[{"data":1,"prerenderedAt":217},["ShallowReactive",2],{"topic-chemical-kinetics":3},{"topic":4,"prev":201,"next":209},{"id":5,"slug":6,"title":7,"shortTitle":8,"lesson":5,"category":9,"description":10,"metaDescription":11,"objectives":12,"conceptSections":34,"workedExample":104,"oerSources":113,"selfStudyQuestions":119,"relatedTopicSlugs":141,"relatedElements":144,"relatedReferences":149,"keyEquations":152,"howTo":169,"conceptProbe":178,"practiceTeaser":182,"gatedContent":190,"seoKeywords":193,"lastAlignmentAudit":199,"objectiveCount":200},20,"chemical-kinetics","Chemical Kinetics","Kinetics","thermodynamics-kinetics","Study reaction rates, rate laws, reaction order, activation energy, and reaction mechanisms.","Learn chemical kinetics: rate laws, reaction order, half-life, activation energy, the Arrhenius equation, and reaction mechanisms. Step-by-step examples.",[13,16,19,22,25,28,31],{"id":14,"text":15},"20.1","Calculate average and instantaneous rates of reaction from concentration-time data and express the stoichiometric rate relationships among reactants and products",{"id":17,"text":18},"20.2","Determine the rate law (rate constant and reaction orders) from experimental initial-rate data",{"id":20,"text":21},"20.3","Use integrated rate laws to relate concentration and time for zero-, first-, and second-order reactions",{"id":23,"text":24},"20.4","Calculate and apply the half-life for first-order reactions",{"id":26,"text":27},"20.5","Use the Arrhenius equation to relate the rate constant to temperature and calculate activation energy",{"id":29,"text":30},"20.6","Identify reactants, products, transition state, activation energy, and the effect of a catalyst from a reaction-energy diagram, and rank reaction rates from diagram features",{"id":32,"text":33},"20.7","Determine the rate law consistent with a proposed reaction mechanism using the steady-state or rate-determining-step approximation",[35,44,54,65,74,83,92,100],{"heading":36,"content":37,"relatedObjectives":38,"deepDive":40},"Reaction Rates","\u003Cp>The \u003Cstrong>reaction rate\u003C/strong> measures how quickly concentrations change over time. For a general reaction aA + bB → cC + dD, the rate can be expressed from any species:\u003C/p>\u003Cp>rate = −(1/a)(Δ[A]/Δt) = −(1/b)(Δ[B]/Δt) = (1/c)(Δ[C]/Δt) = (1/d)(Δ[D]/Δt)\u003C/p>\u003Cp>The negative signs for reactants reflect decreasing concentrations; stoichiometric coefficients ensure the rate is the same regardless of which species is monitored. Rates are always positive and have units of M/s (mol·L\u003Csup>−1\u003C/sup>·s\u003Csup>−1\u003C/sup>).\u003C/p>\u003Cp>The \u003Cstrong>instantaneous rate\u003C/strong> at a given time is the slope of the tangent to a concentration-vs.-time curve at that point. The \u003Cstrong>initial rate\u003C/strong> — measured at t ≈ 0 before products accumulate — is especially useful because it avoids complications from reverse reactions and is the basis of the method of initial rates.\u003C/p>",[39],150,[41],{"label":42,"body":43},"Why this matters: diamonds are forever only kinetically","\u003Cp>Thermodynamics says every diamond should be crumbling into graphite: the conversion is energetically downhill. Diamonds persist because the rate is essentially zero at Earth-surface temperatures; rearranging that carbon lattice has an activation barrier so high the reaction would take longer than the age of the universe. Kinetics, not stability, is what “forever” means.\u003C/p>\u003Cp>The same distinction runs your kitchen: the expiration date on milk is a rate measurement, not a thermodynamic verdict, and rust versus fire are the same iron-plus-oxygen chemistry at wildly different rates. Thermodynamics decides \u003Cem>whether\u003C/em>; kinetics decides \u003Cem>when\u003C/em>, and this topic is about when.