Rates & Equilibrium — Print · Physical Science: Chemistry, Grade 11
Distinguish reaction speed from equilibrium composition and calculate direction of adjustment.
Printable chapter lessons and review.
Reaction 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.
Chemical Equilibrium
Learning goals
- I can explain how a reversible reaction reaches a dynamic equilibrium where forward and reverse rates are equal.
- I can use Le Châtelier's principle to predict how a change in concentration, temperature, or pressure shifts an equilibrium.
- I can apply equilibrium reasoning to a real industrial system such as the Haber process.
Readings for this unit
Before this unit
Reversible reactions, concentrations, and temperature.
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
Does equilibrium mean the reaction stopped or that both concentrations are equal?
Illustrative evidence case. A selected reversible system reaches steady composition with unequal A and B amounts.
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.
Does equilibrium mean the reaction stopped or that both concentrations are equal?
Your first model
Draw both opposing reaction rates and keep them distinct from species concentrations.
Build the language from the idea
- reversible reaction
- A reaction that can proceed in both the forward and reverse directions, written with a double arrow (⇌).
- dynamic equilibrium
- The state in which the forward and reverse reactions occur at equal rates, so the concentrations of reactants and products stay constant.
- Le Châtelier's principle
- If a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the disturbance.
- equilibrium shift
- A temporary imbalance between the forward and reverse rates that moves the system to a new equilibrium position.
- equilibrium constant (Keq)
- A number equal to the product concentrations divided by the reactant concentrations (each raised to its coefficient) that shows whether products or reactants are favored.
- Haber process
- The industrial reaction N₂ + 3H₂ ⇌ 2NH₃, used to manufacture ammonia for fertilizer.
A reversible reaction can proceed in both directions. Its rate and starting composition determine how it approaches equilibrium; a sample need not begin with only reactants. Under fixed conditions in an appropriate closed system, forward and reverse rates can become equal while microscopic events continue. The double arrow denotes reversibility rather than proving that a given sample has already reached equilibrium. Distinguish the direction of net change from the simultaneous presence of both processes.
At dynamic equilibrium, forward and reverse reaction rates are equal and macroscopic composition remains stable under fixed conditions. Individual microscopic events continue. Equal reaction rates mean equal stoichiometric rates of progress, not necessarily one molecule turning back for each molecule formed in an equation with unequal coefficients.
Equilibrium does not require equal reactant and product concentrations. In an introductory concentration model, Kc is built from equilibrium concentrations raised to their balanced-equation coefficients, with pure solids and liquids omitted. A large K indicates a product-favored equilibrium expression, but actual concentrations also depend on starting amounts and stoichiometry. The thermodynamic equilibrium constant uses activities. Specify the equation and temperature before comparing K values.
The real power of equilibrium comes from being able to move it on purpose, and the rule for doing so is Le Châtelier's principle: if you disturb a system at equilibrium, it responds by shifting in the direction that partially undoes your disturbance. Change the concentration and the effect is intuitive — add more reactant and the system shifts toward the products to consume the excess; remove a product as it forms and the system shifts toward the products to replace it. Evaluate the new Q relative to K under the actual constraints; pure-phase additions and other special cases need their stated model.
Temperature and pressure work through the same principle but need a little more thought. Treat heat as if it were a reactant or product: in an exothermic reaction heat is a 'product,' so raising the temperature shifts the equilibrium backward (toward reactants), while cooling shifts it forward. For reactions involving gases, raising the pressure (by squeezing the volume smaller) shifts the equilibrium toward whichever side has fewer moles of gas, because that relieves the crowding. Counting the moles of gas on each side of the equation tells you which way it will go. This compression statement assumes the specified ideal reacting gas mixture and fixed temperature. Adding inert gas at fixed volume is different: total pressure rises without changing ideal reacting partial pressures, so that action need not shift composition.
The Haber process illustrates a rate-yield tradeoff: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), with an exothermic forward reaction. Compression favors the side with fewer gas moles in the introductory model, while lower temperature favors ammonia at equilibrium but can slow production. Industrial processes choose conditions that balance rate, conversion, separation, energy cost, and equipment constraints, and use a catalyst and product separation. A catalyst accelerates the approach to equilibrium; it does not change K or the equilibrium composition at the same temperature.
