Elementary-Step Rate Laws

Introduction

A reaction network's arrows become a kinetic model when each arrow is assigned a flux law. For a genuine elementary collision or transformation, mass-action kinetics relates that flux to the activities of its reacting species. An overall balanced reaction is different: its coefficients do not generally reveal the measured orders. Keeping that distinction prevents a common error when converting mechanisms into differential equations.

Core explanation

For an elementary unimolecular step A → B, a simple law is v₁ = k₁[A], with k₁ in inverse time. For an elementary bimolecular step A + B → C, v₂ = k₂[A][B], with k₂ in concentration⁻¹ time⁻¹. If two identical A molecules collide in a proposed elementary step 2A → D, a conventional law is v₃ = k₃[A]², while the precise numerical definition of k₃ absorbs combinatorial conventions. These orders arise from the elementary event model under suitable dilute-solution or ideal-gas conditions, not from arbitrary overall stoichiometric coefficients.

For a reversible elementary reaction A + B ⇌ C, net flux can be written v = kf[A][B] − kr[C] under ideal mass-action conditions. At equilibrium v = 0, so the concentration-based equilibrium ratio is K = kf/kr for the chosen standard-state convention. At nonideal conditions, activities replace bare concentrations and activity coefficients matter. A network with multiple reversible steps must maintain thermodynamic consistency around cycles; independently fitted constants cannot imply perpetual net motion at equilibrium.

Surface and enzyme mechanisms require additional variables. A surface adsorption step A(g) + → A may have forward flux proportional to A pressure or activity times vacant-site fraction θ , and a reverse flux proportional to θA. A catalysed overall equation A → P may show saturation in A even though each microscopic step follows its own appropriate law. ACS work on microkinetic modelling uses elementary reaction rates with gas species, adsorbates and vacant sites to predict observables.

An empirically measured overall rate law must be determined from data. For 2NO + O₂ → 2NO₂, the overall coefficients do not prove rate = k[NO]²[O₂] unless that overall equation is itself a valid elementary event, which is rarely justified by stoichiometry alone. A multistep mechanism may produce non-integer, negative or condition-dependent apparent orders through intermediate balances and inhibition. Rate-law testing thus helps accept or reject proposed pathways.

Step-by-step reasoning

1. Label every reaction as elementary or lumped, and specify its phase. 2. Write forward flux from relevant reactant activities or site fractions. 3. Include reverse flux where the step is reversible. 4. Multiply fluxes by stoichiometric coefficients to get species rates. 5. Check units, thermodynamic consistency and measured concentration dependence.

Visual explanation

Draw two arrows: A + B → I with v₁ = k₁[A][B], then I → P with v₂ = k₂[I]. Write the net rates d[A]/dt = −v₁, d[B]/dt = −v₁, d[I]/dt = v₁ − v₂ and d[P]/dt = v₂. Add a separate overall arrow A + B → P with a warning that it does not by itself establish a rate law.

Real-world analogy

A warehouse task requiring one worker and one item may proceed roughly in proportion to the availability of both when they meet randomly. But a complete order can involve picking, packing and shipping; the number of packages and workers in the overall job description does not reveal the rate of the full workflow. Elementary and overall chemical equations differ in a similar way.

Real-world example

A substrate reacts on a metal surface. At low concentration, rate rises with substrate concentration because adsorption increases. At high concentration, the rate plateaus as sites saturate. A naive overall first-order law cannot capture both regimes, while elementary adsorption plus surface reaction with a site balance can. Product inhibition may even make the high-concentration trend decline.

Why?

Why check units for every rate constant? Rate has concentration per time. For v = k[A], k must have inverse-time units; for v = k[A][B], k needs concentration⁻¹ time⁻¹. Wrong units often reveal a missing concentration factor or an incorrect interpretation of a fitted parameter.

Common misconception

“Balanced coefficients always equal kinetic orders” is false for overall reactions. “Every arrow in a diagram is elementary” is false if it lumps steps. “At equilibrium both forward and reverse rates are zero” is false; they can be equal and nonzero. “Negative apparent reaction order violates mass action” is false when intermediate coverage or inhibition creates it.

Worked example

For A + B → I with k₁ = 2 L mol⁻¹ s⁻¹, [A] = 0.10 M and [B] = 0.20 M, v₁ = 2 × 0.10 × 0.20 = 0.040 M/s. For I → P with k₂ = 0.5 s⁻¹ and [I] = 0.04 M, v₂ = 0.020 M/s. Thus d[I]/dt = 0.040 − 0.020 = +0.020 M/s, while d[P]/dt = +0.020 M/s. A and B each fall at 0.040 M/s. If an experiment measures P formation at 0.020 M/s, that rate alone does not imply [I] is steady; its calculated pool is currently growing.

Quick check

1. What are units of k for an elementary law v = k[A][B] when v is M/s? Answer: M⁻¹ s⁻¹, equivalent to L mol⁻¹ s⁻¹.

Exam focus

Write elementary mass-action laws and net fluxes with correct units. Use N v or direct production-minus-consumption accounting to obtain species rates. Explain why overall coefficients cannot be used as reaction orders without mechanistic evidence.

Advanced insight

Activities make thermodynamic rate laws more consistent in concentrated or ionic solutions. For surface systems, coverage-dependent barriers can make a nominal rate constant vary with θ. These refinements do not erase the value of simple mass-action models; they specify the domain in which those models approximate the chemistry.

Summary

Elementary-step laws turn a proposed mechanism into quantitative fluxes. Their concentration and site dependencies require mechanistic justification, correct units and reversibility where needed. Overall reaction stoichiometry alone does not determine kinetic order.

Practice questions

1. What is the rate law for an ideal elementary A → B step? Answer: v = k[A]. 2. For A + B ⇌ C, what is the ideal net flux? Answer: v = kf[A][B] − kr[C]. 3. If v₁ makes I at 3 M/s and v₂ consumes I at 2 M/s, what is d[I]/dt? Answer: +1 M/s at that instant. 4. Why might an overall A → P reaction show saturation in A? Answer: It may involve a finite population of catalyst sites that become occupied at high A activity.