\u003C/p>",{"heading":45,"content":46,"relatedObjectives":47,"deepDive":50},"Rate Laws and the Method of Initial Rates","\u003Cp>The \u003Cstrong>rate law\u003C/strong> expresses the rate as a function of reactant concentrations: \u003Cstrong>rate = k[A]\u003Csup>m\u003C/sup>[B]\u003Csup>n\u003C/sup>\u003C/strong>, where k is the \u003Cstrong>rate constant\u003C/strong> and the exponents m and n are \u003Cstrong>reaction orders\u003C/strong>. The overall order equals m + n.\u003C/p>\u003Cp>Reaction orders are found \u003Cem>experimentally\u003C/em>, not from the balanced equation. The \u003Cstrong>method of initial rates\u003C/strong> compares experiments in which one concentration changes while others stay constant:\u003C/p>\u003Cul>\u003Cli>If doubling [A] doubles the rate → first order in A (m = 1).\u003C/li>\u003Cli>If doubling [A] quadruples the rate → second order (m = 2).\u003C/li>\u003Cli>If the rate is unchanged → zero order (m = 0).\u003C/li>\u003C/ul>\u003Cp>Once all orders are known, use the trial named in the problem to solve for k; use Trial 1 if none is named. Values from other trials provide a consistency check, and experimental scatter can make them differ slightly. Pay close attention to the \u003Cstrong>units of k\u003C/strong> — they depend on the overall order (e.g., M\u003Csup>−1\u003C/sup>s\u003Csup>−1\u003C/sup> for second order overall).\u003C/p>",[48,49],151,155,[51],{"label":52,"body":53},"Common mistake: reading the orders off the coefficients","\u003Cp>The balanced equation cannot tell you the rate law. The textbook cautionary tale: NO\u003Csub>2\u003C/sub> + CO → NO + CO\u003Csub>2\u003C/sub> has neat 1:1 coefficients, yet experiment finds rate = k[NO\u003Csub>2\u003C/sub>]\u003Csup>2\u003C/sup> at lower temperatures: second order in NO\u003Csub>2\u003C/sub> and \u003Cem>zero\u003C/em> order in CO. Adding more CO does nothing to the rate, because CO only participates after the slow step is already over.\u003C/p>\u003Cp>Orders come from experiments (the method of initial rates on this page) because they encode the hidden mechanism, not the overall accounting. The one exception arrives later in this topic: an \u003Cem>elementary\u003C/em> step, a single molecular event, does take its orders from its own molecularity.\u003C/p>",{"heading":55,"content":56,"relatedObjectives":57,"deepDive":61},"Integrated Rate Laws and Half-Life","\u003Cp>\u003Cstrong>Integrated rate laws\u003C/strong> relate concentration to time, letting you calculate how much reactant remains or how long to reach a target concentration:\u003C/p>\u003Ctable>\u003Cthead>\u003Ctr>\u003Cth>Order\u003C/th>\u003Cth>Integrated Law\u003C/th>\u003Cth>Linear Plot\u003C/th>\u003Cth>Half-Life\u003C/th>\u003C/tr>\u003C/thead>\u003Ctbody>\u003Ctr>\u003Ctd>Zero\u003C/td>\u003Ctd>[A] = [A]\u003Csub>0\u003C/sub> − kt\u003C/td>\u003Ctd>[A] vs. t\u003C/td>\u003Ctd>t\u003Csub>½\u003C/sub> = [A]\u003Csub>0\u003C/sub> / 2k\u003C/td>\u003C/tr>\u003Ctr>\u003Ctd>First\u003C/td>\u003Ctd>ln[A] = ln[A]\u003Csub>0\u003C/sub> − kt\u003C/td>\u003Ctd>ln[A] vs. t\u003C/td>\u003Ctd>t\u003Csub>½\u003C/sub> = 0.693 / k\u003C/td>\u003C/tr>\u003Ctr>\u003Ctd>Second\u003C/td>\u003Ctd>1/[A] = 1/[A]\u003Csub>0\u003C/sub> + kt\u003C/td>\u003Ctd>1/[A] vs. t\u003C/td>\u003Ctd>t\u003Csub>½\u003C/sub> = 1 / (k[A]\u003Csub>0\u003C/sub>)\u003C/td>\u003C/tr>\u003C/tbody>\u003C/table>\u003Cp>For \u003Cstrong>first-order\u003C/strong> reactions the half-life is \u003Cem>constant\u003C/em> — independent of concentration. This is characteristic of radioactive decay and many drug-elimination processes. For zero and second order, half-life depends on [A]\u003Csub>0\u003C/sub>. To determine order graphically, plot data in all three forms; the one yielding a straight line reveals the order.