Worked examples — Rate of reaction — average vs instantaneous

This page is a drill. We take the ideas from the parent note — average rate, instantaneous rate, the negative sign, and dividing by stoichiometric coefficients — and hammer them against every kind of situation a problem can hand you.

Before we start, one reminder in plain words so no symbol sneaks in undefined:

Because every example below leans on one master equation, we state it here in full so the page is self-contained.

The scenario matrix

Every rate problem falls into one of these cells. The examples below are labelled with the cell they cover.

Cell	Case class	What makes it tricky
A	Reactant disappearing	must apply the negative sign
B	Product appearing	sign is already positive
C	Coefficient $=1$ (degenerate)	dividing by 1 changes nothing — check you notice
D	Coefficient $>1$	must divide to get reaction rate
E	Linking two species' rates	convert A's rate into C's rate
F	Average from a table of data	pure $\Delta /\Delta t$
G	Instantaneous from a tangent slope	derivative = slope of tangent
H	Average vs instantaneous compared	limiting behaviour: rate falls over time
I	Zero / degenerate input	$\Delta t\to 0$, or no change at all
J	Real-world word problem	translate words into $\Delta []/\Delta t$
K	Exam twist (mixed units / hidden coefficient)	unit conversion + stoichiometry together

The nine examples below hit all eleven cells.

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Worked examples

Forecast: Will the reaction rate be bigger or smaller than the raw rate at which ${\text{O}}_2$ vanishes? (Guess before reading.)

1. Raw change in ${\text{O}}_2$. $\Delta [{\text{O}}_2]=0.180-0.240=-0.060\text{M}$. Why this step? $\Delta$ always means final minus initial; the minus tells us ${\text{O}}_2$ is being used up.
2. Rate ${\text{O}}_2$ disappears. $-\frac{\Delta [{\text{O}}_2]}{\Delta t}=-\frac{-0.060}{30.0}=2.0\times 10^{-3}\text{M/s}$. Why this step? The negative sign in the definition turns the falling concentration into a positive speed.
3. Normalise by the coefficient. $r=\frac{1}{3}\times 2.0\times 10^{-3}=6.67\times 10^{-4}\text{M/s}$. Why this step? Three ${\text{O}}_2$ molecules vanish per reaction event, so the reaction itself runs $3\times$ slower than ${\text{O}}_2$ disappears — divide by the coefficient $a=3$, exactly the $-\frac{1}{a}\frac{d[\text{A}]}{dt}$ term of the general equation.

Verify: Reaction rate $6.67\times 10^{-4}$ is smaller than $2.0\times 10^{-3}$, matching the forecast (coefficient $>1$ shrinks the reaction rate). Units: $\text{M/s}$ ✓.

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Forecast: Faster or slower than the reaction rate? (Coefficient of ${\text{O}}_3$ is 2.)

1. Write the link. $r=\frac{1}{c}\frac{d[{\text{O}}_3]}{dt}$, where $c=2$ is the coefficient of the product ${\text{O}}_3$ from the general rate equation stated above. Why this step? The product term $+\frac{1}{c}\frac{d[\text{C}]}{dt}=r$ says a product's rate, divided by its coefficient $c$, equals $r$. No negative sign — products increase, so the slope is already positive.
2. Solve for the formation rate. $\frac{d[{\text{O}}_3]}{dt}=2r=2\times 6.67\times 10^{-4}=1.33\times 10^{-3}\text{M/s}$. Why this step? Two ${\text{O}}_3$ appear per reaction event, so ${\text{O}}_3$ builds up twice as fast as the reaction runs.

Verify: Ratio $\frac{d[{\text{O}}_3]/dt}{-d[{\text{O}}_2]/dt}=\frac{1.33\times 10^{-3}}{2.0\times 10^{-3}}=\frac{2}{3}$, exactly the stoichiometric ratio $\frac{2{\text{O}}_3}{3{\text{O}}_2}$ ✓.

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Forecast: Does dividing by the coefficient do anything here?

