Integrated Rate Law: Zero Order

A drug patch delivers medicine into the bloodstream at a perfectly steady rate for hours. A catalytic converter oxidizes pollutant molecules at a fixed pace no matter how much pollutant remains. Both are governed by the same mathematics: zero-order kinetics. Zero-order reactions matter because they reveal when a reaction's speed is decoupled from reactant concentration. This happens when some other factor, like a saturated catalyst surface or a fixed light intensity, becomes the true bottleneck. Recognizing this pattern lets chemists diagnose hidden mechanisms from simple concentration-time data. This article derives the zero-order integrated rate law from its differential form. You will see the linear concentration-time plot that defines it. Then you will explore the physical origin in surface saturation. Finally, two worked examples show how to extract rate constants and predict reaction completion times.

Key takeaways

A zero-order reaction has rate independent of reactant concentration: rate = $k$ .
The integrated law is $[A] = [A]_0 - kt$ , a straight line with slope $-k$ .
Units of $k$ are concentration per time (e.g., mol L⁻¹ s⁻¹), unlike first-order $k$ .
Zero order typically arises from a saturated catalytic surface, enzyme, or fixed light flux.
The reaction must stop abruptly at $[A] = 0$ ; the linear law cannot go negative.
Key idea: When a physical bottleneck (not the reactant) limits reaction speed, concentration falls linearly with time until reactant is exhausted.

The core insight

A zero-order rate law, rate $= k$ , means the reaction proceeds at a constant, concentration-independent speed because some other component—a saturated catalyst surface, a fixed enzyme population, or a constant photon flux—is the true rate-limiting bottleneck, so plotting $[A]$ versus $t$ gives a straight line that must terminate exactly when the reactant runs out.

The physical picture

Most reactions slow down as reactants are consumed—that is the hallmark of first- and second-order kinetics. Zero-order reactions break this pattern entirely. The rate stays constant regardless of how much reactant remains, right up until the reactant is exhausted, at which point the rate must drop to zero because there is nothing left to react. This behavior almost always signals that concentration is not the limiting factor. Instead, some other fixed resource controls the pace. The classic example is a heterogeneous catalytic reaction where every active site on the catalyst surface is occupied. Adding more reactant molecules just increases the queue waiting for a site; it does not speed up the per-site turnover. Similarly, an enzyme operating at saturation ( $[S] \gg K_M$ in Michaelis-Menten terms) converts substrate at a fixed maximum velocity, $V_{max}$ , independent of substrate concentration. Photochemical reactions limited by a fixed light intensity behave the same way: each photon absorbed triggers one event, and the photon flux—not the reactant concentration—sets the pace. Because the differential rate law is simply rate $= k$ , a constant, integrating with respect to time gives a straight line rather than a curve. This linearity is the single most recognizable experimental signature of zero-order behavior: a plot of $[A]$ against $t$ is a straight line with negative slope, in sharp contrast to the exponential decay of first order or the reciprocal growth of second order.

:::svg[Concentration versus time for a zero-order reaction: a straight line of slope −k, terminating at [A] = 0]

Figure

Mechanistic origin: a saturated catalyst surface gives a constant rate independent of bulk reactant concentration

The core relations

\text{rate} = -\frac{d[A]}{dt} = k

where: $k$ is the zero-order rate constant with units of concentration per time (e.g., mol L⁻¹ s⁻¹).

[A] = [A]_0 - kt

where: $[A]_0$ is the initial concentration, $[A]$ is the concentration at time $t$ .

t_{1/2} = \frac{[A]_0}{2k}

where: $t_{1/2}$ is the half-life, which for zero order depends directly on $[A]_0$ (foreshadowing the order-dependence of half-life).

t_{\text{complete}} = \frac{[A]_0}{k}

where: $t_{\text{complete}}$ is the finite time at which $[A]$ reaches exactly zero.

Derivation AFrom the differential rate law to the linear plot

Goal. Starting from the zero-order differential rate law, derive the integrated law $[A] = [A]_0 - kt$ and identify the finite completion time.

Step 1. Write the differential rate law. For a reaction A → products that is zero order in A, the rate of consumption of A is constant:

-\frac{d[A]}{dt} = k

Here $k > 0$ has units of concentration/time, since rate always carries those units and the right-hand side must match with no concentration dependence.

Step 2. Separate variables. Rearranging,

d[A] = -k\, dt

This step is valid because $k$ contains no $[A]$ dependence—this is precisely what "zero order" means.

Step 3. Integrate both sides. Integrate the left side from $[A]_0$ (at $t=0$ ) to $[A]$ (at time $t$ ), and the right side from $0$ to $t$ :

\int_{[A]_0}^{[A]} d[A]' = -k \int_0^t dt'

Step 4. Evaluate the integrals.

