University Chemistry — Year 1 · Bachelor Year 1

8Chemical Kinetics: Rate Laws

Milk keeps a week in the refrigerator and a day on a summer table. The box of a medicine says “store below $25\,{}^{\circ}\mathrm{C}$ ” and carries an expiry date two or three years ahead, which the manufacturer had no time to observe directly. Equilibrium constants say where a reaction ends; they say nothing of how long it takes to get there. That is the domain of chemical kinetics. This chapter defines the rate of a reaction, the rate laws that express it in terms of concentrations, the integrated forms that predict concentrations at any time, the experimental methods that determine a rate law, and the Arrhenius law that describes the effect of temperature — the tool by which the shelf life of a medicine is predicted in a few weeks of experiments.

You already know

Book 1 (grade 12) introduced the rate of appearance and disappearance of a species, the kinetic factors (concentration, temperature, catalyst) and the half-life of a first-order reaction. The extent $\xi$ of a reaction was defined in Definition 7.6.

A bottle of milk left in the sun next to an open refrigerator. The reactions that spoil it go several times faster at summer temperature than in the cold.

8.1 The rate of a reaction

Definition 8.1 (Rate of reaction, rates of formation and disappearance)

For a reaction $0 = \sum_i \nu_i\mathrm{A}_i$ taking place in a closed reactor of constant volume $V$ , the rate of reaction is

v = \frac{1}{V}\frac{\dd\xi}{\dd t} = \frac{1}{\nu_i}\frac{\dd[\mathrm{A}_i]}{\dd t} \quad\text{(any } i\text{)},

in $\mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{s}^{-1}$ . The rate of formation of a product is $\dd[\mathrm{A}_i]/\dd t$ ; the rate of disappearance of a reactant is $-\dd[\mathrm{A}_i]/\dd t$ .

Proposition 8.2 (One rate for all species)

The rate of reaction does not depend on the species chosen to follow it: for $a\,\mathrm{A} + b\,\mathrm{B} \longrightarrow c\,\mathrm{C} + d\,\mathrm{D}$ ,

v = -\frac1a\frac{\dd[\ce{A}]}{\dd t} = -\frac1b\frac{\dd[\ce{B}]}{\dd t} = \frac1c\frac{\dd[\ce{C}]}{\dd t} = \frac1d\frac{\dd[\ce{D}]}{\dd t} .

Proof. $[\mathrm{A}_i] = (n_{i,0} + \nu_i\xi)/V$ with $V$ constant, so $\dd[\mathrm{A}_i]/\dd t = (\nu_i/V)\,\dd\xi/\dd t = \nu_i v$ . ∎

Example 8.3 (Decomposition of dinitrogen pentoxide)

For $\ce{2N2O5 -> 4NO2 + O2}$ : $v = -\frac12\frac{\dd[\ce{N2O5}]}{\dd t} = \frac14\frac{\dd[\ce{NO2}]}{\dd t} = \frac{\dd[\ce{O2}]}{\dd t}$ . Nitrogen dioxide appears four times as fast as dioxygen, and dinitrogen pentoxide disappears twice as fast as dioxygen appears.

8.2 Rate laws and orders

Definition 8.4 (Rate law, orders, rate constant)

A rate law expresses the rate as a function of the concentrations. A reaction has an order when its rate law takes the form

v = k\,[\mathrm{A}]^{p}\,[\mathrm{B}]^{q} \cdots ;

the exponents $p$ , $q$ , … are the partial orders with respect to A, B, …, their sum is the overall order, and $k$ is the rate constant, which depends on temperature only. Orders are found by experiment; they need not equal the stoichiometric coefficients, and they can be fractional or zero.

Remark 8.5 (Units of $k$ )

Since $v$ is in $\mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{s}^{-1}$ , $k$ is in $\mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{s}^{-1}$ for order 0, $\mathrm{s}^{-1}$ for order 1, $\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$ for order 2, and in general $(\mathrm{mol}/\mathrm{L})^{1-n}\,\mathrm{s}^{-1}$ for overall order $n$ . The unit of $k$ tells the overall order.

Remark 8.6 (Reactions without order)

Not every reaction has an order. The rate of some reactions is a fraction whose denominator contains concentrations, or depends on the concentration of a product. Such rate laws are explained by the mechanism of the reaction (Chapter 9). A reaction may also have an order at the beginning only: the initial rate law.

8.3 Integrated rate laws and half-lives

Definition 8.7 (Half-life)

The half-life $t_{1/2}$ of a reactant is the time after which half of its initial amount has been consumed.

