Methods to determine order — initial rates, integrated method, half-life method: Visual walkthrough

Step 0 — The words, before any symbols

Before a single formula, let us earn every symbol.

We want to know: when I start with a taller tank, does it take longer, shorter, or the same time to drop to half? That single yes/no/how-much is the whole method.

Step 1 — Draw the tank emptying

WHAT. Picture a tank of reactant draining. Mark the start height $[A]_{0}$ and the half-line $[A]_{0} /2$ . The time to go from one to the other is $t_{1/2}$ .

WHY. Everything abstract about "order" becomes visible once we agree that concentration = height and half-life = "time to reach the half-line."

PICTURE.

Step 2 — The one equation everything comes from

WHAT. The rate law: the rule saying how fast the level drops depends on the current height.

$speed the level drops - \frac{d [ A ]}{d t} = rate constant k current height, raised to power n [A]^{n}$

WHY these symbols:

$\frac{d [ A ]}{d t}$ is the derivative — the instantaneous slope of the falling curve, "how much height per second." We use a derivative and not a plain fraction because the slope keeps changing as the tank empties; only calculus captures a moving slope. (See Rate Law and Rate Constant.)
The minus sign is because $[A]$ decreases, so $d [A] / d t$ is negative; we flip it so the left side is a positive speed.
$k$ is the rate constant — a fixed number for the reaction at a given temperature (it hides how easy collisions are; see Arrhenius Equation). It sets the overall scale of speed but not the shape of dependence.
$n$ is the order — the star of the show. It is the power the height is raised to.

PICTURE.

Step 3 — Turn the rule into a level-vs-time law (integrate)

WHAT. The rate law only tells us the slope at each instant. To get the actual height at a later time we must add up all those slopes — that is integration. This gives the integrated rate law.

WHY integrate and not just plug in? Because $[A]$ appears on both sides (the slope depends on the very height that is changing). We cannot solve it in one step; we accumulate the tiny drops. Integration is exactly the tool that answers "given the slope everywhere, what is the curve?"

We do it for each order separately.

n = 0 : n = 1 : n = 2 : [A] = [A]_{0} - k t ln [A] = ln [A]_{0} - k t \frac{1}{[ A ]} = \frac{1}{[ A ] _{0}} + k t

Reading each term:

$[A]$ — height now.
$[A]_{0}$ — height at the start (the constant of integration, fixed by "at $t = 0$ height is $[A]_{0}$ ").
$k t$ — how much "draining budget" the clock has spent. Every law says: start value, minus (or plus) $k t$ , in whatever coordinate makes that order straight.
$ln$ (natural log) appears for $n = 1$ because integrating $1/ [A]$ produces a logarithm — the log is the honest answer, not a trick.
$1/ [A]$ appears for $n = 2$ because integrating $1/ [A]^{2}$ produces $- 1/ [A]$ .

PICTURE.

Step 4 — Put the tank down to half in each law

WHAT. Set $[A] = [A]_{0} /2$ and solve for the time. That solved time is $t_{1/2}$ .

WHY. "Half" is the cleanest checkpoint: it does not care about units of $[A]$ (it is always a factor of 2), so the only things left in the answer are $k$ and $[A]_{0}$ — precisely what we want to compare.

Zero order. $\frac{[ A ] _{0}}{2} = [A]_{0} - k t_{1/2} \Rightarrow k t_{1/2} = \frac{[ A ] _{0}}{2} \Rightarrow t_{1/2} = \frac{[ A ] _{0}}{2 k}$ Term-by-term: $[A]_{0} - \frac{[ A ] _{0}}{2} = \frac{[ A ] _{0}}{2}$ is the height we had to lose; divide by drain speed $k$ to get the time. Taller start ⇒ more to lose ⇒ longer.

First order. $ln \frac{[ A ] _{0} /2}{[ A ] _{0}} = - k t_{1/2} \Rightarrow ln \frac{1}{2} = - k t_{1/2} \Rightarrow - ln 2 = - k t_{1/2} \Rightarrow t_{1/2} = \frac{ln 2}{k} = \frac{0.693}{k}$ Look what vanished: the ratio $\frac{[ A ] _{0} /2}{[ A ] _{0}} = \frac{1}{2}$ cancels $[A]_{0}$ completely. The starting height leaves no trace — see Half-Life of Reactions.