\u003C/p>",[58,59,60],152,153,154,[62],{"label":63,"body":64},"Why this matters: your prescription schedule is a half-life","\u003Cp>Most drugs are cleared from the bloodstream by processes that are approximately first order, and first order has a signature property: the half-life does not care how much you started with. Whether the dose was large or small, the same fraction is gone after each fixed interval, which is precisely why prescriptions say “every 8 hours” rather than “when you feel like it.” Caffeine works the same way, with a half-life around five hours: a quarter of your afternoon espresso is still circulating at midnight.\u003C/p>\u003Cp>Carbon-14 dating, drug dosing, and radioactive-waste planning are all the same integrated first-order rate law wearing different units.\u003C/p>",{"heading":66,"content":67,"relatedObjectives":68,"deepDive":70},"The Arrhenius Equation","\u003Cp>The rate constant k increases with temperature. The \u003Cstrong>Arrhenius equation\u003C/strong> quantifies this:\u003C/p>\u003Cp>\u003Cstrong>k = A · e\u003Csup>−E\u003Csub>a\u003C/sub>/RT\u003C/sup>\u003C/strong>\u003C/p>\u003Cul>\u003Cli>A = frequency (pre-exponential) factor — accounts for collision frequency and molecular orientation\u003C/li>\u003Cli>E\u003Csub>a\u003C/sub> = activation energy (J/mol)\u003C/li>\u003Cli>R = 8.314 J/(mol·K); T in kelvins\u003C/li>\u003C/ul>\u003Cp>The linear form, \u003Cstrong>ln k = −E\u003Csub>a\u003C/sub>/R · (1/T) + ln A\u003C/strong>, gives a straight line when ln k is plotted against 1/T with slope = −E\u003Csub>a\u003C/sub>/R.\u003C/p>\u003Cp>The two-point form \u003Cstrong>ln(k\u003Csub>2\u003C/sub>/k\u003Csub>1\u003C/sub>) = (E\u003Csub>a\u003C/sub>/R)(1/T\u003Csub>1\u003C/sub> − 1/T\u003Csub>2\u003C/sub>)\u003C/strong> lets you calculate E\u003Csub>a\u003C/sub> from rate constants at two temperatures or predict k at a new temperature when E\u003Csub>a\u003C/sub> is known. A useful rule of thumb: for many reactions near room temperature, a 10 °C rise roughly doubles the rate.\u003C/p>",[69],156,[71],{"label":72,"body":73},"Why this matters: your refrigerator is an Arrhenius machine","\u003Cp>For reactions with typical activation energies, a rough but useful rule falls out of the Arrhenius equation: the rate roughly doubles for every 10 °C rise. Run it backwards and you have refrigeration: dropping food from room temperature (about 21 °C) to 4 °C slows spoilage reactions severalfold, and the freezer slows them further still. Nothing about the spoilage chemistry changed; only the fraction of molecular collisions energetic enough to clear E\u003Csub>a\u003C/sub> collapsed.\u003C/p>\u003Cp>The exponential is the point: small temperature changes move rates a lot, which is why fevers are dangerous, why crickets chirp faster on warm nights, and why chemists report k with the temperature attached.\u003C/p>",{"heading":75,"content":76,"relatedObjectives":77,"deepDive":79},"Collision Theory and Energy Diagrams","\u003Cp>The \u003Cstrong>collision model\u003C/strong> states that for a reaction to occur, molecules must: (1) collide, (2) with sufficient kinetic energy (≥ E\u003Csub>a\u003C/sub>), and (3) with proper orientation. Only collisions meeting all three criteria — \u003Cem>effective collisions\u003C/em> — lead to product formation.