Integrated rate laws — half-life t₁ - ₂ for each order: Visual walkthrough

Step 1 — What we are even plotting

WHAT. Draw two axes. The horizontal axis is time $t$ (how long the reaction has run, in seconds). The vertical axis is concentration $[A]$ — the amount of our reactant $A$ crammed into each litre, measured in moles per litre (written $M$ ). As the reaction eats $A$ , the dot representing "how much is left" slides downward as it slides rightward.

WHY. Every half-life story is really a story about this one falling curve. Before we can ask "when does it reach half?", we must agree on what "it" is and where "half" sits on the axis.

PICTURE. The starting height is $[A]_{0}$ — the little "0" means at time zero, before anything happened. The dashed line halfway up is $[A]_{0} /2$ . The half-life $t_{1/2}$ is simply the horizontal distance from the start to where the curve first crosses that dashed line.

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

Step 2 — Three different ways the curve can fall

WHAT. The shape of the falling curve is decided by the rate law — the rule that says how fast $[A]$ drops. First we must pin down what "how fast it drops" means as a symbol. The rate is the steepness of our falling curve at an instant — the drop in height per tiny slice of time. We write this with the derivative: $rate \equiv - \frac{d [ A ]}{d t}$ Read $\frac{d [ A ]}{d t}$ as "the tiny change in $[A]$ divided by the tiny slice of time $d t$ that caused it" — the slope of the curve. Since $[A]$ is falling, that slope is negative, so we stick a minus sign in front to make rate a positive "how fast it disappears" number. We use the derivative and not just "average speed" because the fall speed keeps changing along the curve, and we want its value right at this instant.

Three simple rules give three shapes:

Zero-order: rate is a fixed number $k$ , no matter the height → a straight ramp down.
First-order: rate is proportional to the current height, $k [A]$ → a curve that eases off, never quite hitting zero.
Second-order: rate is proportional to height squared, $k [A]^{2}$ → a curve that plunges early then crawls.

WHY. We introduce $k$ , the rate constant, because we need one number to say "how eager" the reaction is. (Its units differ per order — we return to that.) The dependence on height is what makes each half-life behave differently, so we must see all three side by side first.

PICTURE. Same start height, same axes — three curves. Notice the straight line, the gently-easing exponential, and the steep-then-flat second-order curve. The half-life is where each curve pierces the dashed $[A]_{0} /2$ line.

Step 3 — Zero-order: a straight ramp, and its shrinking half-life

WHAT. Zero-order means $- \frac{d [ A ]}{d t} = k$ : the height drops at a constant rate. "Adding up" (integrating) a constant slope $- k$ from the start gives a straight line: $[A]_{t} = start height [A]_{0} - slope of drop k time t$ Here $[A]_{t}$ just means "the height at time $t$ ". Set that height to half and solve for the time: $\frac{[ A ] _{0}}{2} = [A]_{0} - k t_{1/2} \Rightarrow k t_{1/2} = \frac{[ A ] _{0}}{2} \Rightarrow t_{1/2} = \frac{[ A ] _{0}}{2 k}$

WHY. Because the ramp is straight, "half the height" is reached after covering half the total drop. A straight ramp covers a fixed height in a fixed time — so covering less height (a smaller $[A]_{0}$ ) takes less time. That is why the half-life shrinks as the reaction proceeds.

PICTURE. The red ramp. The first half-life is a wide horizontal span; the next half-life (starting from half-height) is a narrower span — same slope, less height to cover.

Step 4 — First-order: the self-similar curve

WHAT. First-order means $- \frac{d [ A ]}{d t} = k [A]$ : the fall speed is proportional to the current height. To solve this we ask "what function has a slope equal to $- k$ times itself?" We gather all the $[A]$ -stuff on one side and all the $t$ -stuff on the other (this is called separating the variables): $\frac{d [ A ]}{[ A ]} = - k d t$ Now add up (integrate) both sides. The left side sums to $ln [A]$ — because $ln$ is precisely the function whose slope is $1/ [A]$ , so summing $1/ [A]$ slices rebuilds $ln [A]$ . The right side sums to $- k t$ . Matching the two from the start point $[A]_{0}$ gives: $ln [A]_{t} - ln [A]_{0} = - k t ⟹ ln [A]_{t} = ln [A]_{0} - k t$ Undoing the $ln$ (raising $e$ to both sides) turns this into the famous exponential decay: $[A]_{t} = [A]_{0} e^{- k t}$

WHY it comes out exponential. A quantity whose rate of fall is proportional to itself must be exponential — that is the defining property of $e^{- k t}$ : its slope at every point is $- k$ times its own value. No other shape has "slope always proportional to height," so the integration can lead nowhere else.

Now find the half-life. The symbol $ln$ (natural logarithm) answers "how many $e$ -foldings?" and is the exact undo of $e^{(\cdot)}$ ; we use it here because it flattens the curvy exponential into a straight line we can solve. Set height to half: $ln (\frac{[ A ] _{0}}{2}) = ln [A]_{0} - k t_{1/2}$ Split the log: $ln ([A]_{0} /2) = ln [A]_{0} - ln 2$ (a division inside a log becomes a subtraction). The $ln [A]_{0}$ on both sides cancels: $- ln 2 = - k t_{1/2} \Rightarrow t_{1/2} = \frac{ln 2}{k} = \frac{0.693}{k}$

WHY the half-life is constant. Look what vanished: $[A]_{0}$ . The starting height is nowhere in the answer. That is the magic of proportional decay — the curve is self-similar: any patch of it, re-zoomed to its own start height, looks identical. So halving always takes the same time.

