Mechanisms & Rate Laws — Relate a simple mechanism to an observed rate law - Chemistry 2

SECTION 1

Historical Context & Motivation

Chemists have long wanted to know more than just what products a reaction makes — they also want to know how fast it happens and why it proceeds at that speed. A balanced equation tells you the starting materials and products, but it says nothing about the molecular-level pathway the reaction actually follows. The study of chemical kinetics — the science of reaction rates — grew out of the need to control industrial processes, understand biological chemistry, and predict how reactions behave under different conditions.

1850s • First Rate Studies

Ludwig Wilhelmy measured the rate of sucrose inversion in acid, producing one of the first quantitative rate laws — mathematical expressions linking reaction speed to concentration.

1889 • Arrhenius Equation

Svante Arrhenius proposed that reactions need a minimum energy — the activation energy — connecting temperature to reaction speed.

1913 • Bodenstein & Chain Reactions

Max Bodenstein studied the H₂ + Br₂ reaction and showed that its complex rate law required a multi-step reaction mechanism, not a single collision event.

1935 • Transition State Theory

Eyring and Polanyi developed a theoretical framework describing the energy profile of each elementary step, solidifying the link between mechanisms and observed rates.

1960s–Today • Computational Kinetics

Modern computers allow chemists to simulate individual reaction steps, predict rate laws from proposed mechanisms, and design catalysts by understanding step-by-step pathways.

The central question that emerged from this history is straightforward: if a reaction happens through a series of small steps, how do those individual steps determine the overall speed we measure in the lab? Answering that question means learning how to connect a proposed mechanism to an experimentally observed rate law.

SECTION 2

Core Principles & Definitions

Before connecting mechanisms to rate laws, you need a solid grasp of a few key ideas. These terms form the vocabulary of kinetics, and every worked problem in this lesson depends on them. Read through the cards below and then see the analogy that ties them together.

Reaction Mechanism

A reaction mechanism is a proposed sequence of elementary steps that together describe how reactants become products. Each step involves one molecular event — such as two molecules colliding.

Elementary Step

An elementary step is a single molecular-level event. Its rate law can be written directly from its balanced equation: the exponents equal the stoichiometric coefficients.

Rate-Determining Step (RDS)

The rate-determining step is the slowest elementary step in a mechanism. Because everything else waits for it to finish, the RDS controls the overall reaction rate.

Intermediate

A reaction intermediate is a species produced in one step and consumed in a later step. It appears in the mechanism but not in the overall balanced equation.

Molecularity

The molecularity of an elementary step is the number of reactant molecules involved: unimolecular (1), bimolecular (2), or, rarely, termolecular (3).

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KEY TAKEAWAY

Think of a reaction mechanism like driving through a city with multiple traffic lights. The rate-determining step is the longest red light on your route — no matter how fast you drive between the other lights, the slowest one sets your overall travel time. In the same way, the slowest elementary step dictates the rate law you observe in the lab.

SECTION 3

Visual Explanation — Energy Profile of a Two-Step Mechanism

One of the best ways to understand a mechanism is to look at its energy diagram (also called a reaction coordinate diagram). The x-axis represents the progress of the reaction from reactants to products, and the y-axis shows energy. Each elementary step has its own energy barrier — the activation energy for that step. The step with the tallest barrier is the rate-determining step.

The curve shows how energy changes as reactants pass through two transition states. Step 1 (pink label) has the taller barrier (E_a1), making it the rate-determining step. Step 2 (orange label) has a smaller barrier (E_a2) and proceeds quickly once the intermediate forms.

Notice that the intermediate sits in the valley between the two peaks. It is a real species that exists briefly during the reaction, but it is not present at the start or the finish. Because the first peak is taller than the second, the overall rate depends on step 1 alone. This is the fundamental insight: the rate law for the slow step is the rate law for the whole reaction — as long as you express it in terms of reactants, not intermediates.

SECTION 4

Mathematical Framework — Writing Rate Laws from Mechanisms

A key rule in kinetics is that you can write a rate law directly from the equation of an elementary step. For an overall (net) reaction, you cannot do this — the exponents in a rate law must be determined experimentally. But for an elementary step, the exponents equal the stoichiometric coefficients.

