9.5 Free Energy and Equilibrium

Free energy (Gibbs free energy, ΔG° under standard conditions) tells you whether a reaction is thermodynamically favored. If ΔG° < 0 the reaction is favored and products are favored at equilibrium (K > 1); if ΔG° > 0 reactants are favored (K < 1); if ΔG° ≈ 0 K ≈ 1 (CED 9.5.A). The exact ties are ΔG° = −RT ln K and K = e^(−ΔG°/RT). So a large negative ΔG° → large K (products dominate); a large positive ΔG° → tiny K (reactants dominate). For nonstandard conditions use ΔG = ΔG° + RT ln Q to see direction of spontaneous change until Q → K. On the AP exam you’ll use these equations in both multiple-choice and free-response problems to justify spontaneity and relate K, ΔG°, and T (show units with R = 8.314 J·mol−1·K−1). For a focused review check the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr), the whole Unit 9 page (https://library.fiveable.me/ap-chemistry/unit-9), and practice problems (https://library.fiveable.me/practice/ap-chemistry).

A negative ΔG° means the reaction is thermodynamically favored because it tells you products are lower in free energy than reactants under standard conditions, so the system can lower its Gibbs free energy by forming products. Mathematically ΔG° = −RT ln K, so if ΔG° < 0 then ln K > 0 and K > 1—meaning at equilibrium the product concentrations (or pressures) dominate. Physically, spontaneity here means the process is energetically downhill when both enthalpy and entropy contributions (ΔG° = ΔH° − TΔS°) are combined. A negative ΔG° doesn’t say anything about how fast the reaction goes (that’s kinetics), only that product formation is favored thermodynamically. For AP Chem, remember the key links: ΔG° < 0 → K > 1 → products favored; ΔG° ≈ 0 → K ≈ 1; ΔG° > 0 → K < 1 (see Topic 9.5 study guide for more practice) (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr). For more problems, check the AP practice set (https://library.fiveable.me/practice/ap-chemistry).

K and ΔG are two ways of describing the same thermodynamic idea, but they’re different “languages.” K (the equilibrium constant) tells you the ratio of product to reactant concentrations at equilibrium. ΔG° (standard free energy change) tells you the energy “drive” under standard conditions. They’re linked by ΔG° = −RT ln K (so K = e^(−ΔG°/RT)). Quick rules you can use on the exam: if ΔG° < 0 → K > 1 → products favored at equilibrium; if ΔG° > 0 → K < 1 → reactants favored; if ΔG° ≈ 0 → K ≈ 1. That’s because when |ΔG°| ≫ RT, K is very far from 1. Also remember ΔG (nonstandard) = ΔG° + RT ln Q, where Q is the reaction quotient—ΔG tells you the spontaneity right now (not just at standard conditions). For more review and worked examples, check the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and plenty of practice problems (https://library.fiveable.me/practice/ap-chemistry).

Use ΔG° = −RT ln K to connect thermodynamic favorability and equilibrium quickly on the exam. How to use it (step-by-step): 1. Decide what you’re solving for: K or ΔG°. If ΔG° is given and you need K, rearrange: K = e^(−ΔG°/RT). If K is given and you need ΔG°, use ΔG° = −RT ln K. 2. Units: R = 8.314 J·mol−1·K−1, T in K, and ΔG° must be in J·mol−1 (convert kJ → J). 3. Typical T = 298 K unless told otherwise (standard conditions). 4. Interpret the sign/magnitude: ΔG° < 0 → ln K > 0 → K > 1 (products favored). ΔG° ≈ 0 → K ≈ 1. If |ΔG°| >> RT (~2.5 kJ at 298 K), K will be far from 1. 5. On the exam show algebra, unit conversion, numeric substitution, and use your calculator (allowed). For a short refresher and practice, see the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr), the Unit 9 overview (https://library.fiveable.me/ap-chemistry/unit-9), and tons of practice problems (https://library.fiveable.me/practice/ap-chemistry).

