5.1 Reaction Rates

Reaction rate = how fast reactants are converted to products—specifically the change in concentration per unit time (CED 5.1.A.1). You calculate an average rate as Δ[species]/Δt (for a reactant use a negative sign because its concentration falls). For a balanced reaction aA + bB → cC, rates are related by stoichiometry: rate = −(1/a)·Δ[A]/Δt = −(1/b)·Δ[B]/Δt = (1/c)·Δ[C]/Δt Instantaneous rate is the derivative: rate = −(1/a) d[A]/dt. Units are typically M·s⁻¹. The numerical rate depends on experimental conditions (concentration, temperature, surface area, catalyst—CED 5.1.A.3). Determining how rate depends on concentration (rate law, orders, and rate constant k) is Topic 5.2; on the AP exam you may be asked to calculate rates from data or relate rates using stoichiometry. For a focused study, see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

Increasing temperature speeds reactions because molecules move faster, so there are more collisions per second and a larger fraction of those collisions have enough energy to get over the activation energy barrier. Collision theory + transition-state ideas: raising T increases kinetic energy, shifting the Maxwell–Boltzmann distribution so more molecules reach or exceed Ea. That raises the rate constant k (Arrhenius equation: k = A e^(−Ea/RT)), so reaction rates increase even if concentrations don’t change. A useful rule of thumb: many reactions get about 2× faster for each 10 °C rise (depends on Ea). This is exactly what Topic 5.1 expects you to explain (collision theory, activation energy, Arrhenius) for the AP exam (LO 5.1.A / EK 5.1.A.3). For a focused review, see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and try related practice problems (https://library.fiveable.me/practice/ap-chemistry).

Stoichiometry tells you how concentration changes of reactants and products are tied together, so it sets the relative rates of those changes. For a balanced reaction aA + bB → cC, the instantaneous rate R is defined so that -(1/a)·d[A]/dt = -(1/b)·d[B]/dt = (1/c)·d[C]/dt = R. Example: 2A → 3B means rate = -(1/2)d[A]/dt = (1/3)d[B]/dt—if [A] drops by 0.20 M/s, [B] rises by 0.30 M/s (because 0.20·(3/2)=0.30). Important AP point (CED 5.1.A.2): stoichiometry controls these relative concentration-change rates, but it does NOT by itself give the rate law. The rate law (how R depends on [A], [B], etc.) must be determined experimentally or from an elementary mechanism (if the step is elementary, its molecularity often gives the rate law). Also remember the role of the rate-determining step in multi-step mechanisms. For more practice and examples tied to the AP curriculum, see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and the AP practice bank (https://library.fiveable.me/practice/ap-chemistry).

Reaction rate is how fast reactant is turned into product—usually expressed as change in concentration per time (∆[A]/∆t) and tied to the stoichiometry of the balanced equation (CED 5.1.A.1–5.1.A.2). It changes when you change concentration, temperature, surface area, or add a catalyst (CED 5.1.A.3). The rate constant, k, is a proportionality factor in the rate law (rate = k [A]^m [B]^n). k tells you how quickly a reaction proceeds under a specific set of conditions, but k itself doesn’t change with reactant concentrations. k does depend on temperature and activation energy (Arrhenius equation) and on the presence of catalysts. Its numerical value and units depend on the reaction order (zero-, first-, second-order give different units). On the AP exam you should: (1) distinguish measured rates (∆[ ]/∆t) from the fitted parameter k in rate laws, and (2) note k’s temperature/catalyst dependence. For a focused review see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

A catalyst speeds up a reaction by providing an alternate reaction pathway with a lower activation energy (Ea). That means more collisions between reactant molecules have enough energy to get over the barrier, so the rate constant k increases (Arrhenius: k = A e^(-Ea/RT)). Catalysts do this by stabilizing the transition state or by forming temporary intermediates; in heterogeneous catalysis the reaction happens on a surface, in homogeneous catalysis the catalyst is in the same phase as reactants. The key reason a catalyst isn’t “used up” is that it’s regenerated by the end of the reaction cycle—it participates in steps of the mechanism but is present in the same amount before and after, so it changes the rate (kinetics) without changing the position of equilibrium or ΔG° (thermodynamics). For AP-level review of catalysts and rate effects see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and more unit practice at (https://library.fiveable.me/ap-chemistry/unit-5).

