Integrated rate laws — half-life t₁ - ₂ for each order: Question bank

Before we start, one shared vocabulary reminder so no symbol sneaks in undefined:

Definition The quantities in play (and their units)

$[A]_{0}$ ::: the starting concentration of reactant A at the stopwatch's zero, measured in molar, $M$ (moles per litre).
$[A]_{t}$ ::: the concentration left after time $t$ has passed, also in $M$ .
$[A]_{start}$ ::: the concentration at the beginning of whichever half-life interval you are timing right now. For the first half-life it equals $[A]_{0}$ ; for the second it equals $[A]_{0} /2$ ; and so on. It is not always $[A]_{0}$ — that distinction is exactly what several traps below exploit.
$t_{1/2}$ ::: the half-life — the time for $[A]$ to fall to exactly half of $[A]_{start}$ for that interval.
$k$ ::: the rate constant. Its units change with order: zero-order $k$ is $M s^{- 1}$ , first-order $k$ is $s^{- 1}$ , second-order $k$ is $M^{- 1} s^{- 1}$ . See Rate constant k — units and temperature dependence.
The superscript in $t_{1/2}^{(0)}$ , $t_{1/2}^{(1)}$ , $t_{1/2}^{(2)}$ below is just a label naming the reaction order (0, 1, or 2). It is not a power — read " $t_{1/2}^{(2)}$ " out loud as "the second-order half-life."

Formula The three half-lives on one card — and why each looks the way it does

Each formula comes from setting $[A]_{t} = [A]_{start} /2$ in that order's integrated rate law. Here is the one-line "why" for each: $t_{1/2}^{(0)} = \frac{[ A ] _{0}}{2 k} (k : M s^{- 1})$ Zero-order drops in a straight line at fixed speed $k$ : $[A]_{0} - k t$ . Put the left side at half the start, $\frac{[ A ] _{0}}{2} = [A]_{0} - k t_{1/2}$ , so $k t_{1/2} = \frac{[ A ] _{0}}{2}$ — the time is proportional to how much you started with. $t_{1/2}^{(1)} = \frac{l n 2}{k} = \frac{0.693}{k} (k : s^{- 1})$ First-order decays exponentially: $[A]_{0} e^{- k t}$ . Setting $e^{- k t_{1/2}} = \frac{1}{2}$ and taking a logarithm to undo the exponential gives $- k t_{1/2} = ln \frac{1}{2} = - ln 2$ , so $t_{1/2} = \frac{l n 2}{k}$ — the $[A]_{0}$ has vanished, leaving a constant. $t_{1/2}^{(2)} = \frac{1}{k [ A ] _{start}} (k : M^{- 1} s^{- 1})$ Second-order's reciprocal grows linearly: $\frac{1}{[ A ] _{t}} = \frac{1}{[ A ] _{0}} + k t$ . Setting $\frac{1}{[ A ] _{0} /2} = \frac{2}{[ A ] _{0}}$ on the left gives $k t_{1/2} = \frac{2}{[ A ] _{0}} - \frac{1}{[ A ] _{0}} = \frac{1}{[ A ] _{0}}$ — inversely proportional to the concentration at the interval's start. Keep this in view — you should never need to leave the page to check which $k$ units belong to which order. Full derivations live in Integrated rate laws — derivation and graphical analysis.

The two figures below make these three "whys" visual — read the captions, then argue the reveals.

Figure 1 — three shapes of decay. All three curves start at the same concentration ( $1.0 M$ , top-left). The red line is zero-order: a straight diagonal that slams into zero at a finite time. The blue curve is first-order: an exponential that bends and only ever approaches zero. The green curve is second-order: a long lazy tail that decays fastest early (crowded molecules) then crawls. The dashed yellow line marks "half of the start" — where each curve crosses it is that reaction's first half-life.