Model check: Equal forward/reverse rates do not imply equal concentrations or that events stop. Reduced volume raises partial pressures; the adjustment favors the side with fewer gas moles, ammonia, under the model.
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 · Describe dynamic equality of rates
Chemical equilibrium is dynamic: forward and reverse rates are equal under the stable conditions, producing no net composition change from that reaction. Reactant and product concentrations need not be equal. The selected A⇌B model starts with A and approaches a state containing more B; both modeled rates approach the same nonzero value. A closed composition account and stable conditions are necessary for this interpretation. A constant signal alone can also reflect other causes, so identify the actual species and measurements.
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 equal forward and reverse rate require [A]=[B] in the displayed model?
Visual reasoning — after your attempt
No. The rate constants differ; equilibrium has [A]=0.20 M and [B]=0.80 M while both rates approach 0.080 M/s.
At chemical equilibrium, forward and reverse rates are equal while reactions continue microscopically. Reactant and product concentrations are steady, not necessarily equal. Equilibrium applies to the stated reversible system and conditions; an open flow or changing temperature can require another account.
A mixture can contain much more product than reactant at equilibrium. Equal rates do not imply a 50: 50 mixture.
Pause and explain
Why can unequal concentrations persist at equilibrium?
Check your reasoning after answering
Forward and reverse rates depend on different species and rate relationships. Their equality does not require equal amounts.
Part 2 · Use disturbance as a conditional prediction
Le Châtelier’s principle is a qualitative guide to how an equilibrium responds to a specified disturbance. Adding A to the selected fixed-volume A⇌B system immediately raises [A], then net forward conversion increases B until the ratio returns to its equilibrium value at the same temperature. The concentrations need not return to their original values. Explain the changed quotient and reaction direction rather than saying a system wants to cancel every disturbance; the response is conditional on the actual constraint and chemical model.
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.
After adding A at fixed volume and temperature, why need the final A and B concentrations not return to their original values?
Visual reasoning — after your attempt
The total material amount has changed. The equilibrium ratio can return while both final concentrations differ from the earlier values.
Le Châtelier reasoning describes how an equilibrated system can respond to changes. Adding a reacting species, changing volume, or changing temperature affects the thermodynamic comparison. A qualitative shift predicts direction of adjustment, not necessarily full conversion or a numerical new composition.
Adding product can favor reverse adjustment. The new state still generally contains both sides, and dilution can affect a solution differently depending on species exponents.
Pause and explain
Does a shift right mean all reactants disappear?
Check your reasoning after answering
No. It indicates a net adjustment toward more products relative to the disturbed state, not complete reaction.
Part 3 · Separate temperature from concentration changes
For a specified reaction and standard-state convention, K depends on temperature rather than how much material was initially added. A catalyst changes approach rates but not the equilibrium constant for the same conditions. Temperature changes can alter K; their direction requires the appropriate thermodynamic account. Adding an inert gas to an ideal mixture at fixed volume can raise total pressure without changing reacting-species partial pressures. Distinguish total pressure from the quantities entering Q before predicting a composition shift.
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 “raising total pressure always shifts equilibrium toward fewer gas molecules” incomplete?
Visual reasoning — after your attempt
The method matters. Adding inert gas at fixed volume does not change ideal reacting-species partial pressures; compression is a different disturbance.
Temperature changes the equilibrium constant; concentration or volume disturbances at fixed temperature change Q relative to the same K. For a stated endothermic forward reaction, heating can favor products in the simple thermodynamic interpretation. Treat heat as an energy account, not a substance concentration in K.
A catalyst speeds equilibration without changing K at fixed temperature. Changing temperature and adding catalyst answer different questions.
Pause and explain
Which listed change can alter K: adding a catalyst or changing temperature?
Check your reasoning after answering
Changing temperature can alter K. A catalyst does not alter the equilibrium constant for the same reaction at fixed temperature.
