Rate Laws and Measurements

Lessons

Senior 6 · Chemistry

Rate Laws and Measurements

5 lessons

Lesson 4 of 5

Concentration–Time Relationships and Half-Life

25 min

How much is left, and when?

A rate equation tells you the rate at one moment. Often you need something else: how much reactant will be left after an hour, or how long until a concentration falls to a safe level. A pharmacist planning doses asks this about a medicine in the blood, and a stores manager asks it about a bottle of hydrogen peroxide disinfectant. The answer comes from integrating the rate equation, which gives the integrated rate law: concentration as a function of time.

Definition

Integrated rate law: an equation for the concentration of a reactant as a function of time, obtained by integrating the rate equation. Example: for first order, $[\mathrm{A}] = [\mathrm{A}]_0\,e^{-kt}$ .

Half-life, $t_{\frac{1}{2}}$ : the time taken for the concentration of a reactant to fall to half of its value at the start of that interval. Example: $\mathrm{H_2O_2}$ with $k = 1.2 \times 10^{-3}\,\mathrm{s^{-1}}$ has $t_{\frac{1}{2}} = 578\,\mathrm{s}$ .

Pseudo-first-order reaction: a reaction of higher order that behaves as first order because every other reactant (or a catalyst) is in such large excess that its concentration stays practically constant. Example: the acid-catalysed hydrolysis of an ester in water, $\text{rate} = k[\text{ester}][\mathrm{H^{+}}] = k'[\text{ester}]$ with $k' = k[\mathrm{H^{+}}]$ .

Deriving the first-order law

For $\text{rate} = -\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]$ , separate the variables and integrate from $[\mathrm{A}]_0$ at $t = 0$ to $[\mathrm{A}]$ at time $t$ :

\int_{[\mathrm{A}]_0}^{[\mathrm{A}]} \frac{d[\mathrm{A}]}{[\mathrm{A}]} = -k\int_0^t dt \quad\Rightarrow\quad \ln\frac{[\mathrm{A}]}{[\mathrm{A}]_0} = -kt \quad\Rightarrow\quad \ln[\mathrm{A}] = \ln[\mathrm{A}]_0 - kt

At one half-life, $[\mathrm{A}] = \frac{1}{2}[\mathrm{A}]_0$ , so $\ln\frac{1}{2} = -kt_{\frac{1}{2}}$ , which gives $t_{\frac{1}{2}} = \frac{\ln 2}{k}$ . The half-life of a first-order reaction does not depend on the starting concentration: every half-life is the same length (Figure 4.1). The same law, with the decay constant $\lambda$ in place of $k$ , describes radioactive decay (Unit 14).

Figure 4.1: First-order decomposition of hydrogen peroxide (illustrative, $k = 1.2 \times 10^{-3}\,\mathrm{s^{-1}}$ ): three equal half-lives of 578 s.

The three integrated laws

	Zero order	First order	Second order ( $\text{rate} = k[\mathrm{A}]^2$ )
Integrated law	$[\mathrm{A}] = [\mathrm{A}]_0 - kt$	$\ln[\mathrm{A}] = \ln[\mathrm{A}]_0 - kt$	$\frac{1}{[\mathrm{A}]} = \frac{1}{[\mathrm{A}]_0} + kt$
Straight-line plot	$[\mathrm{A}]$ against $t$	$\ln[\mathrm{A}]$ against $t$	$\frac{1}{[\mathrm{A}]}$ against $t$
Gradient	$- k$	$- k$	$+ k$
Half-life	$\frac{[\mathrm{A}]_0}{2k}$	$\frac{\ln 2}{k}$	$\frac{1}{k[\mathrm{A}]_0}$
Successive half-lives	each half the one before	all equal	each double the one before

Figure 4.2: Plot the data three ways; the plot that gives a straight line shows the order.