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

Figure 2 — successive half-life widths. Each coloured block is one successive halving of $[A]$ (block 1 takes you from full to half, block 2 from half to quarter, and so on). Read the widths: the red (zero-order) blocks shrink, the blue (first-order) blocks stay equal, and the green (second-order) blocks double each round. Block width = the time that half-life takes, so the picture literally shows "Down, Flat, Up" for orders 0, 1, 2.

True or false — justify

Every reaction has a fixed half-life you can quote as one number

False. Only first-order reactions have a single constant half-life. Zero- and second-order half-lives depend on how much reactant you currently have.

For a first-order reaction, doubling

[A]_{0}

doubles the half-life

False. First-order

t_{1/2} = \frac{ln 2}{k}

contains no

[A]_{0}

at all, so doubling the start changes nothing.

For a zero-order reaction, starting with more reactant means a longer half-life

True.

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

is directly proportional to

[A]_{0}

; the rate is fixed at

k

, so halving a bigger pile takes proportionally longer.

For a second-order reaction, the second half-life is longer than the first

True. Each interval's half-life is

\frac{1}{k [ A ] _{start}}

, where

[A]_{start}

is the concentration at the start of that interval; since it halves each round, the next

t_{1/2}

doubles.

Radioactive decay having a constant half-life proves it is first-order

True. Constant

t_{1/2}

independent of amount is the unique fingerprint of first-order kinetics — see Radioactive decay — first-order kinetics.

Three successive first-order half-lives leave you at

[A]_{0} /6

False. Halving three times gives

(\frac{1}{2})^{3} = \frac{1}{8}

, i.e.

[A]_{0} /8

, not

[A]_{0} /6

. Half-lives multiply, they don't add up in the denominator.

For a zero-order reaction,

[A]

reaches exactly zero after a finite time and stops

True. The straight-line drop

[A]_{t} = [A]_{0} - k t

hits zero at

t = [A]_{0} / k

and the reaction halts abruptly — unlike first-order, which only approaches zero forever.

For a first-order reaction,

[A]

reaches zero after two half-lives

False. Two first-order half-lives leave

[A]_{0} /4

, and no finite number of halvings ever reaches zero — exponential decay only approaches it.

A larger

k

always means a shorter half-life

True for every order, because

k

sits in the denominator of all three half-life expressions (

\frac{[ A ] _{0}}{2 k}

\frac{l n 2}{k}

\frac{1}{k [ A ] _{0}}

). Faster constant, less time to halve.

Spot the error

"For a second-order reaction

t_{1/2} = \frac{0.693}{k}

Wrong — that is the first-order formula. Second-order is

t_{1/2} = \frac{1}{k [ A ] _{start}}

, which also depends on the concentration at the interval's start.

k = 0.5 s^{- 1}

, second-order, so

t_{1/2} = \frac{1}{k [ A ] _{start}}

The units give it away: a second-order

k

must be

M^{- 1} s^{- 1}

. A bare

s^{- 1}

means someone handed you a first-order constant, so this formula doesn't apply.

"Deriving first-order

t_{1/2}

: setting

[A]_{t} = [A]_{0} /2

gives

ln \frac{1}{2} = - k t_{1/2}

, so

t_{1/2} = - \frac{ln 2}{k}

— a negative time."

The sign slip is dropping that

ln \frac{1}{2} = - ln 2

. The two minus signs cancel:

- ln 2 = - k t_{1/2} \Rightarrow t_{1/2} = + \frac{ln 2}{k}

. Time comes out positive, as it must.

"Since half-life for first-order is constant, concentration drops by a constant amount each half-life."

Confuses fraction with amount. Each half-life removes a constant fraction (half of what remains), so the amount removed shrinks each round:

[A]_{0} \to \frac{[ A ] _{0}}{2} \to \frac{[ A ] _{0}}{4}

removes

\frac{[ A ] _{0}}{2}

then

\frac{[ A ] _{0}}{4}

"Zero-order

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

, so more reactant means faster halving."

