Integrated rate laws — half-life t₁ - ₂ for each order: Worked examples

This page is the practice lab for the half-life topic. The parent note built three formulas. Here we hunt down every kind of question those formulas can be wrapped in — every order, every trick, every degenerate edge — and solve each one out loud.

Before we start, let us pin down the three tools we will keep reaching for, so no symbol appears unexplained.

The scenario matrix

Think of this table as a checklist. Every worked example below is tagged with the cell it fills. When all cells are ticked, you have seen every scenario this topic can throw at you.

Cell	Scenario class	What makes it tricky
C1	Zero-order, forward calc	$t_{1/2}$ depends on $[A]_{0}$ — pick the right starting value each step
C2	First-order, forward calc	must recognise $[A]_{0}$ is irrelevant
C3	Second-order, successive half-lives	each half-life doubles — track the new starting concentration
C4	Zero-order degenerate edge	reaction finishes before you expect — concentration can hit zero
C5	"Number of half-lives" (first-order)	fractional decay $(1/2)^{n}$ , not a straight countdown
C6	Diagnose the order from half-life data	read the trend, not a single number
C7	Units / sign sanity trap	wrong-order $k$ units give nonsense time
C8	Real-world word problem	drug elimination / decay dressed in prose
C9	Exam twist: back-solve for $k$ or $[A]_{0}$	rearrange the formula the other way

Worked examples

C1 — Zero-order forward calculation

Worked example A saturated surface reaction

A gas decomposes on a fully-covered catalyst surface, so it runs zero-order with $k = 0.050 M s^{- 1}$ . Starting at $[A]_{0} = 2.0 M$ , find the first half-life, then the half-life that begins the moment $[A]$ reaches $1.0 M$ .

Forecast: for zero-order, $t_{1/2}$ is proportional to the starting concentration. Since the second stage starts at half the concentration, guess the second half-life is half as long.

First half-life. Why this step? Use $t_{1/2} = [A]_{0} / (2 k)$ with the true start. $t_{1/2} = \frac{2.0}{2 ( 0.050 )} = \frac{2.0}{0.10} = 20 s$
New starting concentration is now $[A] = 1.0 M$ . Why this step? For zero-order each stage's half-life uses that stage's starting value. $t_{1/2}^{'} = \frac{1.0}{2 ( 0.050 )} = \frac{1.0}{0.10} = 10 s$ Verify: units $\frac{M}{M s ^{- 1}} = s$ ✓. The second half-life (10 s) is exactly half the first (20 s) — matching the forecast. Zero-order half-lives shrink.

C2 — First-order forward calculation

C3 — Second-order successive half-lives

Worked example Each half-life doubles

A second-order reaction has $k = 0.54 M^{- 1} s^{- 1}$ and $[A]_{0} = 0.20 M$ . Find the first three successive half-lives.

Forecast: second-order $t_{1/2}$ is inversely proportional to the starting concentration. Each stage halves the concentration, so each half-life should double.

First half-life. Why this step? $t_{1/2} = 1/ (k [A]_{0})$ with $[A]_{0} = 0.20$ . $t_{1/2, 1} = \frac{1}{( 0.54 ) ( 0.20 )} = \frac{1}{0.108} = 9.26 s$
New start $= 0.10 M$ . Why this step? Stage two begins at half the concentration. $t_{1/2, 2} = \frac{1}{( 0.54 ) ( 0.10 )} = \frac{1}{0.054} = 18.5 s$
New start $= 0.050 M$ . Why this step? Stage three begins at half again. $t_{1/2, 3} = \frac{1}{( 0.54 ) ( 0.050 )} = \frac{1}{0.027} = 37.0 s$ Verify: $9.26 \to 18.5 \to 37.0$ — each is double the last ✓. The figure below shows the three orders side by side so you can see the trends.

Figure — Integrated rate laws — half-life t₁ - ₂ for each order

C4 — Zero-order degenerate edge: the reaction can end

Worked example What if it runs out?

The zero-order reaction of C1 ( $k = 0.050 M s^{- 1}$ , $[A]_{0} = 2.0 M$ ). How long until all of A is gone? Is it valid to compute a "third" or "fourth" half-life?

Forecast: zero-order concentration falls in a straight line (constant slope), so it hits zero at a finite time — unlike first/second-order which only approach zero. Half-lives keep shrinking and must stop.

Total time to consume all A. Why this step? Set $[A]_{t} = 0$ in $[A]_{t} = [A]_{0} - k t$ . $0 = 2.0 - 0.050 t ⟹ t = \frac{2.0}{0.050} = 40 s$
Check against half-lives. Why this step? Successive zero-order half-lives are $20, 10, 5, 2.5, \dots$ s; their sum is $20 + 10 + 5 + \dots \to 40$ s. $20 + 10 + 5 + 2.5 + \dots = 20 \cdot \frac{1}{1 - \frac{1}{2}} = 40 s$ Verify: the infinite chain of shrinking half-lives sums to exactly the finite completion time 40 s ✓. Degenerate warning: once $t > 40$ s the formula $[A]_{t} = [A]_{0} - k t$ would give a negative concentration, which is physically impossible — concentration stays at 0. Look at the mint straight line in the figure: it truly touches the axis.

C5 — "Number of half-lives" for first-order

C6 — Diagnose the order from data

Worked example Reading the trend

Two experiments give successive half-lives. Identify the order in each.

Set X: $[A]_{0} = 0.80 M$ : first $t_{1/2} = 50 s$ , second $t_{1/2} = 100 s$ .
Set Y: $[A]_{0} = 0.80 M$ : first $t_{1/2} = 50 s$ , second $t_{1/2} = 50 s$ .

Forecast: constant $\Rightarrow$ first-order; increasing $\Rightarrow$ second-order; decreasing $\Rightarrow$ zero-order. So X looks second-order, Y first-order.

Set X: half-life doubled as concentration halved. Why this step? Doubling matches $t_{1/2} \propto 1/ [A]_{0}$ . $second-order ⟹ k = \frac{1}{t _{1/2} [ A ] _{0}} = \frac{1}{50 \times 0.80} = 0.025 M^{- 1} s^{- 1}$
Set Y: half-life unchanged. Why this step? Constant $t_{1/2}$ is the unique first-order signature. $first-order ⟹ k = \frac{0.693}{50} = 0.01386 s^{- 1}$ Verify: X units $\frac{1}{s \cdot M} = M^{- 1} s^{- 1}$ ✓ (second-order); Y units $s^{- 1}$ ✓ (first-order). Method matches Reaction order determination from experimental data.

C7 — Units / sign sanity trap

C8 — Real-world word problem (drug elimination)

C9 — Exam twist: back-solve for the unknown

Which cell did each example fill?

Recall Quick self-test

Zero-order successive half-lives get shorter or longer? ::: Shorter (they halve each stage). Which order lets you use fraction $= (1/2)^{n}$ safely? ::: First-order only (constant $t_{1/2}$ ). Second-order $k$ has what units? ::: $M^{- 1} s^{- 1}$ . If two successive half-lives are equal, the order is? ::: First-order. Why can a zero-order reaction reach $[A] = 0$ but first-order cannot? ::: Straight-line decay hits the axis at finite time; exponential only approaches zero.