Worked examples — Rate law — order vs molecularity

This page is the drill hall for the parent topic. We will not learn new ideas — we will hit every kind of question this topic can throw at you, one worked example per cell. If a phrase like "rate law" or "molecularity" feels blurry, re-read the parent note first; here we assume you can already read $\text{Rate}=k[\text{A}{]}^m[\text{B}{]}^n$ and know that $m$ is measured, not guessed.

The scenario matrix

#	Case class	What makes it tricky	Example that hits it
C1	Integer order from initial-rates data	reading ratios, taking logs	Example 1
C2	Zero order in one reactant	rate ignores a concentration	Example 2
C3	Fractional order from data	exponent is not a whole number	Example 3
C4	Negative order (inhibition)	more reactant → slower	Example 4
C5	Elementary step: order = molecularity	stoichiometry is the rate law	Example 5
C6	Multi-step: order ≠ stoichiometry	slow step decides everything	Example 6
C7	Degenerate / units of $k$	$k$'s units change with order	Example 7
C8	Real-world word problem	translate words → rate law	Example 8
C9	Exam twist: pseudo-order	one reactant swamped, "hidden" order	Example 9

We work them in order. Each tells you which cell it covers.

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Example 1 — Integer order (Cell C1)

Forecast: Doubling A doubled the rate — guess order 1 in A. Tripling B multiplied rate by 9 — guess order 2 in B. Overall 3.

Figure — ratio bars for A and B. The left pair of bars compares trials 1 and 2: the concentration doubles and the bar exactly doubles → order 1. The right pair compares trials 1 and 3: concentration triples but the bar leaps ×9 → order 2. The height jump versus the width jump is literally the exponent.

Alt-text: two pairs of vertical bars on a white grid. Left pair: blue bars of height 2 and 4 under "[A] x2", labelled order 1. Right pair: orange bars of height 2 and 18 under "[B] x3", labelled order 2.

1. Isolate A using trials 1 and 2 ($[\text{B}]$ fixed): $\frac{4.0\times 10^{-3}}{2.0\times 10^{-3}}={\left(\frac{0.20}{0.10}\right)}^m⟹2=2^m⟹m=1$ Why this step? Holding $[\text{B}]$ constant makes the $[\text{B}]$ factor cancel, so the ratio depends on $[\text{A}]$ alone — that ratio is the order.

2. Isolate B using trials 1 and 3 ($[\text{A}]$ fixed): $\frac{18.0\times 10^{-3}}{2.0\times 10^{-3}}={\left(\frac{0.30}{0.10}\right)}^n⟹9=3^n⟹n=2$ Why this step? Same isolation trick, now for B. $9=3^2$ so $n=2$.

3. Overall order $=m+n=1+2=3$. Why this step? The overall order is defined as the sum of the individual exponents, because it tells you how the rate responds when you scale all concentrations together — multiplying every concentration by $c$ multiplies the rate by $c^m\cdot c^n=c^{m+n}$. So the exponents add.

4. Solve for $k$ from trial 1: $k=\frac{\text{Rate}}{[\text{A}][\text{B}{]}^2}=\frac{2.0\times 10^{-3}}{(0.10)(0.10{)}^2}=2.0{\text{M}}^{-2}{\text{s}}^{-1}$ Why this step? Once exponents are known, the rate law has only one unknown left ($k$); any single trial then fixes it by simple division.

Verify: Predict trial 3 with $k=2.0$: $\text{Rate}=2.0\times 0.10\times 0.30^2=2.0\times 0.10\times 0.09=0.018=18\times 10^{-3}$ ✓. Matches the given value.

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Example 2 — Zero order in one reactant (Cell C2)

Forecast: Trials 1→3 change $[\text{Y}]$ 4-fold yet rate is unchanged — smells like order 0 in Y.

Figure — the flat bar. The picture shows two orange bars of the same height even though $[\text{Y}]$ jumps ×4. A concentration that quadruples while the rate bar stays flat is exactly what "order 0" looks like.

Alt-text: two equal-height orange bars on a white grid, both height 6, under labels "[Y]=0.10" and "[Y]=0.40", annotated "rate unchanged -> order 0".

1. Order in X (trials 1,2, $[\text{Y}]$ fixed): $\frac{12.0}{6.0}={\left(\frac{0.10}{0.05}\right)}^m⟹2=2^m⟹m=1$ Why this step? Standard isolation of X — $[\text{Y}]$ cancels, so the rate ratio is the X-order.

