AP Chemistry Unit 5 · S01 — Reaction rate and its measurement — your review

This is the Unit 5 pilot: the first sequence, Reaction rate and its measurement – 11 lessons, 41 checks and 21 figures, built end to end and QA'd. The rest of Unit 5 waits for your verdict.

Do two things. First, rule on the 11 cards below: one tap each, and every tap sends straight to the inbox.

Second, read the 11 lesson cards (keys and feedback are shown for you) and tap Looks right or Needs work on each. Try it as a student on the course page.

Needs you – 11 rulings

Each card is one build call, made on the record under your option-(b) delegation. The filled button is what the build did; tap it to confirm, tap the other to veto, or use Other.

Keep KIN-085, the reverse activation energy read off a labelled profile?

The call: Adopted — keep it.

Released AP items ask for the reverse barrier off a labelled single-step profile (your 2026-08-06 directive: bias toward what the exam asks).
It costs one minimal-pair read beside KIN-084 and no new teaching assets.
No contrary ruling anywhere in the revision log.

What changes: Nothing — KIN-085 stays, cold-sampled in U5-G3 as registered.

wording change

Unit 5 says “energy released / energy absorbed” and saves the exo/endothermic labels for Unit 6?

The call: Adopted — plain language in Unit 5.

Your jargon-before-use principle (JM-P12, unconditional) outranks the exam-styling pull.
Unit 6 defines the labels and sharpens KIN-086 by retrieval.
The engine's sweep enforces the ban; a released AP item that says “exothermic” is re-worded or held for Unit 6.

What changes: Register rev 1.3 removed the “pending decision D2” tail; Unit 5 gate and test items never use the labels.

Half-life items count whole half-lives only, with no stretch tier?

The call: Adopted — whole half-lives only.

AP items are overwhelmingly whole-half-life counting.
The ln route is already taught at KIN-053/054, so nothing is lost.
A stretch tier is your call to switch on, not the operator's.

What changes: Item specs scope KIN-064 to whole half-lives; the atom statement supports either depth.

Teach the pre-equilibrium substitution once in full, and let items accept the kₒₕₛ form?

The call: Adopted as recommended.

The full substitution is the point of the approximation, so the worked example shows it once.
The AP task is the consistency judgment (KIN-108), not constant bookkeeping.

What changes: A spec-layer note only; no register change.

KIN-015's consequence clause stays register-side rather than extending the shared lexicon?

The call: The original ask was overtaken; the clause stays register-side.

The decisions doc said MECH-KMT-TEMP was missing from the shared lexicon. It is there: Unit 3 v2 declared it, and Units 3, 6 and 9 already use it.
KIN-015 says the canonical sentence verbatim, then adds “so collisions occur more often and transfer more energy”.
The shared sentence stays as Unit 3 declared it; Unit 5 owns its consequence clause.

What changes: Register rev 1.3 registers KIN-015 under MECH-KMT-TEMP, so the verbatim sweep protects the opening.

Arrhenius temperature sensitivity stays inside KIN-091, direction-of-effect items only?

The call: Adopted — option (a), keep it inside KIN-091.

The CED excludes Arrhenius calculations, and the register's exclusions already say so.
The engine's sweep bans the machinery (frequency factor, e^(−Ea/RT)) in student text.

What changes: Items limited to direction-of-effect judgments; no factor-of-change arithmetic anywhere.

question change

Keep lesson 4's find-the-rate question, which overlaps lesson 11's skill?

The call: Kept — logged for you to veto.

KIN-004 names a quantity, and your 7 Oct rule says a fact naming a quantity is tested by finding it.
The item reads a tangent the figure supplies; KIN-011 owns choosing read points and building the slope from a bare curve.
The question reviewer flagged the overlap (W1); the blueprint mandates the item.

What changes: Nothing unless you veto; the item stays in lesson 4.

figure change

Lesson 4's tangent reads 0.0010 M/s, the same number as lesson 3's average. Leave the match?

The call: Left as built — the arithmetic is honest.

The tangent at t = 30 s runs (30 s, 0.76 M) to (90 s, 0.70 M): 0.0010 M/s.
Lesson 3's average over 0–60 s is (0.80 − 0.74) ÷ 60: also 0.0010 M/s.
Both come from the same curve, but a student could read the match as “instantaneous equals average”.

What changes: Nothing if left; re-plumbing the curve makes the two numbers differ.

wording change

Unit 5 keeps the “This option is wrong — you …” feedback opener, against the vault's newer R156?

The call: Unit 5 followed the course contract, as Unit 4 did.

This course's contract opens wrong-option feedback with the verdict and the student's own error.
The vault's newer R156 says reasoning only, no verdict prefix.
Switching mid-course would split the voice between units.

What changes: Your ruling sets the voice for Units 6–9; a switch means a feedback-line pass over Units 4 and 5.

figure change

The solvent water is never tracked. Keep lesson 6's two-curve figure?

The call: The operator overrode the blueprint: two curves, not three.

The blueprint drew [H₂O] as a third curve and quoted “the water changes at +0.0010 M/s”.
Water is the solvent, roughly 55 M; its concentration is not a tracked quantity.
The register's own bracketed examples stop at H₂O₂ and O₂; the blueprint went one species past the fence.

What changes: Lesson 6 plots [H₂O₂] falling and [O₂] rising at 2 : 1 steepness; lesson 7 says once that nobody tracks the solvent.

wording change

Is KIN-015's wording right – the canonical KMT sentence verbatim, plus the collision consequence?

The call: Kept at register rev 1.3.

The canonical sentence: “because Kelvin temperature is proportional to the average kinetic energy of the particles”.
KIN-015 adds: “so collisions occur more often and transfer more energy”.
Registered under mechanismLexiconUse, so the verbatim sweep protects it.

What changes: Nothing unless you edit; the S02+ build takes your wording.

Applied without asking – 12 QA rulings the student test decided

E1 — lesson 4's secant redrawn (it was parallel to the tangent; both reviewers called it Essential).
E2 — the false “no ions form” iodine item cut (the real reaction ionises).
E3 — “measured” wording replaced where nothing was measured.
E4 — lesson 3 now says the “size” of the joining line's slope is the average rate.
E5 — both NO₂ figures' concentrations divided by 10; the key rescaled to 4.0 × 10⁻⁴ M/s.
E7 — the taught-instance halogen swap re-instanced to a visibly different reaction.
E8 — the stem that carried its key's head noun reworded.
E9 — one Yes/No claim item converted to a true claim (all four keyed No).
E10 — small wording and contract fixes (feedback openers, entry hints, two genuine third options).
E11 — two figure label collisions offset.
E12 — every Leave row in both QA reports stands as left.
B1(U5) — in-lesson checks may run 2–4 options per the vault's R113 refinement; gates and tests keep exactly four.
The full records: unit5_build/05_operator_adoption.md, unit5_build/blueprints/ADJUDICATIONS.md, unit5_build/BUILD_STATE.md, unit5_build/qa/.

The QA evidence, one line each

Cold solve (non-author, pre-fix)42/42 keys matched, zero clashes
Question pass (review-questions)1 Essential · 11 Worth fixing · 6 Leave — all adopted as fixes except E6 and the Leave rows (carded above)
Accuracy pass (review-accuracy)1 Essential · 4 Worth fixing · 6 Leave — every number recomputed, all adopted as fixes
Blind read of the fixed items41/41 (qa/BLIND_READ_S01.md)
Spec validator (true U5 sweeps, mutation-tested)0 errors · 31 warnings (option-count rows licensed by R113)
Full clickthrough (headless browser)122 pages · 41 checks driven · 0 failures · 0 console errors
Browser figure sweep20 figures rendered · 0 problems

The 11 lessons

Reviewer view: keys are marked, every feedback line is visible, figures are the course's own renders. Students see none of the green marking.

Your criteria (as you pasted them, 7 Oct 2026 – the doc): lessons are good enough if they are 8/10 on these.

Gut checks: Are we confident this is better than the alternative? Would a student benefit from taking this lesson?

Clear and concise language
Accuracy
No “clangers” (content that causes a problematic misconception)
Appropriate questions
Initial testing
Expansion into harder questions (no exam)
The tasks should use similar language to that of the lesson

L01 Reaction rate L02 Keeping the rate positive L03 Average rate L04 Instantaneous rate L05 Initial rate L06 Rates follow the coefficients L07 One rate for one reaction L08 Converting one species' rate into another's L09 Watching a reaction without stopping it L10 Average rate from a data table L11 Instantaneous rate from a tangent

Lesson 1 of 11 · KIN-001

Reaction rate

Atom: State that the rate of a chemical reaction is the change in the concentration of a reactant consumed or of a product formed per unit time, with units such as M/s [e.g., a rate of 0.10 M/s means the concentration of H₂O₂ falls by 0.10 mol/L each second during its decomposition].

Outcome: State what a reaction rate is: the change in a concentration per unit time, in M/s.

Part 1One number for how fast

A flask holds hydrogen peroxide solution at a concentration of 0.80 M.

