Zero-Order Integrated Rate Law

Introduction

A zero-order rate law says the disappearance rate of A is constant with respect to [A] over a measured range: −d[A]/dt = k. Integrating this relation gives a straight-line decline in concentration and a half-life that depends on the starting concentration. The model is useful for saturated surfaces or limited energy input, but it cannot remain valid after reactant is exhausted.

Core explanation

For a simple A → products process with a normalized one-to-one rate, integrate d[A]/dt = −k from t=0 to t: [A]t = [A]0 − kt. A plot of [A] versus time is a straight line with intercept [A]0 and slope −k. The unit of k is concentration/time, such as mol L⁻¹ s⁻¹. This distinguishes zero order from first order, whose k has inverse-time units.

Half-life occurs when [A]t = [A]0/2. Substitute into the equation: [A]0/2 = [A]0 − kt₁/₂, so t₁/₂ = [A]0/(2k). If starting concentration doubles with the same k, the first half-life doubles. Successive halvings take progressively shorter times because the fixed absolute amount removed per unit time is the same while the amount being halved shrinks.

Suppose [A]0=0.80 M and k=0.020 M s⁻¹. The first half-life is 0.80/(2×0.020)=20 s. From 0.40 M to 0.20 M takes only 0.20/0.020=10 s. A constant half-life would instead be the signature of ideal first-order decay, not zero-order decay.

Zero-order behavior can arise when a catalyst surface is saturated by A. All active sites are occupied, so adding more A does not increase the number of reacting sites. It can also appear in certain photochemical systems when absorbed light supply is limiting. The mechanism need not literally contain zero A molecules; “zero” is a concentration exponent in the observed law.

The integrated line is valid only while the zero-order conditions persist. Extrapolating [A]t = [A]0 − kt beyond t=[A]0/k gives negative concentrations, which are physically impossible. At or before depletion, the model must stop or change. Data showing a straight line over one interval do not prove zero order under every possible concentration or temperature.

If the observed species has a stoichiometric coefficient other than one, the disappearance slope may be a multiple of the normalized reaction rate. The familiar integrated equation assumes −d[A]/dt=k as the explicitly defined species law. State which rate k represents before applying formulas mechanically.

Step-by-step reasoning

1. Confirm rate is independent of [A] in the tested range. 2. Write −d[A]/dt=k and integrate to [A]t=[A]0−kt. 3. Read k from the negative slope of a concentration-time line. 4. Set [A]t=[A]0/2 to derive half-life. 5. Reject extrapolated negative concentrations as outside model validity.

Visual explanation

Draw a straight [A] versus time line from [A]0 downward with slope −k. Mark the time at half the starting concentration, then mark the shorter interval required for the next halving. Extend the line only to zero concentration and shade further extrapolation as invalid.

Real-world analogy

A machine removing exactly two liters from a tank each minute empties a larger tank in proportion to its starting amount. Halving the remaining contents takes less time after the tank is already partly empty. This mirrors constant absolute removal.

Real-world example

A catalytic surface fully occupied by reactant molecules may process material at a nearly constant maximum throughput. Within that saturated range, bulk concentration can fall linearly until the surface is no longer saturated.

Why?

Why is zero-order half-life dependent on initial concentration? The rate removes a fixed concentration amount each unit time. Half of a larger starting amount is a larger amount to remove at the same pace.

Common misconception

“Zero order means no reaction.” It means the observed rate has no dependence on [A] over the stated range; product can still form at a nonzero constant rate.

Worked example

A zero-order reactant starts at 0.60 M and has k=0.015 M min⁻¹. After 12 min, [A] = 0.60−0.015(12)=0.42 M. Its initial half-life is 0.60/(2×0.015)=20 min. The calculated 0.42 M is positive and inside the model's possible range; a later prediction after 50 min would be negative and therefore invalid.

Quick check

1. Which plot is linear for an ideal zero-order A disappearance? Answer: [A] against time, with slope −k.

Exam focus

Use the linear integrated equation, give k in concentration/time units and derive t₁/₂=[A]0/(2k). State the concentration range where the model is valid.

Advanced insight

Zero-order segments can be limiting approximations of saturating rate laws such as rate = Vmax[A]/(K+[A]). When [A] is much larger than K, rate approaches Vmax; when [A] falls, the order changes.

Summary

Zero-order disappearance gives a linear concentration decline and a k with concentration/time units. Its half-life rises with starting concentration. The approximation ends when depletion or loss of saturation changes the rate law.

Practice questions

1. If [A]0=1.0 M and k=0.10 M min⁻¹, what is [A] after 3 min? Answer: 0.70 M from 1.0−0.10(3). 2. What is its initial half-life? Answer: 1.0/(2×0.10)=5.0 min. 3. Why is a predicted [A]=−0.20 M unacceptable? Answer: Concentration cannot be negative; the linear law has been extrapolated beyond reactant depletion or its valid regime.