Integrated Rate-Law Comparisons

Introduction

Initial rates use only the first moments of a reaction. An integrated rate law uses the whole concentration–time curve from a single run. Integrating the differential rate law gives a formula for how [A] falls with time, and each order gives a distinctly shaped curve. Choosing the right axes turns that curve into a straight line, which is the classic way of determining order from one experiment.

Core explanation

Consider a single reactant A → products with rate = −d[A]/dt = k[A]ⁿ.

Zero order (n = 0). The rate is constant, so [A] falls linearly:

[A] = [A]₀ − kt

A plot of [A] against t is a straight line of gradient −k. The concentration reaches zero at a finite time, t = [A]₀/k. Zero-order behaviour usually means a surface or enzyme is saturated, so the rate is set by the number of active sites rather than by [A].

First order (n = 1). Separating variables and integrating gives

ln[A] = ln[A]₀ − kt, or [A] = [A]₀e^(−kt)

A plot of ln[A] against t is linear with gradient −k. The concentration decays exponentially and never quite reaches zero; in each equal time interval the same fraction reacts. Radioactive decay and many unimolecular rearrangements follow this form.

Second order (n = 2). Integration gives

1/[A] = 1/[A]₀ + kt

A plot of 1/[A] against t is linear with gradient +k. The curve of [A] against t starts steeply but develops a long tail, because the rate falls with the square of concentration.

Comparing profiles. If the three reactions start at the same [A]₀ with the same initial rate, the zero-order curve stays straight and finishes first, the first-order curve decays exponentially, and the second-order curve lags furthest behind at late times. The late-time tail is therefore the most sensitive region for distinguishing orders.

Units of k. Zero order: mol dm⁻³ s⁻¹. First order: s⁻¹. Second order: mol⁻¹ dm³ s⁻¹. A value of k is meaningless without its units, which also reveal the order used.

Two-reactant second order. For A + B → P with rate = k[A][B] and unequal starting concentrations, the integrated form is more complex. In practice chemists use a large excess of B so that k′ = k[B]₀ is effectively constant, and treat the decay of A as pseudo-first-order.

Formulae

Zero order: [A] = [A]₀ − kt. First order: ln[A] = ln[A]₀ − kt. Second order: 1/[A] = 1/[A]₀ + kt.

Step-by-step reasoning

1. Record [A] at a series of times over at least two or three half-lives. 2. Make three plots: [A], ln[A] and 1/[A] against t. 3. Identify which plot is straight across the whole range, not only the early part. 4. Check the residuals for a systematic curve rather than random scatter. 5. Take k from the gradient, with the sign and units appropriate to that order.

Visual explanation

Sketch three curves from the same starting point: a straight falling line (zero order), a smooth exponential that flattens towards the axis (first order), and a curve that drops quickly then flattens even more slowly (second order). Beside them, sketch the three matching straight-line transformations.

Real-world analogy

A zero-order reaction is like a ticket barrier letting through one person per second whatever the queue length. First order is like a shop where each customer has a fixed chance of leaving per minute. Second order is like people pairing up to dance: pairs form more slowly as the crowd thins.

Real-world example

Pharmacologists model how a drug's plasma concentration falls after a dose. Most drugs are eliminated with first-order kinetics, giving a straight ln(concentration) plot. Ethanol at typical intoxicating levels is a notable exception: its metabolising enzyme is saturated, so blood alcohol falls roughly linearly, close to zero order.

Why?

Why does only one plot give a straight line? Each linearising function is designed so that its derivative with respect to time is constant for exactly one order. Applying the wrong transformation leaves a time-dependent gradient, which appears as curvature.

Common misconception

"If the first few points of a plot look straight, the order is confirmed." Over the first 10–20% of reaction, all three plots are nearly linear. Data must extend over a large fraction of reaction before curvature reveals the wrong model.

Worked example

Question: A second-order reaction has k = 0.50 mol⁻¹ dm³ s⁻¹ and [A]₀ = 0.20 mol dm⁻³. Find [A] after 30 s.

Reasoning: 1/[A] = 1/0.20 + 0.50 × 30 = 5.0 + 15 = 20 dm³ mol⁻¹. So [A] = 1/20 = 0.050 mol dm⁻³.

Answer: [A] = 0.050 mol dm⁻³, one quarter of the starting value.

Quick check

1. Which plot is linear for a first-order reaction, and what does its gradient equal? Answer: ln[A] against time is linear, and its gradient equals minus the rate constant, −k.

Exam focus

Memorise the three integrated forms, the linear plot for each and the sign of each gradient. Examiners often supply a table and ask you to decide which plot to draw; justify your choice by stating that the chosen plot gives a straight line across all the data.

Advanced insight

Linearising transformations distort experimental error. Taking 1/[A] greatly magnifies the uncertainty in small late-time concentrations, so those points dominate a naive least-squares fit. Fitting the untransformed integrated equation directly by non-linear regression weights the data more honestly and is standard practice in research kinetics.

Summary

Integrating a rate law gives concentration as a function of time. Zero order gives [A] linear in t; first order gives ln[A] linear in t; second order gives 1/[A] linear in t. The correct plot is straight over most of the reaction, and its gradient gives k with order-specific units. Late-time data distinguish the orders most clearly.

Practice questions

1. A plot of 1/[A] against time is straight with gradient 0.040 mol⁻¹ dm³ s⁻¹. State the order and k. Answer: Second order, with k = 0.040 mol⁻¹ dm³ s⁻¹. 2. A first-order reaction has k = 0.010 s⁻¹ and [A]₀ = 1.00 mol dm⁻³. Calculate [A] after 100 s. Answer: [A] = 1.00 × e^(−1.0) = 0.37 mol dm⁻³. 3. Why does a zero-order reaction stop at a definite time while a first-order reaction does not? Answer: The zero-order rate stays constant until A is used up, whereas the first-order rate falls in proportion to [A], so [A] approaches zero only asymptotically. 4. Explain why a large excess of B simplifies the analysis of rate = k[A][B]. Answer: [B] stays almost constant, so k[B] acts as a single constant k′ and A decays by pseudo-first-order kinetics.