PICTURE. Equal horizontal steps of width $t_{1/2}$ each chop the height in half: full → ½ → ¼ → ⅛. The steps are all the same width.

Step 5 — Second-order: the plunge that stalls

WHAT. Second-order means $- \frac{d [ A ]}{d t} = k [A]^{2}$ : the fall speed is proportional to height squared. Separate the variables again — but now the $[A]$ side carries an extra power: $\frac{d [ A ]}{[ A ] ^{2}} = - k d t$ Add up (integrate) both sides. The left side sums to $- \frac{1}{[ A ]}$ — because the function $- 1/ [A]$ has slope $1/ [A]^{2}$ , so summing $1/ [A]^{2}$ slices rebuilds $- 1/ [A]$ . The right side sums to $- k t$ . Matching from the start $[A]_{0}$ : $- \frac{1}{[ A ] _{t}} + \frac{1}{[ A ] _{0}} = - k t ⟹ \frac{1}{[ A ] _{t}} = \frac{1}{[ A ] _{0}} + k t$

WHY the reciprocal straightens it. The natural bookkeeping quantity for an $[A]^{2}$ law is not $[A]$ itself but $1/ [A]$ — because integrating $1/ [A]^{2}$ produces $1/ [A]$ . So while $[A]_{t}$ traces a curvy plunge, its reciprocal $1/ [A]_{t}$ climbs in a perfect straight line of slope $k$ . Taking one-over-the-height is exactly the change of viewpoint that turns the messy curve into a line we can read off.

Now the half-life. Set height to half, so the reciprocal is $1/ [A]_{t} = 1/ ([A]_{0} /2) = 2/ [A]_{0}$ : $\frac{2}{[ A ] _{0}} = \frac{1}{[ A ] _{0}} + k t_{1/2} \Rightarrow k t_{1/2} = \frac{1}{[ A ] _{0}} \Rightarrow t_{1/2} = \frac{1}{k [ A ] _{0}}$

WHY the half-life grows. Squaring the height means the reaction is ferocious when crowded, feeble when sparse — two molecules must find each other to react. A bigger $[A]_{0}$ makes collisions frequent, so the first half-life is short. As $A$ thins out, collisions get rare and each new half-life stretches longer.

PICTURE. The steep plum curve. First half-life = a short span; second half-life = a span twice as long (since height halved, $t_{1/2} = 1/ (k [A])$ doubles).

Step 6 — The units must fit (a degenerate-case guard)

WHAT. Each order forces a different unit on $k$ so that $t_{1/2}$ comes out in seconds: $zero: k = \frac{M}{s}, first: k = \frac{1}{s}, second: k = \frac{1}{M \cdot s}$

WHY. Plug the wrong-unit $k$ into a formula and you get nonsense (a "half-life" measured in $M \cdot s^{2}$ , say). Checking units is a free error-catcher. For second order: $\frac{1}{( M ^{- 1} s ^{- 1} ) ( M )} = \frac{1}{s ^{- 1}} = s$ . ✓

PICTURE. A cancellation ledger showing the second-order units collapsing to seconds.

Step 7 — The degenerate case: what if $k = 0$ ?

WHAT. If the rate constant is zero, the reaction never happens. In every formula the half-life blows up: $[A]_{0} / (2 \cdot 0)$ , $0.693/0$ , $1/ (0 \cdot [A]_{0})$ — all infinite.

WHY. A curve that never falls never reaches half — correctly, "never" $= \infty$ . This is the sanity check that our formulas behave at the boundary. Likewise if $[A]_{0} = 0$ there is nothing to halve, and zero/second-order half-lives are already $0$ or undefined — an empty flask has no chemistry to time.

PICTURE. A flat line at height $[A]_{0}$ that never touches the dashed half-line — the arrow to $t_{1/2}$ runs off to infinity.

The one-picture summary

All three curves, same start, with their half-life spans marked: zero-order's spans shrink, first-order's stay equal, second-order's grow. Read the trend off the picture and you have named the order — this is the diagnostic used in order determination and integrated-rate-law analysis, and it underlies drug-elimination modelling.

Recall Feynman retelling — say it back in plain words

Picture how much reactant is left, dropping as a curve over time. The rate is just the steepness of that fall, $- d [A] / d t$ . Ask one question: how long to fall to half? If the fall speed is a fixed number (zero-order), the curve is a straight ramp and each half-life is shorter than the last — answer $[A]_{0} /2 k$ . If the fall speed is proportional to the height (first-order), separating variables and integrating gives $[A]_{0} e^{- k t}$ ; every halving takes the same time because the shrunk curve is a copy of the original — answer $0.693/ k$ , with the start height cancelling out. If the fall speed goes as height squared (second-order), integrating $1/ [A]^{2}$ gives the reciprocal $1/ [A]$ climbing in a straight line, and each half-life is longer, doubling each round — answer $1/ (k [A]_{0})$ . Guard rails: pick the formula whose $k$ -units give seconds, and if $k = 0$ the curve never falls so the half-life is infinite.

Recall Quick self-test

Zero-order half-life formula ::: $t_{1/2} = [A]_{0} / (2 k)$ — proportional to $[A]_{0}$ , so it decreases. First-order half-life formula ::: $t_{1/2} = 0.693/ k$ — independent of $[A]_{0}$ , constant. Second-order half-life formula ::: $t_{1/2} = 1/ (k [A]_{0})$ — inversely proportional to $[A]_{0}$ , so it increases. What does the symbol "rate" mean? ::: The instantaneous fall speed $- d [A] / d t$ , the slope of the concentration curve. Which order shows constant half-life? ::: First-order only.