Second-Order Integrated Rate Law

Introduction

For the specific second-order law −d[A]/dt = k[A]², the reciprocal of concentration rises linearly with time. Its half-life depends inversely on starting concentration, unlike the zero-order and first-order cases. This formula applies to a one-species squared law, not automatically to every reaction whose overall order happens to be two.

Core explanation

Separate variables: d[A]/[A]² = −k dt. Integrating gives −1/[A]t + 1/[A]0 = −kt, which rearranges to 1/[A]t = 1/[A]0 + kt. A graph of 1/[A] versus time is a straight line with slope +k and intercept 1/[A]0. Since 1/[A] has units M⁻¹, k has units M⁻¹ time⁻¹, such as M⁻¹ s⁻¹.

To derive half-life, substitute [A]t=[A]0/2. Then 1/([A]0/2)=1/[A]0+kt₁/₂, so 2/[A]0−1/[A]0=kt₁/₂. Therefore t₁/₂=1/(k[A]0). Doubling initial concentration halves the initial half-life at fixed k. Each successive halving takes longer: after the concentration has halved, the next half-life is twice the first because its new starting concentration is half as large.

For a concrete case, [A]0=0.20 M and k=0.50 M⁻¹ min⁻¹ give t₁/₂=1/(0.50×0.20)=10 min. To fall from 0.10 M to 0.05 M at the same k takes 1/(0.50×0.10)=20 min. A constant half-life is therefore inconsistent with this simple second-order law.

Second-order rate can arise from an elementary A+A encounter, but it need not. Multi-step processes can yield the same concentration law. Also, a reaction with rate = k[A][B] is second order overall, yet if [A] and [B] begin at unequal concentrations, the integrated expression is not the same as 1/[A]=1/[A]0+kt. The simple reciprocal formula is for −d[A]/dt=k[A]² with the stated species definition.

Data fitting should compare plausible plots rather than assume second order because the balanced equation shows two A molecules. A linear 1/[A] plot supports the model over the measured interval, but changes in temperature, volume or pathway can invalidate extrapolation. Measurement noise grows after reciprocal transformation when concentrations become small, so late-time data can be especially uncertain.

The equation predicts concentration approaches zero asymptotically and never becomes negative. This contrasts with a zero-order linear model that must stop at depletion. In a real reaction, detection limits, reverse reaction or side chemistry may prevent the ideal asymptotic law from describing very long times.

Step-by-step reasoning

1. Confirm the specific species law −d[A]/dt=k[A]². 2. Write 1/[A]t=1/[A]0+kt. 3. Obtain k from the positive slope of a reciprocal plot. 4. Use t₁/₂=1/(k[A]0) for a stated starting concentration. 5. Check that a mixed A–B second-order law is not being substituted into this formula blindly.

Visual explanation

Draw a curved [A] versus time graph and a straight 1/[A] versus time graph. Mark an initial half-life interval and a second interval twice as long for the next halving. Label slope +k and intercept 1/[A]0 on the straight plot.

Real-world analogy

When two members of the same crowd must meet to react, a smaller crowd makes encounters disproportionately rarer. As the crowd shrinks, removing the next half can take longer than removing the first half.

Real-world example

An experiment measuring a reacting solute at regular times may find that [A] itself curves and ln[A] also curves, while 1/[A] forms a line. That pattern supports a squared concentration law over the observed range.

Why?

Why does second-order half-life grow as concentration falls? Rate is proportional to the square of concentration, so halving [A] reduces instantaneous rate to one quarter while only halving the amount to be removed.

Common misconception

“Every overall second-order reaction obeys 1/[A]t=1/[A]0+kt.” That form requires the specific disappearance equation k[A]². A two-reactant k[A][B] law needs its own integration conditions.

Worked example

For [A]0=0.40 M, k=0.25 M⁻¹ min⁻¹ and t=6.0 min, 1/[A]t = 1/0.40 + 0.25(6.0) = 2.5+1.5=4.0 M⁻¹. Thus [A]t=0.25 M. The initial half-life is 1/(0.25×0.40)=10 min, so at 6 min it is reasonable that concentration has not yet reached 0.20 M.

Quick check

1. What is the linear diagnostic plot for rate = k[A]²? Answer: 1/[A] versus time, with positive slope k.

Exam focus

Specify the exact squared law, show reciprocal equation and units, and calculate half-life from the current starting concentration. Do not use the one-species formula for an arbitrary A+B reaction.

Advanced insight

For A+B with unequal initial concentrations, integration can use the fixed stoichiometric difference between [A] and [B], yielding a logarithmic ratio expression. Equal initial concentrations can simplify to a reciprocal form under the correct species-rate definition.

Summary

The squared second-order law gives a linear 1/[A] plot and k units M⁻¹ time⁻¹. Its half-life is 1/(k[A]0) and lengthens as concentration falls. The formula applies only to its stated rate law.

Practice questions

1. What is half-life when k=2.0 M⁻¹ s⁻¹ and [A]0=0.10 M? Answer: 1/(2.0×0.10)=5.0 s. 2. What are the reciprocal-plot slope units when time is seconds? Answer: M⁻¹ s⁻¹, the units of k. 3. Why does a second halving take longer than the first? Answer: At lower starting concentration, the squared-concentration rate is smaller and t₁/₂ increases inversely with [A]0.