1. Average rate of A loss. $-\frac{0.320-0.500}{6.0}=-\frac{-0.180}{6.0}=0.030\text{M/s}$. Why this step? Standard definition with the sign flip.
2. Normalise. Coefficient of A is $a=1$, so $r=\frac{1}{1}\times 0.030=0.030\text{M/s}$. Why this step? This is the trap: dividing by 1 leaves the number unchanged, but you should still consciously check the coefficient rather than skip the step.

Verify: Because $a=1$, $r$ equals the A-loss rate exactly ✓. Units $\text{M/s}$ ✓.

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$t$ (s)	0	20	40	60
$[\text{X}]$ (M)	0.800	0.560	0.400	0.288

Find the average rate over the first interval ($0\to 20\text{s}$) and over the last interval ($40\to 60\text{s}$). Coefficient of X is 1.

Forecast: Which interval has the bigger average rate?

1. First interval. $-\frac{0.560-0.800}{20-0}=-\frac{-0.240}{20}=0.0120\text{M/s}$. Why this step? Average rate is always $\Delta /\Delta t$ over the chosen interval.
2. Last interval. $-\frac{0.288-0.400}{60-40}=-\frac{-0.112}{20}=0.0056\text{M/s}$. Why this step? Same definition, different two rows.

Verify: First interval rate $0.0120>$ last interval rate $0.0056$: the reaction slows down as X is consumed — the expected limiting behaviour ✓.

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Forecast: The tangent falls left-to-right, so its slope is negative — what sign will the rate be?

1. Slope of the tangent line. $\frac{d[\text{A}]}{dt}=\frac{0.28-0.60}{60-20}=\frac{-0.32}{40}=-0.0080\text{M/s}$. Why this step? The instantaneous rate is the derivative, and a derivative is exactly the slope of the tangent line at that point (look at the orange tangent touching the curve at the magenta dot).
2. Apply the sign convention. $r_{\text{inst}}=-\frac{d[\text{A}]}{dt}=-(-0.0080)=0.0080\text{M/s}$. Why this step? Rate is reported positive; the definition's minus cancels the tangent's negative slope.

Verify: Slope negative $\Rightarrow$ rate positive $0.0080\text{M/s}$, matching the forecast ✓. Units: $\frac{\text{M}}{\text{s}}$ ✓.

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Forecast: Which is larger, and why?

1. Average rate. $-\frac{0.25-1.00}{200-0}=-\frac{-0.75}{200}=3.75\times 10^{-3}\text{M/s}$. Why this step? Average over the whole trip is the total $\Delta [\text{A}]$ over total time — the slope of the straight chord in the figure.
2. Compare. $\frac{r_{\text{inst}}(0)}{r_{\text{avg}}}=\frac{8.0\times 10^{-3}}{3.75\times 10^{-3}}=2.13$. Why this step? The initial instantaneous rate is the steep tangent at $t=0$; the average is the shallower chord. The tangent is steeper, so the ratio exceeds 1.

Verify: $r_{\text{inst}}(0)>r_{\text{avg}}$ because the reaction is fastest when $[\text{A}]$ is highest and slows afterward — the chord's slope is a blend of fast-early and slow-late ✓.

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Forecast: Is a zero rate physically allowed? What object is the "limit" in part (b)?

1. (a) No change. $-\frac{0.50-0.50}{\Delta t}=-\frac{0}{\Delta t}=0\text{M/s}$. Why this step? $\Delta [\text{A}]=0$ makes the numerator zero, so the rate is genuinely zero — the reaction has stalled (e.g. equilibrium, or over that window nothing measurable happened).
2. (b) Shrinking window. As $\Delta t\to 0$, $-\frac{\Delta [\text{A}]}{\Delta t}⟶-\frac{d[\text{A}]}{dt}$ at $t=40\text{s}$. Why this step? This IS the definition of the instantaneous rate: the average over a vanishing interval becomes the derivative. You may not put $\Delta t=0$ directly (that gives $0/0$, undefined) — the limit is what we mean.

Verify: (a) gives exactly $0$ ✓. (b) The average rate has a well-defined limit (the derivative) even though the fraction is undefined at $\Delta t=0$ — this is precisely why chemists needed the instantaneous definition ✓.