K tells you which side is favored at equilibrium because ΔG° and K are linked by ΔG° = −RT ln K (or K = e^(−ΔG°/RT)). If K > 1, ln K is positive so ΔG° < 0—the reaction is thermodynamically favored under standard conditions and products are favored at equilibrium. If K < 1, ln K is negative so ΔG° > 0—reactants are favored at equilibrium. If K ≈ 1, ΔG° ≈ 0 and neither side is strongly favored. Remember “much larger or much smaller than RT” (CED 9.5.A.3) means K will deviate strongly from 1; temperature appears in the RT term so K can change with T. For AP review on this exact topic, see the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Think of ΔG° as a “balance” that tells you which side has lower free energy at equilibrium. Mathematically ΔG° = −RT ln K, so: - If ΔG° < 0, then −RT ln K is negative → ln K > 0 → K > 1. That means at equilibrium the product concentrations are bigger than reactants, so products are favored. - If ΔG° > 0, then ln K < 0 → K < 1, so reactants are favored at equilibrium. Physically: a negative ΔG° means forming products lowers the system’s free energy (spontaneous toward products); a positive ΔG° means forming products would raise free energy, so the reaction favors staying on the reactant side. Remember ΔG (nonstandard) also depends on Q: ΔG = ΔG° + RT ln Q, so the current mixture can shift direction until Q = K. This is an AP-CED core connection (9.5.A: ΔG°, K, T). For more review, see the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

Look at K compared to 1. If K > 1, products are favored at equilibrium under standard conditions (ΔG° < 0); if K < 1, reactants are favored (ΔG° > 0); if K ≈ 1, ΔG° ≈ 0 and neither side is strongly favored. That follows directly from ΔG° = −RT ln K (so K = e^(−ΔG°/RT)). Large K (≫1) means −RT ln K is large negative (thermodynamically favored toward products); very small K (≪1) gives a large positive ΔG°. Keep in mind temperature appears in RT, so K can change with T (van ’t Hoff)—a reaction favorable at one T might not be at another. For AP exam practice and quick reminders on using ΔG° ↔ K, see the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).

Temperature affects free energy two ways: through ΔG°’s link to K (ΔG° = −RT ln K) and through the temperature term in ΔG = ΔH − TΔS. Increasing T multiplies ΔS’s contribution: if ΔS > 0, larger T makes −TΔS more negative so ΔG becomes more negative (reaction more thermodynamically favored); if ΔS < 0, larger T makes ΔG more positive (less favored). Because ΔG° = −RT ln K, changing T changes K (and thus equilibrium position); you can see this more quantitatively with the van ’t Hoff relation. Also remember magnitude matters: compare ΔG° to RT (~2.5 kJ·mol−1 at 298 K) to judge if K is near 1 or far from 1. These are exactly the CED connections you’ll need on the AP: use ΔG°, K, and T to justify favorability (Topic 9.5.A). For a quick topic review, check the Fiveable study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and grab extra practice problems at (https://library.fiveable.me/practice/ap-chemistry).

Start from the fundamental relation for free energy under any conditions: ΔG = ΔG° + RT ln Q. At equilibrium ΔG = 0 and the reaction quotient Q equals the equilibrium constant K, so 0 = ΔG° + RT ln K → ΔG° = −RT ln K. Solve for K by exponentiating both sides: K = e^(−ΔG°/RT). That’s where K = e^(−ΔG°/RT) comes from—it’s just the equilibrium special case of the nonstandard free energy equation. Remember ΔG° is for standard states (1 M, 1 atm); the sign and size of ΔG° tell you whether K > 1 (ΔG° < 0, products favored) or K < 1 (ΔG° > 0, reactants favored), which is exactly the CED Essential Knowledge (9.5.A.1–9.5.A.3). For more review and AP-style practice on this topic, see the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and the practice bank (https://library.fiveable.me/practice/ap-chemistry).

Because ΔG° and K are linked by ΔG° = −RT ln K, when ΔG° is very close to zero the right-hand side is near 0 = −RT ln K, so ln K ≈ 0 and therefore K ≈ e^0 = 1. Intuitively, ΔG° ≈ 0 means there’s almost no free-energy advantage for products or reactants under standard conditions, so neither side is strongly favored at equilibrium—about equal amounts of products and reactants, which is what K ≈ 1 means. Remember the size of RT matters: at 298 K, RT ≈ 2.48 kJ·mol⁻¹, so a ΔG° of a few kJ·mol⁻¹ already makes K deviate noticeably from 1. This connection (LO 9.5.A; ΔG° = −RT ln K) is exactly what you should be able to use on the AP exam. For a quick Topic 9.5 review see the study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Use ΔG° = -RT ln K. At 298 K RT ≈ 2.48 kJ·mol⁻¹, so compare ΔG° to that scale: - If ΔG° ≈ 0 (within a few kJ·mol⁻¹) → K ≈ 1 (neither side strongly favored). - If ΔG° is negative and several times RT (e.g., −5.7 kJ ≈ −2.3·RT) → ln K ≈ 2.3 → K ≈ 10. - If ΔG° = −11.4 kJ (≈ −4.6·RT) → K ≈ 100; −17.1 kJ → K ≈ 1000. So “much less than 1” means ΔG° ≫ +RT (positive, favor reactants); “much greater than 1” means ΔG° ≪ −RT (negative, favor products). Quick rule of thumb: divide |ΔG°| by RT (~2.5 kJ at 298 K)—if the result is > ~3–4, K differs from 1 by orders of magnitude. For more practice and AP-aligned explanations, see the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and the unit overview (https://library.fiveable.me/ap-chemistry/unit-9).