2. Order in Y (trials 1,3, $[\text{X}]$ fixed): $\frac{6.0}{6.0}={\left(\frac{0.40}{0.10}\right)}^n⟹1=4^n⟹n=0$ Why this step? $4^n=1$ forces $n=0$: quadrupling Y did nothing, so its exponent must be zero (only $4^0=1$).

3. Rate law: $\text{Rate}=k[\text{X}{]}^1[\text{Y}{]}^0=k[\text{X}]$. Overall order $=m+n=1+0=1$. Why this step? $[\text{Y}{]}^0=1$ so the Y factor disappears; the exponents still add, giving overall order 1.

4. $k$ from trial 1: $k=\frac{6.0\times 10^{-4}}{0.05}=1.2\times 10^{-2}{\text{s}}^{-1}$. Why this step? With Y gone, the law is $\text{Rate}=k[\text{X}]$, so $k$ is just $\text{Rate}/[\text{X}]$.

Verify: A first-order $k$ carries units ${\text{s}}^{-1}$ — this follows from the general units rule $[k]={\text{M}}^{1-n}{\text{s}}^{-1}$ derived in Example 7 (order $n=1$ gives ${\text{M}}^0{\text{s}}^{-1}={\text{s}}^{-1}$) ✓. Predict trial 2: $1.2\times 10^{-2}\times 0.10=1.2\times 10^{-3}=12\times 10^{-4}$ ✓.

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Example 3 — Fractional order (Cell C3)

Forecast: $[\text{A}]$ went up ×4 but rate only ×2. Since $4^{1/2}=2$, guess order $\frac{1}{2}$.

Figure — a small-jump bar. The concentration bar quadruples but the rate bar only doubles. When the height grows less than the width, the exponent is between 0 and 1 — here exactly $\frac{1}{2}$.

Alt-text: two green bars of height 5 and 10 under labels "[A]=1.00" and "[A]=4.00", annotated "conc x4 but rate x2 -> order 0.5".

1. Solve for $m$ using logs — needed because the ratio isn't a nice power of 2: $\frac{10.0}{5.0}={\left(\frac{4.00}{1.00}\right)}^m⟹2=4^m⟹m=\frac{\ln 2}{\ln 4}=\frac{\ln 2}{2\ln 2}=0.5$ Why this step? When the $\text{Rate}$ ratio isn't an obvious power, take $\ln$ of both sides: the unknown exponent falls down out of the power (that is the whole reason logarithms exist), giving the general order formula $m=\frac{\ln ({\text{Rate}}_2/{\text{Rate}}_1)}{\ln ([\text{A}{]}_2/[\text{A}{]}_1)}$. Every integer case in this page is just this formula with a clean answer.

2. Rate law: $\text{Rate}=k[\text{A}{]}^{0.5}$. Why this step? The rate law is just the general form $\text{Rate}=k[\text{A}{]}^m$ with the exponent we solved for ($m=0.5$) slotted in — nothing more; writing it out isolates the single remaining unknown $k$ for the next step.

3. $k$ from trial 1: $k=\frac{5.0\times 10^{-3}}{1.00^{0.5}}=5.0\times 10^{-3}{\text{M}}^{0.5}{\text{s}}^{-1}$. Why this step? Rearrange $\text{Rate}=k[\text{A}{]}^{0.5}$ to $k=\text{Rate}/[\text{A}{]}^{0.5}$ and plug in trial 1.

Verify: Predict trial 2: $5.0\times 10^{-3}\times 4^{0.5}=5.0\times 10^{-3}\times 2=10\times 10^{-3}$ ✓.

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Example 4 — Negative order / inhibition (Cell C4)

Forecast: Trials 1→3 raise $[\text{P}]$ ×4 and the rate drops to a quarter. A rate that falls when a concentration rises means a negative exponent.

Figure — the shrinking bar. Unlike every earlier picture, here raising $[\text{P}]$ makes the rate bar shrink to a quarter. A downward jump is the visual signature of a negative order.

Alt-text: two red bars, the left of height 8 and the right of height 2, under labels "[P]=0.10" and "[P]=0.40", annotated "conc x4 but rate to 1/4 -> order minus 1".

