubility product constant Ksp for calcium hydroxide at 25°C.

1. Convert temperature to Kelvin: T = 25 + 273.15 = 298.15 K.
2. Convert ΔG° to J/mol to match R's units: 39.0 kJ/mol = 39000 J/mol.
3. Rearrange the formula to solve for ln K: $\ln K_{\text{sp}}=-\frac{\Delta G^{\circ }}{RT}$.
4. Plug in values: $\ln K_{\text{sp}}=-\frac{39000}{(8.314)(298.15)}\approx -15.75$.
5. Exponentiate to get Ksp: $K_{\text{sp}}=e^{-15.75}\approx 5.6\times 10^{-7}$, which matches the approximate experimental value.

Exam tip: Always use R = 8.314 J/mol·K for Gibbs free energy calculations, not 0.0821 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 (and how to avoid them)

• 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: If ΔG > 0, 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. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

The reaction A(s) + 2 B(g) ⇌ 3 C(g) has ΔH° = +120 kJ/mol and ΔS° = +145 J/mol·K at 298 K. Which of the following statements is true? A) The reaction is thermodynamically favorable at all temperatures B) The reaction is thermodynamically favorable only at temperatures above 828 K C) The reaction is thermodynamically favorable only at temperatures below 828 K D) The reaction is never thermodynamically favorable at any temperature

Worked Solution: When ΔH and ΔS are both positive, ΔG = ΔH - TΔS is negative (favorable) only when TΔS > ΔH, which occurs at high temperatures. To find the boundary temperature where ΔG = 0, rearrange to T = ΔH/ΔS. Convert ΔS to kJ: 145 J/mol·K = 0.145 kJ/mol·K. Then T = 120 kJ/mol / 0.145 kJ/mol·K ≈ 828 K. For temperatures above 828 K, TΔS > ΔH so ΔG < 0, and the reaction is favorable. The correct answer is B.

Question 2 (Free Response)

The decomposition of dinitrogen tetroxide is ${\text{N}}_2{\text{O}}_4(g)\rightleftharpoons 2{\text{NO}}_2(g)$ at 298 K. Use the given data to answer the questions:

(a) Calculate ΔG° for the reaction using standard Gibbs free energy of formation. (b) Calculate the equilibrium constant K for this reaction at 298 K. (c) Is the reaction thermodynamically favorable under standard conditions? Justify your answer.

Worked Solution: (a) Use the formula $\Delta G_{\text{rxn}}^{\circ }=2\Delta G_f^{\circ }({\text{NO}}_2)-\Delta G_f^{\circ }({\text{N}}_2{\text{O}}_4)$. Plugging in values: $\Delta G^{\circ }=2(51.31)-97.89=102.62-97.89=4.73$ kJ/mol.

(b) Use $\Delta G^{\circ }=-RT\ln K$. Convert ΔG° to J/mol: 4.73 kJ/mol = 4730 J/mol. Rearrange for ln K: $\ln K=-\Delta G^{\circ }/(RT)=-4730/(8.314\times 298)\approx -1.91$. Exponentiate: $K=e^{-1.91}\approx 0.15$.

(c) The reaction is not thermodynamically favorable under standard conditions. ΔG° = +4.73 kJ/mol, which is positive, so the forward reaction is thermodynamically unfavorable.

Question 3 (Application / Real-World Style)

Muscle cells use the coupling of ATP hydrolysis to power muscle contraction. The ΔG for ATP hydrolysis is -30.5 kJ/mol under cellular conditions at 37°C. The conformational change that allows muscle contraction has a ΔG of +20.1 kJ/mol. What is the overall ΔG for the coupled process, and is the overall process thermodynamically favorable?

Worked Solution: Gibbs free energy is additive for coupled reactions, so the total ΔG is the sum of the ΔG values of the individual reactions. We have ΔG₁ (ATP hydrolysis) = -30.5 kJ/mol and ΔG₂ (conformational change) = +20.1 kJ/mol. Calculate total ΔG: ΔG_total = -30.5 + 20.1 = -10.4 kJ/mol. Since ΔG_total is negative, the overall coupled process is thermodynamically favorable, allowing muscle contraction to proceed spontaneously in cells.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational link connecting thermodynamics to two core AP Chemistry topics: chemical equilibrium and electrochemistry. Next you will apply the relationship between ΔG° and K to predict how equilibrium constants change with temperature, a common 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. Follow-on topics:

• Thermodynamics of equilibrium
• Electrochemistry and cell potential
• Entropy and entropy change