Surface area speeds a reaction because it increases the number of reactant particles exposed to one another, so more effective collisions happen per second. In heterogeneous reactions (solid + liquid/gas) only molecules at the solid’s surface can collide with other reactants. Breaking a solid into powder raises its surface area, raising collision frequency and therefore the rate (think powdered CaCO3 reacting faster than a single chunk). This ties to collision theory and activation energy: more collisions → higher chance of collisions that have enough energy and correct orientation to overcome Ea. On the AP Chem CED, this is part of Topic 5.1 (rates depend on surface area; see 5.1.A.3) and connects to the collision model in Topic 5.5. For a quick review, check the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ). For practice applying this idea to graphs and experiments, try problems at Fiveable’s Unit 5 page (https://library.fiveable.me/ap-chemistry/unit-5) or the general practice bank (https://library.fiveable.me/practice/ap-chemistry).

It depends on the reaction’s rate law—doubling concentration doesn’t always give the same change in rate. From the AP CED: the rate = k [A]^m [B]^n, where m and n are orders determined experimentally (not from stoichiometry). So: - Zero-order in A (m = 0): rate unchanged when [A] doubles. - First-order in A (m = 1): rate doubles (×2). - Second-order in A (m = 2): rate quadruples (×4). Collision theory and the Arrhenius equation explain why: increasing [reactant] raises collision frequency, so rate usually increases for positive orders; temperature and catalysts also matter. Always determine the rate law from experiments (initial rates or integrated/differential rate laws)—that’s what AP problems test (Topic 5.2–5.3). For a quick review, see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

Think of concentration like the number of cars on a road: more cars (higher [reactant]) → more chances of bumping into each other (collisions) → usually a faster reaction. Collision theory explains that increasing concentration raises collision frequency and so often increases rate, but how much it changes depends on the reaction’s rate law, which you determine experimentally (AP Exam expects you to know this). Rate law form: rate = k[A]^n(B)^m. The exponents (orders) tell you the effect: - n = 0 (zero order): changing [A] doesn’t change rate. - n = 1 (first order): double [A] → rate doubles. - n = 2 (second order): double [A] → rate quadruples. Remember: stoichiometry tells you how concentrations change with time, but it doesn’t tell you the rate law—you must measure it. For more review on Topic 5.1 and examples, see the Fiveable reaction rates study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and extra practice (https://library.fiveable.me/practice/ap-chemistry).

Think of a concentration vs. time graph as a position vs. time graph for molecules: the slope (rise/run) tells you the rate of change in concentration. Mathematically slope = Δ[A]/Δt (average rate) or the tangent d[A]/dt (instantaneous rate). For reactants the slope is negative (concentration is decreasing); for products it’s positive. The magnitude (absolute value) is the speed of the reaction at that time and you must include stoichiometry: if 2A → B, rate = -1/2 d[A]/dt = d[B]/dt. Units are usually M·s⁻¹. Shapes you’ll see on AP: zero-order = straight line (constant slope); first-order = exponential decay (slope gets less steep over time); second-order = steeper curvature. On the exam, be ready to calculate average slopes from two points or instantaneous slopes from a tangent and to relate rates to stoichiometry (CED 5.1.A). For extra practice, check the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and lots of practice problems (https://library.fiveable.me/practice/ap-chemistry).

Start by deciding whether you need an average rate, instantaneous rate, or the rate law. - Average rate: use Δ[reactant or product]/Δt from experimental concentration (or mass/pressure) vs time. Include stoichiometry: rates of reactants are negative and scaled by coefficients (–1/coeff for reactants, +1/coeff for products) per EK 5.1.A.2. - Instantaneous rate: take the slope of the concentration vs time curve at a point (tangent). That’s what you’ll see on FR questions. - To get the rate law and orders: use initial rates from experiments with different starting concentrations (method of initial rates). Fit rate = k[A]^m[B]^n to find m,n and k (CED keywords: rate law, order, rate constant). - For integrated rate laws: match data to zero-, first-, or second-order linear forms to find k and half-life (use ln for first-order). - Remember: temperature, catalysts, surface area affect k/rate (5.1.A.3). For guided practice and AP-style problems, see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and thousands of practice questions (https://library.fiveable.me/practice/ap-chemistry). Unit 5 appears on the exam (~7–9%); expect both multiple-choice and free-response uses of these skills.

