Units For Rate Constant K Third Order

You're staring at a rate law: rate = k[A]²[B]. Your brain freezes. Think about it: the problem asks for the units of k. Third order overall — but what does that actually mean for the constant?

Yeah. Been there.

What Is a Third-Order Rate Constant Anyway

Let's back up. The rate constant k isn't some universal number you look up in a table. Practically speaking, its value — and critically, its units* — depend entirely on the overall reaction order. That's the sum of the exponents in your rate law.

For a third-order reaction, the overall order is 3. Could be rate = k[A]³. Here's the thing — could be rate = k[A]²[B]. Worth adding: could be rate = k[A][B][C]. So doesn't matter. On the flip side, the sum of exponents is 3. That's what "third order" means.

The rate itself always has units of concentration over time. Usually M/s (molarity per second) or mol·L⁻¹·s⁻¹ if you're being formal.

So k has to carry whatever units make the math work out.

The dimensional analysis shortcut

Write it out:

rate = k × [concentration]³

M/s = k × M³

k = (M/s) / M³ = M⁻²·s⁻¹

That's it. M⁻²·s⁻¹. Or written out: L²·mol⁻²·s⁻¹.

Same thing. Different notation.

Why the Units Trip People Up

Here's the thing — nobody forgets the units for first order. Also, second order? Here's the thing — easy. Even so, s⁻¹. Because of that, m⁻¹·s⁻¹. Still manageable.

Third order is where it gets messy because the exponent on concentration is 3, and suddenly you're dealing with M⁻². Squared inverse molarity. It feels wrong intuitively.

I've seen students write M⁻³·s⁻¹. Because of that, that's fourth order. I've seen M⁻¹·s⁻¹. That's second order. The pattern is simple — subtract the overall order from 1 for the concentration exponent — but under exam pressure, patterns vanish.

Real talk: the units tell you the mechanism

This isn't just bookkeeping. The units of k are a fingerprint. If you determine experimentally that k has units of L²·mol⁻²·s⁻¹, you know* the overall order is 3. That constrains possible mechanisms. A termolecular elementary step? Rare but possible. Consider this: a pre-equilibrium followed by a rate-determining step? More common.

The units are data. Treat them like data.

How to Derive Units for Any Order (Not Just Third)

Don't memorize. Here's the thing — derive. It takes ten seconds and you'll never get it wrong.

Step 1: Write the rate law with k isolated.

k = rate / ([A]ˣ[B]ʸ[C]ᶻ...)

Step 2: Plug in units for rate and concentration.

Rate → M·s⁻¹ (or mol·L⁻¹·s⁻¹) Concentration → M (or mol·L⁻¹)

Step 3: Do the algebra.

Overall order = x + y + z = n

k units = M·s⁻¹ / Mⁿ = M¹⁻ⁿ·s⁻¹

Step 4: Convert if needed.

M = mol·L⁻¹, so M¹⁻ⁿ = (mol·L⁻¹)¹⁻ⁿ = mol¹⁻ⁿ·Lⁿ⁻¹

For n = 3: mol⁻²·L²·s⁻¹ = L²·mol⁻²·s⁻¹

Quick reference table (because sometimes you just want the answer)

Overall Order (n)	Units of k (M-based)	Units of k (SI-ish)
0	M·s⁻¹	mol·L⁻¹·s⁻¹
1	s⁻¹	s⁻¹
2	M⁻¹·s⁻¹	L·mol⁻¹·s⁻¹
3	M⁻²·s⁻¹	L²·mol⁻²·s⁻¹
n	M¹⁻ⁿ·s⁻¹	Lⁿ⁻¹·mol¹⁻ⁿ·s⁻¹

Third order is bolded because that's why you're here.

Common Mistakes That Cost Points

Mixing up overall order vs. individual orders

Rate = k[A]²[B] — the order with respect to A is 2. Which means with respect to B is 1. That said, the units of k depend on the overall* order. Overall is 3. Always.

I've graded exams where a student correctly identified the individual orders, then used 2 instead of 3 for the unit derivation. Wrong answer. Full stop.

Forgetting time units

The "per second" part (s⁻¹) is always* there. Always. So unless your rate data uses minutes, hours, days — then it's min⁻¹, h⁻¹, day⁻¹. Match the time unit in your rate data.

Confusing M⁻² with M²

M⁻² means 1/M². Day to day, it's inverse concentration squared. The negative exponent matters. Not concentration squared. Write it as L²·mol⁻²·s⁻¹ if negative exponents trip you up — same thing, harder to misread.

Using the wrong concentration unit

If your concentrations are in mol/L (M), fine. If they're in mol/m³ (SI), your k units shift. If they're in partial pressures (atm, bar) for gas-phase reactions, k carries pressure units instead.

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Gas-phase third order: k units = atm⁻²·s⁻¹ or bar⁻²·s⁻¹ or Pa⁻²·s⁻¹.

The principle is identical. The concentration unit just changes.

