Methods to determine order — initial rates, integrated method, half-life method: Worked examples

The scenario matrix

Every problem in "find the order" collapses into one of these cells. Below the table, examples E1–E9 each announce which cell they cover. Together they fill every row.

Cell	Method	Situation being stressed	Covered by
C1	Initial rates	one reactant, clean integer order	E1
C2	Initial rates	two reactants, isolation logic	E2
C3	Initial rates	fractional / non-integer order	E3
C4	Integrated	pick the linear plot from raw data	E4
C5	Integrated	degenerate: zero order (constant rate)	E5
C6	Half-life	degenerate: first order ( $t_{1/2}$ constant)	E6
C7	Half-life	second order, $t_{1/2}$ grows/shrinks	E7
C8	Any	real-world word problem (drug clearance)	E8
C9	Any	exam twist: units of $k$ pin the order	E9

E1 — One reactant, clean integer order (cell C1)

Worked example Statement

For $A \to$ products the rate law is $r = k [A]^{n}$ . Two experiments:

Expt	$[A]$ (M)	initial rate (M/s)
1	0.050	$1.5 \times 1 0^{- 4}$
2	0.150	$1.35 \times 1 0^{- 3}$

Find $n$ .

Forecast: $[A]$ tripled. Guess: does the rate go ×3, ×9, or ×27?

Step 1 — rate ratio. $\frac{r _{2}}{r _{1}} = \frac{1.35 \times 1 0 ^{- 3}}{1.5 \times 1 0 ^{- 4}} = 9$ . Why this step? The ratio cancels the unknown $k$ , leaving only the concentration effect.

Step 2 — concentration ratio. $\frac{[ A ] _{2}}{[ A ] _{1}} = \frac{0.150}{0.050} = 3$ . Why? This is the cause; the rate ratio is the effect.

Step 3 — solve $3^{n} = 9$ . Using $n = \frac{ln ( r _{2} / r _{1} )}{ln ([ A ] _{2} / [ A ] _{1} )} = \frac{ln 9}{ln 3} = 2$ . Why the log? Logs pull $n$ down out of the exponent so we can read it off directly.

Answer: $n = 2$ , second order.

Verify: predict $r_{2}$ from $n = 2$ : $r_{2} = r_{1} \cdot 3^{2} = 1.5 \times 1 0^{- 4} \times 9 = 1.35 \times 1 0^{- 3}$ M/s. ✓ Matches the table exactly.

E2 — Two reactants, isolation logic (cell C2)

Worked example Statement

$A + B \to$ products, $r = k [A]^{a} [B]^{b}$ .

Expt	$[A]$	$[B]$	rate (M/s)
1	0.10	0.10	$2.0 \times 1 0^{- 3}$
2	0.30	0.10	$6.0 \times 1 0^{- 3}$
3	0.10	0.30	$1.8 \times 1 0^{- 2}$

Find $a$ , $b$ , and overall order.

Forecast: which reactant matters more — the one where tripling did ×3, or the one where tripling did ×9?

Step 1 — find $a$ using 1→2. Here $[B]$ is fixed, only $[A]$ triples. $\frac{r _{2}}{r _{1}} = \frac{6.0 \times 1 0 ^{- 3}}{2.0 \times 1 0 ^{- 3}} = 3 = 3^{a} \Rightarrow a = 1$ . Why fix $[B]$ ? So the rate change is caused by $[A]$ alone — that isolates $a$ .

Step 2 — find $b$ using 1→3. Now $[A]$ fixed, $[B]$ triples. $\frac{r _{3}}{r _{1}} = \frac{1.8 \times 1 0 ^{- 2}}{2.0 \times 1 0 ^{- 3}} = 9 = 3^{b} \Rightarrow b = 2$ . Why? Same isolation trick, now the roles swap.

Step 3 — overall order $= a + b = 1 + 2 = 3$ . Why add? Overall order is the sum of the individual exponents in $r = k [A]^{a} [B]^{b}$ .

Answer: $a = 1, b = 2$ , overall third order.

Verify: $k = \frac{r _{1}}{[ A ] ^{a} [ B ] ^{b}} = \frac{2.0 \times 1 0 ^{- 3}}{( 0.10 ) ^{1} ( 0.10 ) ^{2}} = \frac{2.0 \times 1 0 ^{- 3}}{1 0 ^{- 3}} = 2.0$ . Predict $r_{3} = k (0.10) (0.30)^{2} = 2.0 \times 0.10 \times 0.09 = 1.8 \times 1 0^{- 2}$ . ✓

E3 — Fractional (non-integer) order (cell C3)

E4 — Integrated method, pick the linear plot (cell C4)

Worked example Statement

Decomposition of $A$ . Data below. Decide the order by testing which of $[A]$ , $ln [A]$ , $1/ [A]$ is linear in $t$ , then get $k$ .

$t$ (s)	$[A]$ (M)
0	1.00
100	0.50
200	0.25
300	0.125

Forecast: $[A]$ halves every 100 s regardless of level. Which order has a constant halving time?

