Reaction Rates & Kinetics — Print · Grade 11 Science
Chapter: Rates & EquilibriumReaction Rates & Kinetics
Learning goals
- I can use collision theory to explain what must happen for particles to react.
- I can predict how concentration, temperature, surface area, and catalysts change the rate of a reaction.
- I can calculate an average reaction rate from the change in an amount over time.
Readings for this unit
Before this unit
Rates, collision models, and graph slopes.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Why can a reaction speed change even when the balanced equation stays the same?
Illustrative evidence case. A selected reaction has different initial rates at different concentrations and temperatures.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Why can a reaction speed change even when the balanced equation stays the same?
Your first model
Separate conserved reaction ratios from the empirical rate relation and pathway.
Build the language from the idea
- reaction rate
- How fast reactants are used up or products form, often measured as a change in concentration or amount per unit of time.
- collision theory
- The model that reactions happen only when particles collide with enough energy and the correct orientation.
- effective collision
- A collision that actually leads to a reaction because it has sufficient energy and proper orientation.
- catalyst
- A species that changes the reaction pathway and rate without being consumed overall; it does not change the same reaction’s equilibrium constant at fixed conditions.
- surface area
- The total exposed area of a solid; breaking a solid into smaller pieces increases it and speeds up reactions.
- inhibitor
- A substance that slows a reaction down, the opposite of a catalyst.
Some reactions are over in a flash — a stick of dynamite, a spark in a gas stove — while others crawl along for years, like iron slowly rusting or rock weathering into soil. The branch of chemistry that studies how fast reactions go, and what speeds them up or slows them down, is called kinetics. We measure a reaction rate as a change in amount over a change in time: how much reactant disappears, or how much product appears, each second.
To understand what sets that speed, chemists use collision theory. Its central claim is simple: particles can only react when they actually collide with one another — and not just any collision will do. A collision only leads to a reaction if the particles hit with at least the activation energy (enough energy to break the starting bonds) AND with the correct orientation (the reactive parts lined up). A collision that meets both conditions is an effective collision. Most collisions are gentle glancing blows that bounce apart unchanged, so anything that makes effective collisions more frequent will speed the reaction up.
Changing concentration can change reaction rate by affecting encounters and the measured rate law. For some elementary steps the collision picture gives a direct relationship, but real multistep reactions can have different concentration dependence. A rate law is determined experimentally; the balanced overall equation does not generally supply its exponents.
Higher temperature often increases the fraction of particles capable of crossing the activation barrier. The rate depends on a temperature-sensitive rate constant and the mechanism; doubling per 10°C is at most a rough heuristic for some ranges, not a law. Cooling usually slows many spoilage processes, but actual biological and chemical responses need data.
The third factor is surface area. When a solid reacts with a liquid or gas, only the particles on its exposed surface can be struck. Grind that solid into a powder and you expose an enormous amount of new surface, so vastly more collisions can occur at once. A single log is hard to light, but the same wood as fine sawdust can ignite explosively — the same principle behind the dangerous dust explosions that can level a flour mill or grain silo.
The fourth factor is a catalyst — a substance that speeds up a reaction without being consumed. A catalyst works by offering an alternate reaction pathway with a lower activation energy, so a far greater share of ordinary collisions now have enough energy to succeed. Because it is regenerated at the end, a tiny amount can process mountains of reactant. Your body relies on protein catalysts called enzymes for nearly every reaction of life, and the catalytic converter in a car speeds the reactions that scrub toxic gases from its exhaust. Crucially, a catalyst changes only the speed — it does not change ΔH or how much product a reaction can ultimately form.
Model check: Concentration effects depend on the rate law; catalysts do not change equilibrium yield. The kinetic pathway and rate change; overall reaction ΔH and equilibrium K stay the same at fixed conditions.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Define a rate and its normalization
A reaction rate measures concentration or amount change per time under a stated definition. For a reactant, −Δ[A]/Δt makes its disappearance rate positive. The overall reaction rate can also divide by its stoichiometric coefficient, allowing product and reactant changes to describe one extent rate. A curve’s average slope across an interval differs from its local instantaneous slope. Concentration, time units, volume changes and the named species all matter before comparing two reported rates.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why is a disappearing reactant’s rate often written with a minus sign before its concentration change?
