Rate of reaction — average vs instantaneous: Visual walkthrough

This page rebuilds the whole idea of reaction rate from a single picture: a curve that falls. Every symbol you meet is earned on that picture before we use it. If you have never seen a slope, a derivative, or the letter $Δ$ , start at line one — you will be fine.

We follow one imaginary reaction the whole way: $2 A ⟶ B$

Step 1 — Draw the thing that changes

WHAT. We plot concentration of A (how many moles of A sit in each litre of solution, written $[A]$ , units mol/L which we shorten to M) on the up-axis, and time $t$ in seconds along the flat-axis. As the reaction runs, A gets used up, so the dots march downward.

Figure — Rate of reaction — average vs instantaneous

WHY. Before you can talk about "how fast," you need a picture of "what is changing." The falling blue curve is the reaction. The steepness of that fall will become our rate — so we draw it first.

PICTURE. Look at the blue curve. High on the left (lots of A at the start), low on the right (A running out). Read the shape: it is steep early, gentle late. Hold that thought — it is the whole reason we will need two kinds of rate.

Step 2 — Pick two moments and connect them

WHAT. Choose two instants: an earlier one $t_{1}$ and a later one $t_{2}$ . Read the curve's height at each: $[A]_{1}$ and $[A]_{2}$ . Draw the straight line joining those two points — the secant line.

WHY. We cannot yet ask "the speed at one instant" — a single point has no steepness. So we cheat: we measure the overall drop between two moments and see how steep the shortcut between them is. That steepness is a speed.

PICTURE. The orange line cuts across the curve like a shortcut. The horizontal run (green) is how much time passed; the vertical drop (red) is how much A vanished.

Step 3 — Steepness = drop ÷ run = average rate

WHAT. The steepness of the orange secant is its rise over its run. Here "rise" is the drop in A and "run" is the time elapsed: $slope of secant \frac{Δ [ A ]}{Δ t} = \frac{[ A ] _{2} - [ A ] _{1} red: how much A dropped}{green: how much time passed t _{2} - t _{1}}$

WHY. "Rate" literally means amount of change per unit time. Divide the change in A by the time it took — that is a per-second speed, exactly like km per hour is distance per time.

PICTURE. The orange triangle: the red side over the green side. A steeper orange line = bigger number = faster average. This ratio is the average rate — the speed of the whole trip between $t_{1}$ and $t_{2}$ , not of any single moment.

Step 4 — Fix the sign so speed is positive

WHAT. Since $Δ [A]$ is negative (A fell), the raw slope is a negative number. We stick a minus sign in front to flip it positive, and we give this positive quantity its own name, $r_{avg}$ — the average rate of reaction measured on A: $name of the positive average speed r_{avg} = makes it positive - \frac{Δ [ A ]}{Δ t}$

So $r_{avg}$ is nothing new — it is exactly the raw secant slope from Step 3, sign-flipped so we can report a clean positive speed.

WHY. A speed should be a positive quantity — "the reaction runs at $1.5 \times 1 0^{- 4}$ M/s," never "minus that." The curve genuinely slopes downhill (negative), but we do not want to keep saying "the rate is negative." So the minus in the definition cancels the downhill sign, and we report a clean positive speed.

PICTURE. Left panel: the raw slope arrow points down-right ( $-$ ). Right panel: after the minus sign, we report the same steepness as a positive magnitude called $r_{avg}$ . Nothing about the reaction changed — only our bookkeeping.

For a product P that is being made, its curve rises, so $Δ [P]$ is already positive — no minus needed: $r_{avg} = + \frac{Δ [ P ]}{Δ t}$

Step 5 — Shrink the interval: the secant becomes a tangent

WHAT. Keep the earlier point fixed and slide the later point toward it. As $t_{2}$ creeps back to $t_{1}$ , the two dots merge and the orange secant tips over into the tangent — the straight line that just grazes the curve at one point.

WHY. The average rate blurs the whole interval into one number. But the curve's steepness changes along the way (recall: steep early, gentle late). To get the speed at one exact moment, we must let the interval shrink to nothing — then the "average over the interval" becomes "the value at the point."

PICTURE. Watch the secants (faint orange) pivot as the gap closes, each one steeper-fitting than the last, until the final green tangent kisses the curve at a single spot. That final line's slope is the instantaneous steepness.

Step 6 — Name the tangent's slope: the derivative → instantaneous rate

WHAT. The slope of that grazing tangent gets its own symbol, $\frac{d [ A ]}{d t}$ , called the derivative. It is the limit from Step 5: $\frac{d [ A ]}{d t} = lim_{Δ t \to 0} secant slope \frac{Δ [ A ]}{Δ t}$ Then we add the same positive-fixing minus from Step 4, and name the result $r_{inst}$ — the instantaneous rate of reaction measured on A: $r_{inst} = - \frac{d [ A ]}{d t}$

WHY. Why a derivative and not just a smaller average? Because a chemist wants the speed at a named instant ("the rate at $t = 50$ s"), not "the average over 45–55 s." The derivative is precisely the tool that answers "steepness at one point," which no finite average can give. The letters $d [A]$ and $d t$ are the infinitely small versions of $Δ [A]$ and $Δ t$ — same triangle from Step 3, shrunk to a point.