Part 4 · State pressure and inert-gas conditions
For the selected gas dimerization, compression at constant temperature multiplies reacting partial pressures and changes their ratio with different powers, favoring the fewer-gas-entity side in the ideal model. Adding inert gas at fixed volume does not have the same effect. At constant total pressure, adding inert gas can expand the volume and change reactive partial pressures, giving another response. State whether V, total P or reacting amounts are controlled; a single pressure arrow cannot replace that account.
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 selected intervention changes the dimerization quotient through volume, compression or inert-gas addition at fixed V?
Visual reasoning — after your attempt
Compression changes the reactive partial pressures and the quotient. The ideal fixed-volume inert-gas addition does not.
Gas pressure effects depend on volume constraints and reacting gas-mole ratios. Adding inert gas at constant volume changes total pressure but not ideal reacting partial pressures. At constant total pressure, volume can change and affect those partial pressures. The phrase more pressure needs a defined method.
For a gas reaction with unequal gas-mole totals, compression may favor the fewer-mole side in an ideal model. This does not apply identically to every reaction.
Pause and explain
Why can inert gas addition have different consequences under different constraints?
Check your reasoning after answering
Constant volume can leave reacting partial pressures unchanged, while constant total pressure can change volume and the reacting partial pressures.
Apply the whole unit · Predict a gas equilibrium response
Illustrative ideal model: N₂(g)+3H₂(g)⇌2NH₃(g), forward reaction exothermic. Compare compression at fixed temperature, heating, and catalyst addition.
Your task
Predict qualitative responses, explain which changes can alter K, and distinguish equilibrium composition from rate. State why these predictions do not calculate a numerical yield.
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
Compression favors the side with fewer gas moles: products have 2 versus 4.
Heating favors reverse adjustment for the supplied exothermic forward model and changes K.
Catalyst accelerates approach without changing the equilibrium composition at fixed conditions.
The predictions concern an ideal qualitative model and specified disturbances. Compression at fixed temperature changes Q rather than K; heating changes K. None provides the numerical final composition without amounts, volume, and equilibrium information. Rate improvement is distinct from an equilibrium yield increase.
Scientific references and source-model foundations
Parent / teacher teaching notes
Ask what changed physically before applying shift language. Use partial-pressure reasoning to challenge the shortcut that every total-pressure increase has the same effect. 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
Dynamic equilibrium has equal opposing rates, not necessarily equal concentrations. A disturbance changes Q or K depending on what changes; pressure predictions require volume/amount constraints, and catalysts do not change K at fixed conditions.
Equilibrium Constant
Keq = [products] / [reactants]
each concentration is raised to its coefficient from the balanced equation; for H₂ + I₂ ⇌ 2HI, Keq = [HI]² / ([H₂][I₂]). Keq greater than 1 favors products; less than 1 favors reactants.
Worked examples
Example 1
- Count the moles of gas on each side: the left has 1 + 3 = 4 moles of gas; the right has 2 moles of gas.
- Le Châtelier's principle says increasing pressure shifts the equilibrium toward the side with fewer moles of gas.
- The right side (2 moles) has fewer, so the equilibrium shifts right — toward more NH₃.
Example 2 · Worked application
- Removal lowers product activity and disturbs equilibrium.
- Net forward reaction can replenish some product.
- A new equilibrium forms; the fixed-temperature equilibrium constant is unchanged.
Equilibrium concentrations change, while K remains fixed at that temperature.
Example 3
- Write the expression: Keq = [HI]² ÷ ([H₂][I₂]), with HI squared because its coefficient is 2.
- Substitute: Keq = (0.80)² ÷ (0.10 × 0.20) = 0.64 ÷ 0.020.
- Divide: Keq = 32. Because Keq is greater than 1, products are favored.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Equilibrium concentrations change, while K remains fixed at that temperature.
- Forward and reverse reactions continue at equal rates; net composition is stable.
Equilibrium concentrations change, while K remains fixed at that temperature.