Worked example 1: a first-order decomposition

Problem. Hydrogen peroxide, $[\mathrm{H_2O_2}]_0 = 0.80\,\mathrm{mol\,dm^{-3}}$ , decomposes with iodide as catalyst; the reaction is first order with $k = 1.2 \times 10^{-3}\,\mathrm{s^{-1}}$ (illustrative). Find (a) the half-life, (b) $[\mathrm{H_2O_2}]$ after 10.0 min, (c) the time for it to fall to $0.10\,\mathrm{mol\,dm^{-3}}$ .

Solution.

(a) $t_{\frac{1}{2}} = \frac{0.693}{1.2 \times 10^{-3}} = 578\,\mathrm{s}$ (9.6 min).

(b) $[\mathrm{H_2O_2}] = 0.80\,e^{-(1.2 \times 10^{-3})(600)} = 0.80\,e^{-0.72} = 0.80 \times 0.487 = 0.39\,\mathrm{mol\,dm^{-3}}$ .

(c) $0.80 \rightarrow 0.40 \rightarrow 0.20 \rightarrow 0.10$ is three half-lives: $3 \times 578 = 1730\,\mathrm{s}$ (about 29 min). By formula: $t = \frac{1}{k}\ln\frac{[\mathrm{A}]_0}{[\mathrm{A}]} = \frac{\ln 8}{1.2 \times 10^{-3}} = 1730\,\mathrm{s}$ .

Worked example 2: which plot is straight?

Problem. Nitrogen dioxide decomposes at a high temperature: $\mathrm{2NO_2(g) \rightarrow 2NO(g) + O_2(g)}$ . Use the data (illustrative) to find the order and $k$ .

$t$ / s	0	60	120	180	240
$[\mathrm{NO_2}]$ / $\mathrm{mol\,dm^{-3}}$	0.0100	0.00755	0.00607	0.00507	0.00436
$\ln[\mathrm{NO_2}]$	−4.605	−4.886	−5.104	−5.284	−5.435
$\frac{1}{[\mathrm{NO_2}]}$ / $\mathrm{dm^3\,mol^{-1}}$	100.0	132.5	164.7	197.2	229.4

Solution. In equal time steps $[\mathrm{NO_2}]$ does not fall by equal amounts (not zero order), and $\ln[\mathrm{NO_2}]$ changes by −0.281, −0.218, −0.180, −0.151 (not first order). $\frac{1}{[\mathrm{NO_2}]}$ rises by about 32.4 in every 60 s, so $\frac{1}{[\mathrm{NO_2}]}$ against $t$ is a straight line: second order.

k = \text{gradient} = \frac{229.4 - 100.0}{240} = 0.54\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}

Its first half-life is $\frac{1}{k[\mathrm{A}]_0} = \frac{1}{0.54 \times 0.0100} = 185\,\mathrm{s}$ ; the next would be twice as long.

Worked example 3: half-lives as an order test

Problem. For a reactant B, the concentration falls from 0.080 to $0.040\,\mathrm{mol\,dm^{-3}}$ in 50 s, and from 0.040 to $0.020\,\mathrm{mol\,dm^{-3}}$ in a further 25 s. State the order and find $k$ .

Solution. The second half-life (25 s) is half the first (50 s), so the reaction is zero order ( $t_{\frac{1}{2}} \propto [\mathrm{B}]_0$ ). The concentration falls at a constant $\frac{0.040}{50} = 8.0 \times 10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ , so $k = 8.0 \times 10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ . Check: $\frac{[\mathrm{B}]_0}{2k} = \frac{0.080}{2 \times 8.0 \times 10^{-4}} = 50\,\mathrm{s}$ .

Warning

Common mistakes

Using $t_{\frac{1}{2}} = \frac{\ln 2}{k}$ for a reaction that is not first order. Only first-order half-lives are independent of the starting concentration.
Taking $\log_{10}$ instead of $\ln$ in the first-order law. If you use $\log_{10}$ , the gradient is $-\frac{k}{2.303}$ .
Measuring a half-life from $t = 0$ each time. Each half-life starts where the previous one ended.
Mixing time units: if $k$ is in $\mathrm{s^{-1}}$ , $t$ must be in seconds.

