7.1 Introduction to Equilibrium

Chemical equilibrium is the state reached in a reversible process when the forward and reverse reactions continue to occur but at equal rates, so you see no net change and the concentrations (or partial pressures) of all species stay constant (dynamic equilibrium—CED 7.1.A.2–7.1.A.3). It can happen for phase changes (evaporation/condensation), dissolution/precipitation, acid–base proton transfer, or redox electron transfer (CED 7.1.A.1). Mechanically, as the forward reaction builds product the reverse reaction speeds up; eventually their rates match. You can show this on concentration-vs-time or rate-vs-time graphs: slopes flatten and level off when equilibrium’s established (CED keywords: reversible process, forward and reverse rates, concentration-time graph). For AP tasks, know that Q (reaction quotient) vs K predicts direction of shifting, and that K (equilibrium constant) describes the fixed ratio at equilibrium. Review Topic 7.1 study guide for examples and graphs (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

Because chemical processes are controlled by both kinetics (rates) and thermodynamics (energy), many reactions can run in both directions. At the molecular level individual collisions convert reactants to products (forward) and products back to reactants (reverse). If both directions are reasonably likely, the system can reach a dynamic equilibrium: forward and reverse rates become equal so concentrations stay constant even though reactions continue (CED 7.1.A.2–7.1.A.3). Whether a reaction “goes both ways” depends on the energy changes and the equilibrium constant K—large K favors products, small K favors reactants, and when K is intermediate you’ll have significant amounts of both. You can predict direction at any moment with Q vs. K (related Topic 7.3). For AP review, see the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2) and practice equilibrium graphs so you can interpret concentration- vs-time and rate-vs-time plots like the exam asks.

Think of equilibrium like a two-way street with the same number of cars going each way: reactions still happen (molecules convert forward and reverse), but the forward rate equals the reverse rate, so there’s no net change in amounts you can measure. At the microscopic level individual reactant molecules become product and some product become reactant, but because those processes balance (equal rates), concentrations and partial pressures stay constant—that’s why no observable change occurs (CED 7.1.A.2–7.1.A.3). You can see this on concentration vs. time graphs: after some time the curves flatten out at constant values even though reactions continue (CED 7.1.A.4). Want a quick refresher? Check the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2) and try practice problems (https://library.fiveable.me/practice/ap-chemistry) to see graphs and rate thinking used on AP-style questions.

A reaction that “goes to completion” means essentially all reactant is converted into product and you end up with (nearly) no reactant left—the process is effectively irreversible under those conditions. By contrast, a reaction that reaches equilibrium is reversible: reactants and products are both present and their concentrations (or partial pressures) stay constant over time because the forward and reverse rates are equal (dynamic equilibrium, CED 7.1.A.2–7.1.A.3). Practically: if Q (reaction quotient) ever equals K, the system is at equilibrium; if Q << K the forward reaction is favored (but not necessarily to absolute completion), and if K is extremely large you may treat the reaction as “going to completion” for calculations. Use concentration-vs-time or rate-vs-time graphs to tell the difference: equilibrium shows flattening with nonzero reactant and product levels, completion shows reactant → ~0. For a quick refresher on CED concepts and example graphs, see the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2). Practice problems are at (https://library.fiveable.me/practice/ap-chemistry).

At equilibrium the concentrations stay constant because the system reaches a dynamic balance: the forward and reverse processes are still happening, but they occur at exactly the same rate, so there’s no net change in amounts of reactants or products (CED 7.1.A.3 and 7.1.A.2). Think of two people passing balls back and forth at equal speed—balls keep moving, but each side’s count stays the same. Mathematically, Q = K at equilibrium, so the ratio of product and reactant concentrations no longer changes. On a concentration-vs-time graph you’ll see curves level off once rates equalize. This is a core AP idea you should be able to explain and sketch (Topic 7.1). For a short review and examples, see the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2). For more practice on rate/equilibrium connections, check the AP practice problems (https://library.fiveable.me/practice/ap-chemistry).

