9.3 Gibbs Free Energy and Thermodynamic Favorability

Gibbs free energy (ΔG°) tells you whether a process is thermodynamically favored under standard conditions (1.0 M, 1.0 atm, pure solids/liquids). Use ΔG° = ΔH° − TΔS° or ΔG°reaction = ΣΔG°f(products) − ΣΔG°f(reactants) to calculate it. If ΔG° < 0 the process is thermodynamically favored (often called “spontaneous” in older texts); if ΔG° > 0 it’s not favored. You need it in AP Chem because it links enthalpy and entropy to predict favorability, shows temperature dependence (use the sign table from the CED), and connects to equilibrium via ΔG° = −RT ln K (so ΔG° predicts K and whether products or reactants dominate). On the exam you’ll be asked to calculate ΔG°, pick signs, or justify favorability using ΔH° and ΔS° (Topic 9.3.A). For a focused review and practice problems tied to the CED, see the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and the Unit 9 page (https://library.fiveable.me/ap-chemistry/unit-9).

Look at the sign (and value) of ΔG°: if ΔG° < 0 the process is thermodynamically favored, if ΔG° > 0 it is not favored, and if ΔG° = 0 the system is at equilibrium (CED 9.3.A.2). If you’re given ΔH° and ΔS° instead, use ΔG° = ΔH° − TΔS° to decide—the signs of ΔH° and ΔS° tell you temperature dependence (e.g., ΔH° < 0 and ΔS° > 0 → favored at all T; ΔH° > 0 and ΔS° < 0 → never favored) (CED 9.3.A.5–6). You can also get ΔG° from formation values: ΔG°rxn = ΣΔG°f(products) − ΣΔG°f(reactants) (CED 9.3.A.3). Finally, ΔG° relates to equilibrium via ΔG° = −RT ln K, so a negative ΔG° means K > 1 (products favored)—a common AP exam link (see Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and practice problems (https://library.fiveable.me/practice/ap-chemistry)).

Short answer: on the AP, they mean the same thermodynamic sign but teachers prefer “thermodynamically favored.” CED says a process is thermodynamically favored when ΔG° < 0. “Spontaneous” historically meant the same, but it’s misleading because it suggests “quick” or “without cause.” Thermodynamically favored just describes direction (ΔG° < 0), not rate—a reaction can be favored but very slow (kinetically inhibited). Also remember ΔG° = ΔH° − TΔS° (so sign can change with T), ΔG° = 0 at equilibrium, and ΔG° = −RT ln K links favorability to K. For AP prep, focus on using ΔG° sign rules and the ΔH°/ΔS° table in the CED; review examples and practice problems in the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and more practice at (https://library.fiveable.me/practice/ap-chemistry).

ΔG < 0 means a process is thermodynamically favored because it tells you the reaction lowers the system’s free energy—energy available to do useful work—under the given conditions. Mathematically ΔG = ΔH − TΔS combines enthalpy and entropy; when ΔG is negative the balance of heat and disorder changes favors product formation. Also ΔG° = −RT ln K links free energy to equilibrium: if ΔG° < 0 then ln K > 0 and K > 1, so products are favored at equilibrium. The AP CED prefers the phrase “thermodynamically favored” instead of “spontaneous” to avoid implying speed; a negative ΔG says nothing about rate (kinetics). If you want to practice these ideas and how signs of ΔH° and ΔS° predict favorability at different T, check the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

Use ΔG = ΔH − TΔS when you know the enthalpy change (ΔH) and entropy change (ΔS) for a process at a specific temperature and want the Gibbs free energy change to decide thermodynamic favorability. Plug T in kelvin (K). If ΔG < 0 the process is thermodynamically favored (CED language prefers “thermodynamically favored” over “spontaneous”); if ΔG > 0 it’s unfavored; ΔG = 0 at equilibrium. For standard-state quantities use ΔG° = ΔH° − TΔS° (reactants/products at 1.0 M, 1.0 atm) or get ΔG° from ΔG°f values or ΔG° = −RT ln K when relevant. You don’t need the calculation when signs give the answer (ΔH < 0 and ΔS > 0 → favored at all T; ΔH > 0 and ΔS < 0 → never favored). This equation is commonly tested on AP free-response and multiple-choice items in Topic 9.3—review the CED criteria and examples in the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO). For extra practice, try problems at (https://library.fiveable.me/practice/ap-chemistry).