You need to know factors that affect reaction rates because they tell you how fast reactants become products and let you predict, control, and explain real reactions—exactly what the CED demands for Topic 5.1. Practically: concentration, temperature, surface area, and catalysts change collision frequency and energy (collision theory), which alters the rate constant k and the rate law you’ll write or use on the exam. Conceptually: understanding activation energy, the Arrhenius relationship, and rate-determining steps helps you connect experimental data to mechanisms and to calculate things like half-lives or k (differential/integrated rate laws). On the AP test, kinetics shows up across multiple-choice and free-response (Unit 5 is 7–9% of the exam), so mastering these factors helps you solve rate-law problems and justify explanations (Science Practices 2 & 5). For a focused review, see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ), the whole Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5), and lots of practice questions (https://library.fiveable.me/practice/ap-chemistry).

Collision theory gives the microscopic why behind reaction rates: only collisions between reactant particles can produce products, and for a collision to be effective it must (1) have sufficient energy to overcome the activation energy and (2) have the correct orientation. So the reaction rate depends on collision frequency (increases with concentration and temperature) and the fraction of collisions with energy ≥ Ea (increases strongly with temperature—Arrhenius equation). Catalysts lower Ea, increasing the fraction of effective collisions without changing thermodynamics. This connects directly to AP CED essentials 5.1.A.1–3 (rates depend on concentration, temperature, catalysts) and links to elementary steps and rate laws (Topics 5.2–5.4). For more practice and quick review of the collision model, see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5). Want problems to test this idea? Try the practice set (https://library.fiveable.me/practice/ap-chemistry).

Smaller particle size means larger total surface area for the same mass of solid. For heterogeneous reactions (solid + gas/liquid), more surface area gives more sites where reactant molecules can collide with the solid—so by collision theory you get more effective collisions per second and a faster rate. Surface area doesn’t change activation energy or the rate constant k, but it increases the frequency of collisions that can reach the activation energy, so the observed reaction rate rises. Think: powder reacts much faster than a single chunk of the same material. This is an AP-Chem Topic 5.1 idea (surface area → rate) and ties to collision theory and heterogeneous catalysis in the CED keywords. For more review, see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

You should’ve watched reactant concentrations fall and product concentrations rise over time—and used those changes to measure the reaction rate (amount converted per time), consistent with CED 5.1.A. Practically you record [reactant] vs. time and note the slope (-d[reactant]/dt) or the slope of [product] vs. time (both tied by stoichiometry). You’d also see that initial rates are larger when reactant concentration is higher (concentration affects rate), and that changes in temperature, surface area, or a catalyst change how fast those concentration curves move. For analysis, different orders give characteristic plots: [A] vs. t linear for zero order, ln[A] vs. t linear for first order (constant half-life), and 1/[A] vs. t linear for second order. Those are the exact ideas AP expects you to use on exam questions (CED 5.1.A.1–3). Review the Topic 5.1 study guide here (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

Some reactions are “instant” and some are slow because of how often and how effectively molecules collide and whether those collisions have enough energy to get over an energy barrier. - Collision theory: molecules must collide with correct orientation and enough energy (the activation energy) to form products. Fast reactions have frequent, effective collisions; slow ones don’t. - Factors that speed reactions: higher concentration (more collisions), higher temperature (more collisions and more molecules exceed EA via the Arrhenius relationship), greater surface area for solids, and catalysts (they lower EA and change the pathway). - Rate laws and the rate constant k quantify this experimentally; k rises with temperature and when a catalyst is used. For AP exam, know how these factors show up in rate laws, reaction mechanisms, and energy profiles (CED Topic 5.1–5.4). If you want a clear walkthrough and practice problems on this, see the Topic 5.1 study guide (https://library.fiveable.me/ap-chemistry/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ) and the Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5). For extra practice, use Fiveable’s AP Chem practice problems (https://library.fiveable.me/practice/ap-chemistry).