Second order. $\frac{1}{[ A ] _{0} /2} = \frac{1}{[ A ] _{0}} + k t_{1/2} \Rightarrow \frac{2}{[ A ] _{0}} - \frac{1}{[ A ] _{0}} = k t_{1/2} \Rightarrow t_{1/2} = \frac{1}{k [ A ] _{0}}$ Here $[A]_{0}$ sits in the denominator: a taller start makes $t_{1/2}$ smaller.

PICTURE.

Step 5 — The fingerprint: three ways the red bar reacts

WHAT. Now do the decisive experiment in a picture: draw each order's curve twice, once from $[A]_{0}$ and once from $2 [A]_{0}$ , and watch what the half-life bar does.

WHY. This is the actual lab move — you cannot see $n$ directly, but you can double the starting concentration and time the halving.

PICTURE.

Step 6 — Compress it into one exponent

WHAT. All three boxed results are secretly one formula. Notice the power of $[A]_{0}$ in each:

$n = 0 : t_{1/2} \propto [A]_{0}^{+ 1}$
$n = 1 : t_{1/2} \propto [A]_{0}^{0}$ (no dependence)
$n = 2 : t_{1/2} \propto [A]_{0}^{- 1}$

The exponent is $1, 0, - 1$ for $n = 0, 1, 2$ — i.e. exactly $1 - n$ .

$t_{1/2} \propto [A]_{0}^{1 - n}$

WHY this matters. One formula covers every order, including fractional ones. And it hands us a direct way to solve for $n$ from two experiments.

Take two starting heights $[A]_{0, 1}, [A]_{0, 2}$ with half-lives $t_{1/2, 1}, t_{1/2, 2}$ . Divide: $\frac{t _{1/2, 2}}{t _{1/2, 1}} = (\frac{[ A ] _{0, 2}}{[ A ] _{0, 1}})^{1 - n}$ Take a logarithm — the tool that pulls an exponent down into a multiplier so we can isolate it: $n = 1 - \frac{ln ( t _{1/2, 2} / t _{1/2, 1} )}{ln ( [ A ] _{0, 2} / [ A ] _{0, 1} )}$ Each piece: the top log measures "how much the halving-time changed," the bottom log measures "how much the start changed," their ratio is the exponent $1 - n$ , and we subtract from $1$ to free $n$ .

Step 7 — The edge and degenerate cases

Never leave the reader in an unshown scenario.

The one-picture summary

Recall Feynman: the walkthrough in plain words

Picture a tank of stuff draining. Order is just "how loudly the drain listens to how full the tank is." If it ignores fullness, order 0; if it listens gently (proportional), order 1; if it listens fiercely (squared), order 2.

We can't peek at the order directly. So we time how long the tank takes to go half empty — that's the half-life. Then the clever move: fill the tank twice as high and time it again.

If the halving time doubles, the drain cared about how much was there — order 0.
If the halving time doesn't budge, the drain didn't care about the extra water at all — order 1 (this is radioactivity's trademark).
If the halving time halves, a fuller tank drains proportionally even faster — order 2.

Package all three into one sentence: $t_{1/2} \propto [A]_{0}^{1 - n}$ . Take a log of two experiments and out pops $n$ . That is the entire half-life method — no memorising, just watching a red bar grow, stay, or shrink.

Recall

Half-life definition

time for

[A]

to fall from

[A]_{0}

[A]_{0} /2

Why integrate the rate law

the slope depends on the changing height, so we add up all the slopes

What cancels

[A]_{0}

for first order

the ratio

\frac{[ A ] _{0} /2}{[ A ] _{0}} = \frac{1}{2}

inside the log

Doubling

[A]_{0}

for order 0 does what to

t_{1/2}

doubles it

Doubling

[A]_{0}

for order 1 does what to

t_{1/2}

nothing (unchanged)

Doubling

[A]_{0}

for order 2 does what to

t_{1/2}

halves it

General half-life dependence

t_{1/2} \propto [A]_{0}^{1 - n}

Order from two half-lives

n = 1 - \frac{ln ( t _{1/2, 2} / t _{1/2, 1} )}{ln ([ A ] _{0, 2} / [ A ] _{0, 1} )}

Why take a log to find

n

it pulls the exponent

1 - n

down where we can isolate it

Order	$t_{1/2}$	Double $[A]_{0} \Rightarrow t_{1/2}$
0	$[A]_{0} /2 k$	doubles ↑
1	$0.693/ k$	unchanged =
2	$1/ k [A]_{0}$	halves ↓