Differential and Integrated Rate Laws Revisited

Introduction

An initial-rate experiment measures how rate depends on concentration near the start of a reaction. A concentration–time curve asks a complementary question: how much reactant remains later? Integration turns a differential rate law into that time prediction. At university level, the equations should be derived from a defined model, checked for units and tested against data rather than memorised as disconnected graph tricks.

Core explanation

Consider one reactant A disappearing in a constant-volume system. Define the positive disappearance rate as r = −d[A]/dt. A zero-order model says r = k₀. Integrating from [A]₀ at t = 0 gives [A] = [A]₀ − k₀t. The concentration falls linearly until the model ceases to apply. Units of k₀ are concentration per time, such as mol L⁻¹ s⁻¹. Its half-life is [A]₀/(2k₀), so the half-life depends on the starting concentration. The zero-order approximation can arise when a catalyst surface is saturated, but the molecular origin needs separate evidence.

A first-order model says r = k₁[A]. Separation gives d[A]/[A] = −k₁dt. Integrating yields ln([A]/[A]₀) = −k₁t, or [A] = [A]₀e^(−k₁t). The ratio inside the logarithm is dimensionless, as it must be. k₁ has units s⁻¹, and t₁/₂ = ln 2/k₁, independent of [A]₀. A straight line of ln[A] versus t has slope −k₁ if the simple model holds over the measured interval. OpenStax Chemistry 2e gives these standard integrated forms.

For a second-order model in one reactant, r = k₂[A]². Separation and integration give 1/[A] − 1/[A]₀ = k₂t. k₂ has units L mol⁻¹ s⁻¹ if concentration is mol L⁻¹. The half-life is 1/(k₂[A]₀), so doubling the initial concentration halves the second-order half-life. A graph of 1/[A] against t should be linear with slope k₂ under the model. This equation is for one-reactant second order, not automatically for a mixed law k[A][B] when both A and B change significantly.

All three results require assumptions. Constant volume allows concentration changes to represent amount changes simply. Constant temperature keeps k stable. A single effective pathway with unchanged kinetic form is assumed across the interval. In a gas reactor changing volume or a solution whose catalyst deactivates, apparent graph curvature may arise without a change in molecular mechanism. Conversely, one straight graph over a narrow range does not prove a unique microscopic route.

The differential form makes the mechanism connection precise. A proposed elementary step A → products gives a first-order rate law under mass-action assumptions; an elementary encounter 2A → products gives a second-order dependence on [A] in a simple homogeneous model. An overall reaction can contain several steps, so its measured order may differ from its stoichiometric coefficients. If a candidate mechanism predicts first order but the data show a clear second-order law under controlled conditions, that candidate is incomplete or incorrect. Agreement makes it plausible, not proven. OpenStax's reaction-mechanism treatment stresses that distinction.

Half-lives can be used without graphing when repeated measurements are reliable. For a first-order process, every halving takes the same time. For a second-order process, halving from a lower concentration takes longer. For a zero-order process, successive halving intervals get shorter as less reactant remains. These patterns are diagnostic only when competing processes and detection limits are controlled.

Numerical fitting should use uncertainty, not only visual straightness. Taking logarithms or reciprocals transforms measurement errors unevenly, especially at low concentrations. Fitting the concentration–time equations directly may be preferable when raw measurement errors are roughly uniform. Residual patterns can reveal a model's failure even if a transformed plot looks approximately linear.

Step-by-step reasoning

1. Define r = −d[A]/dt and state the assumed rate law. 2. Separate concentration and time variables, then integrate with [A] = [A]₀ at t = 0. 3. Check that logarithm arguments are dimensionless and all other units match. 4. Solve for half-life by setting [A] = [A]₀/2. 5. Compare the predicted curve with data over the full useful range. 6. Use mechanism evidence separately; do not equate a fitted order with a unique elementary step.

