Integrated Rate Equations & Half-Life

CHEIntegrated Rate Laws & Calculations

Integrated rate laws relate concentration to time, letting you find $k$ , predict how much reactant is left, or compute the time for any given completion. First-order kinetics dominate ISC numericals, so its integrated form and half-life expression must be memorised cold.

First-order integrated rate law

k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}

[A]_0

initial concentration,

[A]

concentration at time

t

k

s^{-1}

First-order half-life

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

independent of

[A]_0

Zero-order integrated rate law

k = \dfrac{[A]_0 - [A]}{t} \qquad t_{1/2} = \dfrac{[A]_0}{2k}

k

mol\,L^{-1}s^{-1}

;

t_{1/2}\propto[A]_0

Concentration remaining (first order)

[A] = [A]_0\,e^{-kt} = [A]_0\times 10^{-kt/2.303}

after

n

half-lives,

[A]=[A]_0/2^{\,n}

Completion-time relations (first order)

t_{75\%} = 2\,t_{1/2} \qquad t_{99.9\%} = 10\,t_{1/2}

each

t_{1/2}

halves what remains

First-order integrated form: $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ ; a plot of $\log[A]$ vs $t$ is linear with slope $-\dfrac{k}{2.303}$ .
First-order half-life: $t_{1/2} = \dfrac{0.693}{k}$ — independent of initial concentration (the key distinguishing feature of first order).
Zero-order integrated form: $k = \dfrac{[A]_0 - [A]}{t}$ ; half-life $t_{1/2} = \dfrac{[A]_0}{2k}$ , which is directly proportional to initial concentration.
Zero-order $k$ carries units of $mol\,L^{-1}s^{-1}$ (same as rate); a plot of $[A]$ vs $t$ is a straight line of slope $-k$ .
For first order, $t_{75\%} = 2\,t_{1/2}$ and $t_{99.9\%} = 10\,t_{1/2}$ — each successive half-life consumes half of what remains.
Percentage-completion trick: substitute $[A] = (100 - x)$ for $[A]_0 = 100$ , e.g. $90\%$ done means $\dfrac{[A]_0}{[A]} = \dfrac{100}{10} = 10$ .
Derivation outline (first order): $-\dfrac{d[A]}{dt}=k[A]$ ; integrating between $[A]_0$ and $[A]$ gives $\ln\dfrac{[A]_0}{[A]}=kt$ , then convert $\ln$ to $2.303\log$ .
Half-life is obtained by putting $[A]=[A]_0/2$ into the integrated law: first order gives $t_{1/2}=\dfrac{2.303}{k}\log 2=\dfrac{0.693}{k}$ .
The exponential form $[A]=[A]_0e^{-kt}$ shows a first-order reaction is asymptotic — it never reaches $100\%$ completion in finite time, only $99.9\%$ etc.
Zero-order reactions do reach completion in finite time: setting $[A]=0$ gives $t_{\text{complete}}=\dfrac{[A]_0}{k}$ (occurs e.g. on a saturated catalyst surface).
For order $n$ ( $n\neq 1$ ), $t_{1/2}\propto\dfrac{1}{[A]_0^{\,n-1}}$ — so half-life rises with $[A]_0$ for zero order and falls with $[A]_0$ for second order.
Watch units: a first-order $k$ given in $min^{-1}$ yields $t_{1/2}$ in minutes; keep $k$ and $t$ in the same time unit throughout.

Where the marks go

Assuming $t_{1/2}=0.693/k$ for every order — it is constant only for first order; for zero and second order half-life depends on $[A]_0$ .
Using $\ln$ and $\log$ inconsistently — the ISC form has the factor $2.303$ because it uses $\log_{10}$ ; dropping or adding $2.303$ is a frequent slip.
Treating ' $90\%$ completed' as $[A]_0/[A]=90$ instead of $100/10=10$ — always use concentration remaining, not consumed.
Confusing zero-order units ( $mol\,L^{-1}s^{-1}$ ) with first-order units ( $s^{-1}$ ) when identifying the order from a given $k$ .
Forgetting the negative sign / slope direction: a $\log[A]$ vs $t$ plot has slope $-k/2.303$ , not $+k/2.303$ .

How the board asks it

Numericalfirst-order integrated rate law
A first-order reaction is $30\%$ complete in $30\,min$ . Calculate the rate constant $k$ and the time required for the reaction to be $90\%$ complete.
Numericalfirst-order half-life and successive half-lives
The half-life of a first-order reaction is $69.3\,s$ . Calculate its rate constant $k$ , and find the time taken for the concentration to fall to $\dfrac{1}{8}$ of its initial value.
Derive / provederivation of first-order integrated form and half-life
Derive the integrated rate equation $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ for a first-order reaction, and hence obtain the expression for its half-life $t_{1/2}$ .
Give reasonshalf-life dependence on initial concentration
Account for the fact that the half-life of a first-order reaction is independent of the initial concentration, whereas that of a zero-order reaction is directly proportional to it.
Distinguishzero-order vs first-order kinetics
How will you distinguish between a zero-order and a first-order reaction on the basis of their half-life expressions and the units of their rate constant $k$ ?
Numericalzero-order integrated form
For a zero-order reaction, the initial concentration is $0.20\,mol\,L^{-1}$ and $k = 2.0\times10^{-2}\,mol\,L^{-1}s^{-1}$ . Calculate the half-life and the time for the reaction to go to completion.

Practise this topic

PreviousUnits of k and Determining Order from Units NextTime for Given Percentage Completion

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.