Rate law — order vs molecularity: Question bank — Chemistry deep dive

Recall The two definitions you keep needing

Order ::: The experimentally measured exponent on a concentration in the rate law $Rate = k [A]^{m} [B]^{n}$ ; the overall order is $m + n$ . Molecularity ::: The number of reactant particles that collide in one elementary step — a theoretical count from the mechanism, always a positive integer (1, 2, or 3).

Toolbox refresher (so the traps land)

Before the trap questions, three quick build-ups the traps rely on: why the ln appears in first order, what a first-order plot looks like, and why termolecular steps are rare. Skip if you already own these.

Why the logarithm shows up in first order

WHAT we do — separate the two variables. Gather every $[A]$ on the left and every $t$ on the right, so each side has only one variable to integrate: $\frac{d [ A ]}{[ A ]} = - k d t$

WHY this step. Once each side holds a single variable we can integrate them independently — that is the whole reason to "separate."

WHAT integrating gives. Integrate from the start ( $t = 0$ , concentration $[A]_{0}$ ) to time $t$ (concentration $[A]$ ): $\int_{[A]_{0}}^{[A]} \frac{d [ A ]}{[ A ]} = - k \int_{0}^{t} d t$ The integral of $1/ [A]$ is $ln [A]$ — that is precisely where the logarithm is born. This gives: $ln [A] - ln [A]_{0} = - k t ⟹ ln [A] = ln [A]_{0} - k t$

WHAT IT LOOKS LIKE. Compare this to $y = c - k t$ : it is a straight line in $t$ with intercept $ln [A]_{0}$ and slope $- k$ . The figure below plots it.

Figure — Rate law — order vs molecularity

Read the figure: the raw $[A]$ curve (cyan) bends — a curve, not a line — but replotting it as $ln [A]$ (amber) straightens into a line whose slope reads off $- k$ . That straightness is the first-order test.

How the initial-rates method isolates an exponent

Worked example Worked isolation of one order

Data for $A + B \to P$ :

Trial	$[A]_{0}$ (M)	$[B]_{0}$ (M)	Rate (M/s)
1	0.10	0.10	0.030
2	0.20	0.10	0.120
3	0.10	0.30	0.090

Isolate $m$ (order in A): trials 1→2 change only $[A]$ (×2), $[B]$ fixed. So $\frac{0.120}{0.030} = (\frac{0.20}{0.10})^{m} \Rightarrow 4 = 2^{m} \Rightarrow m = 2$ . Isolate $n$ (order in B): trials 1→3 change only $[B]$ (×3), $[A]$ fixed. So $\frac{0.090}{0.030} = (\frac{0.30}{0.10})^{n} \Rightarrow 3 = 3^{n} \Rightarrow n = 1$ . Rate law: $Rate = k [A]^{2} [B]$ , overall order $3$ . Changing one concentration at a time is what makes each ratio collapse to a single unknown exponent.

Why termolecular steps are rare

Read the figure: a two-body meeting (left) only needs two spheres to overlap — common. A three-body meeting (right) needs all three inside the same tiny reaction volume at the same instant — the shaded overlap shrinks dramatically, so the collision probability plummets. This is why nature almost always splits reactions into uni- and bimolecular steps rather than one four-or-more-body collision.

Deriving the units of $k$ by dimensional analysis

True or false — justify

True/False: The overall order of a reaction always equals the sum of the stoichiometric coefficients.

False. That equality holds only for a single elementary step; for a multi-step reaction the order is set by the slow step and often has no link to the balanced coefficients.

True/False: Molecularity can be zero.

False. Molecularity counts colliding particles in one step, so it is at least 1; only order can be zero (rate independent of a concentration).

True/False: Order can be fractional or negative, but molecularity cannot.

True. Order is an empirical fit and can come out as

\frac{1}{2}

(e.g. Br₂ in H₂+Br₂) or negative (an inhibitor); molecularity is a physical collision count, so it must be a positive integer.

True/False: For an elementary step

2 A + B \to P

, the rate law is

Rate = k [A]^{2} [B]

True. In an elementary step the stoichiometry is the collision requirement, so exponents equal the coefficients — but this is licensed only because we were told the step is elementary.

True/False: A zero-order reaction proceeds at a constant rate until a reactant runs out.

True. Rate

= k

is independent of concentration, so

[A]

falls linearly with time (slope

- k

) right up to depletion, where the model breaks.

True/False: The rate constant

k

has the same units for every reaction.

False. The units follow

M^{1 - n} s^{- 1}

for overall order

n

M s^{- 1}

(zero),

s^{- 1}

(first),

M^{- 1} s^{- 1}

(second).