Tip

To find the order from a single run, compare successive half-lives on the concentration–time curve: shrinking → zero order; constant → first order; growing (doubling) → second order.

Chad's Prep: Integrated Rate Laws, Zero, 1st and 2nd Order. Watch which quantity is plotted against time for each order, and how the half-life formulas come from the integrated laws.

✦

Practice

Show that the half-life of a first-order reaction is $\frac{\ln 2}{k}$ .
A first-order reaction has a half-life of 5.0 min. Calculate $k$ in $\mathrm{s^{-1}}$ .
A first-order reaction has $k = 2.5 \times 10^{-3}\,\mathrm{s^{-1}}$ . How long does it take for the reactant to fall from 0.30 to $0.20\,\mathrm{mol\,dm^{-3}}$ ?
In a first-order reaction, $[\mathrm{A}]$ falls from 0.0500 to $0.0125\,\mathrm{mol\,dm^{-3}}$ in 40 min. Find the half-life and $k$ .
A zero-order reaction has $k = 2.0 \times 10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ and $[\mathrm{A}]_0 = 0.050\,\mathrm{mol\,dm^{-3}}$ . Find its half-life and the time for A to be used up completely.
A medicine is removed from the body by first-order kinetics with a half-life of 6 hours. What percentage of a single dose remains after 24 hours?
Explain why the hydrolysis of an ester in a large excess of water, catalysed by acid, behaves as a first-order reaction, and how you would find the true rate constant.

Show answersHide answers

$\ln\frac{[\mathrm{A}]}{[\mathrm{A}]_0} = -kt$ ; put $[\mathrm{A}] = \frac{1}{2}[\mathrm{A}]_0$ : $\ln\frac{1}{2} = -kt_{\frac{1}{2}}$ , so $t_{\frac{1}{2}} = \frac{\ln 2}{k}$ .
$k = \frac{0.693}{300\,\mathrm{s}} = 2.3 \times 10^{-3}\,\mathrm{s^{-1}}$ .
$t = \frac{1}{k}\ln\frac{0.30}{0.20} = \frac{0.405}{2.5 \times 10^{-3}} = 162\,\mathrm{s}$ .
0.0500 → 0.0250 → 0.0125 is two half-lives, so $t_{\frac{1}{2}} = 20\,\mathrm{min}$ ; $k = \frac{0.693}{1200\,\mathrm{s}} = 5.8 \times 10^{-4}\,\mathrm{s^{-1}}$ .
$t_{\frac{1}{2}} = \frac{0.050}{2 \times 2.0 \times 10^{-4}} = 125\,\mathrm{s}$ ; used up when $kt = [\mathrm{A}]_0$ : $t = 250\,\mathrm{s}$ .
24 h is 4 half-lives: $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$ , i.e. 6.25 %.
The concentrations of water and of the $\mathrm{H^{+}}$ catalyst stay effectively constant, so $\text{rate} = k'[\text{ester}]$ with $k' = k[\mathrm{H^{+}}]$ (water absorbed into $k$ ). Measure $k^{'}$ from a $\ln[\text{ester}]$ plot, then divide by $[\mathrm{H^{+}}]$ ; repeating at other $[\mathrm{H^{+}}]$ confirms the order in acid.

Sources and credits

Video: Integrated Rate Laws, Zero, 1st and 2nd Order, 14.5 General Chemistry, Chad's Prep, embedded from YouTube.
The text, tables, graphs, illustrative data, worked examples and Practice items were written and drawn by Ganzaa.

1How much is left, and when?

2Deriving the first-order law

3The three integrated laws

How much is left, and when?

Deriving the first-order law

The three integrated laws