Use the standard-formation values (ΔG°f) and the stoichiometry. For any reaction: ΔG°reaction = Σ [ν · ΔG°f (products)] − Σ [ν · ΔG°f (reactants)] Steps: - Write the balanced equation and note each species’ stoichiometric coefficient ν. - Plug in standard Gibbs free energies of formation (ΔG°f) for each substance (states = standard: pure solids/liquids, 1.0 M solutions, or 1.0 atm gases). - Multiply each ΔG°f by its ν, sum products, subtract the sum for reactants. - Keep units consistent (kJ or J). If you need ΔG° in J, convert kJ → J before adding. Interpretation: - If ΔG°reaction < 0 the reaction is thermodynamically favored (spontaneous) under standard conditions; if > 0 it’s not favored. - You can also relate ΔG° to equilibrium: ΔG° = −RT ln K (use when asked on the AP exam). For an AP-aligned refresher and worked examples, see the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO). For broader Unit 9 review and practice questions, check the unit page (https://library.fiveable.me/ap-chemistry/unit-9) and practice bank (https://library.fiveable.me/practice/ap-chemistry).

ΔH < 0 means the process releases heat (exothermic); ΔS > 0 means disorder increases. Plugging into ΔG = ΔH − TΔS gives a negative number at all temperatures because you’re subtracting a positive TΔS from an already negative ΔH. So ΔG < 0 and the process is thermodynamically favored (the CED explicitly states no ΔG calculation is needed in this case). On the AP exam, recognize this sign pattern quickly: exothermic + increasing entropy → favored at all T (CED 9.3.A.5–6). Also remember ΔG relates to equilibrium (ΔG° = −RT ln K), so a negative ΔG° corresponds to K > 1 (products favored). For a quick review and practice, see the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and try problems at (https://library.fiveable.me/practice/ap-chemistry).

Yes—many reactions are thermodynamically favored at some temperatures but not others. Use ΔG° = ΔH° − TΔS°: the sign of ΔG° depends on temperature when ΔH° and ΔS° have the same sign. If ΔH° > 0 and ΔS° > 0, the reaction becomes favored at high T (TΔS° can outweigh ΔH°). If ΔH° < 0 and ΔS° < 0, it’s favored at low T. If ΔH° < 0 and ΔS° > 0, it’s favored at all T; if ΔH° > 0 and ΔS° < 0, it’s unfavored at all T (CED 9.3.A.5–6). Classic AP examples: freezing of water (favored at low T) and dissolution of NaNO3 (temperature-sensitive). On the exam you may be asked to calculate ΔG° from ΔH° and ΔS° or predict favored conditions from signs (use ΔG° = −RT ln K to link to equilibrium). For a quick review, see the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Freezing water is a classic case where you need both enthalpy and entropy because the two effects compete. When liquid water freezes, ΔH° < 0 (heat is released—freezing is exothermic) but ΔS° < 0 (the system becomes more ordered, so entropy decreases). Which effect wins depends on temperature via ΔG° = ΔH° − TΔS°. At high T the −TΔS° term (negative times negative → positive) can make ΔG° positive (unfavored), while at low T the negative ΔH° dominates and ΔG° becomes negative (favored). For H2O, ΔHfus ≈ 6.01 kJ·mol−1 and ΔSfus ≈ 22.0 J·mol−1·K−1, so ΔG° = 0 at T ≈ ΔH/ΔS ≈ 273 K (0 °C)—above that ice melts, below that freezing is thermodynamically favored. This is exactly what the CED wants you to use: evaluate ΔH° and ΔS° and apply ΔG° = ΔH° − TΔS° (Topic 9.3). For a quick review and practice problems on this idea, see the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and the unit page (https://library.fiveable.me/ap-chemistry/unit-9) or try practice questions (https://library.fiveable.me/practice/ap-chemistry).

Think of ΔG = ΔH − TΔS and make a 2×2 quadrant in your head (or draw it): ΔH on vertical, ΔS on horizontal. Then memorize these four rules: - ΔH < 0, ΔS > 0 → always favored (ΔG always < 0). Both terms help. - ΔH > 0, ΔS < 0 → never favored (ΔG always > 0). Both hurt. - ΔH > 0, ΔS > 0 → favored at high T (TΔS can overcome +ΔH). - ΔH < 0, ΔS < 0 → favored at low T (−TΔS makes ΔG less negative at high T). Memory tricks: call the extremes “both-help” (favored always) and “both-hurt” (never). For the middle cases, think “entropy winner at high T” (if ΔS positive) or “enthalpy winner at low T” (if ΔH negative). If you need the crossover temperature use T = ΔH/ΔS (watch signs) to decide high vs low T. For AP prep, know that when ΔH < 0 & ΔS > 0 or ΔH > 0 & ΔS < 0 you don’t need a ΔG calculation (CED 9.3.A.4–5). For a quick review, check the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and drill problems at the unit page (https://library.fiveable.me/ap-chemistry/unit-9) or practice question bank (https://library.fiveable.me/practice/ap-chemistry).