RATE LAW FOR AN ELEMENTARY STEP

rate = k [A]ᵃ [B]ᵇ

Here a and b are the coefficients of reactants A and B in that elementary step, and k is the rate constant for that step. This rule applies only to elementary steps, never to overall reactions.

Strategy: Three Steps to Derive the Overall Rate Law

Identify the slow (rate-determining) step. Write the rate law for that step using its stoichiometric coefficients.
Check for intermediates. If the rate law contains an intermediate, use a preceding fast equilibrium step to substitute it out in terms of actual reactants.
Compare with experiment. The derived rate law must match the experimentally observed rate law. If it doesn't, the proposed mechanism is wrong.

FAST EQUILIBRIUM APPROXIMATION

K₁ = k₁ / k₋₁ = [Intermediate] / ([Reactant₁][Reactant₂])

If a fast, reversible step precedes the slow step, the forward and reverse rates are approximately equal. The equilibrium constant K₁ lets you solve for the concentration of the intermediate and substitute it into the slow-step rate law.

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Common Mistake

Students often try to write rate laws directly from the overall balanced equation. Remember: the overall equation is a summary — its coefficients do not give you the rate law exponents. Only elementary steps have that direct relationship.

SECTION 5

Detailed Breakdown — Mapping a Mechanism to Its Rate Law

Let's walk through a concrete example mechanism and see exactly how each piece connects to the observed rate law. Consider the reaction between nitrogen dioxide (NO₂) and carbon monoxide (CO):

OVERALL REACTION

NO₂ + CO → NO + CO₂

Experimentally, the observed rate law is: rate = k[NO₂]². Notice that CO does not appear in the rate law at all, even though it is a reactant. This tells us CO is not involved in the rate-determining step.

This diagram shows the two-step mechanism for the NO₂ + CO reaction. Step 1 is the slow RDS — two NO₂ molecules collide. Step 2 is fast and consumes the intermediate NO₃. When we add both steps, NO₃ cancels, leaving the overall equation. The rate law from Step 1 matches the experimental rate law.

There are several important things to notice. First, NO₃ is a reaction intermediate — it is produced in Step 1 and consumed in Step 2, so it never appears in the overall equation. Second, CO only participates in the fast second step, which explains why it doesn't appear in the rate law. Third, when we write the rate law for Step 1 using its coefficients (two molecules of NO₂), we get rate = k₁[NO₂]², matching the experimental observation. This agreement is what makes the proposed mechanism consistent with the data — though it does not prove the mechanism is the only possible one.

SECTION 6

Worked Example — Deriving a Rate Law from a Mechanism

Let's work through a mechanism that involves the fast-equilibrium substitution technique. This is a common exam scenario where the slow step's rate law contains an intermediate that must be eliminated.

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Problem Statement

The overall reaction is: 2 NO + O₂ → 2 NO₂. A proposed mechanism is: Step 1 (fast, reversible): 2 NO ⇌ N₂O₂. Step 2 (slow): N₂O₂ + O₂ → 2 NO₂. Derive the predicted rate law and compare it to the experimental rate law: rate = k[NO]²[O₂].

Deriving the Rate Law for 2 NO + O₂ → 2 NO₂

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Step 1 — Identify the rate-determining step

Step 2 is labeled as slow, so it is the rate-determining step. We write its rate law using the stoichiometric coefficients of that elementary step: rate = k₂[N₂O₂][O₂].

rate = k₂[N₂O₂][O₂]
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Step 2 — Check for intermediates

The species N₂O₂ is a reaction intermediate — it does not appear in the overall equation. We cannot leave it in our final rate law because intermediates are not easy to measure experimentally. We need to express [N₂O₂] in terms of the actual reactants, NO and O₂.
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Step 3 — Use the fast equilibrium to substitute

Since Step 1 is fast and reversible, we can write an equilibrium expression: K₁ = k₁/k₋₁ = [N₂O₂]/[NO]². Solving for the intermediate gives: [N₂O₂] = K₁[NO]².