[A] - [A]_0 = -kt

Rearranging gives the integrated zero-order rate law:

[A] = [A]_0 - kt

This is manifestly linear in $t$ : plotting $[A]$ versus $t$ gives a straight line of slope $-k$ and intercept $[A]_0$ .

Step 5. Recognize the physical constraint. Concentration cannot be negative. The linear law above is only physically meaningful while $[A] \geq 0$ . Setting $[A] = 0$ and solving for $t$ gives the moment the reactant is fully consumed:

0 = [A]_0 - k\,t_{\text{complete}} \quad\Longrightarrow\quad t_{\text{complete}} = \frac{[A]_0}{k}

Beyond this time the rate law simply does not apply—there is no more A to react, and $\text{rate}=0$ (a different physical regime, not a mathematical extension of the line).

Step 6. Extract the half-life. Setting $[A] = [A]_0/2$ :

\frac{[A]_0}{2} = [A]_0 - k\,t_{1/2} \quad\Longrightarrow\quad t_{1/2} = \frac{[A]_0}{2k}

Unlike first order, this half-life depends on the starting concentration—each successive half-life is shorter than the last.

\boxed{[A] = [A]_0 - kt, \qquad t_{\text{complete}} = \frac{[A]_0}{k}}

Worked examples

Example 1 — Foundational: extracting $k$ and predicting completion time. A heterogeneous catalytic decomposition of a pollutant gas A on a saturated platinum surface is monitored. Data show $[A]_0 = 0.500$ mol L⁻¹, and after $t = 40.0$ s, $[A] = 0.320$ mol L⁻¹. The reaction is known (from a separate order-determination experiment) to be zero order in A. Step 1 — Solve for $k$ . Using $[A] = [A]_0 - kt$ :

k = \frac{[A]_0 - [A]}{t} = \frac{0.500 \text{ mol L}^{-1} - 0.320 \text{ mol L}^{-1}}{40.0 \text{ s}} = \frac{0.180 \text{ mol L}^{-1}}{40.0\text{ s}} = 4.50\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}

Step 2 — Predict concentration at $t = 90.0$ s.

[A] = 0.500\ \text{mol L}^{-1} - (4.50\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1})(90.0\ \text{s}) = 0.500 - 0.405 = 0.095\ \text{mol L}^{-1}

Step 3 — Find the completion time.

t_{\text{complete}} = \frac{[A]_0}{k} = \frac{0.500\ \text{mol L}^{-1}}{4.50\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}} = 111\ \text{s}

Sanity check. At 90.0 s (less than 111 s), concentration is a small positive number, 0.095 mol L⁻¹—consistent. At exactly 111 s the reaction should just finish; plugging back, $0.500 - (4.50\times10^{-3})(111) = 0.500 - 0.4995 \approx 0$ . ✓ The half-life is $t_{1/2} = 0.500/(2 \times 4.50\times10^{-3}) = 55.6$ s, sensibly less than the 111 s completion time.

Example 2 — Applied: enzyme saturation kinetics with a detection limit. An enzyme-catalyzed reaction converts substrate S to product under conditions where $[S] \gg K_M$ , so the enzyme is fully saturated and the reaction is zero order in S with $V_{max} = k = 2.50\times10^{-4}$ mol L⁻¹ s⁻¹. The initial substrate concentration is $[S]_0 = 1.00\times10^{-2}$ mol L⁻¹. The analytical instrument used to follow the reaction can only reliably detect substrate concentrations above $5.0\times10^{-4}$ mol L⁻¹ (its detection limit). Step 1 — Find the time at which $[S]$ drops to the detection limit. Using $[S] = [S]_0 - kt$ , solve for $t$ :

t = \frac{[S]_0 - [S]}{k} = \frac{1.00\times10^{-2}\ \text{mol L}^{-1} - 5.0\times10^{-4}\ \text{mol L}^{-1}}{2.50\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}}

t = \frac{9.50\times10^{-3}\ \text{mol L}^{-1}}{2.50\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}} = 38.0\ \text{s}

Step 2 — Check whether zero-order behavior persists that whole time. Zero order via saturation holds only while $[S] \gg K_M$ . Suppose separate experiments show $K_M = 1.0\times10^{-4}$ mol L⁻¹. At the detection limit, $[S] = 5.0\times10^{-4}$ mol L⁻¹, which is 5 times $K_M$ —still reasonably saturating, so treating the whole run as zero order is a fair approximation, though the last moments would show the beginning of a curve-over toward first-order-like behavior as $[S]$ approaches $K_M$ . Step 3 — Compute the true completion time (ignoring the caveat) for comparison.

t_{\text{complete}} = \frac{[S]_0}{k} = \frac{1.00\times10^{-2}\ \text{mol L}^{-1}}{2.50\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}} = 40.0\ \text{s}

Sanity check. The instrument loses the signal at 38.0 s, just 2.0 s before the reaction would nominally finish at 40.0 s—consistent with the detection limit being a small fraction (5%) of $[S]_0$ . This illustrates a common practical issue: apparent "completion" in an experiment is often set by instrument sensitivity, not true zero concentration.