ΔG° (standard Gibbs free energy change) is the free-energy change for a reaction when all reactants and products are in their standard states (pure substances, 1.0 M for solutions, 1.0 atm for gases). It’s the value you use to judge thermodynamic favorability under standard conditions and to get K via ΔG° = −RT ln K (CED 9.3.A.1–3, 9.3.A.5). ΔG is the actual Gibbs free energy change under whatever concentrations/pressures you have right now. They’re related by: ΔG = ΔG° + RT ln Q, where Q is the reaction quotient. So even if ΔG° < 0, ΔG can be positive if Q makes RT ln Q big enough (and vice versa). When Q = K, ΔG = 0 (equilibrium). For AP review, see the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and more unit resources (https://library.fiveable.me/ap-chemistry/unit-9) or practice problems (https://library.fiveable.me/practice/ap-chemistry).

No—a positive ΔG (or ΔG° > 0) means the process is thermodynamically unfavored under the given conditions, not that it “can never” happen. AP wording prefers “thermodynamically favored” instead of “spontaneous.” If ΔG° > 0, the reaction tends to lie toward reactants at standard state (ΔG° = −RT ln K gives a small K), but you can still get product formation by: - changing conditions (concentrations, pressure, temperature) so the reaction quotient Q makes ΔG = ΔG° + RT ln Q negative, - coupling the reaction to a strongly favorable one (e.g., ATP hydrolysis), or - overcoming a kinetic barrier: even if ΔG < 0, a large activation energy can make the reaction very slow. Remember ΔG = ΔH − TΔS tells when temperature can flip favorability (use the ΔH/ΔS sign table in the CED). For more AP-aligned review, see the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Temperature enters directly into ΔG° = ΔH° − TΔS°, so whether a process is thermodynamically favored (ΔG° < 0) can change with T. Use the signs of ΔH° and ΔS°: - ΔH° < 0 and ΔS° > 0: ΔG° is negative at all T (favored always). - ΔH° > 0 and ΔS° < 0: ΔG° is positive at all T (never favored). - ΔH° > 0 and ΔS° > 0: TΔS° can overcome ΔH° at high T → favored at high T. - ΔH° < 0 and ΔS° < 0: negative ΔH° dominates at low T → favored at low T. So to tell how T affects favorability, identify ΔH° and ΔS° (or use ΔG° vs T). For equilibrium connections, remember ΔG° = −RT ln K, so raising T can shift K by changing ΔG°. For AP-style questions, you’ll be asked to predict favored/unfavored from ΔH° and ΔS° or calculate ΔG° at a given T—practice both (see the Topic 9.3 study guide) (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO). For more unit review or practice problems, check the unit page (https://library.fiveable.me/ap-chemistry/unit-9) and the AP practice bank (https://library.fiveable.me/practice/ap-chemistry).

“Spontaneous” can sound like “happens quickly or by itself,” which confuses students. AP prefers “thermodynamically favored” because it ties the idea to ΔG° and equilibrium, not to how fast a process occurs. Thermodynamically favored means ΔG° < 0 (so the products are more stable under those standard conditions); it doesn’t say anything about rate. A classic example: rusting of iron is thermodynamically favored (ΔG° < 0) but can be very slow because of a high activation energy—that’s kinetics, not thermodynamics. On the exam you should state ΔG° = ΔH° − TΔS° (or ΔG° = −RT ln K) when you justify favorability, and avoid implying “spontaneous = fast.” For a quick refresher on how AP frames this, see the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO) and the Unit 9 overview (https://library.fiveable.me/ap-chemistry/unit-9). For extra practice, use the AP problem set collection (https://library.fiveable.me/practice/ap-chemistry).

Step-by-step using the formation (ΔGf°) method: 1. Write the balanced chemical equation for the reaction (standard states: pure solids/liquids, 1.0 M solutions, 1.0 atm gases). 2. Look up standard Gibbs free energies of formation, ΔGf° (kJ·mol⁻¹), for every product and reactant (use tables in your reference or the Topic 9.3 study guide (https://library.fiveable.me/ap-chemistry/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO)). 3. Multiply each ΔGf° by its stoichiometric coefficient. 4. Sum the results for products and for reactants. 5. Calculate ΔG°reaction = ΣΔGf°(products) − ΣΔGf°(reactants). Keep units consistent (kJ or J). 6. Interpret the sign: if ΔG° < 0 the process is thermodynamically favored (CED 9.3.A.2); if ΔG° > 0 it’s unfavored; if ΔG° ≈ 0 it’s at equilibrium. 7. (Optional) Connect to temperature or K: use ΔG° = ΔH° − TΔS° or ΔG° = −RT ln K when needed (CED 9.3.A.5, 9.3.A.3). For extra practice and AP-style questions, try the Unit 9 practice set (https://library.fiveable.me/practice/ap-chemistry) and the unit overview (https://library.fiveable.me/ap-chemistry/unit-9).