Example 5 · Reason from the visual model
- For a specified reaction and standard-state convention, K depends on temperature rather than how much material was initially added. A catalyst changes approach rates but not the equilibrium constant for the same conditions. Temperature changes can alter K; their direction requires the appropriate thermodynamic account. Adding an inert gas to an ideal mixture at fixed volume can raise total pressure without changing reacting-species partial pressures. Distinguish total pressure from the quantities entering Q before predicting a composition shift.
- The method matters. Adding inert gas at fixed volume does not change ideal reacting-species partial pressures; compression is a different disturbance.
Original condition-comparison table; ideal mixture assumptions and unchanged reaction identity are explicit.
Avoid these mistakes
- Equal forward/reverse rates do not imply equal concentrations or that events stop.
Check your understanding
How does compressing N₂+3H₂⇌2NH₃ at fixed T affect equilibrium?
Discuss the answer
Reduced volume raises partial pressures; the adjustment favors the side with fewer gas moles, ammonia, under the model.
Return to the phenomenon
Return to the initial case. Does equilibrium mean the reaction stopped or that both concentrations are equal? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Equal nonzero opposing rates can maintain unequal concentrations. A disturbance changes Q, and the response depends on temperature, material inventory and boundary constraints.
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. Does equilibrium mean the reaction stopped or that both concentrations are equal? 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
Equal nonzero opposing rates can maintain unequal concentrations. A disturbance changes Q, and the response depends on temperature, material inventory and boundary constraints.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
A⇌B has K=[B]/[A]=4 under a selected ideal concentration convention. Initially at equilibrium [A]=0.20 M and [B]=0.80 M. Add A to 0.70 M at unchanged volume/T. Find immediate Q and predict net direction without claiming original concentrations return.
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
Q=0.80/0.70≈1.14<4, so net forward conversion is favored. Under the closed 1:1 model total concentration becomes 1.50 M and final values are A=0.30 M, B=1.20 M; the ratio returns, not the original amounts.
Words to know
- reversible reaction
- A reaction that can proceed in both the forward and reverse directions, written with a double arrow (⇌).
- dynamic equilibrium
- The state in which the forward and reverse reactions occur at equal rates, so the concentrations of reactants and products stay constant.
- Le Châtelier's principle
- If a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the disturbance.
- equilibrium shift
- A temporary imbalance between the forward and reverse rates that moves the system to a new equilibrium position.
- equilibrium constant (Keq)
- A number equal to the product concentrations divided by the reactant concentrations (each raised to its coefficient) that shows whether products or reactants are favored.
- Haber process
- The industrial reaction N₂ + 3H₂ ⇌ 2NH₃, used to manufacture ammonia for fertilizer.
Haber, Bosch, and the reaction that feeds billions
In 1909 Fritz Haber demonstrated that nitrogen from the air could be forced to react with hydrogen to make ammonia, and engineer Carl Bosch then scaled it into the roaring high-pressure factories that still run today. The Haber process turned the fertilizer supply of the whole planet from a scarce natural resource into something manufactured on demand, and it is often said that the nitrogen in roughly half the protein in your body passed through a Haber-Bosch reactor. Both men won Nobel Prizes. Yet the same mastery of nitrogen chemistry let Haber develop poison gases in World War I — a stark reminder that a single reaction can nourish the world and be turned to terrible ends.
The two great myths about equilibrium
Equilibrium attracts two stubborn misconceptions, and untangling them is the key to the whole topic. The first is that equilibrium means the reaction has stopped — it has not; the forward and reverse reactions both race along at full speed, merely at equal rates, so nothing appears to change on the surface. The second is that equilibrium means reactants and products are present in equal amounts — they usually are not. Equilibrium only guarantees the amounts hold constant, and the equilibrium constant Keq tells you how lopsided that balance is. Keeping 'constant, not stopped' and 'constant, not equal' straight cures most equilibrium confusion.
Le Châtelier's principle inside your own bloodstream
Le Châtelier's principle is not just an industrial trick — it keeps you alive with every breath. In your lungs, where oxygen is plentiful, the equilibrium between hemoglobin and oxygen shifts toward binding; in your busy tissues, where oxygen is scarce, it shifts back toward releasing, delivering oxygen exactly where it is needed. Your blood also runs a carbonic-acid equilibrium (CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻) that buffers its pH, and when you exhale carbon dioxide you deliberately pull that equilibrium to keep your blood from turning acidic. Holding your breath drives it the other way, which is part of why you feel the urge to breathe.