[N₂O₂] = K₁[NO]²
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Step 4 — Substitute into the slow-step rate law

Replace [N₂O₂] in the rate law from Step 1 of this solution: rate = k₂ × K₁[NO]² × [O₂]. We can combine k₂ × K₁ into a single observed rate constant k: rate = k[NO]²[O₂].

rate = k[NO]²[O₂]
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Step 5 — Compare with experiment

The experimentally observed rate law is rate = k[NO]²[O₂]. Our derived rate law matches perfectly, so the proposed mechanism is consistent with experimental data. The reaction is second order in NO and first order in O₂, giving an overall third-order reaction.

SECTION 7

Strengths & Limitations of Mechanism-Based Rate Laws

Connecting mechanisms to rate laws is a powerful technique, but it has important boundaries. Understanding what this method can and cannot do will help you avoid common pitfalls on exams and in real chemistry.

Strengths	Limitations
Explains why the rate law has certain exponents, even when they don't match the overall equation's coefficients.	A consistent mechanism is not proof that it is correct — other mechanisms may produce the same rate law.
Predicts how changing concentration of a specific reactant affects the rate, aiding process control in industry.	Assumes the fast-equilibrium or steady-state approximation is valid, which may break down under extreme conditions.
Identifies intermediates and transition states, guiding catalyst design.	Cannot determine the actual values of rate constants — those must be measured experimentally.
Reveals which species participate in the slow step, explaining why some reactants are absent from the rate law.	Complex mechanisms with many steps can be difficult to analyze without advanced mathematics (steady-state approximation).

Comparing strengths and limitations of deriving rate laws from mechanisms

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KEY TAKEAWAY

Think of a proposed mechanism like a suspect's alibi in a detective show. If the alibi matches the evidence (the experimental rate law), the suspect stays in the running. But matching evidence doesn't prove guilt — another suspect might have the same alibi. In chemistry, a mechanism that predicts the correct rate law is consistent but not proven; additional experiments are needed to build confidence.

SECTION 8

Connection to Advanced Theory — The Steady-State Approximation

In this lesson you've used the fast-equilibrium approximation to eliminate intermediates. In more advanced courses, you'll encounter the steady-state approximation (SSA), which handles cases where the first step is not neatly reversible. The SSA assumes the concentration of the intermediate remains roughly constant over most of the reaction because it is consumed as fast as it is formed.

Feature	Fast-Equilibrium Method (This Lesson)	Steady-State Approximation (Advanced)
When to use	When a fast, reversible step precedes the slow step	When intermediate formation and consumption rates are similar, even without a clear equilibrium
Key assumption	Fast step reaches equilibrium quickly	d[intermediate]/dt ≈ 0 (rate of change is near zero)
Math complexity	Simple algebra — solve the equilibrium expression for the intermediate	More involved algebra — set the intermediate's net rate of formation to zero and solve
Typical course level	AP Chemistry / Honors Chemistry 2	College physical chemistry or chemical engineering

Fast-equilibrium vs. steady-state approaches to eliminating intermediates

For now, the fast-equilibrium method covers most problems you'll see in high school and AP Chemistry. As you move into college-level physical chemistry, the steady-state approximation will become your primary tool for handling more complex mechanisms with multiple intermediates.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why you can write a rate law directly from an elementary step's equation, but you cannot do the same for an overall balanced equation.

Show answer

An elementary step describes a single molecular event — the molecules shown actually collide in one step, so the probability of the event depends directly on the concentrations raised to their stoichiometric coefficients. An overall equation, however, may summarize multiple elementary steps. Its coefficients reflect the net stoichiometry, not any single collision event, so the exponents in the rate law must be determined by experiment. This is why a reaction like 2 NO₂ + F₂ → 2 NO₂F might have a rate law that is first order in F₂ even though its coefficient is 1, depending on the mechanism.

Question 2: BASIC CALCULATION

Consider the one-step mechanism: H₂ + I₂ → 2 HI. If this is truly a single elementary step, what is the rate law? What is the overall order?