Theorem 8.8 (Integrated rate laws)

For a reaction $a\mathrm{A} \rightarrow \text{products}$ of rate $v = -\frac1a\frac{\dd[\ce{A}]} {\dd t} = k[\ce{A}]^n$ , with $[\ce{A}](0) = a_0$ :

order	integrated law	linear plot	half-life
0	$[\ce{A}] = a_0 - akt$	$[\ce{A}]$ against $t$	$a_0/(2ak)$
1	$[\ce{A}] = a_0\,\eu^{-akt}$	$\ln[\ce{A}]$ against $t$	$\ln2/(ak)$
2	$\dfrac{1}{[\ce{A}]} = \dfrac{1}{a_0} + akt$	$1/[\ce{A}]$ against $t$	$1/(ak\,a_0)$

Proof. Write $c = [\ce{A}]$ , so $\dd c/\dd t = -akc^n$ . Order 0: $\dd c/\dd t = -ak$ , hence $c = a_0 - akt$ (valid until $c = 0$ ). Order 1: $\dd c/\dd t = -akc$ is a first-order linear differential equation whose solution is $c = a_0\eu^{-akt}$ . Order 2: $\dd c/c^2 = -ak\,\dd t$ ; integrating from 0 to $t$ , $-1/c + 1/a_0 = -akt$ , so $1/c = 1/a_0 + akt$ . Half-lives: set $c = a_0/2$ in each law and solve for $t$ . ∎

Proposition 8.9 (The half-life test)

The half-life is independent of the initial concentration if and only if the reaction is of order 1. It is proportional to $a_0$ for order 0 and to $1/a_0$ for order 2. A first-order reaction loses half of what is left in each successive half-life: after $m$ half-lives a fraction $2^{-m}$ remains.

Proof. For order $n \ne 1$ , separating variables gives $t_{1/2} = \dfrac{2^{n-1} - 1}{(n-1)\,ak\,a_0^{n-1}}$ , which depends on $a_0$ unless $n = 1$ ; for $n = 1$ , $t_{1/2} = \ln2/(ak)$ . For order 1, $c(t + t_{1/2}) = c(t)/2$ at every $t$ , since $\eu^{-ak(t + t_{1/2})} = \eu^{-akt}/2$ . ∎

Three reactions with the same initial concentration (1.00\, mol/ L) and the same initial rate (0.050\, mol\, L-1\, min-1). Top: each integrated law becomes a straight line in its own coordinates. Bottom: the concentrations; the half-lives are 10, 13.9 and 20 minutes. — Three reactions with the same initial concentration ( $1.00\,\mathrm{mol}/\mathrm{L}$ ) and the same initial rate ( $0.050\,\mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{min}^{-1}$ ). Top: each integrated law becomes a straight line in its own coordinates. Bottom: the concentrations; the half-lives are 10, 13.9 and 20 minutes.

8.4 Finding a rate law

Method 8.10 (The integral method)

To test an order with measurements $[\ce{A}](t)$ : plot $[\ce{A}]$ , $\ln[\ce{A}]$ and $1/[\ce{A}]$ against $t$ ; the plot that is a straight line gives the order, and its slope gives $k$ (slope $-ak$ , $-ak$ or $+ak$ respectively).

Method 8.11 (The half-life method)

Measure the half-life for several initial concentrations: constant, order 1; proportional to $a_0$ , order 0; proportional to $1/a_0$ , order 2. In a single experiment, compare the times to go from $a_0$ to $a_0/2$ and from $a_0/2$ to $a_0/4$ : equal times mean order 1.

Definition 8.12 (Initial rate)

The initial rate $v_0$ is the rate at $t = 0$ , read as the slope of the tangent at the origin of the curve $[\ce{A}](t)$ , divided by the stoichiometric coefficient. At that moment the concentrations are the known initial ones and no product interferes.

Method 8.13 (The method of initial rates)

For $v_0 = k[\ce{A}]_0^p[\ce{B}]_0^q$ : run experiments that change $[\ce{A}]_0$ alone; then $\ln v_0 = \ln(k[\ce{B}]_0^q) + p\ln[\ce{A}]_0$ , and the slope of $\ln v_0$ against $\ln[\ce{A}]_0$ is $p$ . In practice, doubling $[\ce{A}]_0$ multiplies $v_0$ by $2^p$ . Do the same for B.