Visual explanation

Place three concentration–time curves side by side: straight descending line for zero order, exponential fall for first order, and slower long tail for second order. Below them, show their diagnostic straight plots: [A] versus t, ln[A] versus t, and 1/[A] versus t. Mark slope signs and units. Draw a note that a graph is a test of a model only over the conditions where data were gathered.

Real-world analogy

Imagine a reservoir draining at a fixed volume per minute, at a rate proportional to how much water remains, or at a rate proportional to the square of that amount. These three rules give different emptying curves and half-life patterns. Chemical rate laws similarly specify how the instantaneous rate changes as reactant disappears. The analogy does not imply that water drainage follows the same molecular mechanisms.

Real-world example

A drug or pollutant may degrade approximately first order over a useful concentration range. An analyst measures concentration at several times and tests whether an exponential model predicts later samples. If temperature changes during storage, a single constant k may fail. The rate-law model is valuable for prediction only when its conditions match the intended use.

Why?

Why is a first-order half-life independent of starting concentration? At every moment the rate is proportional to the amount remaining, so reducing the concentration slows the absolute loss proportionally. The same fraction is lost over equal time intervals. In zero- and second-order models, the fraction lost per interval changes differently, making half-life depend on the initial concentration.

Common misconception

“Second order always means two reactant molecules collide in one step.” A one-reactant [A]² law can be consistent with a bimolecular step, but complex mechanisms may produce the same dependence. Another mistake is to integrate r = k[A]² as if the result were ln[A] = −kt; the reciprocal form belongs to this second-order case.

Worked example

A first-order reaction begins at [A]₀ = 0.200 mol L⁻¹ with k = 0.100 min⁻¹. At t = 10.0 min, [A] = 0.200e^(−0.100 × 10.0) = 0.0736 mol L⁻¹. Its half-life is ln 2/0.100 = 6.93 min. The exponent is dimensionless because min⁻¹ times min cancels. A 10-minute measurement near 0.0736 mol L⁻¹ supports the model at that point but cannot alone distinguish it from all possible alternative rate laws.

Quick check

1. Which straight-line plot tests a simple second-order disappearance law r = k[A]²? Answer: Plot 1/[A] against time; its slope should be +k under the stated assumptions.

Exam focus

Write the differential equation before selecting an integrated formula. State whether the second-order case is [A]² or a mixed [A][B] law. Include initial concentration and units in every calculation. Use the half-life dependence on [A]₀ as a diagnostic, and explain that order is experimentally established for an overall reaction rather than copied from a balanced equation.

Advanced insight

When a reaction follows A + B → products with r = k[A][B] and both concentrations change, the integrated result depends on their initial difference. If one reactant is in large excess, its concentration may be approximately constant, giving pseudo-first-order behaviour for the other reactant with k obs = k[B] approx. The underlying bimolecular step has not become truly unimolecular; only the experimental concentration range makes the simpler time law effective.

Summary

Integrating a rate law converts instantaneous disappearance into a concentration–time prediction. Zero order gives a linear concentration decline, first order an exponential decline, and one-reactant second order a reciprocal relation. Their k units and half-life dependencies differ. These equations are models valid under stated conditions; matching a time curve can reject incompatible mechanisms but cannot uniquely prove a molecular pathway.

Practice questions

1. What is the first-order integrated law using a dimensionless logarithm? Answer: ln([A]/[A]₀) = −kt, equivalently [A] = [A]₀e^(−kt). 2. If a second-order k has units L mol⁻¹ s⁻¹, what units does k[A]² have? Answer: mol L⁻¹ s⁻¹, the required rate units. 3. How does doubling [A]₀ affect the half-life for r = k[A]²? Answer: It halves t₁/₂ because t₁/₂ = 1/(k[A]₀). 4. Why is a straight transformed plot alone insufficient to prove a mechanism? Answer: Different multistep mechanisms can share an effective rate law, and transformed-data errors or limited ranges can hide deviations.