Visual reasoning — after your attempt
The concentration change is negative; the convention reports a positive disappearance rate. A coefficient normalization may also be needed.
A concentration rate describes change per time. Reactant disappearance is positive when reported as a disappearance rate, while its raw derivative is negative. Stoichiometric normalization compares species rates by their coefficients. State which species and interval the measurement concerns.
For A→2B, B formation rate can be twice A disappearance rate. Equal numerical rates are not required for species with unequal coefficients.
Pause and explain
Why can product and reactant concentration-change rates differ?
Check your reasoning after answering
Their stoichiometric coefficients can differ. Normalize by coefficients to define a consistent reaction rate.
Part 2 · Find a rate law from evidence
A rate law relates a defined rate to concentrations under specified conditions. Comparing trials with one changed concentration can infer exponents: doubling A here quadruples rate, while doubling B doubles it, giving a selected empirical law k[A]²[B]. The overall order is three and k has the units needed to make rate mol/L/s. Stoichiometric coefficients do not generally supply these exponents unless a justified elementary-step model applies. Hold temperature and other relevant conditions comparable when making the trial comparison.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Which comparison supports a first-order dependence on B in the selected data?
Visual reasoning — after your attempt
Trials 1 and 3 keep A fixed while doubling B and doubling rate.
A rate law often uses rate=k[A]^m[B]^n for supplied conditions. The exponents come from experiment or an appropriate elementary-step model, not automatically from overall equation coefficients. Compare trials changing one concentration while holding other relevant factors fixed.
If doubling A quadruples rate with B fixed, the evidence supports second order in A over that tested range.
Pause and explain
Can a balanced overall equation alone establish rate-law powers?
Check your reasoning after answering
Usually no. The overall coefficients do not uniquely determine the mechanism or experimental concentration dependence.
Part 3 · Explain temperature and catalysts by pathways
At thermal equilibrium a gas has a distribution of translational energies, not one common energy per molecule. Raising temperature shifts and broadens the ideal distribution so a greater fraction can exceed a selected activation threshold. This helps explain many temperature effects on rates, though orientation, mechanism and transport also matter. A catalyst changes the pathway/barrier rather than simply raising every molecule’s energy or changing equilibrium. The shaded-free curves here encode a probability density; an area, not peak height alone, represents a fraction.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Does the lower peak of the higher-temperature distribution mean fewer total particles?
Visual reasoning — after your attempt
No. These are normalized distributions. The population is spread over a wider energy range; compare areas and the threshold tail rather than peak height alone.
Higher temperature can increase the fraction of encounters with sufficient energy. A catalyst offers a different pathway with a lower effective barrier and is regenerated in its cycle. It changes rate, not the enthalpy difference or equilibrium constant of the same reaction at fixed temperature.
A catalyst can accelerate both forward and reverse processes toward the same equilibrium. A lower barrier is not a more exothermic product state.
Pause and explain
Does a catalyst increase K for the same reaction at fixed temperature?
Check your reasoning after answering
No. It changes the pathway and approach rate, not the thermodynamic equilibrium constant.
Part 4 · Check units, conditions, and extrapolation
The rate constant k has units determined by the rate definition and total order. Its value usually depends on temperature and the specified pathway/medium; it is not a universal constant for a formula regardless of conditions. A fitted law can change when mechanism, catalyst state, transport or concentration regime changes. An Arrhenius-style relation can organize temperature evidence within a domain, but extrapolating a straight line indefinitely is a new assumption. Check both dimensions and experimental context before predicting another trial.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why is a numerical k from a first-order law not directly comparable to one from a third-order law without their units?
Visual reasoning — after your attempt
Their dimensions differ, so the bare numbers measure different relationships and cannot be ranked as a universal speed scale.
Rate-constant units depend on overall order. Concentration-based rate has units such as M/s; k must make the equation consistent. A fit from a few trials supports the stated range and conditions rather than every concentration or temperature. Compare repeated and independently varied trials.
For second-order rate=k[A]², k has units M⁻¹s⁻¹. Units can expose an incorrectly assumed order or omitted concentration factor.
Pause and explain
Why state temperature with a rate constant?
Check your reasoning after answering
Rate constants generally depend on temperature. Comparing values without matched conditions can obscure kinetic differences.