\u003C/p>\u003Cp>An \u003Cstrong>energy diagram\u003C/strong> (reaction-coordinate diagram) plots potential energy vs. reaction progress:\u003C/p>\u003Cul>\u003Cli>The peak is the \u003Cstrong>transition state\u003C/strong> (activated complex) — a fleeting, high-energy arrangement of atoms that cannot be isolated.\u003C/li>\u003Cli>The height from reactants to the peak is the \u003Cstrong>activation energy (E\u003Csub>a\u003C/sub>)\u003C/strong>.\u003C/li>\u003Cli>The difference between reactant and product energy levels is \u003Cstrong>ΔH\u003C/strong>. If products are lower, the reaction is exothermic; if higher, endothermic.\u003C/li>\u003C/ul>\u003Cp>Multi-step mechanisms show multiple peaks on the diagram — one per elementary step — with valleys between them representing reaction \u003Cstrong>intermediates\u003C/strong>. In a multi-step diagram, the step with the \u003Cstrong>highest peak\u003C/strong> (largest E\u003Csub>a\u003C/sub>) is the rate-determining step; between two pathways for the same overall reaction, the one with the lower E\u003Csub>a\u003C/sub> proceeds faster.\u003C/p>",[69,78],157,[80],{"label":81,"body":82},"Why the bookshelf does not burst into flame","\u003Cp>Paper sitting in air is far downhill from CO\u003Csub>2\u003C/sub> and water thermodynamically, yet libraries are not fire hazards by default. The activation-energy hill is the guardian: at room temperature, essentially no collisions between cellulose and O\u003Csub>2\u003C/sub> carry enough energy to clear it. A match changes the local story: its flame pays the activation toll for a small region, and the heat released by that first burning patch pays the toll for its neighbors, a self-sustaining chain that no longer needs the match.\u003C/p>\u003Cp>Ignition temperature, in kinetics language, is the temperature at which the heat-release rate finally outruns the heat-loss rate. The energy diagram in this section is a map of that hill.\u003C/p>",{"heading":84,"content":85,"relatedObjectives":86,"deepDive":88},"Catalysts","\u003Cp>A \u003Cstrong>catalyst\u003C/strong> increases a reaction rate by providing an \u003Cem>alternative pathway\u003C/em> with a \u003Cem>lower activation energy\u003C/em>. It participates in intermediate steps but is regenerated and not consumed overall. On an energy diagram, the catalysed path has a lower peak than the uncatalysed one.\u003C/p>\u003Cp>Two main categories exist:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Homogeneous catalysts\u003C/strong> — present in the same phase as the reactants (e.g., H\u003Csup>+\u003C/sup> catalysing ester hydrolysis in solution).\u003C/li>\u003Cli>\u003Cstrong>Heterogeneous catalysts\u003C/strong> — in a different phase, typically a solid surface on which gaseous or dissolved reactants adsorb, react, and desorb (e.g., the iron catalyst in the Haber–Bosch process, catalytic converters in vehicles).\u003C/li>\u003C/ul>\u003Cp>Biological catalysts (\u003Cstrong>enzymes\u003C/strong>) are highly specific proteins that lower E\u003Csub>a\u003C/sub> dramatically. Crucially, catalysts do \u003Cem>not\u003C/em> change ΔH or shift the equilibrium position — they only affect \u003Cem>how fast\u003C/em> equilibrium is reached.\u003C/p>",[87],158,[89],{"label":90,"body":91},"Why this matters: catalysts in your exhaust pipe and your cells","\u003Cp>The catalytic converter under your car is a platinum-group catalyst that gives exhaust pollutants (CO, unburned fuel, nitrogen oxides) a cheaper reaction path, converting the large majority of them to CO\u003Csub>2\u003C/sub>, N\u003Csub>2\u003C/sub>, and water in the fraction of a second they spend passing through. It emerges unchanged, catalyzing the same reactions for years.