Definition 8.14 (Isolation method, apparent order)

In the isolation method, all reactants but one are put in large excess (ten times or more). Their concentrations stay practically constant, and $v = k[\ce{A}]^p[\ce{B}]^q \approx k'[\ce{A}]^p$ with $k' = k[\ce{B}]_0^q$ : the reaction behaves as if of apparent order $p$ with the apparent rate constant $k'$ . This is also called degeneracy of the order.

Reading an initial rate: the tangent at the origin (red) of a first-order curve meets the time axis at t_0 = 20\, min, so v_0 = 1.00\, mol/ L/20\, min = 0.050\, mol\, L-1\, min-1. — Reading an initial rate: the tangent at the origin (red) of a first-order curve meets the time axis at $t_0 = 20\,\mathrm{min}$ , so $v_0 = 1.00\,\mathrm{mol}/\mathrm{L}/20\,\mathrm{min} = 0.050\,\mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{min}^{-1}$ .

In the lab — Following a reaction in time

A rate law is built from concentrations measured while the reaction runs. When a reactant or a product is coloured, a spectrophotometer records the absorbance, proportional to its concentration (Beer–Lambert, Chapter 17); when ions appear or disappear, a conductimeter follows the conductivity; when a gas forms in a closed vessel, a manometer follows the pressure. Otherwise samples are taken at known times, the reaction is stopped (quenched) by cooling or dilution, and each sample is titrated. The temperature of the vessel is held constant by a thermostatic bath, because $k$ depends strongly on it.

8.5 Temperature: the Arrhenius law

Definition 8.15 (Activation energy)

The activation energy $E_a$ of a reaction of rate constant $k(T)$ is defined by

\frac{\dd \ln k}{\dd T} = \frac{E_a}{RT^2} ,

in $\mathrm{J}/\mathrm{mol}$ . When $E_a$ is independent of $T$ , integrating gives $k = A\,\eu^{-E_a/RT}$ , where the constant $A$ , of the unit of $k$ , is the pre-exponential factor.

Theorem 8.16 (Arrhenius law)

For most reactions, over a moderate range of temperature, the rate constant follows the Arrhenius law

k(T) = A\,\eu^{-E_a/RT} ,

with $A$ and $E_a > 0$ independent of $T$ : $\ln k$ is a linear function of $1/T$ of slope $-E_a/R$ .

Proof. Admitted at this level. ∎

Remark 8.17 (The meaning of $E_a$ )

The law is empirical. Its interpretation — only the molecular collisions that carry at least an energy of the order of $E_a$ lead to reaction, and the fraction of such collisions varies as $\eu^{-E_a/RT}$ — is sketched with the energy profiles of the next chapter and made quantitative by the theories of reaction rates in the Year 3 volume.

Method 8.18 (Determining $E_a$ )

From rate constants at several temperatures, plot $\ln k$ against $1/T$ (in kelvin): the slope is $-E_a/R$ , the intercept $\ln A$ . With two temperatures only,

E_a = \frac{R\,\ln(k_2/k_1)}{1/T_1 - 1/T_2} .

Example 8.19 (A rule of thumb)

For $E_a = 53\,\mathrm{kJ}/\mathrm{mol}$ , going from 25 to $35\,{}^{\circ}\mathrm{C}$ multiplies $k$ by $\exp\!\big(\frac{53000}{8.314}(\frac{1}{298.15} - \frac{1}{308.15})\big) = 2.0$ . The familiar rule “ten degrees more, twice as fast” holds near room temperature for reactions of activation energy around $50\,\mathrm{kJ}/\mathrm{mol}$ ; for larger $E_a$ the factor is larger.

Arrhenius plot of the degradation of a medicine (weekend problem): three measured rate constants (red) lie on a line of slope -E_a/R; extrapolated to 25\, C, it gives the shelf life at room temperature. — Arrhenius plot of the degradation of a medicine (weekend problem): three measured rate constants (red) lie on a line of slope $-E_a/R$ ; extrapolated to $25\,{}^{\circ}\mathrm{C}$ , it gives the shelf life at room temperature.

History — Arrhenius, 1889

The Swedish chemist Svante Arrhenius, studying the inversion of sucrose by acids in 1889, proposed that only a small fraction of the molecules, those in an “activated” state, react, and that this fraction grows with temperature as $\eu^{-E/RT}$ . The idea, controversial at first, became the foundation of chemical kinetics. Arrhenius received the Nobel Prize in Chemistry in 1903, for his theory of electrolytic dissociation.

8.6 Exercises

Exercise 8.1 ★

For $\ce{2N2O5 -> 4NO2 + O2}$ at constant volume, nitrogen dioxide appears at $2.0 \times 10^{-4}\,\mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{s}^{-1}$ . Give the rate of reaction, the rate of formation of dioxygen and the rate of disappearance of $\ce{N2O5}$ .