True/False: Molecularity is defined for the overall reaction H₂+Br₂→2HBr.

False. That reaction is not a single elementary step, so molecularity is undefined for it; only its individual elementary steps have a molecularity.

True/False: Doubling every concentration of a second-order reaction doubles the rate.

False. If overall order is 2, doubling all concentrations multiplies the rate by

2^{2} = 4

, not 2.

Spot the error

"The reaction is

2 NO + O_{2} \to 2 NO_{2}

, so it must be termolecular and third order."

Error: molecularity and order are being read straight off the coefficients. Whether it is truly termolecular depends on the mechanism, and the order must be measured — the coefficients alone prove neither.

"Order w.r.t. B came out to 0, so B is not a reactant."

Error: zero order means the rate does not depend on [B], not that B is absent. B may still be consumed and appear after a fast pre-equilibrium or as a saturating species.

"Since molecularity is always ≥1, the reaction can never be zero order."

Error: this confuses the two ideas. Molecularity (a mechanism count) and order (a measured exponent) are independent — an enzyme-saturated or surface-catalysed step can measure as zero order while every elementary step still has molecularity ≥1.

"The slow step is

NO_{2} + F_{2} \to NO_{2} F + F

, so Rate

= k [NO_{2}]^{2} [F_{2}]

Error: the exponents should match the slow step's own species, giving Rate

= k [NO_{2}] [F_{2}]

. The extra NO₂ appears only in a fast later step and never enters the rate law.

"A fractional order like

\frac{1}{2}

means half a molecule collides."

Error: half a molecule is meaningless. A fractional order is an effective exponent produced by combining several elementary steps (e.g. a dissociation equilibrium), not a picture of one collision.

"For

2 A + B \to C

we always get Rate

= k [A] [B]

because only one A is in the slow step."

Error: you cannot assume the slow step contains one A — that is precisely what experiment must decide. The rate law follows the measured slow step, whatever it turns out to be.

Why questions

Why is molecularity greater than 3 essentially never observed?

The chance of four or more particles arriving at the same point at the same instant with the right orientation is vanishingly small, so nature routes reactions through sequences of uni- and bimolecular steps instead.

Why can order not be predicted from the balanced equation for a multi-step reaction?

The balanced equation only enforces mass balance; it says nothing about which collisions actually control the speed. The order is fixed by the slowest elementary step, which the overall equation hides.

Why does the slow step determine the overall rate law?

It is the bottleneck — the whole reaction cannot outrun its slowest stage, so the rate law inherits the concentrations that appear in that step, just as a pipeline's throughput is set by its narrowest section.

Why do we plot

ln [A]

vs time for a suspected first-order reaction?

Integrating

d [A] / d t = - k [A]

gives

ln [A] = ln [A]_{0} - k t

, which is a straight line in

t

; a straight

ln

plot therefore confirms first order and its slope gives

- k

Why does the initial-rates method change only one concentration at a time?

Holding the others fixed isolates a single reactant's influence, so the rate ratio depends on that one exponent alone and the order can be read off directly.

Why is order an experimental quantity while molecularity is theoretical?

Order is whatever exponent makes the measured rate data fit, so it can only come from the lab; molecularity is deduced from a proposed mechanism's collision count, so it lives in theory until the mechanism is verified.

Edge cases

For an intermediate F that is consumed by a fast step, why does [F] not appear in the rate law?

It is destroyed as fast as it forms, so its concentration never builds up; only species present in the rate-determining slow step survive into the observable rate law.

What is the molecularity of a reaction whose mechanism is unknown?

Undefined. Molecularity attaches to specific elementary steps; without a known mechanism there are no elementary steps to count.

If a reaction is truly elementary, what is the relationship between its order and molecularity?

They coincide numerically — the exponents equal the coefficients, so overall order equals the molecularity of that single step (e.g. both 3 for a termolecular step).

What happens to a zero-order rate law as

[A] \to 0

The model predicts a constant rate

k

even at zero concentration, which is unphysical; in reality the zero-order behaviour breaks down near depletion (the catalyst or surface is no longer saturated) and the reaction slows.

Can two reactions with the same balanced equation have different orders?

Yes. Identical stoichiometry can hide different mechanisms with different slow steps, so their measured rate laws — and therefore orders — can differ entirely.

Is a first-order rate law proof that the reaction happens in one unimolecular step?

No. First order is consistent with a unimolecular step but can also arise from a multi-step mechanism; order alone never uniquely fixes the mechanism.