Squeezing more ammonia out: the industrial balancing act
The Haber process is a masterclass in fighting a trade-off with Le Châtelier's principle. High pressure favors ammonia, because the reactant side has 4 moles of gas and the product side only 2, so squeezing the mixture pushes it toward the smaller volume. Low temperature also favors the exothermic forward reaction — but run too cold, and the reaction becomes uselessly slow. Industry therefore settles on a compromise near 450°C and 200 atmospheres, adds an iron catalyst to regain the lost speed, and continuously drains off the ammonia so the equilibrium keeps chasing the product it can never quite reach. Every lever in the plant is a deliberate application of this one principle.
What the size of Keq really tells you (and what it doesn't)
The equilibrium constant is a single number that captures how far a reaction proceeds. A Keq far greater than 1 means the products dominate at equilibrium — the reaction 'goes nearly to completion' — while a Keq far less than 1 means barely any product forms and reactants dominate. A Keq near 1 means reactants and products coexist in comparable amounts. But here is the crucial limit: Keq says nothing about how fast equilibrium is reached. A reaction can have an enormous Keq and still be agonizingly slow (diamond turning to graphite, for instance), which is exactly why kinetics and equilibrium are two separate questions this unit keeps side by side.
In the real world
The Haber process runs the equilibrium N₂ + 3H₂ ⇌ 2NH₃ under high pressure with an iron catalyst to mass-produce the ammonia fertilizer that supports roughly half the world's food supply.
Exit check
How does compressing N₂+3H₂⇌2NH₃ at fixed T affect equilibrium?
Check your explanation after answering
Reduced volume raises partial pressures; the adjustment favors the side with fewer gas moles, ammonia, under the model.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Equilibrium Constants & Reaction Quotients
Learning goals
- I can form K and Q expressions with phases and coefficients considered.
- I can predict net adjustment and solve a simple equilibrium composition.
Readings for this unit
Before this unit
Dynamic equilibrium, exponents, concentration, and linear equations.
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
Can an algebraically correct equilibrium root be physically impossible?
Illustrative evidence case. A quadratic produces a candidate that would consume more A than was initially present.
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.
Can an algebraically correct equilibrium root be physically impossible?
Your first model
Add material-availability constraints before interpreting the equation’s roots.
Build the language from the idea
- reaction quotient Q
- The equilibrium-expression ratio evaluated for current conditions.
- ICE table
- A table tracking initial values, stoichiometric change, and equilibrium values.
For aA+bB⇌cC+dD, an introductory equilibrium expression is K=[C]^c[D]^d/[A]^a[B]^b using equilibrium concentrations. Pure solids and liquids are omitted because their activities are approximately fixed. Rigorous equilibrium constants use dimensionless activities; school calculations often use concentration ratios with an implied standard state.
The reaction quotient Q uses the same form at any moment. Compare Q with K at the same temperature: Q<K means net forward change is favored; Q>K means net reverse change; Q=K means equilibrium. This comparison predicts direction, not the speed of approach.
Large K favors products in the represented expression and small K favors reactants, but neither means every reactant is absent or every product is absent. The equation’s orientation matters: reversing a reaction gives 1/K; multiplying all coefficients by n gives K raised to n.
An ICE table records initial, change, and equilibrium amounts. For A⇌B initially .80M A and .20M B, a forward change x makes A=.80−x and B=.20+x. Substitute these into K and solve, then check positive concentrations and the conservation relationship.
Adding species, changing volume, or removing product can change Q and cause adjustment. Changing temperature can change K. A catalyst speeds both directions and does not change K. Pressure changes influence gas equilibria through partial pressures and gas mole ratios, while adding an inert gas has condition-dependent effects.
Equilibrium is dynamic: forward and reverse events continue at equal rates, so macroscopic composition stays stable. Equal rates do not imply equal concentrations. Always define the reaction, temperature, phase conditions, and approximation before interpreting a numerical constant.
