Chemistry · General chemistry II · Concept

Integrated rate laws and half-life

An integrated rate law gives a reactant’s concentration at any time from its starting concentration and $k$ . For a first-order reaction $[A] = {[A]}_{0} e^{- k t}$ , and the half-life, $0.693 / k$ , is the same however much is left, which is why radioactive decay is described by half-lives.

A concrete case

A medicine that halves every six hours

Many drugs leave the blood by first-order kinetics. Say a patient’s blood holds $80 mg / L$ of a drug whose half-life is $6.0 hours$ . After $6 hours$ , $40 mg / L$ is left, after $12 hours$ , $20 mg / L$ , and after $18 hours$ , $10 mg / L$ .

Each 6-hour stretch removes half of whatever is there. The first removes $40 mg / L$ and the fourth only $5 mg / L$ . The level never reaches zero in a fixed number of steps.

At the start the level falls by about $9.2 mg / L$ per hour. At that pace the drug would be gone in under $9 hours$ , yet after $9 hours$ about $28 mg / L$ is still there. The rate shrinks as the drug runs down, so a rate times a time cannot give the amount left. The integrated rate law can.

A first-order drug with a 6.0-hour half-life
Time $(h)$	Level $(mg / L)$	Removed in the last $6 h$ $(mg / L)$
$0$	$80$	–
$6$	$40$	$40$
$12$	$20$	$20$
$18$	$10$	$10$
$24$	$5$	$5$

The rule

The three integrated rate laws

Each law is for one reactant, $A \to products$ , whose rate law is $rate = k {[A]}^{n}$ with $n = 0$ , $1$ or $2$ . ${[A]}_{0}$ is the concentration at the start, $t = 0$ , and $[A]$ is the concentration at time $t$ . $k$ must use the same time unit as $t$ .

For a first-order reaction, $ln [A]$ falls in a straight line with time, $ln [A] = ln {[A]}_{0} - k t$ . Undoing the logarithm gives the exponential form, $[A] = {[A]}_{0} e^{- k t}$ . For second order it is $1 / [A]$ that rises in a straight line, and for zero order $[A]$ itself falls in a straight line until A runs out.

ln [A] = ln {[A]}_{0} - k t, [A] = {[A]}_{0} e^{- k t}

\begin{array}{c} ln [A] = ln {[A]}_{0} - k t \\ [A] = {[A]}_{0} e^{- k t} \end{array}

Integrated rate laws for one reactant
Order	Rate law	Integrated law	Half-life	Units of $k$
$0$	$rate = k$	$[A]$ $= {[A]}_{0} - k t$	${[A]}_{0} \div 2 k$	$M s^{- 1}$
$1$	$rate$ $= k [A]$	$ln [A]$ $= ln {[A]}_{0} - k t$	$0.693 \div k$	$s^{- 1}$
$2$	$rate$ $= k {[A]}^{2}$	$1 / [A]$ $= k t + 1 / {[A]}_{0}$	$1 \div (k {[A]}_{0})$	$M^{- 1} s^{- 1}$

Half-life

The half-life, $t_{1 / 2}$ , is the time for $[A]$ to fall to half its value. For a first-order reaction it is $ln 2 \div k$ , which is $0.693 \div k$ , and it does not depend on how much A there is. After $n$ half-lives the fraction left is ${(\frac{1}{2})}^{n}$ .

The other orders have half-lives that change as the reaction goes. A second-order half-life, $1 \div (k {[A]}_{0})$ , doubles each time, because ${[A]}_{0}$ halves. A zero-order half-life, ${[A]}_{0} \div 2 k$ , halves each time.

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

Why it works

Why first order means a constant half-life

In a first-order reaction the rate is proportional to how much A is left. In each minute the same fraction of the remaining A reacts, whether there is a lot or a little. Losing a fixed fraction in each equal time is exponential decay, like a balance that shrinks by a fixed percentage every year.

Calculus makes this exact. The rate law, $- d [A] / d t = k [A]$ , separates into $d [A] / [A] = - k d t$ . Integrating from time $0$ to time $t$ gives $ln [A] - ln {[A]}_{0} = - k t$ , the first-order law.

The half-life comes from the same law. Put $[A] = {[A]}_{0} / 2$ and ${[A]}_{0}$ cancels, leaving $ln 2 = k t_{1 / 2}$ . Nothing in that equation depends on the starting amount, so every halving takes the same time.

A second-order reaction slows faster, because its rate goes as ${[A]}^{2}$ . Integrating $- d [A] / d t = k {[A]}^{2}$ gives $1 / [A] = k t + 1 / {[A]}_{0}$ . Its half-life, $1 / (k {[A]}_{0})$ , grows as ${[A]}_{0}$ shrinks, so each halving takes twice as long as the one before.

