The first two articles in this series leaned hard on a symbol I never actually explained: $-r_A$. Every design equation, every reactor sizing, every selectivity argument bottoms out in that one term. So where does it come from?

Short answer: you measure it. And that’s the single most misunderstood thing in reaction engineering, because the notation makes it look like something you can derive from the balanced equation. You can’t.

The trap: order is not stoichiometry

Here’s the reflex almost every student has. You see a reaction:

$$2A + B \rightarrow \text{products}$$

and you write down $-r_A = k\,C_A^2\,C_B$, reading the exponents straight off the coefficients. It feels right. It is, in general, wrong.

Reaction order is empirical. The exponents in a rate law are determined by experiment, not by the balanced equation. A reaction with a 2 in front of $A$ may well turn out to be first order in $A$ — or half order, or zero order, or something with no integer in sight.

Why? Because a balanced equation is bookkeeping. It tells you what goes in and what comes out; it says nothing about how the reaction actually happens. And the rate is set by the how — the actual molecular collisions, usually a sequence of elementary steps, one of which is slower than the rest and throttles everything. The stoichiometry is the summary; the rate law is the mechanism’s fingerprint.

This is where molecularity and order part ways, and exams love the distinction:

MolecularityOrder
What it isHow many molecules collide in one elementary stepThe exponent in the measured rate law
Where it comes fromTheory / mechanismExperiment
Possible valuesOnly 1, 2, or (rarely) 3Anything: 0, ½, 1, 1.5, 2, even negative
Applies toA single elementary step onlyThe overall observed reaction

There’s exactly one case where you can read the exponents off the equation: when the reaction is genuinely elementary — a single molecular event, exactly as written. Then order matches molecularity and the coefficients are the exponents. But you only know a reaction is elementary because someone told you or proved it, never because it looks simple.

If a problem says “the reaction is elementary,” that sentence is doing enormous work — it’s granting you permission to write the rate law from the stoichiometry. Without it, you need data.

Anatomy of a rate law

The general power-law form you’ll use 90% of the time:

$$-r_A = k\,C_A^{\alpha}\,C_B^{\beta}$$

A detail that quietly costs people points: the units of $k$ depend on the overall order. They have to, because $-r_A$ always has units of mol/(L·s) no matter what. Work backwards and for overall order $n$ you get $k$ in units of $(\text{mol/L})^{1-n}\,\text{s}^{-1}$:

Overall orderUnits of $k$Sanity check
0mol·L⁻¹·s⁻¹$k$ is the rate
1s⁻¹pure inverse time
2L·mol⁻¹·s⁻¹needs a concentration to cancel
Free error-check: if someone hands you $k$ with units of L·mol⁻¹·s⁻¹, the reaction is second order overall. The units of $k$ tell you the order. Use this to catch mistakes in seconds.

Integrated rate laws: the shapes that give it away

Knowing $-r_A = kC_A^n$ tells you the rate right now. To get concentration as a function of time, integrate the batch mole balance $\dfrac{dC_A}{dt} = -kC_A^n$. Three cases dominate coursework:

Zero order ($n=0$) — the rate doesn’t care about concentration at all. Straight-line decay:

$$C_A = C_{A0} - kt$$

First order ($n=1$) — the classic exponential decay:

$$C_A = C_{A0}\,e^{-kt} \qquad\Longleftrightarrow\qquad \ln C_A = \ln C_{A0} - kt$$

Second order ($n=2$) — a long, slow tail:

$$\frac{1}{C_A} = \frac{1}{C_{A0}} + kt$$

Look at the right-hand forms. Each one is a straight line in a different variable — and that’s not a coincidence, it’s the entire experimental strategy. Plot your data three ways; whichever one straightens out tells you the order.

ZERO ORDER C_A time plot C_A vs t FIRST ORDER ln C_A time plot ln C_A vs t SECOND ORDER 1 / C_A time plot 1/C_A vs t

The integral method in one picture. Transform your data three ways and see which one gives a straight line — that’s your order. Note the second-order plot slopes up, because $1/C_A$ grows as $C_A$ falls.

The half-life fingerprint

There’s a faster tell, and it’s beautiful. Half-life $t_{1/2}$ is the time to consume half the reactant. Set $C_A = C_{A0}/2$ in each integrated law and see what pops out:

OrderHalf-lifeDepends on $C_{A0}$?
Zero$t_{1/2} = \dfrac{C_{A0}}{2k}$Yes — proportional to it
First$t_{1/2} = \dfrac{\ln 2}{k}$No — totally independent
Second$t_{1/2} = \dfrac{1}{k\,C_{A0}}$Yes — inversely

The first-order row is the famous one. A first-order reaction takes the same time to go from 100% to 50% as it does from 50% to 25%, or 25% to 12.5% — forever. That constant halving is the signature of exponential decay, and it’s why radioactive decay (first order) has a meaningful, quotable half-life while a second-order reaction doesn’t really.

Diagnostic shortcut: run the reaction twice at different starting concentrations. If half-life doesn’t budge → first order. If doubling $C_{A0}$ doubles the half-life → zero order. If doubling $C_{A0}$ halves it → second order.

Worked example: finding the order from data

You run a liquid-phase batch experiment and measure $C_A$ over time:

$t$ (min)010203040
$C_A$ (mol/L)2.0001.2200.7440.4540.277

Test the three shapes. Are the successive drops constant (zero order)? $2.000 \to 1.220$ is a drop of 0.780; $1.220 \to 0.744$ is 0.476. Not constant — not zero order.

Now try first order: take logs and look for a constant slope.

$$\ln(2.000)=0.693,\quad \ln(1.220)=0.199,\quad \ln(0.744)=-0.296,\quad \ln(0.454)=-0.790,\quad \ln(0.277)=-1.284$$

The successive differences are $-0.494$, $-0.495$, $-0.494$, $-0.494$ — dead constant over every 10-minute interval. That’s a straight line on a $\ln C_A$ vs. $t$ plot, so the reaction is first order, and the slope gives $k$:

$$k = -\text{slope} = \frac{0.494}{10\ \text{min}} \approx 0.0494\ \text{min}^{-1}$$

Cross-check with the half-life fingerprint: $t_{1/2} = \ln 2 / k = 0.693/0.0494 \approx 14.0$ min. And sure enough, the data crosses $C_A = 1.000$ (half of 2.000) at right about 14 minutes. Two independent methods agreeing is how you know you’ve got it.

When the exponent isn’t a nice number

Real rate laws get weird, and it’s worth seeing why so you’re not thrown when it happens.

That last one is a good reminder that “order” is a description of behavior in a regime, not a fundamental property of the reaction. It’s a fitted parameter, and fitted parameters have domains of validity.

Why this matters for the reactors

Now loop back. In Part 1 the whole PFR-beats-CSTR-on-volume argument rested on the rate falling as conversion rises — which is exactly a statement that $n > 0$. In Part 2, the selectivity rule (high concentration favors the higher-order pathway) is a statement about the relative orders of two rate laws. Everything in reactor design is downstream of the exponents.

Which means: get the order wrong and every number after it is wrong. Size a reactor assuming first order when it’s really second, and you won’t just be a little off — you’ll be off in a direction that gets worse at high conversion, precisely where you were trying to operate. The rate law is the foundation. Measure it, don’t assume it.

⚡ Put it into practice

Run the integrated rate laws

Plug in $k$, $C_{A0}$, and $t$ for zero, first, and second order — and check half-lives against your data.

Batch Reaction Calculators →