Lesson 2 · 25 min

Integrated rate laws and half-lives

How does concentration fall with time? Plot the data three ways — [A], ln[A] and 1/[A] against t. Only one is straight, and it gives you the order, the rate constant and the half-life.

In this lesson

Write the integrated rate laws for zero-, first- and second-order reactions
Identify the order from plots of [A], ln[A] and 1/[A] against t, and get k from the slope
Calculate half-lives and explain why only a first-order half-life is independent of the starting concentration
Judge a straight-line fit by its residuals, not only by R²

In 1921 Daniels and Johnston followed the decomposition of gaseous $\ce{N2O5}$ : at 45 °C, however much $\ce{N2O5}$ the flask held to begin with, half of it was gone after about 23 minutes, and half of the rest after another 23. The liver clears ethanol from the blood the opposite way: roughly the same amount disappears every hour, whatever is left. Two very different countdowns, two different rate laws.

Predict first

A first-order reaction takes 23 minutes to fall from 0.020 M to 0.010 M. How long does it take to fall from 0.010 M to 0.005 M?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

11.5 minutes 23 minutes 46 minutes Cannot tell without k

From rate law to concentration–time curve

The rate law $-d[\mathrm{A}]/dt = k[\mathrm{A}]^n$ is a differential equation: it says how fast the concentration is falling right now. Integrating it over time gives concentration as a function of time — the integrated rate law. Each of the three common orders has its own way of turning the curve into a straight line:

\begin{aligned} n=0:&\quad [\mathrm{A}] = [\mathrm{A}]_0 - kt, &\quad t_{1/2} &= \frac{[\mathrm{A}]_0}{2k}\\ n=1:&\quad \ln[\mathrm{A}] = \ln[\mathrm{A}]_0 - kt, &\quad t_{1/2} &= \frac{\ln 2}{k}\\ n=2:&\quad \frac{1}{[\mathrm{A}]} = \frac{1}{[\mathrm{A}]_0} + kt, &\quad t_{1/2} &= \frac{1}{k[\mathrm{A}]_0} \end{aligned}

Each left-hand side is linear in t: the matching plot is a straight line whose slope gives k.

$n$: reaction order
$[\mathrm{A}]_0$: starting concentration
$t_{1/2}$: half-life: time for the concentration to halve

The fingerprints of the three orders
Order	Straight plot	Slope	Successive half-lives
0	[A] vs t	−k	get shorter (halving each time)
1	ln[A] vs t	−k	constant
2	1/[A] vs t	+k	get longer (doubling each time)

Worked example

How long does N₂O₅ last?

At 45 °C, $\ce{N2O5}$ decomposes with $k = 4.98\times10^{-4}\ \mathrm{s^{-1}}$ (first order), starting from $[\ce{N2O5}] = 0.0200$ M. Find the half-life, the concentration after 1 hour, and the time to fall to 10 % of the start.

Testing the zero-, first- and second-order linearisations

Which plot is straight?

$\ce{2NO2(g)} \mathrel{\ce{->}} \ce{2NO(g)} + \ce{O2(g)}$ — Two NO₂ molecules must collide to react, so the rate goes as [NO₂]².

Measured data (13)

t / s	[A] / M
0	0.00998
100	0.00644
200	0.00473
300	0.00363
400	0.00308
500	0.00264
600	0.00253
700	0.00208
800	0.00177
900	0.00166
1000	0.0016
1100	0.00156
1200	0.00136

Zero order: [A] vs t

R² = 0.7100

3 runs of residual signs

First order: ln[A] vs t

R² = 0.9124

3 runs of residual signs

Second order: 1/[A] vs t

R² = 0.9901

6 runs of residual signs

Switch to NO₂ decomposing at 300 °C, then to ethanol in the blood: guess which plot is straight first, then look at the residuals — curved residuals (long runs of one sign) mean the wrong choice. “Measure again” draws a fresh set of measurement noise.

Second-order half-lives keep getting longer: for $\ce{2NO2} \mathrel{\ce{->}} \ce{2NO} + \ce{O2}$ at 300 °C, $k = 0.54\ \mathrm{L\,mol^{-1}\,s^{-1}}$ , and from 0.0100 M the first half-life is 185 s, the second 370 s and the third 741 s — the lower the concentration, the harder it is for two $\ce{NO2}$ molecules to meet. A zero-order rate does not depend on concentration: ethanol in the blood is processed by saturated alcohol dehydrogenase at about 0.15 g/L per hour, so its half-life shrinks as the level falls. Zero-order behaviour is almost always a sign of saturation, which returns in the lesson on catalysis.

Concept check

For a second-order reaction A → products, which plot is a straight line?

[A] against t ln[A] against t 1/[A] against t [A]² against t

Try the calculation

After 30 minutes at 45 °C ( $k = 4.98\times10^{-4}\ \mathrm{s^{-1}}$ ), what percentage of the $\ce{N2O5}$ is left?

Try the calculation

A blood ethanol level of 0.80 g/L falls at a zero-order rate of 0.15 g L⁻¹ h⁻¹. How many hours until it reaches 0.20 g/L?

Key takeaways

Zero order: [A] falls linearly; first order: ln[A] falls linearly; second order: 1/[A] rises linearly.
Only the first-order half-life, ln 2/k, is independent of the starting concentration.
To find the order, plot the data three ways and pick the one whose residuals scatter at random.
Zero-order behaviour usually signals saturation (an enzyme, a catalyst surface).

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Data and references

F. Daniels, E. H. Johnston, “The thermal decomposition of gaseous nitrogen pentoxide. A monomolecular reaction”, J. Am. Chem. Soc. 43, 53–71 (1921).
A. W. Jones, “Evidence-based survey of the elimination rates of ethanol from blood with applications in forensic casework”, Forensic Sci. Int. 200, 1–20 (2010).
R. H. Petrucci et al., General Chemistry: Principles and Modern Applications (Pearson), chapter on chemical kinetics (NO₂ decomposition at 300 °C).
P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 17B (integrated rate laws).