The hydrogen peroxide slowly decomposes into water and oxygen gas.

After 60 s, the concentration of hydrogen peroxide has fallen to 0.74 M.

A chemist calls this reaction slow.

But "slow" is a feeling, not a number.

How can the chemist put one number on how fast the hydrogen peroxide disappears?

Part 2Change per unit time

You've already seen molarity: the moles of solute per liter of solution, M = $\frac{n}{V}$ .

The chemist watches one substance's concentration as time passes.

The hydrogen peroxide's concentration dropped by 0.06 M in 60 s.

Divide the change by the time: 0.06 M ÷ 60 s = 0.0010 M/s.

That change in concentration per unit time, 0.0010 M/s here, is called the reaction rate.

A rate of 0.0010 M/s means the molarity falls by 0.0010 mol/L each second.

So M/s and mol/L·s name the same unit.

A rate can track a reactant being consumed or a product being formed.

Every rate question that follows starts from this one number.

What the student is expected to be able to do

State that the rate of a reaction is the change in a reactant's or product's concentration per unit time, with units of M/s.

Part 3Check what you just learned

Check 1 · numeric entry

Dinitrogen pentoxide, N₂O₅, decomposes in a sealed flask. Its concentration falls from 0.50 M to 0.44 M over 30 s. What is the rate of consumption of N₂O₅? Give your answer to two significant figures.

Key: 0.0020 M/s

On a correct answerCorrect. The concentration fell by 0.06 M in 30 s, so the rate is 0.06 M ÷ 30 s = 0.0020 M/s.

Mapped wrong answers

0.06	This is the concentration change alone. A rate is the change per unit time: divide the 0.06 M by the 30 s to get 0.0020 M/s.
-0.0020	Right size, wrong sign. The rate of consumption says how fast the N₂O₅ disappears, reported as a positive number: 0.0020 M/s.
0.015	This comes from dividing the final concentration, 0.44 M, by the 30 s. Divide the change, 0.50 M − 0.44 M = 0.06 M, by the time instead: 0.0020 M/s.
0.0147	This comes from dividing the final concentration, 0.44 M, by the 30 s. Divide the change, 0.50 M − 0.44 M = 0.06 M, by the time instead: 0.0020 M/s.

Worked steps shown after a second miss

Write down the values from the question

change in concentration = 0.50 M − 0.44 M = 0.06 M

time = 30 s

Write down the equation

rate = change in concentration ÷ time

Substitute in the values, and calculate

rate = 0.06 M ÷ 30 s

rate = 0.0020 M/s

Check 2 · numeric entry

Iodine, I₂, forms in a sealed flask as hydrogen iodide decomposes. The concentration of I₂ rises from 0 M to 0.030 M over 60 s. What is the rate of formation of I₂? Give your answer to two significant figures.

Key: 0.00050 M/s

On a correct answerCorrect. The concentration of I₂ rose by 0.030 M in 60 s, so the rate of formation is 0.030 M ÷ 60 s = 0.00050 M/s.

Mapped wrong answers

0.030	This is the concentration change alone. Divide it by the 60 s: 0.030 M ÷ 60 s = 0.00050 M/s.
1.8	This comes from multiplying 0.030 M by 60 s. A rate is the change divided by the time: 0.030 ÷ 60 = 0.00050 M/s.
2000	This comes from dividing the time by the concentration change, 60 ÷ 0.030. A rate is the change divided by the time: 0.00050 M/s.

Worked steps shown after a second miss

Write down the values from the question

change in concentration = 0.030 M − 0 M = 0.030 M

time = 60 s

Write down the equation

rate = change in concentration ÷ time

Substitute in the values, and calculate

rate = 0.030 M ÷ 60 s

rate = 0.00050 M/s

Check 3 · multiple choice

In reaction A, the reactant's concentration falls by 0.20 M. In reaction B, the reactant's concentration falls by 0.05 M. Can you tell from these two changes alone which reaction had the higher rate?

Yes

This option is wrong — you compared the concentration changes without their times. A rate is change per unit time: 0.05 M in one second beats 0.20 M spread over an hour.

No KEY

Correct. A rate is a concentration change per unit time. Neither change names its time, so the two rates cannot be compared.

Check 4 · multiple choice

A reaction's change in concentration per unit time, with units such as M/s, is called what?

The reaction rate KEY

Correct. The change in a reactant's or product's concentration per unit time is the reaction rate.

The molarity

This option is wrong — molarity is a concentration itself, the moles of solute per liter. The change per unit time is the reaction rate.

The mole ratio

This option is wrong — a mole ratio compares two species' coefficients and carries no time. The change per unit time is the reaction rate.

Judge this card on:

Does the headache land before the answer?
Is each reveal line one idea?
Does every check test this lesson's atom (under the title)?
Is any key wrong?
Is each figure right – labels, shapes, the point it makes?

Your own criteria sit above the lesson cards.

Note (saves in this browser as you type)

Lesson 2 of 11 · KIN-002

Keeping the rate positive

Atom: State that a reaction rate is reported as a positive quantity, so a reactant's rate of consumption is written with a minus sign as rate = $- \frac{Δ [reactant]}{Δ t}$ because the reactant's concentration change is itself negative [e.g., rate = $- \frac{Δ [H₂O₂]}{Δ t}$ for the decomposition of H₂O₂].

Outcome: State why a reactant's consumption is written rate = $- \frac{Δ [reactant]}{Δ t}$ , with the rate reported positive.

Part 1A rate that seems negative

In the first lesson's flask, the concentration of hydrogen peroxide dropped from 0.80 M to 0.74 M in 60 s.

Compute the change: 0.74 M minus 0.80 M equals −0.06 M.

Divide by 60 s, and the rate seems to come out at −0.0010 M/s.

A negative rate sounds as if the reaction had reversed, turning water and oxygen back into hydrogen peroxide.

But the hydrogen peroxide is still decomposing: the flask bubbles out oxygen the whole time.

How can the rate stay positive when the reactant's concentration change is negative?

Part 2The built-in minus sign

Square brackets name a concentration: [H₂O₂] stands for the hydrogen peroxide's concentration, in M.

The Greek letter Δ (delta) names a change: Δ[H₂O₂] is the change in [H₂O₂], and Δt is the change in time.

Chemists report every reaction rate as a positive quantity.

The reaction consumes the reactant.

So, while the reaction is consuming reactant, the reactant's concentration change is negative.

The definition carries a built-in minus sign: rate = $- \frac{Δ [reactant]}{Δ t}$ .

The minus sign flips the negative change into a positive rate.

For the flask: rate = −(−0.0010 M/s) = +0.0010 M/s — the same positive number lesson one found.

A product's concentration change is already positive.

So a product's rate needs no minus sign.

The table compares a consumed reactant with a formed product: the sign of $\frac{Δ [X]}{Δ t}$ , where the minus sign goes, and the sign of the reported rate.

Where the minus sign goes
	Consumed reactant	Formed product
Sign of $\frac{Δ [X]}{Δ t}$	negative	positive
Rate expression	rate = $- \frac{Δ [X]}{Δ t}$	rate = $\frac{Δ [X]}{Δ t}$
Sign of the reported rate	positive	positive

What the student is expected to be able to do

State that a reaction rate is reported as a positive quantity, so a reactant's consumption is written rate = $- \frac{Δ [reactant]}{Δ t}$ .

Part 3Check what you just learned

Check 1 · multiple choice

Sulfur dioxide is consumed in the reaction 2 SO₂ + O₂ → 2 SO₃. While the SO₂ is being consumed, what is the sign of $\frac{Δ [SO₂]}{Δ t}$ ?

Negative KEY

Correct. The reaction consumes the SO₂, so [SO₂] falls, and a falling concentration makes $\frac{Δ [SO₂]}{Δ t}$ negative.

Positive

This option is wrong — you gave the sign of the reported rate. The reaction consumes the SO₂, so [SO₂] falls, and $\frac{Δ [SO₂]}{Δ t}$ is negative.

Check 2 · multiple choice

Chlorine, Cl₂, is consumed as it reacts with hydrogen in a flask. Which expression gives chlorine's rate of consumption as a positive quantity?

rate = $\frac{Δ [Cl₂]}{Δ t}$

This option is wrong — you left the minus sign out. $\frac{Δ [Cl₂]}{Δ t}$ is negative for a consumed reactant, so this rate would come out negative.

rate = $- \frac{Δ [Cl₂]}{Δ t}$ KEY

Correct. $\frac{Δ [Cl₂]}{Δ t}$ is negative because the Cl₂ is consumed; the built-in minus sign flips it into a positive rate.

rate = $- \frac{Δ t}{Δ [Cl₂]}$

This option is wrong — you inverted the fraction. A rate is a concentration change divided by a time interval, with the minus sign in front: rate = $- \frac{Δ [Cl₂]}{Δ t}$ .