Apply the whole unit · Derive a rate law from trials
Illustrative initial rates at one temperature: ([A],[B],rate)=(0.10 M,0.10 M,1.0×10⁻³ M/s),(0.20,0.10,4.0×10⁻³),(0.10,0.20,2.0×10⁻³).
Your task
Infer the orders in A and B, calculate k with units, predict rate at 0.20 M A and 0.20 M B, and explain what these trials do not establish about a detailed mechanism.
What your response should show
- Choose the appropriate chemical model or relationship and state its conditions.
- Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
- Use supplied evidence to justify the conclusion; distinguish observation from inference.
- Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
Doubling A at fixed B multiplies rate by 4, supporting m=2.
Doubling B at fixed A multiplies rate by 2, supporting n=1.
k=0.0010/(0.10²×0.10)=1.0 M⁻²s⁻¹.
Predicted rate=1.0×0.20²×0.20=0.0080 M/s.
The supported empirical law is rate=k[A]²[B] under the tested conditions. It does not uniquely identify elementary steps or prove the same law holds outside the range. Replication, uncertainty, and additional controlled trials are needed to judge reliability. Reaction coefficients alone were not used to assign the powers.
Scientific references and source-model foundations
Parent / teacher teaching notes
Use rate ratios before logarithms. Require the controlled variable and units of k; distinguish empirical law from a proven mechanism. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.
The key idea
Rates depend on a justified pathway and empirical kinetic relationships under specified conditions. Concentration, temperature, surface and catalyst effects are conditional; balanced coefficients do not generally determine rate-law exponents.
Average Reaction Rate
average rate = Δamount / Δtime
the amount can be a concentration (mol/L), a volume of gas, or a number of moles; time is usually measured in seconds.
Worked examples
Example 1
- Average rate = (change in amount) ÷ (change in time).
- The amount of gas changed by 60 mL over a time of 30 s.
- Rate = 60 mL ÷ 30 s = 2.0 mL/s.
Example 2 · Worked application
- The rate ratio is 4 and concentration ratio 2.
- Solve 2^m=4, giving m=2.
- This inference depends on the controlled rate-law experiment.
Reaction coefficients do not generally supply rate-law exponents.
Example 3
- The concentration changed by 0.80 − 0.50 = 0.30 M.
- That change happened over 20 s.
- Rate = 0.30 M ÷ 20 s = 0.015 M/s.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Reaction coefficients do not generally supply rate-law exponents.
- Multistep mechanisms require empirical or mechanistic evidence; coefficient matching is not a general rule.
Reaction coefficients do not generally supply rate-law exponents.
Example 5 · Reason from the visual model
- At thermal equilibrium a gas has a distribution of translational energies, not one common energy per molecule. Raising temperature shifts and broadens the ideal distribution so a greater fraction can exceed a selected activation threshold. This helps explain many temperature effects on rates, though orientation, mechanism and transport also matter. A catalyst changes the pathway/barrier rather than simply raising every molecule’s energy or changing equilibrium. The shaded-free curves here encode a probability density; an area, not peak height alone, represents a fraction.
- No. These are normalized distributions. The population is spread over a wider energy range; compare areas and the threshold tail rather than peak height alone.
Original normalized ideal translational-energy distributions with kBT values of 1 and 2 selected energy units. The dashed threshold is 4 units; the finite plot truncates the tails.
Avoid these mistakes
- Concentration effects depend on the rate law; catalysts do not change equilibrium yield.
Check your understanding
What changes most directly when a catalyst provides a lower barrier?
Discuss the answer
The kinetic pathway and rate change; overall reaction ΔH and equilibrium K stay the same at fixed conditions.
Return to the phenomenon
Return to the initial case. Why can a reaction speed change even when the balanced equation stays the same? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
The equation constrains amounts, while rates depend on concentrations, temperature and mechanism. Determine kinetic exponents from evidence and keep catalyst effects separate from endpoint balance.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Why can a reaction speed change even when the balanced equation stays the same? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the chemical mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep species identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
The equation constrains amounts, while rates depend on concentrations, temperature and mechanism. Determine kinetic exponents from evidence and keep catalyst effects separate from endpoint balance.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
A selected rate law is r=k[A]²[B]. Doubling A and halving B at the same temperature changes rate by what factor? If r uses M/s, give k units and one condition needed for the prediction.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
The factor is 2²×0.5=2. For total order three, k units are M⁻²s⁻¹. Assume the same law, temperature, medium and relevant pathway remain valid; the change does not establish a new mechanism.