Solution

Solution of Exercise 8.1.

$v = \frac14 \times 2.0 \times 10^{-4} = 5.0 \times 10^{-5}\,\mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{s}^{-1}$ ; dioxygen forms at the same rate, $5.0 \times 10^{-5}\,$ ; $\ce{N2O5}$ disappears at $2v = 1.0 \times 10^{-4}\,\mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{s}^{-1}$ .

Exercise 8.2 ★

Give the unit of the rate constant for overall orders 0, 1, 2 and 3, concentrations in $\mathrm{mol}/\mathrm{L}$ and times in seconds.

Solution

Solution of Exercise 8.2.

Order 0: $\mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{s}^{-1}$ ; order 1: $\mathrm{s}^{-1}$ ; order 2: $\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$ ; order 3: $\mathrm{L}^{2}\,\mathrm{mol}^{-2}\,\mathrm{s}^{-1}$ .

Exercise 8.3 ★

A first-order reaction $\mathrm{A} \longrightarrow \mathrm{B}$ has $k = 3.0 \times 10^{-3}\,\mathrm{s}^{-1}$ . Compute its half-life and the fraction of $\ce{A}$ left after $10\,\mathrm{min}$ .

Solution

Solution of Exercise 8.3.

$t_{1/2} = \ln2/k = 0.693/(3.0 \times 10^{-3}) = 231\,\mathrm{s}$ . After $600\,\mathrm{s}$ , $\eu^{-kt} = \eu^{-1.8} = 0.165$ : 16.5 % left.

Exercise 8.4 ★

When the initial concentration of a reactant is halved, its half-life doubles. What is the order? What if the half-life is halved?

Solution

Solution of Exercise 8.4.

A half-life proportional to $1/a_0$ means order 2. If halving $a_0$ halves the half-life, $t_{1/2} \propto a_0$ : order 0.

Exercise 8.5 ★★

The concentration of a reactant $\ce{A}$ ( $\mathrm{A} \longrightarrow \mathrm{P}$ ) is measured: at $t = 0$ , 10, 20, 30 and $40\,\mathrm{min}$ , $[\ce{A}] = 0.100$ , 0.0741, 0.0549, 0.0407 and $0.0301\,\mathrm{mol}/\mathrm{L}$ . Determine the order and the rate constant.

Solution

Solution of Exercise 8.5.

$[\ce{A}]$ is not linear in $t$ (losses 0.0259, 0.0192, 0.0142, 0.0106). $\ln[\ce{A}] = -2.303$ , $-2.602$ , $-2.902$ , $-3.202$ , $-3.503$ : it drops by 0.300 every $10\,\mathrm{min}$ , a straight line. Order 1, $k = 0.0300\,\mathrm{min}^{-1}$ .

Exercise 8.6 ★★

For $\mathrm{A} + \mathrm{B} \longrightarrow \mathrm{P}$ , $v_0 = 2.0 \times 10^{-6}\,\mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{s}^{-1}$ for $[\ce{A}]_0 = [\ce{B}]_0 = 0.010\,\mathrm{mol}/\mathrm{L}$ ; $8.0 \times 10^{-6}\,$ when $[\ce{A}]_0$ is doubled; $4.0 \times 10^{-6}\,$ when $[\ce{B}]_0$ is doubled. Find the partial orders, the rate law and the rate constant.

Solution

Solution of Exercise 8.6.

Doubling $[\ce{A}]_0$ multiplies $v_0$ by $4 = 2^2$ : order 2 in $\ce{A}$ . Doubling $[\ce{B}]_0$ multiplies it by 2: order 1 in $\ce{B}$ . $v = k[\ce{A}]^2[\ce{B}]$ with $k = 2.0 \times 10^{-6}/(0.010^2 \times 0.010) = 2.0\,\mathrm{L}^{2}\,\mathrm{mol}^{-2}\,\mathrm{s}^{-1}$ .

Exercise 8.7 ★★

The rate law of $\mathrm{A} + \mathrm{B} \longrightarrow \mathrm{P}$ is $v = k[\ce{A}][\ce{B}]$ . With $[\ce{B}]_0 = 0.50\,\mathrm{mol}/\mathrm{L}$ and $[\ce{A}]_0 = 5.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ , $\ce{A}$ disappears with first-order kinetics and an apparent constant of $0.012\,\mathrm{s}^{-1}$ . Explain and compute $k$ .

Solution

Solution of Exercise 8.7.