Check 3 · multiple choice

In the reaction 2 SO₂ + O₂ → 2 SO₃, a student computes $\frac{Δ [SO₂]}{Δ t}$ = −0.0040 M/s and says the negative sign shows the reaction is going in reverse, turning SO₃ back into SO₂ and O₂. Is the student correct?

Yes

This option is wrong — you read the minus sign as a direction. A forwards reaction consumes its reactant, so $\frac{Δ [SO₂]}{Δ t}$ is negative even while the reaction goes forwards.

No KEY

Correct. A forwards reaction consumes its reactant, so a reactant's $\frac{Δ [X]}{Δ t}$ is negative. The sign marks consumption, not direction.

Check 4 · numeric entry

In the reaction 2 SO₂ + O₂ → 2 SO₃, [SO₂] falls from 0.060 M to 0.036 M over 120 s. What is $\frac{Δ [SO₂]}{Δ t}$ over this stretch, with its sign? Give your answer to two significant figures.

Key: -0.00020 M/s

On a correct answerCorrect. Δ[SO₂] = 0.036 M − 0.060 M = −0.024 M, over Δt = 120 s: −0.00020 M/s. The negative sign marks the falling concentration.

Sign retry on 0.00020Right size, wrong sign.

\frac{Δ [SO₂]}{Δ t}

is the change in [SO₂], and [SO₂] fell. Write the change with its minus sign.

Mapped wrong answers

-0.024	This is the concentration change alone. Divide it by the 120 s: −0.024 M ÷ 120 s = −0.00020 M/s.
0.024	This is the size of the concentration change alone. Divide by the 120 s, and keep the minus sign, because [SO₂] fell: −0.00020 M/s.
0.0003	This comes from dividing the final concentration, 0.036 M, by the time. Use the change: (0.036 − 0.060) M ÷ 120 s = −0.00020 M/s.
-0.0003	This comes from dividing the final concentration, 0.036 M, by the time. Use the change: (0.036 − 0.060) M ÷ 120 s = −0.00020 M/s.

Worked steps shown after a second miss

Write down the values from the question

Δ[SO₂] = 0.036 M − 0.060 M = −0.024 M

Δt = 120 s

Write down the equation

$\frac{Δ [SO₂]}{Δ t}$ = Δ[SO₂] ÷ Δt

Substitute in the values, and calculate

$\frac{Δ [SO₂]}{Δ t}$ = −0.024 M ÷ 120 s

$\frac{Δ [SO₂]}{Δ t}$ = −0.00020 M/s

Judge this card on:

Does the headache land before the answer?
Is each reveal line one idea?
Does every check test this lesson's atom (under the title)?
Is any key wrong?
Is each figure right – labels, shapes, the point it makes?

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Lesson 3 of 11 · KIN-003

Average rate

Atom: State that an average rate is a concentration change divided by the finite time interval over which the change happens [e.g., the average rate over the first 60 s of a decomposition].

Outcome: State what an average rate is, and name the interval it covers.

Part 1Fast at the start, slower later

The flask from lesson one loses 0.06 M of hydrogen peroxide in the first 60 s.

Over the next 60 s, the flask loses only 0.04 M more.

The hydrogen peroxide disappears quickly in the first minute and more slowly in the second.

So one overall number cannot honestly describe the whole 120 s.

Any fair rate must name the stretch of time it covers.

What does a rate taken over one chosen stretch of time actually tell the chemist?

Part 2The average over a stated stretch

Pick a stretch of time, say the first 60 s.

Divide the concentration change by the length of that stretch.

The answer is the rate averaged over that whole stretch.

Chemists call this number the average rate.

For the flask's first 60 s: average rate = $- \frac{Δ [H₂O₂]}{Δ t}$ = −(−0.06 M) ÷ 60 s = 0.0010 M/s.

Over the next 60 s, the average rate is only 0.04 M ÷ 60 s ≈ 0.00067 M/s.

On the graph of [H₂O₂] against time, the points at 0 s and 60 s are marked and joined with a straight line.

The size of that joining line's slope is the average rate over the 60 s.

An average rate hides any speeding up or slowing down inside its interval.

So a stated average rate always names its interval.

What the student is expected to be able to do

State that an average rate is a concentration change divided by the finite time interval over which the change happens.

Part 3Check what you just learned

Check 1 · numeric entry

Nitrosyl chloride, NOCl, decomposes in a sealed flask. [NOCl] is 0.500 M at t = 0 and 0.420 M at t = 100 s. What is the average rate of consumption of NOCl between t = 0 and t = 100 s? Give your answer to two significant figures.

Key: 0.00080 M/s

On a correct answerCorrect. [NOCl] fell by 0.080 M over the 100 s stretch, so the average rate is 0.080 M ÷ 100 s = 0.00080 M/s.

Mapped wrong answers

0.080	This is the concentration change over the interval. Divide it by the 100 s: 0.00080 M/s.
-0.00080	Right size, wrong sign. A rate of consumption is reported as a positive quantity: 0.00080 M/s.
0.0042	This comes from dividing the final concentration, 0.420 M, by the time. Use the change: (0.500 − 0.420) M ÷ 100 s = 0.00080 M/s.

Worked steps shown after a second miss

Write down the values from the question

Δ[NOCl] = 0.420 M − 0.500 M = −0.080 M

Δt = 100 s

Write down the equation

average rate = $- \frac{Δ [NOCl]}{Δ t}$

Substitute in the values, and calculate

average rate = −(−0.080 M) ÷ 100 s

average rate = 0.00080 M/s

Check 2 · multiple choice

A lab report says a reaction's average rate was 0.0025 M/s. Which further detail must the report give for that number to carry its meaning?

The total volume of the reacting solution

This option is wrong — concentration already accounts for volume: molarity is moles per liter. What the average still hides is its time stretch.

The mass of reactant present in the flask

This option is wrong — the rate tracks the concentration change, not the amount present. What the average still hides is its time stretch.

The stretch of time the rate was averaged over KEY

Correct. An average rate always names its interval: 0.0025 M/s over one minute and 0.0025 M/s over an hour describe different experiments.

Check 3 · multiple choice

A concentration change divided by the stretch of time over which the change happened is called what?

The molarity

This option is wrong — molarity is a single concentration, not a change over time. A change divided by its time stretch is the average rate.

The mole ratio

This option is wrong — a mole ratio compares coefficients and has no time in it. A change divided by its time stretch is the average rate.

The average rate KEY

Correct. The concentration change divided by the finite time interval it happened over is the average rate over that interval.

Judge this card on:

Does the headache land before the answer?
Is each reveal line one idea?
Does every check test this lesson's atom (under the title)?
Is any key wrong?
Is each figure right – labels, shapes, the point it makes?

Your own criteria sit above the lesson cards.

Note (saves in this browser as you type)

Lesson 4 of 11 · KIN-004

Instantaneous rate

Atom: State that an instantaneous rate is the rate at one moment, given by the slope of the line tangent to the concentration–time curve at that instant [e.g., the tangent slope at t = 30 s on an [H₂O₂]–time curve].

Outcome: State what an instantaneous rate is: the slope of the tangent to the concentration–time curve at that instant.

Part 1The rate at one moment

The average rate over the first 60 s blends the fast start with the slower finish.

A chemist wants the rate at exactly t = 30 s.

Shrink the interval around 30 s, and the average settles toward one steady number.

But an interval of zero seconds would divide the change by zero.

How can anyone read a rate at a single instant?

Part 2The tangent

Here is the curve of [H₂O₂] against time.

At t = 30 s, the figure lays a straight line that just touches the curve and matches its steepness there.

That line is called the tangent.

The slope of the tangent is the rate at that single instant.

Chemists call this slope the instantaneous rate.

Rise over run turns the curve's steepness at one point into one readable number.

Q: What is the instantaneous rate of consumption of H₂O₂ at t = 30 s, from the drawn tangent?

Read two points on the tangent

(30 s, 0.76 M)(the touch point)

(90 s, 0.70 M)(far along the tangent)

Write down the equation

slope = rise ÷ run

Substitute in the values, and calculate

slope = (0.70 M − 0.76 M) ÷ (90 s − 30 s)

slope = −0.06 M ÷ 60 s = −0.0010 M/s

instantaneous rate = −(−0.0010) = 0.0010 M/s(lesson two's rule: a consumption rate is reported positive)

Where the curve is steep, the tangent is steep, and the instantaneous rate is large.

Where the curve is flat, the tangent is flat, and the instantaneous rate is small.

What the student is expected to be able to do

State that an instantaneous rate is the rate at one moment, given by the slope of the line tangent to the concentration–time curve at that instant.

Part 3Check what you just learned

Check 1 · multiple choice

The graph shows [NO₂] falling as NO₂ reacts away in a sealed flask. Two straight lines, J and K, are drawn on the curve. Which line's slope gives the instantaneous rate at t = 40 s?

Line J KEY

Correct. Line J touches the curve at t = 40 s and matches its steepness there — the tangent. Its slope is the instantaneous rate at 40 s.