\frac{d [A]}{[A]} = - k d t \Rightarrow ln [A] - ln {[A]}_{0} = - k t

\begin{array}{c} \frac{d [A]}{[A]} = - k d t \\ \Rightarrow ln [A] - ln {[A]}_{0} = - k t \end{array}

The method

Solving an integrated rate law

1. Find the order. The problem may state it, or the units of $k$ give it: $M s^{- 1}$ for zero order, $s^{- 1}$ for first and $M^{- 1} s^{- 1}$ for second.

2. Put $k$ and the time in the same unit.

3. Write the law for that order, and solve it for the unknown before putting in numbers.

4. Substitute with units. In the first-order law, $k t$ must come out as a pure number.

5. Finish the algebra. Take $e$ to the power for $[A]$ , or invert $1 / [A]$ in a second-order problem.

6. Check the answer against the half-lives.

Counting half-lives

When a time is a whole number of half-lives, skip the exponential. Divide the time by the half-life to get $n$ , the number of half-lives, and the amount left is ${(\frac{1}{2})}^{n}$ of the start. Three half-lives leave one eighth, and ten leave about a thousandth.

Going backward works the same way. If $80.0 mg$ has fallen to $5.00 mg$ , the ratio is $16 = 2^{4}$ , so four half-lives have passed. For a fraction that is not a power of $\frac{1}{2}$ , use $k$ and the logarithmic form.

m = m_{0} {(\frac{1}{2})}^{n}, n = \frac{t}{t_{1 / 2}}

Where it breaks

The laws assume one reactant sets the rate, the temperature stays constant and the reaction does not run back. Near equilibrium the reverse reaction slows the loss of A, and the laws predict too little A.

Subtracting $k t$ is the zero-order law only. For the first-order example below, $0.880 - (0.0410) (30.0)$ gives $- 0.35$ , which is not a concentration at all. The true answer is $0.257 M$ .

A zero-order law stops at zero. $[A] = {[A]}_{0} - k t$ turns negative after $t = {[A]}_{0} / k$ , when the reaction has simply ended.

The second-order law in the table is for $rate = k {[A]}^{2}$ . A reaction with $rate = k [A] [B]$ follows it only when A and B start at the same concentration and are used one for one.

A constant half-life belongs to first order alone. Using $0.693 / k$ for a second-order reaction gives one number where each half-life is really twice the one before.

Radioactive dating assumes the sample began with the same share of the isotope that living things have, and lost none of it except by decay. Carbon-14 dating fades out near $50, 000 years$ , almost nine half-lives, when too little is left to measure.

Check your answer

Count half-lives. Divide the time by the half-life and halve the start that many times. $30.0 \min$ with $t_{1 / 2} = 16.9 \min$ is just under two half-lives, so a little more than a quarter should be left. $0.257 M$ is $29$ percent of $0.880 M$ .

Check the direction and the range. $[A]$ must fall, and it must stay between $0$ and ${[A]}_{0}$ . A concentration above the start means a lost minus sign in the exponent.

Check the units. $k t$ has no units in a first-order problem, so $k$ in $\min^{- 1}$ needs $t$ in minutes. In a second-order problem $k t$ comes out in $M^{- 1}$ , the units of $1 / [A]$ .

In a Chemistry box, write each given on its own row, such as ${[A]}_{0} = 0.880 M$ , $k = 0.0410 \min^{- 1}$ and $t = 30.0 \min$ . Then write the law, then the substitution. In the first example’s box the Checker reads the substitution as “Substitution matches row 4 with the values defined above.” A half-life worked out under $t_{1 / 2} = ln 2 / k$ reads ✓ the same way.

Whole half-lives have their own rows. $n = 37.2 h \div 12.4 h$ $= 3.00$ reads ✓, and the Checker says “The number of half-lives is the time over the half-life.”

The row $m = 80.0 mg \div 8$ $= 10.0 mg$ then reads “After n half-lives, m = m₀ × (½)ⁿ.”

For a time that is not a whole number of half-lives, write the power with the time in its exponent, $m = m_{0} {(\frac{1}{2})}^{t / t_{1 / 2}}$ , and $m_{0}$ , $t_{1 / 2}$ and $t$ on rows of their own. With $80.0 mg$ , $t_{1 / 2} = 12.4 h$ and $t = 30.0 h$ , $m = (80.0 mg) {(1 / 2)}^{30.0 / 12.4}$ $= 15.0 mg$ reads ✓, and $20.0 mg$ reads ✗. Using $k$ and $[A] = {[A]}_{0} e^{- k t}$ works too.