A concentration vs. time graph shows how reactant and product amounts change until they level off—that flat part is equilibrium (concentrations constant, dynamic equilibrium: forward and reverse rates equal). Read it like this: - At t = 0 note starting concentrations. - Steep slope = fast net change (fast net forward or reverse reaction); shallow slope = slow net change. - When lines flatten to horizontal, no observable change occurs—that’s equilibrium (CED 7.1.A.2–7.1.A.3). The y-values at the plateau are the equilibrium concentrations. - If both reactant and product curves are present, where they stop tells you which side is favored (large product [eq] means K large). - Instantaneous slopes before plateau give relative net rates; equal magnitudes of forward and reverse rates aren’t shown directly but are implied once concentrations stop changing. - If you disturb the system (add/remove species, change T or V), curves will shift to new plateaus (Le Châtelier idea, Topic 7.2–7.5). For practice interpreting these graphs (and AP-style questions tied to CED Topic 7.1), check the unit study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).

You can see reversible processes all the time. Easy, observable examples: water evaporating from a puddle while vapor condenses on a cold window (vapor–liquid equilibrium); a sugar or salt solution that dissolves when you stir but re-forms crystals as it cools or evaporates (dissolution ↔ precipitation); a sponge or activated charcoal that soaks up a gas and then releases it when heated (absorption ↔ desorption). On the chemical side, acid–base proton transfers (H+ hopping between molecules) and redox electron transfers are reversible at equilibrium in many buffered or electrochemical systems. Remember: at equilibrium you won’t see net change because forward and reverse rates are equal—it’s dynamic, not static (CED 7.1.A.2–7.1.A.3). Tracking concentration or pressure vs. time shows curves leveling off when equilibrium is reached. For more examples and AP-aligned notes, check the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Equilibrium happens because the forward and reverse processes keep changing concentrations until their rates depend on those concentrations in exactly the same way—so the rate of product formation equals the rate of reactant formation. Early on, one direction is faster (because reactant or product concentrations favor that rate), so concentrations shift. As those concentrations change, the rates change too; eventually they reach values that make the two rates identical. That steady state is dynamic: molecules still react both ways, but there’s no net change in concentrations (CED 7.1.A.2–7.1.A.3). Mathematically, rateforward = ratereverse when the reaction quotient Q equals the equilibrium constant K, which is why Q and K predict direction (see Topic 7.3). On the AP exam you may be asked to interpret concentration- or rate-vs-time graphs showing this establishment of equilibrium (Topic 7.1; review: https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2). For more practice problems, check https://library.fiveable.me/practice/ap-chemistry.

Short answer: products turning back into reactants doesn’t contradict what you learned—many reactions are reversible. At the microscopic level molecules keep colliding and reacting both ways. If the forward and reverse reactions happen at different rates at first, concentrations change; when those rates become equal, the system reaches a dynamic equilibrium. That means you still have both reactants and products present, but their concentrations stay constant because forward = reverse (CED 7.1.A.2–7.1.A.3). Think of evaporation/condensation or salt dissolving/precipitating—you can see both directions happening even though macroscopically nothing appears to change. Kinetics explains it: some product molecules have enough energy (and the right orientation) to collide and reform reactants, so the reverse pathway is real. Use concentration-vs-time or rate-vs-time graphs to see equilibrium being established (CED keywords: reversible process, forward and reverse rates, dynamic equilibrium). For a clear AP-level walkthrough, check the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2). For more practice problems across Unit 7, see (https://library.fiveable.me/practice/ap-chemistry).

Static equilibrium means nothing is happening at all—no forward or reverse process occurs (e.g., a cold, inert solid sitting on a table). Dynamic equilibrium means opposing processes continue, but at equal rates so there’s no net change you can observe. In chemistry that latter idea is the one AP expects: at equilibrium reactants and products coexist and their concentrations or partial pressures stay constant because the forward and reverse reaction rates are equal (CED 7.1.A.2–7.1.A.3). Quick examples: a sealed container with liquid water + vapor at a fixed T shows dynamic vapor↔liquid exchange (molecules evaporate and condense at equal rates). A truly “static” situation would be when neither process is occurring. For more AP-aligned explanation and graphs of concentration vs. time, check the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2). For extra practice, hit Fiveable’s AP Chem practice set (https://library.fiveable.me/practice/ap-chemistry).