Limiting cases & common traps

Situation	What happens	Common mistake
$t < [A]_0/k$	Linear decay, rate constant $= k$	Assuming rate changes with $[A]$
$t \geq [A]_0/k$	Reaction is over, $[A]=0$ , rate $=0$	Extending the line to negative $[A]$
High reactant concentration (saturating regime)	Zero order is a good approximation	Assuming zero order holds at all concentrations
Low reactant concentration (near depletion)	Real reactions often crossover to first order as the bottleneck resource is no longer saturated	Ignoring the crossover and misreading the tail of the plot
Units of $k$	mol L⁻¹ time⁻¹	Confusing with first-order $k$ (units of time⁻¹)

The chemist behind it

Zero-order kinetics as a formal concept grew out of the study of heterogeneous catalysis in the early twentieth century, especially surface reactions on metals. Irving Langmuir's work on gas adsorption and surface reaction rates provided the physical mechanism—the Langmuir isotherm—that explains why a catalytic reaction rate can plateau and become independent of gas-phase concentration once the surface is saturated. This surface-science foundation, later extended into the Langmuir–Hinshelwood mechanism for surface-catalyzed reactions, gave zero-order kinetics its clearest physical justification and remains the standard explanation taught alongside enzyme saturation kinetics.

Origins · 1 minIrving Langmuir and how this was discovered

Irving Langmuir

1881–1957 · American chemist whose adsorption isotherm explains why surface-catalyzed reactions become zero order at high pressure

:::quiz[A reaction has $[A]_0 = 0.80$ mol/L and follows zero-order kinetics with $k = 0.020$ mol L⁻¹ s⁻¹. What is $[A]$ at $t = 20$ s?]

0.80 mol/L
0.40 mol/L ✓
0.20 mol/L
0.60 mol/L

:::faq[Why is the plot of $[A]$ versus $t$ linear for zero order, but curved for first order?] The differential rate law sets the shape. For zero order, $-d[A]/dt = k$ is a constant, so integrating gives a straight line, $[A] = [A]_0 - kt$ . For first order, $-d[A]/dt = k[A]$ depends on concentration itself, so the rate slows as $[A]$ falls, producing an exponential decay curve rather than a straight line. The linearity test (plotting $[A]$ vs. $t$ ) is therefore a fast diagnostic for zero order.

Ask the tutor: How does the half-life of a zero-order reaction compare to first- and second-order half-lives as concentration changes?

Can a zero-order reaction really go on forever without slowing down?

No. The linear law $[A] = [A]_0 - kt$ predicts $[A]$ hitting exactly zero at $t = [A]_0/k$ , and physically the reaction simply stops there—concentration cannot go negative. In real saturated-catalyst or saturated-enzyme systems, as $[A]$ approaches very low values the surface or enzyme is no longer kept saturated, so the kinetics typically curve over toward first order shortly before this true endpoint.

How can I tell experimentally that a reaction is zero order rather than some other order?

Plot $[A]$ versus $t$ for several trial orders: zero order gives a straight line for $[A]$ vs. $t$ , first order gives a straight line for $\ln[A]$ vs. $t$ , and second order gives a straight line for $1/[A]$ vs. $t$ . Whichever plot is most linear (highest $R^2$ ) identifies the order. This method is developed fully in the discussion of methods for determining reaction order.

Does the half-life formula for zero order behave like the first-order half-life?

No, and this is a key distinguishing feature. For zero order, $t_{1/2} = [A]_0/(2k)$ , which depends directly on the starting concentration—higher $[A]_0$ means a longer half-life. For first order, the half-life is $\ln2/k$ , completely independent of concentration. Comparing how half-life scales with $[A]_0$ is one of the cleanest ways to distinguish reaction orders experimentally.

Is zero order common in solution-phase chemistry, or mostly a surface phenomenon?

It appears wherever a fixed external resource, not reactant concentration, controls the rate. Classic cases include heterogeneous catalysis on saturated surfaces, enzyme kinetics at substrate saturation, and photochemical reactions limited by constant light intensity. It is less common for simple bimolecular solution reactions without such a bottleneck, since those typically follow first- or second-order kinetics instead.