“Standard conditions” for ΔG° means the reactants and products are in their standard states: pure solids or liquids, gases at 1.00 atm, and all solutes at 1.00 M. By convention ΔG° values are usually tabulated at 298 K (25°C), so most AP problems assume T = 298 K unless told otherwise. ΔG° is the change in Gibbs free energy under those standard states—it’s the value that links directly to K via ΔG° = -RT ln K (so ΔG° < 0 means K > 1 under standard conditions). For formation reactions, ΔGf° is the free energy change to make 1 mol of a compound from elements in their standard states. Review this in the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and try practice questions (https://library.fiveable.me/practice/ap-chemistry)—AP free-response and multiple-choice often expect you to assume 1 M/1 atm and 298 K unless the problem states a different temperature.

Good question. We use the natural log (ln) because the relationships come from exponential factors with base e—specifically K = e^(−ΔG°/RT) and ΔG° = −RT ln K (CED 9.5.A.2). In thermodynamics and statistical mechanics, probabilities and Boltzmann factors are naturally expressed as e^(−E/kT), so the equilibrium constant K emerges from sums/products of those exponentials. Using ln keeps units consistent: RT has units of energy, so −RT ln K gives energy (ΔG°). If you used log base 10 you’d need an extra factor (ln K = 2.303 log10 K), which just adds an unnecessary constant. For AP exam work, memorize ΔG° = −RT ln K and how sign/magnitude of ΔG° compares to RT to estimate K (CED 9.5.A.3–4). For a quick review, check the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

Short answer: Yes—but you need to be precise about which ΔG you mean. If ΔG° (standard free energy) is positive, products are not favored under standard conditions (K < 1). However the actual free energy change at the moment of reaction is ΔG = ΔG° + RT ln Q. A reaction with ΔG° > 0 can proceed spontaneously if the reaction quotient Q is small enough that ΔG becomes negative. Also you can make a non-spontaneous step occur in real systems by coupling it to a very favorable process (e.g., ATP hydrolysis) or by doing work on the system—both change the net ΔG for the overall process. Note that kinetics matters: even if ΔG < 0 the reaction may be very slow without a catalyst (activation energy issue). For AP exam terms, remember K = e^(−ΔG°/RT) and ΔG = ΔG° + RT ln Q (Topic 9.5). For extra review and practice problems, see the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and AP practice sets (https://library.fiveable.me/practice/ap-chemistry).

Think of K and ΔG° as two ways to say the same thing: where equilibrium sits and whether products are thermodynamically favored. For FRQs you should do three things clearly and quickly: - Use the key equations: ΔG° = −RT ln K and K = e^(−ΔG°/RT). Plug in R = 8.314×10^(−3) kJ·mol^(−1)·K^(−1) (or 8.314 J) and T in K. Show units. - Interpret sign and size: ΔG° < 0 → K > 1 → products favored; ΔG° ≈ 0 → K ≈ 1; ΔG° > 0 → K < 1 → reactants favored. For estimates, compare |ΔG°| to RT (~2.5 kJ/mol at 298 K) to say if K deviates strongly from 1. - If conditions aren’t standard, use ΔG = ΔG° + RT ln Q (or ΔG = ΔG° + RT ln(Q) rearranged) to judge spontaneity now (ΔG < 0 means reaction proceeds toward products). On the exam, show the math, state the qualitative conclusion, and justify with one sentence linking sign/magnitude to K (CED 9.5.A.1–4). For extra practice, see the Topic 9.5 study guide (https://library.fiveable.me/ap-chemistry/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr) and hundreds of practice problems (https://library.fiveable.me/practice/ap-chemistry).