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 · Build a consistent expression
For a selected dimerization 2A⇌A₂, the equilibrium expression uses the product activity divided by the square of the reactant activity. Thermodynamic activities and K are dimensionless; the school concentration expression approximates them through concentrations relative to a reference c°. Pure solid and liquid activities are often set to one in the stated standard-phase approximation, not because those substances cease to matter physically. Changing coefficients or reversing a reaction changes the associated constant convention.
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 the A activity squared in the selected expression?
Visual reasoning — after your attempt
Its stoichiometric coefficient is two in the specified equilibrium reaction, so the mass-action expression includes that power.
For aA+bB⇌cC+dD, K uses product activities raised to c,d divided by reactant activities raised to a,b. Introductory Kc calculations approximate with concentrations referenced to standard concentration. Pure solids and liquids have effectively constant activity and are omitted from the simplified expression.
For A⇌2B, the concentration form is [B]²/[A]. Omitting a pure solid does not mean it plays no chemical role.
Pause and explain
Why are coefficients used as powers in an equilibrium expression?
Check your reasoning after answering
The thermodynamic expression reflects the balanced reaction stoichiometry. This equilibrium rule differs from assigning experimental rate-law orders.
Part 2 · Compare Q with K at the same temperature
The reaction quotient Q has the same form as K but is evaluated for the current mixture. Comparing Q and K at the same temperature predicts the thermodynamic net direction toward equilibrium under the model. Q<K favors forward conversion, while Q>K favors reverse conversion. Neither comparison determines the speed or how long equilibration takes. Use the same reaction direction, exponent pattern and normalized-activity convention for both quantities or the numerical comparison loses its meaning.
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.
What net direction is predicted for the selected mixture with Q=8 and K=4?
Visual reasoning — after your attempt
Net reverse conversion toward A is favored under the stated reaction convention and conditions; no timescale is provided.
Q has the same expression using the current mixture. Q<K supports net forward adjustment, Q>K reverse, and Q=K equilibrium in the model. The comparison concerns actual composition and a specified equation orientation. Reversing the equation changes both K and Q reciprocally.
A small product ratio relative to K can favor forward reaction even if products are already present. Direction is not decided by which side looks more abundant.
Pause and explain
What does Q>K indicate?
Check your reasoning after answering
The current product-to-reactant activity ratio exceeds the equilibrium value, so reverse adjustment is favored for the equation as written.
Part 3 · Use an equilibrium-change table
For dimerization, producing x concentration units of A₂ consumes 2x of A. The ICE table therefore differs from a one-to-one isomerization. Substitution into the selected K equation can yield more than one algebraic root; reject any root that makes a species concentration negative or exceeds available material. The displayed K=2 case gives x≈0.3048 M and [A]≈0.3904 M. Substitute the accepted values back into both the equilibrium ratio and atom-equivalent conservation to check the result.
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 must x be no greater than 0.500 M for the selected initial 1.00 M A inventory?
Visual reasoning — after your attempt
Each A₂ requires two A units, so producing more than 0.500 M A₂ would require a negative remaining A amount.
Write initial quantities, changes linked by coefficients, and equilibrium values. Substitute equilibrium values into K, not initial values unless the sample is already equilibrated. Reject algebraic roots producing negative concentrations or violating the available matter.
For A⇌B starting only with A, changes are −x,+x. For A⇌2B, B changes by +2x; the coefficient cannot be dropped from the table.
Pause and explain
Why must an algebraic root be checked chemically?
Check your reasoning after answering
Mathematics can supply roots incompatible with nonnegative concentrations or material constraints. Only physically consistent roots describe the stated mixture.
Part 4 · Interpret the magnitude without claiming a rate
The magnitude of K summarizes a particular equilibrium activity ratio under its convention. A large value often supports strong product favorability for a suitable starting inventory, but it does not itself give a universal percent conversion independent of concentrations and stoichiometry. A small K does not mean no product forms, and K near one does not require every concentration to be equal. K is an equilibrium quantity, not a kinetic rate constant despite both often being called constants.