\u003C/p>\u003Cp>Enzymes do the same job for biochemistry, with accelerations that can reach factors of 10\u003Csup>17\u003C/sup>: reactions that would take millions of years uncatalyzed complete in milliseconds. Life is not made of unusual reactions; it is made of ordinary reactions with extraordinary catalysts.\u003C/p>",{"heading":93,"content":94,"relatedObjectives":95,"deepDive":96},"Reaction Mechanisms and the Rate-Determining Step","\u003Cp>A \u003Cstrong>reaction mechanism\u003C/strong> proposes a sequence of elementary steps whose sum gives the overall balanced equation. Each elementary step’s rate law can be written directly from its molecularity:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Unimolecular:\u003C/strong> A → products, rate = k[A]\u003C/li>\u003Cli>\u003Cstrong>Bimolecular:\u003C/strong> A + B → products, rate = k[A][B]\u003C/li>\u003C/ul>\u003Cp>The \u003Cstrong>rate-determining step (RDS)\u003C/strong> is the slowest step — it acts as a bottleneck. The overall rate law matches the RDS’s rate law, after substituting for any intermediates using equilibrium assumptions from faster preceding steps.\u003C/p>\u003Cp>A valid mechanism must satisfy two tests: (1) the elementary steps sum to the overall equation, and (2) the predicted rate law matches the experimentally observed one. \u003Cstrong>Intermediates\u003C/strong> are produced in one step and consumed in a later step — they never appear in the overall equation and should not appear in the final rate law. When the fast pre-equilibrium step is a dissociation (such as Cl\u003Csub>2\u003C/sub> ⇌ 2Cl), solving the equilibrium expression for the intermediate introduces a square root, so half-integer orders such as [Cl\u003Csub>2\u003C/sub>]\u003Csup>1/2\u003C/sup> can appear in the predicted rate law.\u003C/p>\u003Cp>\u003Cstrong>Identifying catalysts vs intermediates in a multi-step mechanism.\u003C/strong> Both species appear inside the mechanism but cancel out of the overall equation, yet they show opposite ordering of appearance:\u003C/p>\u003Cul>\u003Cli>A \u003Cstrong>catalyst\u003C/strong> appears first as a \u003Cem>reactant\u003C/em> in an earlier step and reappears as a \u003Cem>product\u003C/em> in a later step — it enters, drives the chemistry, and exits unchanged. A reaction usually has at most one catalyst.\u003C/li>\u003Cli>An \u003Cstrong>intermediate\u003C/strong> appears first as a \u003Cem>product\u003C/em> in an earlier step and is then consumed as a \u003Cem>reactant\u003C/em> in a later step — it is created and destroyed inside the mechanism. A mechanism can have several intermediates.\u003C/li>\u003C/ul>\u003Cp>Operational test: write down each species that does not appear in the overall balanced equation, then mark each occurrence in the mechanism as “reactant” or “product.” First-as-reactant-then-product = catalyst; first-as-product-then-reactant = intermediate.\u003C/p>",[78,87],[97],{"label":98,"body":99},"The toll booth sets the traffic flow","\u003Cp>Picture a three-lane highway with one single-lane toll booth. Cars per hour through the whole system equals cars per hour through the booth; widening the fast stretches changes nothing. Reaction mechanisms work identically: the overall rate is the rate of the slowest elementary step, and speeding up an already-fast step is wasted effort.\u003C/p>\u003Cp>This is why the rate law is a window into the mechanism: it reflects only the species involved up to and including the rate-determining step. It is also why catalysts are designed against the slow step specifically; a catalyst that accelerates a fast step is a new lane painted next to the same old toll booth.