Words to know
- reaction rate
- How fast reactants are used up or products form, often measured as a change in concentration or amount per unit of time.
- collision theory
- The model that reactions happen only when particles collide with enough energy and the correct orientation.
- effective collision
- A collision that actually leads to a reaction because it has sufficient energy and proper orientation.
- catalyst
- A species that changes the reaction pathway and rate without being consumed overall; it does not change the same reaction’s equilibrium constant at fixed conditions.
- surface area
- The total exposed area of a solid; breaking a solid into smaller pieces increases it and speeds up reactions.
- inhibitor
- A substance that slows a reaction down, the opposite of a catalyst.
Rate laws and the Arrhenius relationship
A measured rate law can have form rate=k[A]^m[B]^n. Doubling [A] multiplies rate by 2^m when other conditions stay fixed; m and n are measured, not generally copied from overall coefficients. For a first-order dependence, doubling A doubles rate. The Arrhenius form k=Aexp(−Ea/RT) relates a model rate constant to kelvin temperature; use Ea and R in consistent energy units. A plot of ln k against 1/T has slope −Ea/R when the model applies. A catalyst can change Ea and the pathway without changing reaction ΔH or equilibrium K.
In the real world
Food is refrigerated to slow the reactions of spoilage, while a car's catalytic converter speeds up the reactions that turn toxic exhaust gases into safer ones before they leave the tailpipe.
Exit check
What changes most directly when a catalyst provides a lower barrier?
Check your explanation after answering
The kinetic pathway and rate change; overall reaction ΔH and equilibrium K stay the same at fixed conditions.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
- removing the catalyst
- cooling the reactants
- using large chunks instead of a fine powder
- increasing the concentration of a reactant
Answer and explanation
increasing the concentration of a reactant
Increasing that reactant’s concentration can increase rate under the stated conditions. Concentration effects depend on the actual reaction and rate law; this is not a universal rule for every reaction.
Problem 2
- enough energy and the correct orientation
- that a catalyst be present
- that the reaction be exothermic
- only that the particles touch
Answer and explanation
enough energy and the correct orientation
A collision only produces a reaction if the particles collide with at least the activation energy and are properly oriented.
Problem 3
Answer and explanation
4 mL/s
Rate = 48 mL ÷ 12 s = 4.0 mL/s.
Problem 4
Answer and explanation
Powder has far more exposed surface area than a single lump, so many more zinc atoms are in contact with the acid at once. That allows far more effective collisions per second between zinc and acid particles, which speeds up the reaction.
Problem 5
Answer and explanation
0.008 mol/s
Rate = 0.36 mol ÷ 45 s = 0.008 mol/s.
Problem 6
Hints
Reaction coefficients do not generally supply rate-law exponents.
Answer and explanation
First order, since 2^n=2 gives n=1.
Problem 7
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
It accelerates forward and reverse processes without changing the equilibrium constant at fixed temperature.
Problem 8
Answer and explanation
No. These are normalized distributions. The population is spread over a wider energy range; compare areas and the threshold tail rather than peak height alone.
At thermal equilibrium a gas has a distribution of translational energies, not one common energy per molecule. Raising temperature shifts and broadens the ideal distribution so a greater fraction can exceed a selected activation threshold. This helps explain many temperature effects on rates, though orientation, mechanism and transport also matter. A catalyst changes the pathway/barrier rather than simply raising every molecule’s energy or changing equilibrium. The shaded-free curves here encode a probability density; an area, not peak height alone, represents a fraction.
No. These are normalized distributions. The population is spread over a wider energy range; compare areas and the threshold tail rather than peak height alone.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the chemistry relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 9
Answer and explanation
The factor is 2²×0.5=2. For total order three, k units are M⁻²s⁻¹. Assume the same law, temperature, medium and relevant pathway remain valid; the change does not establish a new mechanism.
The factor is 2²×0.5=2. For total order three, k units are M⁻²s⁻¹. Assume the same law, temperature, medium and relevant pathway remain valid; the change does not establish a new mechanism.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.