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 K=10⁶ prove the reaction will be fast?
Visual reasoning — after your attempt
No. It describes equilibrium favorability under the specified convention; the activation pathway and rate constants are separate evidence.
Large K indicates a product-favored equilibrium ratio for the equation and conditions, not a fast reaction or complete conversion at every starting amount. K changes with equation scaling and temperature. A dimensionless activity definition prevents confusing unit conventions with a changed physical prediction.
Multiplying all equation coefficients by two squares the corresponding K. The chemical system is the same but the expression convention changed.
Pause and explain
Does a large K guarantee immediate completion?
Check your reasoning after answering
No. It concerns equilibrium composition ratios. Kinetics, starting amounts, and conditions determine timescale and actual conversion.
Apply the whole unit · Solve a simple equilibrium and direction test
Illustrative model: A⇌B,Kc=4.0 at fixed temperature; initially [A]=0.50 M and [B]=0. In another mixture [A]=0.10 M and [B]=0.20 M. Use the supplied ideal concentration approximation.
Your task
Calculate the first mixture’s equilibrium concentrations, verify K, and use Q to predict direction in the second. Explain why these values cannot establish reaction speed.
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
First mixture: equilibrium A=0.50−x,B=x.
x/(0.50−x)=4.0 gives x=0.40 M.
Equilibrium A=0.10 M,B=0.40 M; ratio=4.0.
Second mixture: Q=0.20/0.10=2.0<K, so forward adjustment is favored.
The first table conserves the supplied total 0.50 M in the fixed-volume 1: 1 model and gives nonnegative values. The second mixture has a different total but the same K at the stated temperature. Equilibrium ratios and Q comparisons do not establish how quickly the system responds; rate evidence is separate.
Scientific references and source-model foundations
Parent / teacher teaching notes
Use a physically simple 1: 1 table before quadratics. Require a nonnegative-root check, a K substitution check, and a direction statement tied to the written equation. 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
Compare Q with K to predict adjustment; equilibrium means equal opposing rates, not equal amounts.
Worked examples
Example 1
- (.20+x)/(.80−x)=3.
- .20+x=2.40−3x; x=.55.
- A=.25, B=.75; ratio=3.
Example 2 · Worked application
- Q_c=[B]/[A]=1.0/0.50=2.
- Q<K, so products are relatively deficient.
- Net forward reaction occurs until equilibrium is approached.
Compare Q and K for the same expression and temperature.
Example 3
- Compare .5 with 4.
- Products are low relative to equilibrium ratio.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Compare Q and K for the same expression and temperature.
- It must increase the product-to-reactant ratio, so the net change is forward.
Compare Q and K for the same expression and temperature.
Example 5 · Reason from the visual model
- For dimerization, producing x concentration units of A₂ consumes 2x of A. The ICE table therefore differs from a one-to-one isomerization. Substitution into the selected K equation can yield more than one algebraic root; reject any root that makes a species concentration negative or exceeds available material. The displayed K=2 case gives x≈0.3048 M and [A]≈0.3904 M. Substitute the accepted values back into both the equilibrium ratio and atom-equivalent conservation to check the result.
- Each A₂ requires two A units, so producing more than 0.500 M A₂ would require a negative remaining A amount.
Original calculated hypothetical dimerization using ideal normalized concentrations and no volume change; A and A₂ conserve the same atomic building units.
Avoid these mistakes
- Use equilibrium concentrations for K and current concentrations for Q.
Check your understanding
Why need not equilibrium concentrations be equal?
Discuss the answer
Equal forward and reverse rates can occur with unequal concentrations because their rate relationships differ.
Return to the phenomenon
Return to the initial case. Can an algebraically correct equilibrium root be physically impossible? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Reject negative species amounts and verify stoichiometric conservation. Only a physically admissible root can represent the stated mixture, even if both roots solve an algebraic rearrangement.
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. Can an algebraically correct equilibrium root be physically impossible? 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
Reject negative species amounts and verify stoichiometric conservation. Only a physically admissible root can represent the stated mixture, even if both roots solve an algebraic rearrangement.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
For 2A⇌A₂ under the normalized ideal convention, K=2.00, [A]=0.400 M and [A₂]=0.160 M. Compute Q, predict net direction, and state the maximum A₂ concentration if only an initial 0.800 M A inventory were available at fixed volume.