Lifetime, flooding and dating

A first-order process also has a mean lifetime, $τ = 1 / k$ , the time for $[A]$ to fall to $1 / e$ of its value, about $37$ percent. $τ$ is $1.443$ times the half-life. IUPAC asks that the two never be confused, because both are quoted for radioactive isotopes and excited molecules.

Many reactions with two reactants are run with one in large excess. Its concentration hardly changes, so the reaction follows the first-order law in the other reactant, with an observed $k$ that contains the excess concentration. Chemists call this pseudo-first order.

Carbon-14 has two half-lives in use. Decay arithmetic, here and in most courses, uses $5730 years$ , the mean of three measurements adopted at the $1962$ radiocarbon conference in Cambridge, and the current evaluation is $5700 \pm 30 years$ . Radiocarbon laboratories report a conventional radiocarbon age with Libby’s older $5568 years$ instead, by an international agreement of 1977, so that every published age sits on one scale. That age is then calibrated against tree rings of known age, which also corrects for changes in the atmosphere’s carbon-14 over time.

Common mistakes

Using $0.693 / k$ for a zero- or second-order reaction. For the second-order example, $0.693 / k$ gives $1.28 M s$ , not even a time, while the first half-life is $1 / (k {[A]}_{0}) = 185 s$ . Only a first-order half-life is free of the starting concentration.
Mixing time units. With $k = 0.0410 \min^{- 1}$ , a time of $30.0 \min$ goes in as $30.0 \min$ , not as $1800 s$ . $k t$ has to be a pure number, and minutes cancel only against minutes.
Dropping the minus sign in the exponent. $(0.880 M) e^{+ 1.23}$ $= 3.01 M$ , more than the reaction started with. The law is $[A] = {[A]}_{0} e^{- k t}$ , because a reactant can only decrease.
Subtracting $k t$ in a first-order problem. $0.880 - (0.0410) (30.0)$ $= - 0.35$ , which cannot be a concentration. ${[A]}_{0} - k t$ is the zero-order law, and a first-order reactant falls by a fraction, not a fixed amount.
Dividing by $2 n$ instead of halving $n$ times. After $3$ half-lives, $80.0 mg$ becomes $80.0 \div 2^{3} = 10.0 mg$ , not $80.0 \div 6 = 13.3 mg$ . Each half-life halves what the one before left.
Stopping at $1 / [A]$ in a second-order problem. $1 / [A] = 154 M^{- 1}$ is the reciprocal, not the concentration. Inverting it gives $[A] = 0.00649 M$ .
Using $log$ , base $10$ , where the laws need $ln$ . $log (0.880 / 0.110) \div 0.0410$ gives $22.0 \min$ , but the time is $50.7 \min$ . The integrated laws come from $e$ , so they need natural logarithms.

Key terms

Integrated rate law: An equation for concentration versus time, derived from the rate law: $ln [A] = - k t + ln {[A]}_{0}$ for first order, $[A] = {[A]}_{0} - k t$ for zero order and $1 / [A] = k t + 1 / {[A]}_{0}$ for second order. Here $k$ is the rate constant, $t$ is the time and ${[A]}_{0}$ is the starting concentration. Each order gives its own straight-line plot.
Half-life: The time it takes for an amount to fall to half its starting value. For an amount that decays as $A_{0} e^{- k t}$ with $k > 0$ , it is $ln 2 \div k$ , whatever the starting amount. For a first-order reaction, $[A] = {[A]}_{0} e^{- k t}$ , so $t_{1 / 2} = ln 2 \div k$ , with $k$ the rate constant.
Reaction order: The exponent on a concentration in the rate law (the order in that reactant), or the sum of all the exponents (the overall order). Orders come from experiments, not from the balanced equation.
Rate law: An equation linking reaction rate to reactant concentrations, such as $rate = k {[A]}^{2} [B]$ . The units of the rate constant $k$ depend on the overall order.
Radioactivity: The decay of an unstable nucleus into another nucleus, with the emission of particles such as alpha or beta particles or of gamma radiation.
Radiocarbon dating: Estimating the age of once-living material from how much carbon-14 (half-life about $5730 years$ ) remains compared with living organisms.

Work through an example

A compound A decomposes in solution by first-order kinetics with $k = 0.0410 \min^{- 1}$ . A solution starts at ${[A]}_{0} = 0.880 M$ . Find $[A]$ after $30.0 \min$ , the half-life, and the time at which $[A]$ has fallen to $0.110 M$ .

Use the first-order integrated rate law →

Count half-lives in radioactive decay →

Use the second-order integrated rate law →

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