Step 1 — test zero order ( $[A]$ vs $t$ ). Drops $0.50, 0.25, 0.125$ — not equal steps. Not linear. Why? Zero order falls by a fixed amount each interval, which this isn't.

Step 2 — test first order ( $ln [A]$ vs $t$ ). $ln$ values: $0, - 0.693, - 1.386, - 2.079$ . Steps of $- 0.693$ each — perfectly even! Linear. ✓ Why? Equal $ln$ steps in equal times is the signature of first order.

Step 3 — get $k$ . Slope $= \frac{- 0.693 - 0}{100 - 0} = - 6.93 \times 1 0^{- 3}$ s⁻¹. Since slope $= - k$ , $k = 6.93 \times 1 0^{- 3}$ s⁻¹. Why the minus? Concentration falls, so the line slopes down; $k$ is the magnitude.

Answer: first order, $k = 6.93 \times 1 0^{- 3}$ s⁻¹. See the three-plot comparison below.

Verify: first-order law $[A] = [A]_{0} e^{- k t}$ : at $t = 200$ , $[A] = 1.00 e^{- 6.93 \times 1 0^{- 3} \times 200} = e^{- 1.386} = 0.250$ M. ✓

Figure — Methods to determine order — initial rates, integrated method, half-life method

E5 — Degenerate case: ZERO order (cell C5)

Worked example Statement

A reaction on a saturated catalyst surface. Data:

$t$ (s)	$[A]$ (M)
0	0.80
100	0.60
200	0.40
300	0.20

Find the order and $k$ .

Forecast: the drop is $0.20$ M every 100 s — always the same amount, not the same fraction. Which order does that?

Step 1 — spot the constant drop. $Δ [A] = - 0.20$ M per $100$ s, every interval. Why care? A fixed drop per fixed time means the rate $- \frac{d [ A ]}{d t}$ is constant — that's the definition of zero order, $r = k [A]^{0} = k$ .

Step 2 — confirm $[A]$ vs $t$ is the straight line. Points $0.80, 0.60, 0.40, 0.20$ fall on a line. Zero order confirmed. Why this plot? From $[A] = [A]_{0} - k t$ , only plotting $[A]$ itself gives a straight line for $n = 0$ .

Step 3 — get $k$ . $k = - slope = - \frac{0.20 - 0.80}{300 - 0} = - \frac{- 0.60}{300} = 2.0 \times 1 0^{- 3}$ M/s. Why units M/s? For zero order $k$ has the units of rate itself.

Answer: zero order, $k = 2.0 \times 1 0^{- 3}$ M/s.

Verify (limiting behaviour): zero-order reactions end — $[A]$ hits 0 at $t = [A]_{0} / k = 0.80/2.0 \times 1 0^{- 3} = 400$ s. After that the rate must drop to 0 (no $A$ left); the law only holds while $[A] > 0$ . Sanity ✓.

E6 — Degenerate case: FIRST order via half-life (cell C6)

Worked example Statement

For a reaction, $[A]_{0} = 0.40$ M gives $t_{1/2} = 200$ s. When you rerun with $[A]_{0} = 0.80$ M, $t_{1/2}$ is still $200$ s. Find the order and $k$ .

Forecast: the half-life ignored the starting concentration. Which single order does that? (Hint: radioactivity behaves this way.)

Step 1 — read the fingerprint. $t_{1/2}$ independent of $[A]_{0}$ $\Rightarrow$ order $= 1$ . Why? Among all orders, only $n = 1$ gives $t_{1/2} = \frac{0.693}{k}$ , which has no $[A]_{0}$ in it.

Step 2 — general check with $n = 1 - \frac{ln ( t _{1/2, 2} / t _{1/2, 1} )}{ln ([ A ] _{0, 2} / [ A ] _{0, 1} )} = 1 - \frac{ln ( 200/200 )}{ln ( 0.80/0.40 )} = 1 - \frac{ln 1}{ln 2} = 1 - 0 = 1$ . Why this formula? It converts the how-t-half-changes-with-concentration slope into the order for any $n \neq = 1$ , and gracefully returns $1$ when there's no change.

Step 3 — get $k$ . $k = \frac{0.693}{t _{1/2}} = \frac{0.693}{200} = 3.47 \times 1 0^{- 3}$ s⁻¹.

Answer: first order, $k = 3.47 \times 1 0^{- 3}$ s⁻¹.

Verify: after one half-life, $[A]$ should be $0.20$ M from $[A]_{0} = 0.40$ : $[A] = 0.40 e^{- 3.47 \times 1 0^{- 3} \times 200} = 0.40 e^{- 0.693} = 0.40 \times 0.500 = 0.200$ M. ✓

E7 — Second order, $t_{1/2}$ changes with $[A]_{0}$ (cell C7)

Worked example Statement

$[A]_{0} = 0.10$ M gives $t_{1/2} = 250$ s. $[A]_{0} = 0.25$ M gives $t_{1/2} = 100$ s. Find order and $k$ .

Forecast: starting concentration went up $2.5 \times$ , and $t_{1/2}$ went down. Inverse relation — smell order 2.