Line K

This option is wrong — line K passes through the curve at 20 s and again at 80 s, so its slope averages the change over that stretch. It does not match the curve's steepness at 40 s.

Check 2 · multiple choice

A student reads a concentration–time curve. At t = 90 s the curve sits at a height of 0.25 M. The student says the instantaneous rate at 90 s is therefore 0.25 M/s. Is the student correct?

Yes

This option is wrong — you read the curve's height, which is a concentration in M. The instantaneous rate is the slope of the tangent at 90 s, in M/s.

No KEY

Correct. The curve's height at 90 s is a concentration, 0.25 M. The instantaneous rate at 90 s is the slope of the tangent there: how fast that concentration is changing.

Check 3 · numeric entry

The graph shows [NO₂] against time, with the tangent at t = 50 s drawn and two points on the tangent marked. What is the instantaneous rate of consumption of NO₂ at t = 50 s? Give your answer to two significant figures. Scientific notation such as 1.2e-4 works.

Key: 4.0e-4 M/s

On a correct answerCorrect. The tangent drops from 0.040 M at 25 s to 0.020 M at 75 s: a rise of −0.020 M over a run of 50 s, so the rate is 4.0 × 10⁻⁴ M/s.

Mapped wrong answers

-4.0e-4	Right size, wrong sign. A rate of consumption is reported positive; the tangent's negative slope marks the falling concentration.
6.0e-4	This comes from dividing the curve's height at 50 s, 0.030 M, by the 50 s. Use the tangent's two marked points: rise −0.020 M over run 50 s.
2.7e-4	This comes from dividing one marked point's concentration, 0.020 M, by its clock time, 75 s. Use rise over run between the two marked points instead.

Worked steps shown after a second miss

Read the two marked points on the tangent

(25 s, 0.040 M)

(75 s, 0.020 M)

Write down the equation

slope = rise ÷ run

Substitute in the values, and calculate

slope = (0.020 M − 0.040 M) ÷ (75 s − 25 s) = −4.0 × 10⁻⁴ M/s

rate of consumption = 4.0 × 10⁻⁴ M/s

Check 4 · multiple choice

The rate at one single moment, read as the slope of the tangent at that instant, is called what?

The average rate

This option is wrong — an average rate covers a stretch of time between two clock readings. The rate at one single moment is the instantaneous rate.

The instantaneous rate KEY

Correct. The slope of the tangent at one instant is the instantaneous rate at that instant.

The concentration at that instant

This option is wrong — the curve's height at an instant is a concentration, in M. The rate at that instant is the slope of the tangent there, in M/s.

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Lesson 5 of 11 · KIN-005

Initial rate

Atom: State that the initial rate is the instantaneous rate at the moment the reaction begins, at t = 0 [e.g., the slope of the tangent drawn at t = 0 on a concentration–time curve].

Outcome: State what the initial rate is: the instantaneous rate at the moment the reaction begins, t = 0.

Part 1The moment of mixing

At the moment of mixing, the chemist knows every concentration exactly.

The chemist chose those concentrations and poured them into the flask.

Sixty seconds later, the reaction has already consumed part of the reactant.

The concentrations now sitting in the flask are whatever the reaction left behind.

Which moment's rate belongs to the mixture the chemist actually made?

Part 2The rate at t = 0

The rate at the moment a reaction begins deserves its own name.

At that moment, the concentrations are exactly the ones the chemist mixed.

Chemists call the instantaneous rate at t = 0 the initial rate.

On the concentration–time curve, the initial rate is the slope of the tangent drawn at t = 0.

This flask's curve is steepest at the start, so its tangent at t = 0 is its steepest tangent.

The three named rates differ only in where the rate is read.

An average rate covers a stated stretch of time.

An instantaneous rate sits at one single instant.

The initial rate is the instantaneous rate at the first instant of all, t = 0.

Whole experiments rest on comparing the initial rates of different mixtures.

What the student is expected to be able to do

State that the initial rate is the instantaneous rate at the moment the reaction begins, at t = 0.

Part 3Check what you just learned

Check 1 · multiple choice

A chemist watches one reaction for 90 s, then divides the whole concentration change over those 90 s by the 90 s. Which rate has the chemist computed? Choose the most exact name.

The average rate KEY

Correct. A concentration change divided by a finite stretch of time, here the first 90 s, is the average rate over that stretch.

The instantaneous rate

This option is wrong — an instantaneous rate belongs to one single moment, read from a tangent. Dividing a change by a 90 s stretch averages over the whole stretch.

The initial rate

This option is wrong — the initial rate is a tangent's slope at the single instant t = 0. Dividing a change by a 90 s stretch averages over the whole stretch.

Check 2 · multiple choice

A chemist draws the tangent to a concentration–time curve at t = 45 s and computes its slope. Which rate has the chemist computed? Choose the most exact name.

The average rate

This option is wrong — no stretch of time was averaged. A tangent's slope belongs to one single instant: the instantaneous rate at 45 s.

The instantaneous rate KEY

Correct. The slope of the tangent at t = 45 s is the instantaneous rate at that instant.

The initial rate

This option is wrong — the initial rate sits at t = 0, the moment of mixing. This tangent sits at t = 45 s, so its slope is the instantaneous rate there.

Check 3 · multiple choice

A chemist draws the tangent to a concentration–time curve at t = 0 and computes its slope. Which rate has the chemist computed? Choose the most exact name.

The average rate

This option is wrong — no stretch of time was averaged; a tangent's slope belongs to one instant. At t = 0, that slope has its own name: the initial rate.

The instantaneous rate

This option is wrong — the slope at t = 0 is indeed an instantaneous rate, but the instantaneous rate at t = 0 has a more exact name: the initial rate.

The initial rate KEY

Correct. The tangent's slope at t = 0 is the instantaneous rate at the moment the reaction begins — the initial rate.

Check 4 · multiple choice

Chemists give the instantaneous rate at the moment a reaction begins, t = 0, its own name. What is that name?

The initial rate KEY

Correct. The instantaneous rate at t = 0 is called the initial rate.

The average rate

This option is wrong — an average rate covers a stretch of time. The rate at the single moment t = 0 is the initial rate.

The starting molarity

This option is wrong — a molarity is a concentration, not a rate. The rate at the moment of mixing is the initial rate.

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Lesson 6 of 11 · KIN-006

Rates follow the coefficients

Atom: State that the rates of change of reactant and product concentrations are related by the ratio of their coefficients in the balanced chemical equation [e.g., in 2 H₂O₂ → 2 H₂O + O₂, H₂O₂ is consumed twice as fast as O₂ is produced].

Outcome: State how the balanced equation's coefficients relate the species' rates of change.

Part 1The classmate's question

In lesson one's flask, the following reaction occurs: 2 H₂O₂ → 2 H₂O + O₂.

The hydrogen peroxide disappears at 0.0010 M/s.

A classmate asks how fast the oxygen appears.

Nobody measured the oxygen.

Does the balanced equation already answer the classmate's question?

Part 2The coefficients answer it

Check 1 · multiple choice

In the reaction N₂ + 3 H₂ → 2 NH₃, how many moles of H₂ are consumed for every 1 mol of N₂ consumed?

1 mol

This option is wrong — you matched the two reactants one-for-one. The coefficients say 3 H₂ for every 1 N₂.

2 mol

This option is wrong — 2 is the coefficient of the NH₃ formed. The H₂ : N₂ mole ratio is 3 : 1.

3 mol KEY

Correct. The coefficients give the mole ratio: 3 mol of H₂ are consumed for every 1 mol of N₂.

You've already seen that a balanced equation's coefficients give the mole ratio between every pair of species.

In the flask's reaction, every 2 mol of hydrogen peroxide consumed makes 1 mol of oxygen.

So the oxygen appears at half the speed the hydrogen peroxide disappears.

Half of 0.0010 M/s is 0.0005 M/s: the oxygen's concentration rises at 0.0005 M/s.

The species' rates of change stand in the same ratio as their coefficients.

The graph shows the two tracked species in one flask: [H₂O₂] falls twice as steeply as [O₂] rises.

The reaction also makes water, but that water joins the solvent, so nobody tracks its concentration.

Measure one species, and the coefficients turn that one rate into every other species' rate — no second measurement needed.

What the student is expected to be able to do

State that the rates of change of reactant and product concentrations are related by the ratio of their coefficients in the balanced equation.

Part 3Check what you just learned

Check 2 · multiple choice

In N₂ + 3 H₂ → 2 NH₃, both reactants are consumed. Does [H₂] fall faster than [N₂] falls, slower, or at the same speed?

Faster KEY

Correct. Every 1 mol of N₂ consumed takes 3 mol of H₂ with it, so [H₂] falls three times as fast as [N₂].

Slower

This option is wrong — you inverted the ratio. The coefficients say 3 H₂ per 1 N₂, so the H₂ is consumed three times as fast.

At the same speed

This option is wrong — the same speed would need equal coefficients. The coefficients say 3 H₂ per 1 N₂, so [H₂] falls three times as fast.

