Integrated Rate Laws

5.3 Integrated Rate Laws

Integrated Rate Laws

Three graphs plot [A] vs t, ln[A] vs t, and 1/[A] vs t simultaneously. The straightest line reveals reaction order, confirmed by live R² values for each plot.

[A] vs tln[A] vs t1/[A] vs tR² Order ID

Topic 5.3

Concentration Changes Over Time

Identify the rate law expression of a chemical reaction using data that show how the concentrations of reaction species change over time.

Reaction order can be inferred from a graph of concentration versus time by asking which plot is linear. Each order has exactly one linearizing transformation, and all three integrated rate laws are on the AP equation sheet:

Order	Linear plot	Integrated law	Slope
Zeroth	[A] vs t	[A]_t − [A]₀ = −kt	−k
First	ln[A] vs t	ln[A]_t − ln[A]₀ = −kt	−k
Second	1/[A] vs t	1/[A]_t − 1/[A]₀ = kt	+k

Note the sign: only the second-order plot has a positive slope, because 1/[A] grows as [A] shrinks.

Half-life is a critical parameter for first-order reactions specifically, because for first order — and only first order — the half-life is constant, independent of starting concentration:

t_1/2 = 0.693 / k

This constancy is itself a diagnostic: if successive halvings take equal time, the reaction is first order. (For zeroth order each successive half-life gets shorter; for second order each gets longer.)

EK 5.3.A.6 names the canonical application: radioactive decay is first order. After n half-lives, the fraction remaining is (1/2)ⁿ.

Key points

Find the order by finding which plot is linear: [A], ln[A], or 1/[A] versus time.
Slope is −k for zeroth and first order, +k for second order.
Constant half-life ⇔ first order. This is a diagnostic, not just a formula.
After n half-lives, (1/2)ⁿ of the original remains.

Equations

$[A]_{t} - [A]_{0} = - k t$
on the exam sheetZeroth order.
$ln [A]_{t} - ln [A]_{0} = - k t$
on the exam sheetFirst order.
$\frac{1}{[ A ] _{t}} - \frac{1}{[ A ] _{0}} = k t$
on the exam sheetSecond order.
$t_{1/2} = \frac{0.693}{k}$
on the exam sheetHalf-life — first order only. Do not apply it to other orders.
- $t_{1/2}$ half-life
- $k$ first-order rate constant (s⁻¹)

Common mistakes

Applying t₁/₂ = 0.693/k to a non-first-order reaction. The equation sheet groups it with kinetics generally, but it is first-order-only.
Sign of the slope. A positive slope on a 1/[A] plot is second order, not an error.
Confusing ln with log. The integrated first-order law uses natural log.
Mixing up half-life and reaction completion. A reaction is never mathematically "finished" in first-order kinetics.

Worked example

The decomposition of N₂O₅ is first order with k = 5.2 × 10⁻³ s⁻¹ at 65 °C. (a) Find the half-life. (b) If [N₂O₅]₀ = 0.400 M, what is the concentration after 300. s? (c) How long until only 12.5% remains?

(a) Half-life.
t₁/₂ = 0.693 / (5.2 × 10⁻³ s⁻¹) = 133 s

(b) Concentration at 300 s.
ln[A]ₜ = ln[A]₀ − kt = ln(0.400) − (5.2 × 10⁻³)(300.)
ln[A]ₜ = −0.916 − 1.56 = −2.476
[A]ₜ = e^(−2.476) = 0.084 M

(c) Time to 12.5%.
12.5% = 1/8 = (1/2)³, which is exactly three half-lives.
t = 3 × 133 s = 399 s

Part (c) needed no logarithms at all — recognizing the power of one-half is the fast route.

Full notes for topic 5.3 →