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

Definition The symbols we will use everywhere

$t$ — time, measured from the moment the reaction starts ( $t = 0$ ), in seconds ( $s$ ), minutes, hours, or days. Every formula on this page tracks how concentration changes as $t$ grows.
$[A]_{0}$ — the starting concentration of reactant A (how much we begin with, i.e. its value at $t = 0$ ), in mol per litre, written $M$ .
$[A]_{t}$ — the concentration left after time $t$ (the subscript $t$ just says "measured at time $t$ ").
$k$ — the rate constant, the number that sets the speed. Its units depend on the order (see Rate constant k — units and temperature dependence).
$t_{1/2}$ — the half-life, the special value of $t$ at which the concentration has fallen to exactly half of what it was.
$ln$ — the natural logarithm, the tool that "undoes" the exponential $e^{- k t}$ . We use it only for first order because only first order gives an exponential decay.

Level 1 — Recognition

Goal: can you name the order from its fingerprint, and read the right formula?

Exercise 1.1

A reaction is monitored and its successive half-lives are measured: 40 s, 40 s, 40 s, 40 s. What is the reaction order?

Recall Solution

A half-life that never changes as the reaction runs is the unique signature of first order.

Why not zero order? Zero-order half-lives shrink over time (each depends on the current concentration, which is falling).
Why not second order? Second-order half-lives grow over time (each is inversely proportional to the current concentration).
Only first order keeps $t_{1/2} = \frac{0.693}{k}$ — no concentration inside it, so nothing to change.

Answer: first order.

Exercise 1.2

For a zero-order reaction, write down the half-life formula and state in one sentence what happens to $t_{1/2}$ if you start with a larger $[A]_{0}$ .

Recall Solution

$t_{1/2} = \frac{[ A ] _{0}}{2 k}$ Since $[A]_{0}$ sits on top, a larger starting concentration gives a larger half-life — they are directly proportional. Physically: a zero-order reaction consumes reactant at a fixed rate $k$ , so twice as much reactant takes twice as long to lose half of it.

Exercise 1.3

A first-order reaction has $k = 0.0250 s^{- 1}$ . Just from the units of $k$ , confirm the order, then find $t_{1/2}$ .

Recall Solution

Units of $k$ are $s^{- 1}$ (just inverse time). That is exactly the first-order unit — so this is first order. $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0250 s ^{- 1}} = 27.7 s$

Level 2 — Application

Goal: plug numbers in correctly, keeping units honest.

Exercise 2.1

A first-order decomposition has $k = 1.15 \times 1 0^{- 3} s^{- 1}$ . Find its half-life in seconds and in minutes.

Recall Solution

$t_{1/2} = \frac{0.693}{1.15 \times 1 0 ^{- 3} s ^{- 1}} = 602.6 s$ In minutes: $602.6 s \div 60 = 10.04 min$ .

Exercise 2.2

A zero-order surface reaction has $k = 0.020 M s^{- 1}$ and $[A]_{0} = 0.50 M$ . Find $t_{1/2}$ , and check the units cancel to seconds.

Recall Solution

$t_{1/2} = \frac{[ A ] _{0}}{2 k} = \frac{0.50 M}{2 ( 0.020 M s ^{- 1} )} = \frac{0.50 M}{0.040 M s ^{- 1}} = 12.5 s$ Unit check: $\frac{M}{M s ^{- 1}} = \frac{1}{s ^{- 1}} = s$ . ✓ The M's cancel, leaving seconds.

Exercise 2.3

A second-order reaction has $k = 0.35 M^{- 1} s^{- 1}$ and $[A]_{0} = 0.80 M$ . Find the first half-life.

Recall Solution

$t_{1/2} = \frac{1}{k [ A ] _{0}} = \frac{1}{( 0.35 M ^{- 1} s ^{- 1} ) ( 0.80 M )} = \frac{1}{0.28 s ^{- 1}} = 3.57 s$ Unit check: $M^{- 1} s^{- 1} \times M = s^{- 1}$ ; one over that is $s$ . ✓

Level 3 — Analysis

Goal: reason about how successive half-lives behave, and read them backwards.