Check 3 · multiple choice

In N₂ + 3 H₂ → 2 NH₃, ammonia forms while hydrogen is consumed. Does [NH₃] rise faster than [H₂] falls, slower, or at the same speed?

Faster

This option is wrong — faster would need NH₃'s coefficient to beat H₂'s. The coefficients say 2 NH₃ form for every 3 H₂ consumed.

Slower KEY

Correct. The coefficients say 2 mol of NH₃ form for every 3 mol of H₂ consumed, so [NH₃] rises at two-thirds of the speed [H₂] falls.

At the same speed

This option is wrong — the same speed would need equal coefficients. The coefficients say 2 NH₃ per 3 H₂, so the NH₃ appears more slowly.

Check 4 · multiple choice

In the reaction 2 CO + O₂ → 2 CO₂, the CO is consumed at 0.0080 M/s. At what rate is the O₂ consumed?

0.0040 M/s KEY

Correct. The coefficients say 1 mol of O₂ per 2 mol of CO, so the O₂ is consumed at half the CO's rate: 0.0040 M/s.

0.0080 M/s

This option is wrong — you matched the two reactants one-for-one. The coefficients say 1 O₂ per 2 CO, so the O₂ rate is half the CO's.

0.016 M/s

This option is wrong — you doubled instead of halving. The ratio of O₂ to CO is 1 : 2, so the O₂ is consumed at half the CO's rate.

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Lesson 7 of 11 · KIN-007

One rate for one reaction

Atom: State that dividing each species' rate of concentration change by its coefficient in the balanced equation gives the same single reaction rate, written rate = $- \frac{1}{a} \frac{Δ [A]}{Δ t}$ = $\frac{1}{c} \frac{Δ [C]}{Δ t}$ [e.g., rate = $- \frac{1}{2} \frac{Δ [H₂O₂]}{Δ t}$ = $\frac{Δ [O₂]}{Δ t}$ for 2 H₂O₂ → 2 H₂O + O₂].

Outcome: State the one-rate expression: rate = $- \frac{1}{a} \frac{Δ [A]}{Δ t}$ = $\frac{1}{c} \frac{Δ [C]}{Δ t}$ .

Part 1Two speeds, one reaction

The following reaction occurs in lesson one's flask: 2 H₂O₂ → 2 H₂O + O₂.

The hydrogen peroxide changes at −0.0010 M/s.

The oxygen changes at +0.0005 M/s.

Nobody tracks the water: the water formed just joins the solvent.

Two chemists watching different species would quote two different speeds for the same reaction.

So which single number is the rate of the reaction?

Part 2Divide by the coefficient

Divide each species' rate of change by that species' own coefficient.

Fix the sign with lesson two's rule: report the rate positive, so a consumed reactant's change takes the minus sign.

Every species then yields the same single number.

From the hydrogen peroxide: −(−0.0010 M/s) ÷ 2 = 0.0005 M/s.

From the oxygen: (+0.0005 M/s) ÷ 1 = 0.0005 M/s.

The table applies the same two steps to both species.

Two species, one shared rate
Species	$\frac{Δ [X]}{Δ t}$	Divide by its coefficient	Rate of the reaction
H₂O₂	−0.0010 M/s	−(−0.0010) ÷ 2	0.0005 M/s
O₂	+0.0005 M/s	+0.0005 ÷ 1	0.0005 M/s

Chemists write the shared number as one expression.

a	the coefficient of reactant A in the balanced equation
c	the coefficient of product C in the balanced equation
$\frac{Δ [A]}{Δ t}$	reactant A's rate of concentration change (negative)
$\frac{Δ [C]}{Δ t}$	product C's rate of concentration change (positive)

One reaction now has one rate, whichever species the laboratory happened to watch.

What the student is expected to be able to do

State that dividing each species' rate of concentration change by its coefficient gives the same single reaction rate, written rate = $- \frac{1}{a} \frac{Δ [A]}{Δ t}$ = $\frac{1}{c} \frac{Δ [C]}{Δ t}$ .

Part 3Check what you just learned

Check 1 · multiple choice

For the reaction 2 N₂O₅ → 4 NO₂ + O₂, which expression gives the rate of the reaction from each species?

rate = $\frac{Δ [N₂O₅]}{Δ t}$ = $\frac{Δ [NO₂]}{Δ t}$ = $\frac{Δ [O₂]}{Δ t}$

This option is wrong — you ignored the coefficients and the sign. The N₂O₅'s change must be divided by 2 and sign-flipped, and the NO₂'s divided by 4.

rate = $- \frac{1}{2} \frac{Δ [N₂O₅]}{Δ t}$ = $\frac{1}{4} \frac{Δ [NO₂]}{Δ t}$ = $\frac{Δ [O₂]}{Δ t}$ KEY

Correct. Each species' change is divided by its own coefficient, and the consumed reactant's term carries the minus sign.

rate = −2 $\frac{Δ [N₂O₅]}{Δ t}$ = 4 $\frac{Δ [NO₂]}{Δ t}$ = $\frac{Δ [O₂]}{Δ t}$

This option is wrong — you multiplied by the coefficients instead of dividing. Dividing each change by its own coefficient is what makes every species give the same number.

rate = $- \frac{1}{2} \frac{Δ [N₂O₅]}{Δ t}$ = $- \frac{1}{4} \frac{Δ [NO₂]}{Δ t}$ = $\frac{Δ [O₂]}{Δ t}$

This option is wrong — you put a minus sign on the NO₂ term, but NO₂ is a formed product. Only the consumed reactant's term needs the minus sign.

Check 2 · multiple choice

Two chemists watch the same flask of reacting N₂O₅: one measures the N₂O₅, the other measures the O₂. A student says the two chemists will compute the same reaction rate, because each will divide the species' $\frac{Δ [X]}{Δ t}$ by its own coefficient and fix the sign. Is the student correct?

Yes KEY

Correct. Dividing each species' $\frac{Δ [X]}{Δ t}$ by its own coefficient, with the sign fixed, turns both measurements into the same single number: the rate of the reaction.

This option is wrong — you stopped at the raw $\frac{Δ [X]}{Δ t}$ values, which do differ between species. Dividing each by its own coefficient, sign fixed, gives both chemists the same rate.

Check 3 · numeric entry

In 2 N₂O₅ → 4 NO₂ + O₂, a measurement gives $\frac{Δ [NO₂]}{Δ t}$ = +0.024 M/s. What is the rate of the reaction? Give your answer to two significant figures.

Key: 0.0060 M/s

On a correct answerCorrect. rate =

\frac{1}{4} \frac{Δ [NO₂]}{Δ t}

= 0.024 ÷ 4 = 0.0060 M/s. Any species, divided by its own coefficient, gives this same number.

Mapped wrong answers

0.024	This is $\frac{Δ [NO₂]}{Δ t}$ itself. Divide it by NO₂'s coefficient, 4: 0.024 ÷ 4 = 0.0060 M/s.
0.096	This comes from multiplying by the coefficient. Divide by it instead: 0.024 ÷ 4 = 0.0060 M/s.
0.012	This comes from dividing by 2, the N₂O₅'s coefficient. Divide the NO₂'s change by NO₂'s own coefficient, 4: 0.0060 M/s.

Worked steps shown after a second miss

Write down the values from the question

$\frac{Δ [NO₂]}{Δ t}$ = +0.024 M/s

coefficient of NO₂ = 4

Write down the equation

rate = $\frac{1}{4} \frac{Δ [NO₂]}{Δ t}$

Substitute in the values, and calculate

rate = 0.024 ÷ 4

rate = 0.0060 M/s

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Lesson 8 of 11 · KIN-008

Converting one species' rate into another's

Atom: Calculate the rate of change of one species' concentration from the measured rate of change of another species using the ratio of their coefficients in the balanced equation [e.g., given $\frac{Δ [O₂]}{Δ t}$ = +0.10 M/s in 2 H₂O₂ → 2 H₂O + O₂, find $\frac{Δ [H₂O₂]}{Δ t}$ = −0.20 M/s].

Outcome: Calculate one species' rate of concentration change from another species' measured rate.

Part 1Which species gets the 2?

A question supplies only the oxygen's rate of change: $\frac{Δ [O₂]}{Δ t}$ = +0.10 M/s.

The question asks for the hydrogen peroxide's rate of change.

Nobody measured the hydrogen peroxide.

Lesson six said the hydrogen peroxide's and the oxygen's rates stand in a 2 : 1 ratio.

Which species gets multiplied by 2?

And which sign does the answer carry?

Part 2Ratio first, then sign

Multiply the known rate by the mole ratio of the wanted species to the known species.

Then set the sign: negative for a consumed reactant, positive for a formed product.

Here the ratio is 2 mol of H₂O₂ per 1 mol of O₂.

Q: In 2 H₂O₂ → 2 H₂O + O₂, a measurement gives $\frac{Δ [O₂]}{Δ t}$ = +0.10 M/s. What is $\frac{Δ [H₂O₂]}{Δ t}$ ?