PICTURE. Three tangents drawn at early, middle, late times. The early one is steepest (fastest reaction), the late one is nearly flat (reaction crawling). This single picture shows why $r_{inst}$ keeps dropping as A runs out — connects directly to Collision theory: fewer A molecules → fewer collisions → gentler slope.

Step 7 — Make everyone agree: divide by stoichiometric coefficients

WHAT. In $2 A \to B$ , two A vanish for every one B made — so the A-curve falls twice as fast as the B-curve rises. If we just quoted slopes, "the rate" would depend on which species you watched. To get one honest number, divide each species' rate by its coefficient. This normalization applies to both flavours of rate — the average one $r_{avg}$ and the instantaneous one $r_{inst}$ — because it is a fact about the stoichiometry, not about how big or small the time interval was. Written with derivatives (the instantaneous case): $r = per unit of A, halved - \frac{1}{2} \frac{d [ A ]}{d t} = per unit of B + \frac{1}{1} \frac{d [ B ]}{d t}$ and identically with $Δ$ 's in place of $d$ 's for the average case. The single normalized number is written $r$ — the reaction rate — with no "avg/inst" tag because the normalization is the same operation either way; you simply feed it whichever slope (secant or tangent) you measured.

WHY. The balanced equation is a recipe: 2 spoons of A per 1 spoon of B. The reaction "happens" once per recipe, not once per molecule. Dividing $\frac{d [ A ]}{d t}$ by its coefficient 2 converts "per A molecule" into "per reaction," so A, B, and any other species all report the same $r$ .

PICTURE. Two curves: A plunging steeply (coefficient 2) and B rising gently (coefficient 1). After we halve A's slope, the two arrows shrink/grow to the same length — the shared reaction rate $r$ . This is the seed of the Rate law and order of reaction, where $r$ is tied to concentrations.

Step 8 — The degenerate cases (so nothing surprises you)

WHAT & WHY & PICTURE, three edge cases the general picture must survive:

A perfectly straight line. If $[A]$ falls in a straight line, its tangent is the line everywhere — so average rate = instantaneous rate at every point. Left panel: secant and tangent coincide. (Rare in real chemistry, but the maths still works.)
The flat tail ( $t \to$ large). As A is nearly gone, the curve flattens: tangent slope $\to 0$ , so $r_{inst} \to 0$ . Middle panel: the reaction never un-happens; it just stops. The rate approaches zero but stays $\geq 0$ .
A product's rising curve. Watching B instead of A, the slope is positive, so we use $+ \frac{d [ B ]}{d t}$ — no minus. Right panel: same reaction, mirror-image curve, opposite sign. This is why the definition splits into " $-$ for reactants, $+$ for products."

Recall Check the edge cases

For a product being formed, do we use a minus sign? ::: No — a product's concentration rises, so its slope is already positive; we write $+ \frac{d [ P ]}{d t}$ . As a reaction nears completion, what happens to the instantaneous rate? ::: The curve flattens, its tangent slope goes to zero, so $r_{inst} \to 0$ . When are average and instantaneous rate exactly equal? ::: Only when the concentration–time graph is a straight line, so its tangent matches its secant everywhere.

Worked check — putting the pictures to numbers

The one-picture summary

Everything on one canvas: the falling blue $[A]$ curve; an orange secant (whose slope, sign-flipped, is $r_{avg}$ ); a green tangent at one instant (whose slope, sign-flipped, is $r_{inst}$ ); the shrinking-interval arrow linking them; and the " $\div a$ " tag showing the final normalization to the single reaction rate $r$ .

Recall Feynman retelling — say it in plain words

I draw how much A there is as a curve that slides downhill over time. To ask "how fast," I first pick two moments and lay a straight shortcut between the curve's heights there — its steepness (drop divided by time) is the average speed of the whole stretch. The curve slopes downhill, giving a negative number, but a speed should be positive, so the definition tucks in a minus sign to flip it. Now I want the speed at one exact instant, not a whole stretch — so I slide the two points together until the shortcut grazes the curve at a single spot: that grazing line is the tangent, and its steepness is the instantaneous rate (same minus-sign flip). The curve is steep at the start and lazy at the end, which is exactly why the instant-speed keeps dropping — fewer molecules, fewer collisions. Finally, because two A vanish per one B in $2 A \to B$ , A's curve falls twice as fast as B's rises; to make everyone quote the same number, I divide each slope by its recipe coefficient. Do that, and reactant or product, they all agree on one honest reaction rate $r$ .

Where this leads: once $r$ is defined, we ask what controls its size — that is the Rate law and order of reaction; solving those gives the Integrated rate laws and Half-life of reactions; the why-it-slows lives in Collision theory and the Arrhenius equation.