Show answer

Since the reaction occurs in a single elementary step, we can write the rate law directly from the coefficients: rate = k[H₂][I₂]. Both reactants have a coefficient of 1, so the reaction is first order in H₂ and first order in I₂. The overall order is 1 + 1 = 2 (second order overall). This is a bimolecular elementary step because two molecules (one H₂ and one I₂) collide in a single event.

Question 3: INTERMEDIATE

A proposed mechanism for the decomposition of ozone (2 O₃ → 3 O₂) is: Step 1 (fast, reversible): O₃ ⇌ O₂ + O. Step 2 (slow): O₃ + O → 2 O₂. Derive the predicted rate law. Identify the intermediate.

Show answer

The slow step is Step 2, so we start with rate = k₂[O₃][O]. The species O (atomic oxygen) is the intermediate — it is produced in Step 1 and consumed in Step 2. To eliminate it, use the fast-equilibrium expression from Step 1: K₁ = [O₂][O]/[O₃], so [O] = K₁[O₃]/[O₂]. Substituting gives rate = k₂ × K₁ × [O₃]²/[O₂], or rate = k[O₃]²[O₂]⁻¹. This is second order in O₃ and negative first order in O₂, meaning that increasing O₂ concentration actually slows the reaction by shifting the Step 1 equilibrium back toward O₃.

Question 4: APPLIED

In the atmosphere, NO₂ reacts with fluorine gas according to the overall equation: 2 NO₂ + F₂ → 2 NO₂F. The observed rate law is rate = k[NO₂][F₂]. Propose a two-step mechanism consistent with this rate law, clearly identifying the slow step and any intermediates.

Show answer

One consistent mechanism is: Step 1 (slow): NO₂ + F₂ → NO₂F + F. Step 2 (fast): NO₂ + F → NO₂F. The slow step gives rate = k₁[NO₂][F₂], which matches the observed rate law. F (fluorine atom) is the intermediate — produced in Step 1 and consumed in Step 2. Adding both steps: 2 NO₂ + F₂ → 2 NO₂F, with F canceling out. The fact that the overall equation has a coefficient of 2 for NO₂ but the rate law is only first order in NO₂ makes sense because only one NO₂ molecule participates in the rate-determining step.

Question 5: CRITICAL THINKING

A student proposes two different mechanisms for a reaction A + 2 B → C + D. Mechanism I has a slow first step: A + B → X (slow), then X + B → C + D (fast). Mechanism II has a fast first step: A + B ⇌ X (fast), then X + B → C + D (slow). Both must produce the overall equation. Do both mechanisms predict the same rate law? Explain your reasoning thoroughly.

Show answer

No, they predict different rate laws. In Mechanism I, the slow step is A + B → X, giving rate = k₁[A][B] — first order in A and first order in B (second order overall). In Mechanism II, the slow step is X + B → C + D, giving rate = k₂[X][B]. Since X is an intermediate, we use the fast-equilibrium from Step 1: K = [X]/([A][B]), so [X] = K[A][B]. Substituting: rate = k₂K[A][B][B] = k[A][B]². This is first order in A and second order in B (third order overall). The two mechanisms predict different dependencies on [B], so an experiment varying [B] while holding [A] constant could distinguish between them. This illustrates why mechanism proposals must always be tested against experimental rate data.

SUMMARY

Lesson Summary

Summary

A reaction mechanism is a proposed sequence of elementary steps that describes how reactants become products at the molecular level. Each elementary step has a rate law whose exponents equal its stoichiometric coefficients. The rate-determining step (RDS) — the slowest step — controls the overall reaction rate. To derive the observed rate law, write the rate law for the RDS; if it contains a reaction intermediate, use the fast-equilibrium approximation from a preceding step to substitute the intermediate's concentration in terms of actual reactants.

A mechanism is consistent with experimental data if its predicted rate law matches the observed rate law and its elementary steps add up to the overall balanced equation. However, consistency does not prove a mechanism is correct — it only means the proposal has not been disproven. The molecularity of each step (unimolecular, bimolecular, or termolecular) describes how many reactant molecules participate in that single event. Master these connections, and you'll be able to move between the microscopic world of molecular collisions and the macroscopic rate data measured in the lab.