Write down the values from the question

$\frac{Δ [O₂]}{Δ t}$ = +0.10 M/s(the known species)

wanted: $\frac{Δ [H₂O₂]}{Δ t}$

Multiply by the mole ratio of the wanted species to the known species

ratio = 2 mol H₂O₂ : 1 mol O₂(from the coefficients)

size = 0.10 × 2 = 0.20 M/s

Set the sign

H₂O₂ is a consumed reactant, so the sign is negative

$\frac{Δ [H₂O₂]}{Δ t}$ = −0.20 M/s

Two moves, the ratio and then the sign, answer every convert-the-rate question on the AP exam.

What the student is expected to be able to do

Calculate one species' rate of concentration change from another species' measured rate, using the coefficient ratio and the sign rule.

Part 3Check what you just learned

Check 1 · numeric entry

In the reaction 2 NO + O₂ → 2 NO₂, a measurement gives $\frac{Δ [NO]}{Δ t}$ = −0.062 M/s. What is $\frac{Δ [O₂]}{Δ t}$ , with its sign? Give your answer to two significant figures.

Key: -0.031 M/s

On a correct answerCorrect. The ratio of O₂ to NO is 1 : 2, so the size is 0.062 ÷ 2 = 0.031 M/s; the O₂ is consumed, so the sign is negative.

Sign retry on 0.031Right size, wrong sign. The O₂ is a consumed reactant, so its concentration falls:

\frac{Δ [O₂]}{Δ t}

is negative.

Mapped wrong answers

-0.124	You multiplied by 2. The ratio of O₂ to NO is 1 : 2, so the O₂ changes at half the NO's speed: −0.031 M/s.
-0.062	You carried the NO's value across unchanged. The ratio of O₂ to NO is 1 : 2, so halve it: −0.031 M/s.
0.062	Two fixes needed: the ratio of O₂ to NO is 1 : 2, so halve the 0.062, and the O₂ is consumed, so the sign is negative: −0.031 M/s.
0.124	You multiplied by 2 and dropped the sign. The ratio of O₂ to NO is 1 : 2, and the O₂ is consumed: −0.031 M/s.

Worked steps shown after a second miss

Write down the values from the question

$\frac{Δ [NO]}{Δ t}$ = −0.062 M/s

wanted: $\frac{Δ [O₂]}{Δ t}$

ratio = 1 mol O₂ : 2 mol NO

Multiply by the mole ratio of the wanted species to the known species

size = 0.062 × (1/2) = 0.031 M/s

Set the sign

O₂ is a consumed reactant, so the sign is negative

$\frac{Δ [O₂]}{Δ t}$ = −0.031 M/s

Check 2 · numeric entry

In the reaction 2 NO + O₂ → 2 NO₂, a measurement gives $\frac{Δ [NO]}{Δ t}$ = −0.062 M/s. What is $\frac{Δ [NO₂]}{Δ t}$ , with its sign? Give your answer to two significant figures.

Key: 0.062 M/s

On a correct answerCorrect. The ratio of NO₂ to NO is 2 : 2, one for one, so the size stays 0.062 M/s; the NO₂ is a formed product, so the sign is positive.

Mapped wrong answers

-0.062	Right size, wrong sign. The NO₂ is a formed product, so its concentration rises: $\frac{Δ [NO₂]}{Δ t}$ is positive.
0.031	You halved the value. The ratio of NO₂ to NO is 2 : 2 — one for one — so the size stays 0.062 M/s.
0.124	You doubled the value. The ratio of NO₂ to NO is 2 : 2 — one for one — so the size stays 0.062 M/s.

Worked steps shown after a second miss

Write down the values from the question

$\frac{Δ [NO]}{Δ t}$ = −0.062 M/s

wanted: $\frac{Δ [NO₂]}{Δ t}$

ratio = 2 mol NO₂ : 2 mol NO

Multiply by the mole ratio of the wanted species to the known species

size = 0.062 × (2/2) = 0.062 M/s

Set the sign

NO₂ is a formed product, so the sign is positive

$\frac{Δ [NO₂]}{Δ t}$ = +0.062 M/s

Check 3 · numeric entry

Ozone decomposes by the reaction 2 O₃ → 3 O₂. A measurement gives $\frac{Δ [O₃]}{Δ t}$ = −0.040 M/s. What is $\frac{Δ [O₂]}{Δ t}$ , with its sign? Give your answer to two significant figures.

Key: 0.060 M/s

On a correct answerCorrect. The ratio of O₂ to O₃ is 3 : 2, so the size is 0.040 × (3/2) = 0.060 M/s; the O₂ is a formed product, so the sign is positive.

Mapped wrong answers

-0.060	Right size, wrong sign. The O₂ is a formed product here, so its concentration rises: $\frac{Δ [O₂]}{Δ t}$ is positive.
0.027	You used the ratio upside down, 2/3. The wanted species is O₂, coefficient 3; the known is O₃, coefficient 2: 0.040 × (3/2) = +0.060 M/s.
0.0267	You used the ratio upside down, 2/3. The wanted species is O₂, coefficient 3; the known is O₃, coefficient 2: 0.040 × (3/2) = +0.060 M/s.
0.040	You carried the O₃'s size across unchanged. The ratio of O₂ to O₃ is 3 : 2: 0.040 × (3/2) = +0.060 M/s.

Worked steps shown after a second miss

Write down the values from the question

$\frac{Δ [O₃]}{Δ t}$ = −0.040 M/s

wanted: $\frac{Δ [O₂]}{Δ t}$

ratio = 3 mol O₂ : 2 mol O₃

Multiply by the mole ratio of the wanted species to the known species

size = 0.040 × (3/2) = 0.060 M/s

Set the sign

O₂ is a formed product, so the sign is positive

$\frac{Δ [O₂]}{Δ t}$ = +0.060 M/s

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Lesson 9 of 11 · KIN-009

Watching a reaction without stopping it

Atom: Identify experimental methods for monitoring reactant or product amounts as a reaction proceeds [e.g., tracking the absorbance of a colored reactant, the pressure of an evolved gas, or the electrical conductivity of a solution as ions form].

Outcome: Identify the instrument that can follow a given reaction as it happens.

Part 1Watching without stopping

Every rate in this sequence so far came from concentrations measured at several times.

But a chemist cannot reach into the flask mid-reaction and count molecules.

Pulling one sample out and testing it costs the chemist minutes.

The mixture keeps reacting for every one of those minutes.

Which instruments can watch a concentration change without stopping the reaction?

Part 2Three properties an instrument can read

Check 1 · multiple choice

A calibration line is built from a set of standard solutions. What does the line let a chemist turn an absorbance reading into?

A volume

This option is wrong — volumes come from glassware, not from the line. The calibration line maps an absorbance onto a concentration.

A molar mass

This option is wrong — molar mass comes from the formula, not from the line. The calibration line maps an absorbance onto a concentration.

A concentration KEY

Correct. The calibration line turns a measured absorbance into the colored species' concentration, read within the standards' range.

Check 2 · multiple choice

Partway through a reaction, a solution's electrical conductivity is rising. What does the rise show is happening in the solution?

More charged particles are becoming free to move KEY

Correct. A material carries current through charged particles that are free to move, so a rising conductivity means more mobile ions.

More molecules of every kind are dissolving

This option is wrong — dissolved particles with no charge carry no current. Conductivity rises only when mobile charged particles become more plentiful.

Electrons have begun to flow through the water itself

This option is wrong — electrons do not travel through the solution. The current in a solution is carried by ions, so a rising conductivity means more mobile ions.

Pick a property that follows one species' concentration.

Read that property continuously as the reaction happens.

A spectrophotometer reads the falling absorbance of a mixture whose colored reactant is being consumed.

You've already seen the calibration line from Unit 3: built from standard solutions, it turns each absorbance into the colored species' concentration, read within the standards' range.

A pressure gauge reads the rising pressure of a sealed flask while a reaction makes more gas.

Seal the first lesson's flask, and its oxygen would raise the pressure exactly that way.

A conductivity probe reads the rising conductivity of a mixture while a reaction creates ions.

You've already seen why: a material carries an electric current only when it holds charged particles that are free to move.

One clear case: tert-butyl chloride reacts with water, and the products are tert-butyl alcohol and hydrochloric acid — H⁺(aq) and Cl⁻(aq) ions.

Written once with formulas: (CH₃)₃CCl + H₂O → (CH₃)₃COH + H⁺(aq) + Cl⁻(aq).

Each new pair of ions carries current, so the conductivity climbs as the ions accumulate.

The table compares the three methods: the property each reads, the instrument that reads it, and a reaction it suits.

Three ways to watch a reaction as it happens
Property the probe reads	Instrument	Example reaction
Absorbance, falling	Spectrophotometer	Crystal violet, a purple dye, fades in a basic solution
Gas pressure, rising	Pressure gauge	2 H₂O₂ → 2 H₂O + O₂ in a sealed flask
Conductivity, rising	Conductivity probe	tert-Butyl chloride reacts with water, releasing H⁺ and Cl⁻

Each method turns a reaction in progress into a concentration–time table a chemist can read.