1. Order in S (trials 1,2): $\frac{16.0}{8.0}={\left(\frac{0.40}{0.20}\right)}^m⟹2=2^m⟹m=1$ Why this step? Isolate S: $[\text{P}]$ is fixed and cancels, so the rate ratio is the S-order.

2. Order in P (trials 1,3, $[\text{S}]$ fixed). Let the symbol $p$ denote the order with respect to P (the exponent on $[\text{P}]$ in the rate law). Apply the same general log method from Example 3, now with an answer below zero: $\frac{2.0}{8.0}={\left(\frac{0.40}{0.10}\right)}^p⟹0.25=4^p⟹p=\frac{\ln 0.25}{\ln 4}=-1$ Why this step? We introduce a named exponent $p$ for P exactly as $m,n$ named the exponents for A and B; then it is the log-isolation formula $p=\frac{\ln ({\text{Rate}}_3/{\text{Rate}}_1)}{\ln ([\text{P}{]}_3/[\text{P}{]}_1)}$ from Example 3 — the only difference is the numerator $\ln 0.25<0$, so the order comes out negative. A negative order is legal in the empirical rate law: it encodes inhibition.

3. Rate law: $\text{Rate}=k[\text{S}][\text{P}{]}^{-1}=\frac{k[\text{S}]}{[\text{P}]}$. Overall order $=m+p=1+(-1)=0$. Why this step? The exponents add just as before; here they cancel to a net overall order of 0, meaning scaling both S and P together leaves the rate unchanged.

4. $k$ from trial 1: $k=\frac{\text{Rate}\cdot [\text{P}]}{[\text{S}]}=\frac{8.0\times 10^{-3}\times 0.10}{0.20}=4.0\times 10^{-3}{\text{s}}^{-1}$. Why this step? Rearrange $\text{Rate}=k[\text{S}]/[\text{P}]$ to solve for $k$, then plug in trial 1.

Verify: Predict trial 3: $\frac{4.0\times 10^{-3}\times 0.20}{0.40}=2.0\times 10^{-3}$ ✓.

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Example 5 — Elementary step: order = molecularity (Cell C5)

Forecast: One NO plus one O₃ collide, so guess $\text{Rate}=k[\text{NO}][{\text{O}}_3]$, molecularity 2, order 2.

Figure — one collision. Two particles (one NO, one O₃) approach and touch at a single point. Because exactly two particles must meet, the collision frequency is proportional to $[\text{NO}]\times [{\text{O}}_3]$ — the picture is the rate law.

Alt-text: two circles labelled NO (blue) and O3 (orange) with arrows pointing toward a shared collision point marked with a green star, annotated "2 particles meet -> Rate proportional to [NO][O3]".

1. Use the elementary rule: for an elementary step, exponents equal stoichiometric coefficients. $\text{Rate}=k[\text{NO}{]}^1[{\text{O}}_3{]}^1$ Why this step? In one collision the frequency of NO-meets-O₃ events is proportional to $[\text{NO}]\times [{\text{O}}_3]$ — the stoichiometry literally is the collision requirement.

2. Molecularity = number of species colliding = $1+1=2$ (bimolecular, i.e. two particles meet at once). Why this step? We just count the reactant particles in the one step: one NO + one O₃ = 2, which is the definition of molecularity above.

3. Order = $m+n=1+1=2$. Here order equals molecularity — but only because the step is elementary. Why this step? Add the exponents from step 1 to get overall order; for an elementary step those exponents came straight from the collision count, so the two numbers coincide.

Verify: If we doubled both concentrations, rate should ×4: ${\left(\frac{2[\text{NO}]}{[\text{NO}]}\right)}^1{\left(\frac{2[{\text{O}}_3]}{[{\text{O}}_3]}\right)}^1=2\times 2=4$ ✓, consistent with overall order 2.

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Example 6 — Multi-step: order ≠ stoichiometry (Cell C6)

Figure — the bottleneck gate. The diagram below draws the two-step mechanism as a pipe. On the left, reactants ${\text{NO}}_2+{\text{F}}_2$ pour toward a narrow red gate (the slow step 1). Only a trickle passes per second. On the right, whatever passes hits the wide green box (fast step 2), which clears instantly. The point the picture makes visually: the rate of the whole pipe is fixed by the narrow gate alone — widening the fast box changes nothing. That is why only step-1 reactants appear in the rate law.