Exercise 3.1

A second-order reaction starts at $[A]_{0} = 0.40 M$ with $k = 0.50 M^{- 1} s^{- 1}$ . (a) First half-life? (b) Concentration after it? (c) Second half-life? (d) What is the ratio (second half-life)/(first half-life)?

Recall Solution

(a) $t_{1/2, 1} = \frac{1}{k [ A ] _{0}} = \frac{1}{( 0.50 ) ( 0.40 )} = \frac{1}{0.20} = 5.0 s$ (b) After one half-life, $[A] = \frac{0.40}{2} = 0.20 M$ . (c) Now treat $0.20 M$ as the new starting concentration: $t_{1/2, 2} = \frac{1}{k [ A ]} = \frac{1}{( 0.50 ) ( 0.20 )} = \frac{1}{0.10} = 10.0 s$ (d) Ratio $= 10.0/5.0 = 2$ . Each successive second-order half-life doubles, because the concentration inside $\frac{1}{k [ A ]}$ halved.

What the figure below shows (alt text: a concentration-vs-time decay curve in black, with two red double-headed horizontal arrows on the time axis marking the first half-life stretch from 0 to 5 s and the second from 5 to 15 s; dashed guide lines drop from the concentration levels 0.40, 0.20, 0.10 M to the curve): the black curve is $[A]$ falling over time $t$ . The two red double-headed arrows mark the two half-life stretches on the time axis. Notice the second red arrow (the $0.20 \to 0.10 M$ leg) is twice as wide as the first (the $0.40 \to 0.20 M$ leg): as the concentration gets thinner, molecules collide less often, so each successive halving takes longer. This visually confirms the ratio of 2 you just computed.

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

Exercise 3.2

A first-order reaction has $t_{1/2} = 25 s$ . Starting from $1.00 M$ , what concentration remains after 100 s? Answer without a calculator by counting half-lives.

Recall Solution

$100 s \div 25 s per half-life = 4$ half-lives. Each half-life multiplies the concentration by $\frac{1}{2}$ : $1.00 \to 0.50 \to 0.25 \to 0.125 \to 0.0625 M$ After 4 half-lives, $[A] = 1.00 \times (\frac{1}{2})^{4} = \frac{1}{16} = 0.0625 M$ .

Exercise 3.3

Given a zero-order reaction with $[A]_{0} = 1.0 M$ , the first half-life is 20 s. Find the second half-life (time to go from $0.5 M$ to $0.25 M$ ).

Recall Solution

From the first half-life, find $k$ : $t_{1/2} = \frac{[ A ] _{0}}{2 k} \Rightarrow k = \frac{[ A ] _{0}}{2 t _{1/2}} = \frac{1.0}{2 ( 20 )} = 0.025 M s^{- 1}$ . The second half-life uses the new starting concentration $0.5 M$ : $t_{1/2, 2} = \frac{[ A ]}{2 k} = \frac{0.5}{2 ( 0.025 )} = \frac{0.5}{0.05} = 10 s$ The zero-order second half-life is half the first — half-lives shrink. Contrast this with second order, where they grow.

Level 4 — Synthesis

Goal: combine order-diagnosis with calculation, and cross tools.

Exercise 4.1

Two experiments on the same reaction:

Exp 1: $[A]_{0} = 1.00 M$ , first $t_{1/2} = 100 s$ .
Exp 2: $[A]_{0} = 2.00 M$ , first $t_{1/2} = 50 s$ . Determine the order and then the rate constant $k$ .

Recall Solution

Diagnose: doubling $[A]_{0}$ (×2) halved $t_{1/2}$ (×½). That inverse relationship $t_{1/2} \propto 1/ [A]_{0}$ is the signature of second order.

(Zero order would have doubled $t_{1/2}$ ; first order would leave it unchanged.) Find $k$ using either experiment. From Exp 1: $t_{1/2} = \frac{1}{k [ A ] _{0}} \Rightarrow k = \frac{1}{t _{1/2} [ A ] _{0}} = \frac{1}{( 100 ) ( 1.00 )} = 0.0100 M^{- 1} s^{- 1}$ Check with Exp 2: $k = \frac{1}{( 50 ) ( 2.00 )} = 0.0100 M^{- 1} s^{- 1}$ . ✓ Consistent. This is the half-life method of finding order.