What the student is expected to be able to do

Identify experimental methods for monitoring reactant or product amounts as a reaction proceeds: absorbance of a colored species, pressure of an evolved gas, or conductivity as ions form.

Part 3Check what you just learned

Check 3 · multiple choice

Heated in a sealed flask, solid calcium carbonate decomposes into solid calcium oxide and carbon dioxide gas. No species is colored, and no solution is present. Which instrument lets a chemist follow the reaction as it happens?

A spectrophotometer, reading the mixture's absorbance

This option is wrong — nothing in the flask is colored, so there is no absorbance change to read. The CO₂ gas raises the sealed flask's pressure; read that.

A pressure gauge, reading the flask's gas pressure KEY

Correct. The reaction releases CO₂ gas into a sealed flask, so the pressure climbs as the CO₂ accumulates, and the gauge reads it continuously.

A conductivity probe, reading the solution's conductivity

This option is wrong — a conductivity probe needs a solution with mobile ions, and this flask holds two solids and a gas. Read the rising gas pressure instead.

Check 4 · multiple choice

In a colorless solvent, orange-brown bromine reacts with excess cyclohexene, and the mixture's color fades. No gas forms and no ions form. Which instrument lets a chemist follow the reaction as it happens?

A spectrophotometer, reading the mixture's absorbance KEY

Correct. The bromine's orange-brown color fades as the bromine is consumed, so the mixture's absorbance falls, and the spectrophotometer reads it continuously.

A pressure gauge, reading the flask's gas pressure

This option is wrong — no gas forms, so a sealed flask's pressure would not change. The color is the property that changes: read the absorbance.

A conductivity probe, reading the solution's conductivity

This option is wrong — no ions form, so the conductivity does not change. The color is the property that changes: read the absorbance.

Check 5 · multiple choice

Benzoyl chloride reacts with water, releasing H⁺(aq) and Cl⁻(aq) ions as it goes. The mixture stays colorless, and no gas leaves it. Which instrument lets a chemist follow the reaction as it happens?

A spectrophotometer, reading the mixture's absorbance

This option is wrong — the mixture stays colorless, so there is no absorbance change to read. Each new H⁺ and Cl⁻ pair raises the conductivity; read that.

A pressure gauge, reading the flask's gas pressure

This option is wrong — no gas leaves the mixture, so there is no pressure change to read. Each new H⁺ and Cl⁻ pair raises the conductivity; read that.

A conductivity probe, reading the solution's conductivity KEY

Correct. Every reacting molecule releases one H⁺ ion and one Cl⁻ ion, so the solution's conductivity climbs as the ions accumulate.

Check 6 · multiple choice

In a sealed flask, gaseous cyclobutane decomposes, each molecule splitting into two ethene molecules, so the moles of gas climb. Every species is colorless, and no solution is present. Which instrument lets a chemist follow the reaction as it happens?

A spectrophotometer, reading the mixture's absorbance

This option is wrong — every species is colorless, so there is no absorbance change to read. The moles of gas climb in a sealed flask, so read the rising pressure.

A pressure gauge, reading the flask's gas pressure KEY

Correct. One molecule becomes two, so the moles of gas climb, and in a sealed flask the pressure climbs with them.

A conductivity probe, reading the solution's conductivity

This option is wrong — a gas-filled flask holds no ionic solution to probe. The moles of gas climb, so read the rising pressure.

Judge this card on:

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Does every check test this lesson's atom (under the title)?
Is any key wrong?
Is each figure right – labels, shapes, the point it makes?

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Lesson 10 of 11 · KIN-010

Average rate from a data table

Atom: Calculate the average rate over a specified interval from concentration–time data given as a table or a graph [e.g., find $- \frac{Δ [crystal violet]}{Δ t}$ between t = 0 and t = 120 s from a data table].

Outcome: Calculate the average rate over a stated interval from a concentration–time table or graph.

Part 1Four rows from a real fade

Crystal violet is a purple dye that slowly loses its color in a basic solution.

A spectrophotometer watched the fade, reading the dye's falling absorbance — lesson nine's method.

The chemist now has a table of the dye's concentration at 0 s, 60 s, 120 s and 180 s.

Crystal violet concentration during the fade
Time (s)	[crystal violet] (M)
0	5.00 × 10⁻⁵
60	4.35 × 10⁻⁵
120	3.80 × 10⁻⁵
180	3.32 × 10⁻⁵

Here is an exam question about this table: what is the average rate of consumption of the dye between t = 0 and t = 120 s?

Which two rows of the table answer the question?

Part 2Two rows answer the question

Take the concentration at each end of the stated interval.

Subtract the two concentrations to get the change in concentration.

Subtract the two times to get the change in time.

Divide the change in concentration by the change in time.

The dye is a consumed reactant.

A consumed reactant's concentration change is negative.

So the rate takes lesson two's built-in minus sign: average rate = $- \frac{Δ [crystal violet]}{Δ t}$ .

Q: What is the average rate of consumption of crystal violet between t = 0 and t = 120 s?

Write down the values from the table

[crystal violet] at 0 s = 5.00 × 10⁻⁵ M(the interval's first row)

[crystal violet] at 120 s = 3.80 × 10⁻⁵ M(the interval's last row)

Δt = 120 s − 0 s = 120 s

Write down the equation

average rate = $- \frac{Δ [crystal violet]}{Δ t}$

Substitute in the values, and calculate

average rate = −(3.80 × 10⁻⁵ − 5.00 × 10⁻⁵) ÷ 120

average rate = 1.20 × 10⁻⁵ ÷ 120

average rate = 1.0 × 10⁻⁷ M/s

The answer is lesson three's average rate, now computed from real laboratory rows.

What the student is expected to be able to do

Calculate the average rate over a specified interval from concentration–time data supplied as a table or a graph.

Part 3Check what you just learned

Check 1 · numeric entry

Methylene blue concentration during its fade
Time (s)	[methylene blue] (M)
0	8.00 × 10⁻⁶
90	6.35 × 10⁻⁶
180	4.85 × 10⁻⁶
270	3.65 × 10⁻⁶

The blue dye methylene blue loses its color as it reacts in a basic solution. The table gives its concentration at four times. What is $\frac{Δ [methylene blue]}{Δ t}$ between t = 0 and t = 180 s, with its sign? Give your answer to three significant figures. Scientific notation such as 1.23e-6 works.

Key: -1.75e-8 M/s

On a correct answerCorrect. Δ[methylene blue] = 4.85 × 10⁻⁶ − 8.00 × 10⁻⁶ = −3.15 × 10⁻⁶ M, over Δt = 180 s: −1.75 × 10⁻⁸ M/s.

Sign retry on 1.75e-8Right size, wrong sign. The dye is consumed, so its concentration change is negative: −1.75 × 10⁻⁸ M/s.

Mapped wrong answers

-1.5e-8	You took the rows at 90 s and 270 s. The stem asks between t = 0 and t = 180 s: (4.85 − 8.00) × 10⁻⁶ M ÷ 180 s = −1.75 × 10⁻⁸ M/s.
-3.15e-6	This is the concentration change alone. Divide it by the 180 s: −1.75 × 10⁻⁸ M/s.
-1.61e-8	You averaged over the whole table, 0 to 270 s. The stem asks between t = 0 and t = 180 s: (4.85 − 8.00) × 10⁻⁶ M ÷ 180 s = −1.75 × 10⁻⁸ M/s.

Worked steps shown after a second miss

Write down the values from the table

[methylene blue] at 0 s = 8.00 × 10⁻⁶ M

[methylene blue] at 180 s = 4.85 × 10⁻⁶ M

Δt = 180 s

Write down the equation

$\frac{Δ [methylene blue]}{Δ t}$ = (final − initial) ÷ Δt

Substitute in the values, and calculate

$\frac{Δ [methylene blue]}{Δ t}$ = (4.85 × 10⁻⁶ − 8.00 × 10⁻⁶) ÷ 180

$\frac{Δ [methylene blue]}{Δ t}$ = −1.75 × 10⁻⁸ M/s

Check 2 · numeric entry

Allura red concentration during its fade
Time (s)	[allura red] (M)
0	6.40 × 10⁻⁵
50	5.90 × 10⁻⁵
100	5.46 × 10⁻⁵
150	5.08 × 10⁻⁵

The red food dye allura red loses its color as it reacts with bleach. The table gives its concentration at four times. What is the average rate of consumption of the dye between t = 50 s and t = 150 s? Give your answer to two significant figures. Scientific notation such as 1.2e-7 works.

Key: 8.2e-8 M/s

On a correct answerCorrect. The dye fell from 5.90 × 10⁻⁵ M at 50 s to 5.08 × 10⁻⁵ M at 150 s: 0.82 × 10⁻⁵ M ÷ 100 s = 8.2 × 10⁻⁸ M/s.