\u003C/p>",{"heading":101,"content":102,"relatedObjectives":103},"Kinetics Problem Decision Framework and Common Mistakes","\u003Cp>Select the right kinetics tool based on what you need to find:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>Finding the rate law?\u003C/strong> Use the method of initial rates: compare experiments where only one concentration changes at a time.\u003C/li>\u003Cli>\u003Cstrong>Finding concentration at a later time?\u003C/strong> Use the integrated rate law for the correct order (zero, first, or second).\u003C/li>\u003Cli>\u003Cstrong>Finding half-life?\u003C/strong> Use the half-life expression for the correct order. Note: only first-order half-life is independent of concentration.\u003C/li>\u003Cli>\u003Cstrong>Finding activation energy?\u003C/strong> Use the Arrhenius equation with data at two temperatures.\u003C/li>\u003Cli>\u003Cstrong>Connecting mechanism to rate law?\u003C/strong> The rate law is determined by the slow (rate-determining) step. The overall rate law must be consistent with the proposed mechanism.\u003C/li>\u003C/ol>\u003Cp>Common mistakes: assuming reaction order from stoichiometric coefficients (order must be determined experimentally), using the wrong integrated rate law for the reaction order, confusing rate with rate constant (k changes with temperature; rate depends on both k and concentrations), and forgetting that a catalyst lowers activation energy but does not change ΔH or the equilibrium constant.\u003C/p>",[39,48,58,59,60,49,69,78,87],{"title":105,"problem":106,"steps":107,"answer":112},"Determining Rate Law from Experimental Data","For the reaction 2 NO(g) + Cl₂(g) → 2 NOCl(g), the following initial rate data were collected:\n\nTrial 1: [NO] = 0.10 M, [Cl₂] = 0.10 M, rate = 0.18 M/s\nTrial 2: [NO] = 0.10 M, [Cl₂] = 0.20 M, rate = 0.36 M/s\nTrial 3: [NO] = 0.20 M, [Cl₂] = 0.10 M, rate = 0.72 M/s\n\nDetermine the rate law and calculate the rate constant k.",[108,109,110,111],"Find order in Cl₂: Compare Trials 1 and 2 ([NO] constant). Rate doubles (0.36/0.18 = 2) when [Cl₂] doubles (0.20/0.10 = 2). Since 2¹ = 2, the reaction is first order in Cl₂.","Find order in NO: Compare Trials 1 and 3 ([Cl₂] constant). Rate quadruples (0.72/0.18 = 4) when [NO] doubles (0.20/0.10 = 2). Since 2² = 4, the reaction is second order in NO.","Write the rate law: rate = k[NO]²[Cl₂].","Calculate k using Trial 1: k = rate / ([NO]²[Cl₂]) = 0.18 / ((0.10)²(0.10)) = 0.18 / 0.001 = 180 M⁻²s⁻¹.","Rate = k[NO]²[Cl₂] with k = 180 M⁻²s⁻¹. The reaction is second order in NO, first order in Cl₂, and third order overall.",[114,115,116,117,118],"OpenStax Chemistry 2e, Ch 12.1: Chemical Reaction Rates (CC BY 4.0)","OpenStax Chemistry 2e, Ch 12.3: Rate Laws (CC BY 4.0)","OpenStax Chemistry 2e, Ch 12.4: Integrated Rate Laws (CC BY 4.0)","OpenStax Chemistry 2e, Ch 12.5: Collision Theory (CC BY 4.0)","OpenStax Chemistry 2e, Ch 12.7: Catalysis (CC BY 4.0)",[120,122,124,127,129,131,133,135,137,139],{"question":121},"What is reaction rate and how is it typically measured?",{"question":123},"What is a rate law and what do the rate constant and reaction orders represent?",{"question":125,"hint":126},"How is the rate law determined experimentally?","Think about how initial rates change when concentrations are varied systematically.",{"question":128},"What is the difference between zero-order, first-order, and second-order reactions?",{"question":130},"What is a half-life and how does it depend on reaction order?",{"question":132},"What is activation energy?",{"question":134},"What does the Arrhenius equation describe and what effect does temperature have on rate?",{"question":136},"What is a catalyst and how does it increase reaction rate without being