Reaction Order and Units of k

Introduction

Reaction order describes how measured rate responds to concentration. The rate itself always has concentration-per-time units in the usual fixed-volume definition, but the rate constant k takes different units for different orders. Dimensional analysis gives a quick and rigorous check on kinetic calculations.

Core explanation

For rate = k[A]ᵐ[B]ⁿ, the overall order is q=m+n. If concentration is represented by M and time by s, rate has units M s⁻¹. The concentration product has units Mᵠ. Therefore k must have units M^(1−q) s⁻¹. This formula works for zero, first, second and higher orders as long as the chosen rate law is a concentration power law.

For zero order, rate = k, so k has units M s⁻¹. For first order, rate = k[A], so k has units s⁻¹. For second order, rate = k[A]² or k[A][B], so k has units M⁻¹ s⁻¹. For third order, k has units M⁻² s⁻¹. A bare number such as k=0.20 is incomplete unless its units and temperature are reported.

Partial orders need not be integers. If rate = k[A]^(1/2)[B], overall order is 1.5 and k units are M⁻0.5 s⁻¹. Negative order can occur in inhibition: rate = k[A]/[I] has total apparent order zero when A is +1 and inhibitor I is −1; k then has M s⁻¹ units in that specific expression. The unit formula still applies, though the mechanistic interpretation requires care.

Overall order differs from the total number of reactant molecules in the balanced equation. For an overall reaction 2A+B → products, its measured law might be rate = k[A][B], giving second order, or some other expression. Only for a verified elementary step does molecularity directly guide the simple exponents. Thus order is a kinetic property, not a stoichiometric coefficient total.

The numerical value of k depends on the concentration and time units chosen. A first-order k of 0.60 min⁻¹ equals 0.010 s⁻¹. Both describe the same physical rate constant if conversion is correct. A second-order value stated in L mol⁻¹ s⁻¹ is equivalent to M⁻¹ s⁻¹ because M means mol L⁻¹. Comparing two published numbers without unit conversion can create a false difference.

Temperature and reaction medium affect k. A particular k value has meaning only for specified conditions and rate-law form. If a catalyst or solvent changes the mechanism, even the apparent order may change. Units can catch an arithmetic error but cannot prove the chosen kinetic model is correct.

Step-by-step reasoning

1. Write the measured rate-law concentration factors. 2. Add their exponents to obtain overall order q. 3. Write rate units as M s⁻¹ and concentration product units as Mᵠ. 4. Divide to obtain k units M^(1−q) s⁻¹. 5. Convert any time or concentration units before comparing numerical k values.

Visual explanation

Draw a row of four boxes: order 0 → M s⁻¹, order 1 → s⁻¹, order 2 → M⁻¹ s⁻¹, order 3 → M⁻² s⁻¹. Above them write [k]=[rate]/[concentration]ᵠ. Underline that the rate-unit numerator is unchanged while denominator power grows.

Real-world analogy

If a recipe's output per second is divided by one ingredient amount, the proportionality constant has one set of units; dividing by the square of an ingredient amount gives different units. Kinetic k units record exactly what concentration factors were divided out.

Real-world example

Two reports of a first-order decomposition may list k as 0.12 min⁻¹ and 0.0020 s⁻¹. Unit conversion shows they are the same value, so the numbers alone should not be used to claim different reaction speeds.

Why?

Why does k have s⁻¹ for a first-order reaction? Rate contributes M s⁻¹ and one concentration factor contributes M; dividing cancels concentration, leaving inverse time.

Common misconception

“All rate constants have units s⁻¹.” That is true for a simple first-order concentration law, not for zero-, second- or other-order laws.

Worked example

For rate = k[A]²[B] at rate units M s⁻¹, overall order is 2+1=3. The concentration factor has units M³, so k units are (M s⁻¹)/M³ = M⁻² s⁻¹. If measured rate is 0.016 M s⁻¹ at [A]=0.20 M and [B]=0.10 M, k=0.016/(0.20²×0.10)=4.0 M⁻² s⁻¹. Substituting the units returns M s⁻¹.

Quick check

1. What are k units for a second-order law when time is seconds? Answer: M⁻¹ s⁻¹, equivalent to L mol⁻¹ s⁻¹.

Exam focus

Show the exponent sum and the unit division explicitly. Keep overall order separate from equation coefficients, and report k with both units and conditions.

Advanced insight

In activities-based kinetic laws, concentrations can be divided by a standard concentration to make factors dimensionless, shifting where units appear by convention. Introductory concentration laws use the familiar order-dependent k units.

Summary

Overall order is the sum of concentration exponents. For a rate law of order q, k has units M^(1−q) s⁻¹ when rate is in M s⁻¹. This dimensional relationship checks calculations but does not establish the mechanism.

Practice questions

1. What is overall order of rate = k[A]^(1/2)[B]? Answer: 1.5, the sum 0.5+1. 2. What units does a zero-order k have? Answer: M s⁻¹. 3. Convert 0.30 min⁻¹ to s⁻¹. Answer: Divide by sixty seconds per minute: 0.30/60 = 0.0050 s⁻¹. The physical first-order rate constant is unchanged by this unit conversion.