Backwards. More

[A]_{0}

makes

t_{1/2}

bigger (longer), because the fixed rate must chew through a proportionally larger amount.

"I measured two half-lives, both 40 s, so I'll call it second-order."

Constant successive half-lives mean first-order, not second. For second-order the second half-life would have been double the first.

"For second-order,

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

works for every successive half-life."

Only for the first one. Each later interval must use

[A]_{start}

— the concentration at that interval's beginning — not the original

[A]_{0}

; the second half-life uses

[A]_{0} /2

, giving double the time.

t_{1/2} = \frac{1}{k [ A ] _{start}}

, so if

[A]_{start} = 0

the half-life is infinite — the reaction never happens."

[A]_{start} = 0

there is nothing to react; "half of zero" happens instantly and the formula's division-by-zero is just a signal that the question is meaningless, not that time is infinite.

Why questions

Why is the first-order half-life independent of concentration, while the other orders aren't?

Because first-order rate is proportional to

[A]

: as concentration drops, the rate drops in lock-step, so the fraction-halving time never changes. Zero-order has a fixed rate (mismatched to a shrinking amount); second-order rate falls even faster than

[A]

Why does

ln 2

appear in the first-order half-life but nowhere else?

Because first-order decay is exponential, and undoing an exponential needs a logarithm; setting

[A]_{t} = [A]_{0} /2

produces

ln \frac{1}{2} = - ln 2

, and that lone minus sign flips to give the positive

t_{1/2} = ln 2/ k

. See the derivation in Integrated rate laws — derivation and graphical analysis.

Why do second-order half-lives keep doubling as the reaction proceeds?

A second-order reaction needs two molecules to meet; as

[A]

falls, encounters get rarer roughly with the square, so each successive halving takes twice as long.

Why is measuring successive half-lives a good way to find reaction order?

The trend (constant, decreasing, or increasing) is a fingerprint no other order shares, so a couple of stopwatch readings distinguish orders without curve-fitting — see Reaction order determination from experimental data.

Why do pharmacologists love the first-order half-life so much?

Because a constant half-life lets them predict drug concentration at any time from one number, regardless of dose — the backbone of Pharmacokinetics — drug elimination.

Why can't we use "number of half-lives" shortcuts for zero- or second-order reactions?

Because the half-life changes each interval for those orders, so "3 half-lives =

\frac{1}{8}

" only works when every half-life is identical, i.e. first-order only.

Edge cases

What happens to a zero-order reaction's half-life as it nears completion?

It keeps shrinking with the remaining

[A]

, and the whole reaction stops dead the instant reactant hits zero (the rate stays

k

right up to the wall).

Does a first-order reaction ever truly reach

[A] = 0

No — exponential decay halves forever and only approaches zero. In practice we call it "done" after ~7 half-lives (

< 1%

left), but mathematically it never touches zero.

For a second-order reaction, what is the half-life when

[A]_{start}

is enormous?

Very short:

t_{1/2} = \frac{1}{k [ A ] _{start}}

shrinks as

[A]_{start}

grows, because crowded molecules collide constantly. The opposite limit (tiny

[A]_{start}

) gives an enormous half-life.

If you observe the first half-life doubling on the second interval, but tripling on a third, is it still simple second-order?

No — clean second-order doubles each time (a fixed geometric pattern). A shifting pattern signals a mixed or higher-order mechanism, not textbook second-order.

What order gives a half-life that is constant in amount removed per unit time rather than per half-life?

Zero-order: reactant disappears at a steady

k

mol per litre per second, a straight-line drop, so "amount removed per second" is fixed even though the half-life itself shrinks.

Can two reactions have the same

k

but wildly different half-lives?

Yes — if they are different orders (or different

[A]_{0}

for orders 0 and 2), because the half-life formula mixes

k

with

[A]_{0}

differently for each order.

Recall One-line self-test

Which single word describes each order's half-life trend as the reaction runs? ::: Zero-order = decreasing, first-order = constant, second-order = increasing.