Alt-text: a flow diagram; a blue "NO2 + F2" reactant box feeds through a narrow red "SLOW step 1" gate labelled Rate = k1[NO2][F2] into a wide green "FAST step 2" box, then to product P. Grey arrows show a few particles passing the gate and instant clearance afterward.

Forecast: The stoichiometry $2{\text{NO}}_2+{\text{F}}_2$ tempts you toward order 3. But only the slow step counts — guess order 2.

1. Rate = rate of the slow step (step 1 is the bottleneck): $\text{Rate}=k_1[{\text{NO}}_2][{\text{F}}_2]$ Why this step? As the figure shows, the slow step is a narrow gate. No matter how fast step 2 runs, overall throughput is capped by the gate, so only the slow step's reactants set the rate.

2. Read off orders: order in NO₂ $=1$, order in F₂ $=1$, overall $=1+1=2$. Why this step? The slow-step law is elementary, so its exponents are the coefficients (both 1); add them for overall order 2.

3. Compare to stoichiometry: sum of coefficients $=2+1=3\ne 2$. The extra NO₂ enters only in the fast step, so it never appears in the rate law. Why this step? This is the whole lesson of the cell: overall order comes from the mechanism's slow step, not from the balanced equation.

4. Molecularity check: step 1 is bimolecular (2), step 2 is bimolecular (2). Individual molecularities are integers (2 each), yet the overall order (2) matching step-1 molecularity is a coincidence of this mechanism, not a rule. Why this step? We count colliding particles in each elementary step (2 and 2) to show that "molecularity" is a per-step count, distinct from the overall order we measured — the two only happen to agree here.

Verify: The intermediate F is produced in step 1 and consumed in step 2 at equal rates in steady state, so it never accumulates and cannot appear in the observed law — consistent with $\text{Rate}=k_1[{\text{NO}}_2][{\text{F}}_2]$. (Numerically: doubling only $[{\text{NO}}_2]$ doubles rate; doubling only $[{\text{F}}_2]$ doubles rate — both order 1.)

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Example 7 — Degenerate case: units of $k$ (Cell C7)

Forecast: Rate always has units ${\text{M s}}^{-1}$, so $k$ must absorb whatever $[\text{A}{]}^n$ leaves behind.

1. General rule: from $\text{Rate}=k[\text{A}{]}^n$, solve for $[k]$ by matching units on both sides: $[k]=\frac{[\text{Rate}]}{[\text{A}{]}^n}=\frac{{\text{M s}}^{-1}}{{\text{M}}^n}={\text{M}}^{1-n}{\text{s}}^{-1}$ Why this step? Both sides of any physical equation must carry the same units; the left side is ${\text{M s}}^{-1}$, so $k$ must supply exactly ${\text{M}}^{1-n}{\text{s}}^{-1}$ to make $k[\text{A}{]}^n$ land on ${\text{M s}}^{-1}$.

2. Plug in each order:

Order $n$	$[k]={\text{M}}^{1-n}{\text{s}}^{-1}$
$0$	${\text{M s}}^{-1}$
$1$	${\text{s}}^{-1}$
$2$	${\text{M}}^{-1}{\text{s}}^{-1}$
$0.5$	${\text{M}}^{0.5}{\text{s}}^{-1}$

Verify: Cross-check with earlier examples — Example 1 (order 3) gave ${\text{M}}^{-2}{\text{s}}^{-1}$: formula gives ${\text{M}}^{1-3}{\text{s}}^{-1}={\text{M}}^{-2}{\text{s}}^{-1}$ ✓. Example 3 (order 0.5) gave ${\text{M}}^{0.5}{\text{s}}^{-1}$ ✓.

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Example 8 — Real-world word problem (Cell C8)

Forecast: "Time to fade" is inverse to rate. Faster rate ⇒ shorter time. Bleach ×2 halves time ⇒ doubles rate ⇒ order 1 in B. Dye ×2 doesn't change time ⇒ order 0 in D.

1. Translate words to rate: fade time $t\propto 1/\text{Rate}$.

   • Bleach ×2 → $t$ halves → Rate ×2 → order in B: $2=2^b\Rightarrow b=1$.
   • Dye ×2 → $t$ same → Rate same → order in D: $1=2^d\Rightarrow d=0$. Why this step? We can't measure $k$ from words, but ratios of rate are exactly what orders need — and rate ratios are the inverse of time ratios.