You know a reaction reached equilibrium in the lab when measurable properties stop changing with time—even though forward and reverse reactions are still happening (dynamic equilibrium). Practically, you look for constant values: concentrations (or absorbance via a spectrophotometer), partial pressure of gases, mass loss/gain, conductivity, or pH that level off on a concentration-vs-time or pressure-vs-time graph. If repeated measurements taken over several half-lives show no net change, [Reactants] and [Products] are effectively constant (CED 7.1.A.2–7.1.A.3). Common checks: spectrophotometric absorbance steady within experimental error, gas pressure constant on a manometer, or titration of aliquots showing the same composition over time. You can also compare Q to K: if Q = K, the system is at equilibrium (Topic 7.3). For a quick AP review, see the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Because equilibrium is dynamic: the forward and reverse reactions keep happening, but they occur at exactly the same rate. That means molecules are still converting back and forth, yet the net amounts of reactants and products don’t change over time—so you see “no observable changes.” In AP terms (CED 7.1.A.2–7.1.A.3), concentrations or partial pressures are constant at equilibrium even though microscopic events continue. Think of it like a busy elevator: people get on and off at equal rates, so the number of people inside stays the same. You can show this on concentration-vs-time or rate-vs-time graphs where curves flatten once rates match. For more review on dynamic equilibrium and practice graphs, see the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2) and try problems at (https://library.fiveable.me/practice/ap-chemistry).

Yes—equilibrium can happen with physical processes as well as chemical reactions. The CED explicitly lists phase and physical equilibria (evaporation ⇄ condensation, absorption ⇄ desorption, dissolution ⇄ precipitation) as reversible processes that reach a dynamic equilibrium: forward and reverse rates equal, so concentrations/partial pressures stay constant (no observable change) (CED 7.1.A.1–3). On the AP exam you may be asked to interpret concentration/pressure vs. time or rate vs. time graphs to show equilibrium establishment (CED 7.1.A.4). So when you see “equilibrium” think dynamic, reversible, and not only bond-making/breaking—it includes vapor–liquid, solubility, and adsorption equilibria. For quick review, check the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

There’s no single answer—“how long” depends on kinetics, not equilibrium itself. Equilibrium is reached when forward and reverse rates are equal and concentrations (or partial pressures) stop changing (CED 7.1.A.2–3). The time to get there depends on rate constants, initial concentrations/pressures, temperature, presence of a catalyst, and whether the process is homogeneous or heterogeneous (surface area, diffusion). Examples: simple gas-phase or proton-transfer equilibria can establish in milliseconds–seconds; many aqueous acid–base equilibria are essentially instantaneous; dissolution/precipitation or heterogeneous reactions can take minutes, hours, or longer. On the AP exam you’ll often use concentration- vs-time or rate-vs-time graphs to identify when equilibrium is reached (CED 7.1.A.4). Want practice estimating/reading those graphs? Check the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2) and more problems at (https://library.fiveable.me/practice/ap-chemistry).

As a reversible process approaches equilibrium, the forward rate decreases (because reactant concentration drops) and the reverse rate increases (because product concentration rises) until they become equal. At equilibrium both rates are nonzero but identical, so there’s continuous microscopic reaction in both directions with no net change in concentrations or partial pressures (dynamic equilibrium, CED 7.1.A.3). On a rate-vs-time graph you’ll see the forward-rate curve falling and the reverse-rate curve rising and then overlapping at the equilibrium rate; on a concentration-vs-time graph concentrations flatten out (CED 7.1.A.2, 7.1.A.4). This idea is tested on the AP exam—know that equal rates, not zero rates, define equilibrium. For a concise review, see the Topic 7.1 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-equilibrium/study-guide/dzIPBIOsEPKoTL4VKEH2); for more unit review and practice, check Unit 7 (https://library.fiveable.me/ap-chemistry/unit-7) and AP Chem practice problems (https://library.fiveable.me/practice/ap-chemistry).