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
Q=0.160/(0.400)²=1.00<2.00, so net forward is favored. From an initial 0.800 M A-only inventory, conservation limits A₂ to at most 0.400 M. The equilibrium value need not reach that maximum, and no rate is given.
Words to know
- reaction quotient Q
- The equilibrium-expression ratio evaluated for current conditions.
- ICE table
- A table tracking initial values, stoichiometric change, and equilibrium values.
In the real world
Use the chapter investigation to connect this unit to measured evidence and a practical decision.
Exit check
Why need not equilibrium concentrations be equal?
Check your explanation after answering
Equal forward and reverse rates can occur with unequal concentrations because their rate relationships differ.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Answer and explanation
3.0 mL/s
Average rate = change in amount ÷ change in time = 75 mL ÷ 25 s = 3.0 mL/s.
Problem 2
Answer and explanation
Heating makes particles collide more often, and — more importantly — a larger fraction of collisions carry at least the activation energy, so more collisions are effective.
Higher temperature gives particles more kinetic energy. They collide more frequently, and a much larger share of those collisions now have enough energy to clear the activation-energy barrier. Both effects increase the number of effective collisions per second, so the rate rises.
Problem 3
Answer and explanation
0.015 M/s
The concentration changed by 0.90 − 0.30 = 0.60 M over 40 s. Average rate = 0.60 M ÷ 40 s = 0.015 M/s.
Problem 4
Answer and explanation
Toward the products
By Le Châtelier's principle, the system responds to the extra reactant by shifting in the direction that consumes it — toward the products.
Problem 5
Answer and explanation
Keq = 4
Keq = [HI]² ÷ ([H₂][I₂]) = (0.40)² ÷ (0.10 × 0.40) = 0.16 ÷ 0.040 = 4.0. Because Keq is greater than 1, products are favored.
Problem 6
Answer and explanation
Increase the pressure (the product side has fewer gas moles, 2 vs 4, so higher pressure favors NH₃) and lower the temperature (the forward reaction is exothermic, so cooling favors it). Continuously removing NH₃ would also help.
By Le Châtelier's principle: (1) Raising the pressure shifts the equilibrium toward the side with fewer moles of gas; the left has 1 + 3 = 4 moles and the right has 2, so higher pressure favors ammonia. (2) Because the forward reaction is exothermic, heat behaves like a product, so lowering the temperature shifts the equilibrium forward, making more NH₃. (Removing ammonia as it forms would also pull the equilibrium to the right.)
Problem 7
Answer and explanation
The kinetic pathway and rate change; overall reaction ΔH and equilibrium K stay the same at fixed conditions.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 8
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.
Problem 9
Answer and explanation
Reduced volume raises partial pressures; the adjustment favors the side with fewer gas moles, ammonia, under the model.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 10
Answer and explanation
Q=0.80/0.70≈1.14<4, so net forward conversion is favored. Under the closed 1:1 model total concentration becomes 1.50 M and final values are A=0.30 M, B=1.20 M; the ratio returns, not the original amounts.
Q=0.80/0.70≈1.14<4, so net forward conversion is favored. Under the closed 1:1 model total concentration becomes 1.50 M and final values are A=0.30 M, B=1.20 M; the ratio returns, not the original amounts.
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.
Problem 11
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
0.2
1/5=.2.
Problem 12
Answer and explanation
Net reverse change lowers the product-to-reactant expression toward K.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 13
Answer and explanation
Q=0.160/(0.400)²=1.00<2.00, so net forward is favored. From an initial 0.800 M A-only inventory, conservation limits A₂ to at most 0.400 M. The equilibrium value need not reach that maximum, and no rate is given.
Q=0.160/(0.400)²=1.00<2.00, so net forward is favored. From an initial 0.800 M A-only inventory, conservation limits A₂ to at most 0.400 M. The equilibrium value need not reach that maximum, and no rate is given.
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.