2. Rate law: $\text{Rate}=k[\text{B}{]}^1[\text{D}{]}^0=k[\text{B}]$. Overall order $=b+d=1+0=1$. Why this step? Add the two exponents; the $[\text{D}{]}^0$ factor is 1 and drops out, leaving overall order 1.

Verify: Sanity — the reaction is limited by how much bleach is present, not dye, which matches "dye is just along for the ride." Doubling B and halving the observed time are consistent with first order in B ($2\times \text{rate}\Rightarrow \frac{1}{2}\times \text{time}$) ✓.

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Example 9 — Exam twist: pseudo-first order (Cell C9)

Forecast: Water is in huge excess, so $[{\text{H}}_2\text{O}]$ is effectively constant. A constant folded into $k$ makes the law appear first order in ester only.

1. Justify the "constant water" approximation: the ester is dilute, say $\sim 0.1\text{M}$, while water is $55.5\text{M}$. Even if all the ester reacts, at most $\sim 0.1\text{M}$ of water is consumed — a change of under $0.2\%$. Why this step? This is the key approximation of the whole example. A quantity that changes by less than a fraction of a percent over the run can be treated as a fixed number, not a variable — that is exactly what lets us fold it into the rate constant in the next step.

2. Absorb the constant into a new constant $k^{\prime }$: since $[{\text{H}}_2\text{O}]$ is (approximately) fixed, define $k^{\prime }\equiv k[{\text{H}}_2\text{O}]$: $\text{Rate}=k[\text{Ester}]\underset{\approx \text{constant}}{\underbrace{[{\text{H}}_2\text{O}]}}=(k[{\text{H}}_2\text{O}])[\text{Ester}]=k^{\prime }[\text{Ester}]$ Why this step? Once $[{\text{H}}_2\text{O}]$ is a number, $k[{\text{H}}_2\text{O}]$ is just another number; the form of the law collapses to $\text{Rate}=k^{\prime }[\text{Ester}]$, which is what a first-order law looks like.

3. Name it: this is a pseudo-first-order reaction — its true order (and molecularity of the elementary collision) is 2, but the observed order is 1 because water's dependence is hidden inside $k^{\prime }$. Why this step? Reading the collapsed law $\text{Rate}=k^{\prime }[\text{Ester}]$ at face value gives observed order 1; the "pseudo" flags that a real second reactant is lurking in $k^{\prime }$.

4. Relate $k^{\prime }$ to the true $k$ numerically. Suppose the true $k=2.0\times 10^{-4}{\text{M}}^{-1}{\text{s}}^{-1}$. Then $k^{\prime }=k[{\text{H}}_2\text{O}]=2.0\times 10^{-4}{\text{M}}^{-1}{\text{s}}^{-1}\times 55.5\text{M}=1.11\times 10^{-2}{\text{s}}^{-1}$ Why this step? Substituting the fixed $[{\text{H}}_2\text{O}]$ into $k^{\prime }=k[{\text{H}}_2\text{O}]$ both gives the number and shifts the units: the $\text{M}$ in $55.5\text{M}$ cancels one ${\text{M}}^{-1}$ in $k$, turning second-order units ${\text{M}}^{-1}{\text{s}}^{-1}$ into first-order units ${\text{s}}^{-1}$.

Verify: Units: ${\text{M}}^{-1}{\text{s}}^{-1}\times \text{M}={\text{s}}^{-1}$ ✓ (matches Example 7's first-order units). Magnitude $2.0\times 10^{-4}\times 55.5=1.11\times 10^{-2}$ ✓.

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Recall Self-test — cover the answers

Doubling A quadruples rate: order in A? ::: 2 (since $4=2^2$) Quadrupling B leaves rate unchanged: order in B? ::: 0 (since $1=4^0$) Rate falls to $1/4$ when P is quadrupled: order in P? ::: $-1$ (since $\frac{1}{4}=4^{-1}$) What does $1\text{M}$ mean? ::: $1{\text{mol L}}^{-1}$ — one mole of solute per litre of solution Units of $k$ for overall order $n$? ::: ${\text{M}}^{1-n}{\text{s}}^{-1}$ In a multi-step reaction, which step sets the rate law? ::: the slow (rate-determining) step Can molecularity be 0, fractional, or negative? ::: No — only $1,2,3$; order can be any of those Why does an ester hydrolysis look first order? ::: water is in vast excess, so $[{\text{H}}_2\text{O}]$ is folded into $k^{\prime }$ (pseudo-first order)