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Forecast: Which sign, and roughly how many $\text{mmol/L}$ vanish per hour?

1. Identify the change. $\Delta [\text{urea}]=7.0-25.0=-18.0\text{mmol/L}$. Why this step? Urea is being removed, so treat it exactly like a disappearing reactant.
2. Average removal rate. $-\frac{\Delta [\text{urea}]}{\Delta t}=-\frac{-18.0}{4.0}=4.5\text{mmol/L per hour}$. Why this step? Same definition — the "reaction" here is physical removal, but the maths of $\Delta /\Delta t$ is identical.

Verify: $4.5\text{mmol/L/hr}\times 4.0\text{hr}=18.0\text{mmol/L}$ removed, matching $25.0-7.0$ ✓. Positive rate for a disappearing quantity ✓.

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Forecast: Two traps hide here — a unit conversion (min→s) and two different coefficients (2 and 3). Predict whether ${\text{H}}_2$ vanishes faster or slower than ${\text{NH}}_3$ forms.

1. Convert units. $1.2\times 10^{-2}\frac{\text{M}}{\text{min}}\times \frac{1\text{min}}{60\text{s}}=2.0\times 10^{-4}\text{M/s}$. Why this step? All rates must share a time unit before we combine them. $1\text{min}=60\text{s}$.
2. Reaction rate from ${\text{NH}}_3$. $r=\frac{1}{c}\frac{d[{\text{NH}}_3]}{dt}=\frac{1}{2}\times 2.0\times 10^{-4}=1.0\times 10^{-4}\text{M/s}$, using the product coefficient $c=2$. Why this step? Coefficient of ${\text{NH}}_3$ is 2, so divide by 2 to get the normalised reaction rate.
3. Rate ${\text{H}}_2$ is consumed. $-\frac{d[{\text{H}}_2]}{dt}=3r=3\times 1.0\times 10^{-4}=3.0\times 10^{-4}\text{M/s}$. Why this step? The reactant term $-\frac{1}{a}\frac{d[{\text{H}}_2]}{dt}=r$ with $a=3$ rearranges to $-\frac{d[{\text{H}}_2]}{dt}=3r$; three ${\text{H}}_2$ vanish per reaction event.

Verify: $\frac{-d[{\text{H}}_2]/dt}{d[{\text{NH}}_3]/dt}=\frac{3.0\times 10^{-4}}{2.0\times 10^{-4}}=\frac{3}{2}$, exactly the coefficient ratio $\frac{3{\text{H}}_2}{2{\text{NH}}_3}$ ✓. ${\text{H}}_2$ disappears faster than ${\text{NH}}_3$ forms, as forecast ✓.

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Recall Which cell is which? (all eleven)

Cell A — reactant disappearing needs the negative sign. ::: A Cell B — a product appearing already has a positive sign (no minus). ::: B Cell C — the degenerate case where the coefficient equals 1 (dividing changes nothing but you still check it). ::: C Cell D — coefficient greater than 1 means you must divide by the coefficient to get the reaction rate. ::: D Cell E — linking two species means converting one species' rate into another via their coefficient ratio. ::: E Cell F — average rate from a data table is pure Δ[ ]/Δt over an interval. ::: F Cell G — instantaneous rate equals the slope of the tangent line to the curve. ::: G Cell H — comparing average and instantaneous shows the rate falls over time (tangent steeper than chord). ::: H Cell I — as $\Delta t\to 0$ the average rate becomes the derivative (instantaneous rate). ::: I Cell J — a word problem just means translate the words into a Δ[ ]/Δt calculation. ::: J Cell K — an exam twist bundles a unit conversion together with a hidden stoichiometric coefficient. ::: K

Connections

• Once you can measure rates, you ask how they depend on concentration → Rate law and order of reaction.
• Integrating the rate law over time → Integrated rate laws and Half-life of reactions.
• Why rates fall as reactant runs out, at the molecular level → Collision theory and Reaction mechanisms.
• How temperature turbo-charges rates → Arrhenius equation; how Catalysts speed them without being consumed.