Mapped wrong answers

-8.2e-8	Right size, wrong sign. A rate of consumption is reported as a positive quantity: 8.2 × 10⁻⁸ M/s.
8.8e-8	You averaged over the whole table, 0 to 150 s. The stem asks between 50 s and 150 s: (5.90 − 5.08) × 10⁻⁵ M ÷ 100 s = 8.2 × 10⁻⁸ M/s.
8.2e-7	Check the powers of ten: the change is 0.82 × 10⁻⁵ M, and dividing by 100 s gives 8.2 × 10⁻⁸ M/s.

Worked steps shown after a second miss

Write down the values from the table

[allura red] at 50 s = 5.90 × 10⁻⁵ M

[allura red] at 150 s = 5.08 × 10⁻⁵ M

Δt = 100 s

Write down the equation

average rate = $- \frac{Δ [allura red]}{Δ t}$

Substitute in the values, and calculate

average rate = −(5.08 × 10⁻⁵ − 5.90 × 10⁻⁵) ÷ 100

average rate = 8.2 × 10⁻⁸ M/s

Check 3 · numeric entry

Sulfuryl chloride, SO₂Cl₂, decomposes in a sealed flask. The graph shows its concentration against time, with two points marked on gridline crossings. What is the average rate of consumption of SO₂Cl₂ between t = 100 s and t = 300 s? Give your answer to two significant figures. Scientific notation such as 1.2e-4 works.

Key: 2.0e-4 M/s

On a correct answerCorrect. The graph reads 0.08 M at 100 s and 0.04 M at 300 s: a change of 0.04 M over 200 s, so 2.0 × 10⁻⁴ M/s.

Mapped wrong answers

-2.0e-4	Right size, wrong sign. A rate of consumption is reported as a positive quantity: 2.0 × 10⁻⁴ M/s.
1.3e-4	You divided by 300 s, the second point's clock time. The interval is 100 s to 300 s, so Δt = 200 s: 0.04 M ÷ 200 s = 2.0 × 10⁻⁴ M/s.
0.04	This is the concentration change read from the graph. Divide it by the 200 s interval: 2.0 × 10⁻⁴ M/s.

Worked steps shown after a second miss

Read the two marked points from the graph

(100 s, 0.08 M)

(300 s, 0.04 M)

Write down the equation

average rate = $- \frac{Δ [SO₂Cl₂]}{Δ t}$

Substitute in the values, and calculate

average rate = −(0.04 − 0.08) M ÷ 200 s

average rate = 2.0 × 10⁻⁴ M/s

Judge this card on:

Does the headache land before the answer?
Is each reveal line one idea?
Does every check test this lesson's atom (under the title)?
Is any key wrong?
Is each figure right – labels, shapes, the point it makes?

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Note (saves in this browser as you type)

Lesson 11 of 11 · KIN-011

Instantaneous rate from a tangent

Atom: Calculate an instantaneous rate, including the initial rate, by drawing a tangent to a concentration–time curve and computing the tangent's slope [e.g., estimate the initial rate from the tangent drawn at t = 0].

Outcome: Calculate an instantaneous rate, including the initial rate, from a drawn tangent's slope.

Part 1An exam question no rows can answer

Lesson one's flask has been measured at four times; here is its data table.

Hydrogen peroxide concentration in lesson one's flask
Time (s)	[H₂O₂] (M)
0	0.80
30	0.76
60	0.74
120	0.70

Here is an exam question about this table: what is the initial rate of consumption of the H₂O₂?

Lesson five named that rate: the initial rate is the instantaneous rate at the moment the reaction begins, t = 0.

Any pair of table rows gives an average rate: a change spread over a stretch of time.

But the question asks for the rate at the single instant t = 0.

So no pair of rows can answer it.

Only the curve through the rows can answer it.

How does a line drawn on that curve turn into a rate in M/s?

Part 2Draw the tangent, read the slope

Draw the tangent at the asked instant: lesson four's straight line that touches the curve there and matches its steepness.

Pick two easy-to-read points that sit far apart on the tangent line.

Compute the slope from those two points: rise over run.

Lesson two's rule sets the sign: a rate is reported positive, so the built-in minus sign flips a reactant's negative slope.

The final line states the rate in M/s.

Here is the curve through the table's four rows, with the tangent drawn at t = 0 and two read points marked.

Q: What is the initial rate of consumption of H₂O₂ in the flask?

Read the two marked points on the tangent

(0 s, 0.80 M)(the touch point at t = 0)

(100 s, 0.65 M)(far along the tangent)

Write down the equation

slope = rise ÷ run

Substitute in the values, and calculate

slope = (0.65 M − 0.80 M) ÷ (100 s − 0 s)

slope = −0.15 M ÷ 100 s = −0.0015 M/s

initial rate = −(−0.0015) = 0.0015 M/s(lesson two's built-in minus sign)

That answers the exam question: the initial rate the table's rows could not give is 0.0015 M/s.

The tangent's slope at t = 0 is the initial rate the AP exam asks for by name.

What the student is expected to be able to do

Calculate an instantaneous rate, including the initial rate, by drawing a tangent to a concentration–time curve and computing the tangent's slope.

Part 3Check what you just learned

Check 1 · multiple choice

A student computes a drawn tangent's slope from two points, using rise over run. Where must both points sit?

On the tangent line KEY

Correct. The slope being computed is the tangent line's, so both read points must sit on the tangent line — far apart, on easy-to-read crossings.

On the curve

This option is wrong — the curve's steepness keeps changing, so two curve points give the average rate over the stretch between them, not the tangent's slope.

One on the tangent line and one on the curve

This option is wrong — a mixed pair gives the slope of neither the tangent nor the curve. Both read points must sit on the tangent line itself.

Check 2 · numeric entry

Cyclopropane gas slowly rearranges into propene in a sealed flask. The graph shows [cyclopropane] against time; the tangent at t = 100 s is drawn, with two points on the tangent marked on gridline crossings. What is $\frac{Δ [cyclopropane]}{Δ t}$ at t = 100 s, with its sign? Give your answer to two significant figures. Scientific notation such as 1.2e-4 works.

Key: -7.5e-4 M/s

On a correct answerCorrect. The tangent reads 0.45 M at 0 s and 0.15 M at 400 s: a rise of −0.30 M over a run of 400 s gives −7.5 × 10⁻⁴ M/s.

Sign retry on 7.5e-4Right size, wrong sign. The cyclopropane is consumed, so its concentration change is negative: −7.5 × 10⁻⁴ M/s.

Mapped wrong answers

-0.30	This is the rise alone. Divide it by the run, 400 s: −7.5 × 10⁻⁴ M/s.
-0.0030	You divided the rise by the touch point's clock time, 100 s. The run is the time between the two marked points: 400 s − 0 s = 400 s.
0.0030	You divided the rise by the touch point's clock time, 100 s, and dropped the sign. The run is 400 s, and the change is negative: −7.5 × 10⁻⁴ M/s.

Worked steps shown after a second miss

Read the two marked points on the tangent

(0 s, 0.45 M)

(400 s, 0.15 M)

Write down the equation

slope = rise ÷ run

Substitute in the values, and calculate

slope = (0.15 M − 0.45 M) ÷ (400 s − 0 s)

$\frac{Δ [cyclopropane]}{Δ t}$ = −7.5 × 10⁻⁴ M/s

Check 3 · numeric entry

Azomethane gas decomposes in a sealed flask. The graph shows its concentration against time; the tangent at t = 0 is drawn, with two points on the tangent marked on gridline crossings. What is the initial rate of consumption of azomethane? Give your answer to two significant figures. Scientific notation such as 1.2e-4 works.

Key: 1.6e-4 M/s

On a correct answerCorrect. The tangent reads 0.032 M at 0 s and 0.008 M at 150 s: a rise of −0.024 M over a run of 150 s, so the initial rate is 1.6 × 10⁻⁴ M/s.

Mapped wrong answers

-1.6e-4	Right size, wrong sign. The question asks the initial rate of consumption, which is reported positive: 1.6 × 10⁻⁴ M/s.
2.1e-4	This comes from dividing the tangent's starting concentration, 0.032 M, by the 150 s. Use rise over run between the two marked points: 0.024 M ÷ 150 s.
0.024	This is the rise alone. Divide it by the run, 150 s: 1.6 × 10⁻⁴ M/s.

Worked steps shown after a second miss

Read the two marked points on the tangent

(0 s, 0.032 M)

(150 s, 0.008 M)

Write down the equation

slope = rise ÷ run

Substitute in the values, and calculate

slope = (0.008 M − 0.032 M) ÷ (150 s − 0 s) = −1.6 × 10⁻⁴ M/s

initial rate = 1.6 × 10⁻⁴ M/s

Judge this card on:

Does the headache land before the answer?
Is each reveal line one idea?
Does every check test this lesson's atom (under the title)?
Is any key wrong?
Is each figure right – labels, shapes, the point it makes?

Your own criteria sit above the lesson cards.

Note (saves in this browser as you type)

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