Practical Tips That Actually Work

1. Carry units through every calculation

Not just at the end. Every step. If you're solving for k from experimental data:

k = rate / ([A]²[B])

Plug in numbers with units*:

rate = 2.So naturally, 4 × 10⁻³ M/s [A] = 0. 10 M [B] = 0.

k = (2.Because of that, 4 × 10⁻³ M/s) / ((0. 10 M)²(0.20 M)) = (2.In practice, 4 × 10⁻³ M/s) / (0. 0020 M³) = 1.

The units fall out automatically. You don't need to derive them separately if you never drop them.

2. Check your k value for sanity

A third-order k in M⁻²·s⁻¹... what's a "normal" magnitude?

Depends wildly on the reaction. But if you get k = 10¹² M⁻²·s⁻¹

If you get k = 10¹² M⁻²·s⁻¹, you’re probably looking at a highly efficient, diffusion‑controlled elementary step rather than a typical elementary third‑order elementary reaction. In practice, elementary third‑order collisions are rare in solution; they become more common in the gas phase where molecules move faster and can encounter each other more often. For a gas‑phase reaction of the type

[ \mathrm{A + B + C ;\xrightarrow{k}; products} ]

the rate law is still

[ \text{rate}=k[\mathrm{A}][\mathrm{B}][\mathrm{C}] ]

but now the concentration unit is expressed in pressure (atm, bar, or Pa). Consequently the units of k will be atm⁻²·s⁻¹, bar⁻²·s⁻¹, or Pa⁻²·s⁻¹. If you plug in pressures of order 1 atm, you’ll often obtain a k in the 10⁻¹⁰ to 10⁻⁸ range for slower reactions, and values approaching 10⁻⁴ atm⁻²·s⁻¹ for very fast, barrierless encounters. This leads to a value of 10¹² M⁻²·s⁻¹ is therefore an outlier in solution chemistry and usually signals that the reaction proceeds via a pre‑equilibrium or that the measured rate constant has been inadvertently multiplied by a large factor (e. That said, g. , a stoichiometric coefficient or a conversion error).

Quick sanity‑check checklist

Units of rate – Are they consistent with the concentration units you used? If you measured rate in mol·L⁻¹·min⁻¹, convert minutes to seconds before extracting k.
Overall order – Add the exponents of all concentration terms in the rate law. That sum is the only thing that determines the exponent on the concentration unit in k.
Negative exponents – Remember that a negative exponent means “per unit concentration.” For a third‑order reaction you will have a denominator of concentration squared, not a numerator.
Dimensional consistency – Multiply the units of k by the units of the concentration terms raised to their powers; the product should give you the units of rate. If they don’t match, you’ve made an algebraic slip.
Magnitude expectations –
- Solution (aqueous): k for a third‑order elementary step is typically 10⁻³–10⁰ M⁻²·s⁻¹. Anything larger than 10² M⁻²·s⁻¹ usually indicates a non‑elementary mechanism or a data‑processing error.
- Gas phase: k values range from 10⁻⁴ (atm⁻²·s⁻¹) for slow, high‑energy collisions to 10⁻⁶ (atm⁻²·s⁻¹) for diffusion‑limited encounters. Values above 10⁻² (atm⁻²·s⁻¹) are uncommon unless the reaction is barrierless.

A worked‑out example (no repetition)

Consider the gas‑phase reaction

[ \mathrm{NO_2 + CO + Cl_2 ;\longrightarrow; NO + CO_2 + Cl_2} ]

with the experimentally determined rate law

[ \text{rate}=k[\mathrm{NO_2}][\mathrm{CO}][\mathrm{Cl_2}] ]

Suppose you measured an initial rate of 4.Here's the thing — 5 × 10⁻⁴ atm·s⁻¹ when each partial pressure was 0. 15 atm.

[ k=\frac{4.Even so, 5\times10^{-4}\ \text{atm·s}^{-1}}{(0. 15\ \text{atm})(0.15\ \text{atm})(0.Even so, 15\ \text{atm})} =\frac{4. Worth adding: 5\times10^{-4}}{3. That's why 375\times10^{-3}}\ \text{atm}^{-2}! \text{s}^{-1} \approx 0.13\ \text{atm}^{-2}!

The resulting k value sits comfortably within the expected range for a third‑order gas‑phase elementary step, illustrating how the unit‑based sanity check instantly tells you whether the number you obtained is plausible.

Final take‑away

Deriving the units of a rate constant is not a rote exercise; it is a diagnostic tool that forces you to confront the order of the reaction, the units of concentration (or pressure), and the consistency of your data. By carrying units through every

calculation and applying the quick sanity-check checklist above, you turn a potential source of error into a powerful self-correcting mechanism. Still, when the units of your derived rate constant match the expected dimensions for the overall reaction order—whether in solution or in the gas phase—you gain immediate confidence that your rate law, experimental data, and mathematical manipulations are all in agreement. Conversely, a mismatch signals that something has gone awry, whether in measurement, unit conversion, or algebraic rearrangement, allowing you to catch and correct mistakes before they propagate further. In short, treating unit analysis as an integral part of kinetic problem-solving—not merely a final formality—ensures both accuracy and deeper conceptual understanding.