Exercise 4.2

A radioactive isotope decays by first-order kinetics with a half-life of 8.0 days. A patient is given a dose; what fraction remains after 24 days, and what is $k$ in $day^{- 1}$ ?

Recall Solution

$24 days \div 8.0 = 3$ half-lives, so fraction remaining $= (\frac{1}{2})^{3} = \frac{1}{8} = 0.125$ . Rate constant: $k = \frac{0.693}{t _{1/2}} = \frac{0.693}{8.0 day} = 0.0866 day^{- 1}$ This is the reasoning behind drug elimination schedules.

Exercise 4.3

A first-order reaction is 75% complete in 60 minutes. Find $k$ and the half-life. (Hint: 75% complete means $[A]_{t} = 0.25 [A]_{0}$ .)

Recall Solution

$75%$ gone leaves $25% = \frac{1}{4}$ , i.e. two half-lives (each halving: $1 \to \frac{1}{2} \to \frac{1}{4}$ ). So $2 t_{1/2} = 60 min \Rightarrow t_{1/2} = 30 min$ . $k = \frac{0.693}{30 min} = 0.0231 min^{- 1}$ Cross-check via the integrated law $ln [A]_{t} = ln [A]_{0} - k t$ : $k = \frac{1}{t} ln \frac{[ A ] _{0}}{[ A ] _{t}} = \frac{1}{60} ln 4 = \frac{1.386}{60} = 0.0231 min^{- 1} ✓$

Level 5 — Mastery

Goal: full mixed reasoning, degenerate cases, and unit thinking.

Exercise 5.1

An enzyme reaction runs at zero order while substrate is plentiful, with $k = 0.10 M s^{- 1}$ and $[A]_{0} = 1.0 M$ . It stays zero order until $[A]$ drops to $0.20 M$ , below which it switches to first order with a new rate constant $k^{'} = 0.50 s^{- 1}$ (the prime symbol $^{'}$ just labels this as the second-phase rate constant, distinct from the zero-order $k$ above; note its units $s^{- 1}$ confirm it is first order). How long, in total, to reach $0.10 M$ ?

Recall Solution

Phase 1 (zero order, $1.0 \to 0.20 M$ ). We need the zero-order integrated law, so let us build it first.

Zero order means the rate is constant: the concentration loses $k$ units of M every second. Why: by definition of zero order the rate does not depend on $[A]$ , so it is just the fixed number $k$ .
Constant loss rate means the concentration falls in a straight line: after time $t$ we have removed $k \times t$ , so $[A]_{t} = [A]_{0} - k t$ (This is the same law quoted in the integrated rate law note.)
Rearranging for the time to fall from $[A]_{0}$ to a target $[A]_{t}$ : subtract, then divide by $k$ : $t = \frac{[ A ] _{0} - [ A ] _{t}}{k} = \frac{1.0 - 0.20}{0.10} = 8.0 s$ Phase 2 (first order, $0.20 \to 0.10 M$ ): this is exactly one halving ( $0.20 \to 0.10$ ), so we use the first-order half-life with $k^{'}$ : $t_{2} = t_{1/2} = \frac{0.693}{k ^{'}} = \frac{0.693}{0.50} = 1.386 s$ Total: $t = 8.0 + 1.386 = 9.39 s$ . This mixed behaviour is why order can appear to change over the course of one reaction.

Exercise 5.2

For a first-order reaction, prove that the time to fall to $\frac{1}{n}$ of the initial value is $t = \frac{ln n}{k}$ , and use it to find the time to reach $10%$ remaining when $k = 0.020 s^{- 1}$ .