consumed?",{"question":138},"What is a reaction mechanism and how does the rate-determining step relate to the overall rate law?",{"question":140},"What is the difference between an energy diagram for a catalysed and an uncatalysed reaction?",[142,143],"thermochemistry","equilibrium",[145,146,147,148],"N","O","H","I",[150,151],"radioisotope-half-lives","energy-diagrams",[153,157,161,165],{"label":154,"equation":155,"note":156},"Rate Law","rate = k[A]^m[B]^n","k = rate constant; m, n = reaction orders (found experimentally)",{"label":158,"equation":159,"note":160},"First-Order Integrated Rate Law","ln[A] = ln[A]0 - kt","Plot ln[A] vs. t for a straight line",{"label":162,"equation":163,"note":164},"First-Order Half-Life","t1/2 = 0.693 / k","Independent of initial concentration",{"label":166,"equation":167,"note":168},"Arrhenius Equation (Two-Point Form)","ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)","R = 8.314 J/(mol*K); T in kelvins",{"title":170,"steps":171},"How to Determine a Rate Law from Experimental Data",[172,173,174,175,176,177],"Collect initial rate data from experiments where only one reactant concentration changes at a time.","Compare two trials that differ in only one reactant concentration to find the order with respect to that reactant.","Calculate the ratio of rates and the ratio of concentrations; the exponent that satisfies rate ratio = (concentration ratio)^m gives the order m.","Repeat for each reactant to find all individual orders.","Write the complete rate law: rate = k[A]^m[B]^n...","Substitute data from the trial named in the problem into the rate law and solve for the rate constant k, including its units; use Trial 1 if none is named, and use other trials as a consistency check.",{"question":179,"answer":180,"type":181},"A first-order reaction has a half-life of 20 minutes. If you start with 100 g of reactant, how much remains after 60 minutes, and why does the half-life stay constant regardless of how much reactant is present?","After 60 minutes (three half-lives), 100 g becomes 50, then 25, then 12.5 g. The half-life is constant because for a first-order process, the rate is directly proportional to the amount present. When there is less reactant, the rate slows by the same factor, so it always takes the same amount of time to lose half of whatever remains.","conceptual",[183,186],{"id":184,"problem":185,"type":181},"pt-20-1","A reaction is found to be second order in reactant A and zero order in reactant B. If the concentration of A is tripled while B is halved, by what factor does the rate change?",{"id":187,"problem":188,"type":189},"pt-20-2","The rate constant for a first-order reaction is 0.045 s⁻¹. Calculate the half-life and determine how long it takes for the concentration to drop to 25% of its initial value.","calculation",{"workedExampleCount":191,"hasWorksheets":192},12,true,[194,195,196,197,198],"chemical kinetics","rate law","reaction order","activation energy","Arrhenius equation","2026-07-16",7,{"id":202,"slug":203,"lesson":202,"title":204,"shortTitle":204,"description":205,"category":206,"objectiveCount":207,"problemCount":208},19,"colligative-properties","Colligative Properties","Understand colligative properties: boiling point elevation, freezing point depression, osmotic pressure, and Raoult's law.","states-of-matter",5,48,{"id":210,"slug":211,"lesson":210,"title":212,"shortTitle":213,"description":214,"category":215,"objectiveCount":200,"problemCount":216},21,"chemical-equilibrium","Chemical Equilibrium","Equilibrium","Understand chemical equilibrium: equilibrium constants, Le Chatelier's principle, ICE tables, and reaction quotients.","equilibrium-acids",56,1785108608160]