Step 1 — general formula. $n = 1 - \frac{ln ( t _{1/2, 2} / t _{1/2, 1} )}{ln ([ A ] _{0, 2} / [ A ] _{0, 1} )} = 1 - \frac{ln ( 100/250 )}{ln ( 0.25/0.10 )}$ . Why? We don't yet trust "it looks inverse"; the formula gives the exact $n$ .

Step 2 — crunch. $ln (0.4) = - 0.9163$ , $ln (2.5) = 0.9163$ . So $n = 1 - \frac{- 0.9163}{0.9163} = 1 - (- 1) = 2$ . Why does it land on exactly 2? Because $t_{1/2} \propto [A]_{0}^{1 - n} = [A]_{0}^{- 1}$ , so its ratio is the reciprocal of the concentration ratio — logs of reciprocals are exact negatives.

Step 3 — get $k$ from $t_{1/2} = \frac{1}{k [ A ] _{0}}$ : using the first data point, $k = \frac{1}{t _{1/2} [ A ] _{0}} = \frac{1}{250 \times 0.10} = 0.040$ M⁻¹s⁻¹. Why units M⁻¹s⁻¹? Second-order $k$ must carry these units so that $k [A]^{2}$ comes out as M/s.

Answer: second order, $k = 0.040$ M⁻¹s⁻¹.

Verify: predict the second half-life for $[A]_{0} = 0.25$ : $t_{1/2} = \frac{1}{0.040 \times 0.25} = \frac{1}{0.010} = 100$ s. ✓ Matches the table.

E8 — Real-world word problem: drug clearance (cell C8)

Worked example Statement

A patient is given a drug. Blood tests show its concentration halves every $4$ hours, no matter the starting dose. Starting at $80$ mg/L, when does it drop to $10$ mg/L, and what is the elimination constant $k$ ?

Forecast: "halves every 4 h regardless of dose" — you've seen this fingerprint one example ago. What order?

Step 1 — identify order. Constant $t_{1/2}$ independent of dose $\Rightarrow$ first order (typical drug pharmacokinetics). Why? Same reasoning as E6 — only $n = 1$ has dose-free half-life.

Step 2 — count half-lives to reach $10$ mg/L. $80 \to 40 \to 20 \to 10$ : that's $3$ halvings. Time $= 3 \times 4 = 12$ h. Why count halvings? First order drops by a fixed fraction (½) each half-life, so successive halvings chain multiplicatively.

Step 3 — get $k$ . $k = \frac{0.693}{t _{1/2}} = \frac{0.693}{4} = 0.173$ h⁻¹. Why? First-order half-life relation, solved for $k$ .

Answer: first order, reaches $10$ mg/L after $12$ h, $k = 0.173$ h⁻¹.

Verify: $[A] = 80 e^{- 0.173 \times 12} = 80 e^{- 2.079} = 80 \times 0.125 = 10.0$ mg/L. ✓

E9 — Exam twist: units of $k$ reveal the order (cell C9)

Worked example Statement

An exam gives only this: for a single-reactant reaction, $k = 3.0 \times 1 0^{- 2} mol L^{- 1} s^{- 1}$ . State the order — no rate data provided.

Forecast: you have no concentrations, no rates... but the units of $k$ are a hidden fingerprint. Can you decode them?

Step 1 — recall the units rule. From $r = k [A]^{n}$ , rate has units $mol L^{- 1} s^{- 1}$ always, so $k$ has units $(mol L^{- 1})^{1 - n} s^{- 1}$ . Why? $[A]^{n}$ carries $(mol L^{- 1})^{n}$ ; to make the product equal rate's units, $k$ must supply $(mol L^{- 1})^{1 - n}$ .

Step 2 — match. Given $k$ has $mol L^{- 1}$ to the power $+ 1$ . Set $1 - n = 1 \Rightarrow n = 0$ . Why does this pin it down? Each order gives a unique power on the concentration unit: $n = 0 \to (M)^{+ 1}$ , $n = 1 \to (M)^{0}$ (just s⁻¹), $n = 2 \to (M)^{- 1}$ .

Answer: zero order.

Verify (unit table sanity):

$n$	power $1 - n$	units of $k$
0	$+ 1$	mol L⁻¹ s⁻¹
1	$0$	s⁻¹
2	$- 1$	L mol⁻¹ s⁻¹

The given units sit in the $n = 0$ row. ✓ Consistent with the Rate Law and Rate Constant convention.

Recall check

Recall Which cells stress the degenerate orders?

Zero order (C5, E9) and first order (C6, E8) — the two "special" cases where behaviour looks unusually simple. C5/E5 shows ::: a constant drop per unit time (rate independent of $[A]$ ). C6/E6 shows ::: a constant half-life independent of $[A]_{0}$ .

Recall The fastest single-clue shortcuts

No data, only $k$ 's units? ::: match the power of the concentration unit — $n = 1 - (that power)$ . $t_{1/2}$ doesn't change with dose? ::: first order. Constant amount lost per equal time? ::: zero order. Doubling $[A]$ makes rate ×4? ::: second order.