Recall Solution

Start from the first-order integrated law $ln [A]_{t} = ln [A]_{0} - k t$ . Why: it links concentration to time for first order. Set $[A]_{t} = [A]_{0} / n$ (the target "one $n$ -th" condition): $ln \frac{[ A ] _{0}}{n} = ln [A]_{0} - k t$ Apply the log-of-a-quotient rule $ln (\frac{a}{b}) = ln a - ln b$ (the logarithm turns division into subtraction — that is exactly why logs are the right tool here). With $a = [A]_{0}$ , $b = n$ : $ln [A]_{0} - ln n = ln [A]_{0} - k t$ Subtract $ln [A]_{0}$ from both sides. Why: it appears identically on both sides, so removing it isolates the terms that actually depend on $t$ and $n$ : $- ln n = - k t$ Multiply both sides by $- 1$ (to clear the minus signs), then divide by $k$ (to isolate $t$ ): $k t = ln n \Rightarrow t = \frac{l n n}{k}$ Sanity check: setting $n = 2$ gives $t = \frac{ln 2}{k}$ , the familiar half-life. ✓ Numerical application — time to reach $10%$ remaining. " $10%$ remaining" means $[A]_{t} = \frac{1}{10} [A]_{0}$ , so $n = 10$ : $t = \frac{l n 10}{k} = \frac{l n 10}{0.020 s ^{- 1}} = \frac{2.3026}{0.020} = 115.1 s$

Exercise 5.3 (degenerate / limiting case)

A zero-order reaction with $[A]_{0} = 0.60 M$ and $k = 0.15 M s^{- 1}$ . (a) When does the reactant run out completely ( $[A] = 0$ )? (b) Explain why the second-order and first-order laws never allow $[A]$ to reach exactly zero, but zero order does.

Recall Solution

(a) $[A]_{t} = [A]_{0} - k t = 0 \Rightarrow t = \frac{[ A ] _{0}}{k} = \frac{0.60}{0.15} = 4.0 s$ . Note this equals $2 t_{1/2}$ since $t_{1/2} = [A]_{0} /2 k = 2.0 s$ . (b) First order gives $[A]_{t} = [A]_{0} e^{- k t}$ ; the exponential $e^{- k t}$ shrinks toward 0 but is never exactly 0 for finite $t$ — so first order approaches, never reaches, empty. Second order gives $1/ [A]_{t} = 1/ [A]_{0} + k t$ , which grows to infinity only as $t \to \infty$ , so $[A]_{t} \to 0$ only in the infinite-time limit. Zero order is the exception: its straight-line decay $[A]_{0} - k t$ hits zero at the finite time $[A]_{0} / k$ , then the model must stop (concentration can't go negative).

What the figure below shows (alt text: three decay curves starting from the same point; two black curves — a solid first-order exponential and a dashed second-order curve — both bend and flatten toward but never touch the time axis; one red straight line for zero order slopes down and hits the time axis at a marked red dot labelled t = [A]0 / k): the two black curves (first order, solid; second order, dashed) both bend and flatten, forever approaching the time-axis without touching it — visual proof they never truly empty. The red straight line (zero order) crashes into the axis at a single finite point, marked with a red dot at $t = [A]_{0} / k$ : that is the true "finish line" only zero order has.

Exercise 5.4

A drug is eliminated by first-order kinetics with $t_{1/2} = 4.0 h$ . To stay effective, its plasma level must not fall below $25%$ of the peak dose. What is the maximum safe dosing interval before the next dose is needed?

Recall Solution

$25% = \frac{1}{4} = (\frac{1}{2})^{2}$ , i.e. exactly two half-lives (each halving: $100% \to 50% \to 25%$ ). $t = 2 \times t_{1/2} = 2 \times 4.0 = 8.0 h$ So the drug should be redosed at most every 8.0 hours. This is the everyday logic behind real dosing schedules.

Recall Self-test — cover the answers

Constant half-life means order ::: first order Zero-order half-life formula ::: $t_{1/2} = [A]_{0} /2 k$ Second-order half-life formula ::: $t_{1/2} = 1/ (k [A]_{0})$ If doubling $[A]_{0}$ halves $t_{1/2}$ , the order is ::: second order "75% complete" means the remaining fraction is ::: 0.25 (one quarter) Time to reach $1/ n$ for first order ::: $t = ln n / k$ The only order whose $[A]$ reaches exactly zero at finite time ::: zero order