$\ce{B}$ is in hundred-fold excess: its concentration stays $0.50\,\mathrm{mol}/\mathrm{L}$ , and $v = k'[\ce{A}]$ with $k' = k[\ce{B}]_0$ . Hence $k = 0.012/0.50 = 0.024\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$ .

Exercise 8.8 ★★

A rate constant is $1.0 \times 10^{-3}\,\mathrm{s}^{-1}$ at $300\,\mathrm{K}$ and $5.0 \times 10^{-3}\,\mathrm{s}^{-1}$ at $320\,\mathrm{K}$ . Compute the activation energy and the rate constant at $310\,\mathrm{K}$ .

Solution

Solution of Exercise 8.8.

$E_a = R\ln5/(1/300 - 1/320) = 8.314 \times 1.609/(2.083 \times 10^{-4}) = 64.2\,\mathrm{kJ}/\mathrm{mol}$ . At $310\,\mathrm{K}$ : $\ln k = \ln(1.0 \times 10^{-3}) + \frac{E_a}{R}\big(\frac{1}{300} - \frac{1}{310}\big) = -6.908 + 0.831$ , $k = 2.3 \times 10^{-3}\,\mathrm{s}^{-1}$ .

Exercise 8.9 ★★

The reaction $2\,\mathrm{A} \longrightarrow \mathrm{P}$ has the rate law $v = -\frac12\frac{\dd[\ce{A}]} {\dd t} = k[\ce{A}]^2$ with $k = 0.50\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$ . Starting from $[\ce{A}]_0 = 0.020\,\mathrm{mol}/\mathrm{L}$ , compute the half-life and the concentration after $100\,\mathrm{s}$ .

Solution

Solution of Exercise 8.9.

Here $a = 2$ : $1/[\ce{A}] = 1/a_0 + 2kt$ , $t_{1/2} = 1/(2k a_0) = 1/(2 \times 0.50 \times 0.020) = 50\,\mathrm{s}$ . At $100\,\mathrm{s}$ : $1/[\ce{A}] = 50 + 100 = 150\,\mathrm{L}/\mathrm{mol}$ , $[\ce{A}] = 6.7 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ .

Exercise 8.10 ★★★

For a first-order reaction, show that the time needed to consume a fraction $f$ of the reactant is $t_f = -\ln(1 - f)/(ak)$ . Compare the times for 50 %, 75 %, 90 % and 99.9 %, in half-lives.

Solution

Solution of Exercise 8.10.

$[\ce{A}] = a_0(1 - f) = a_0\eu^{-akt_f}$ gives $t_f = -\ln(1-f)/(ak)$ , and in half-lives $t_f/t_{1/2} = -\ln(1-f)/\ln2$ : 1 for 50 %, 2 for 75 %, 3.32 for 90 %, 9.97 for 99.9 %: about ten half-lives for the reaction to be “finished”.

Exercise 8.11 ★★★

A skin patch releases a drug at a constant rate of $0.40\,\mathrm{mg}/\mathrm{h}$ as long as its reservoir, initially $10\,\mathrm{mg}$ , is not empty. Write the law of the amount left, give its order, its half-life and the time after which the patch is exhausted. Why do zero-order kinetics suit a drug patch?

Solution

Solution of Exercise 8.11.

$m(t) = 10 - 0.40\,t$ ( $\mathrm{mg}$ , $\mathrm{h}$ ): order 0. Half-life $10/(2 \times 0.40) = 12.5\,\mathrm{h}$ ; empty after $25\,\mathrm{h}$ . The dose delivered per hour does not depend on what is left: a steady supply, which is the purpose of a patch.

Exercise 8.12 ★★★

Show that the rule “ten degrees more, twice as fast” between $298.15\,\mathrm{K}$ and $308.15\,\mathrm{K}$ corresponds to $E_a \approx 53\,\mathrm{kJ}/\mathrm{mol}$ . What factor does the same activation energy give between $0\,{}^{\circ}\mathrm{C}$ and $10\,{}^{\circ}\mathrm{C}$ ?

Solution

Solution of Exercise 8.12.

$E_a = R\ln2/(1/298.15 - 1/308.15) = 8.314 \times 0.693/(1.088 \times 10^{-4}) = 52.9\,\mathrm{kJ}/\mathrm{mol}$ . Between 273.15 and $283.15\,\mathrm{K}$ : $\exp\big(\frac{52900}{8.314}(\frac{1}{273.15} - \frac{1}{283.15})\big) = 2.3$ : the same activation energy gives a larger factor at lower temperature.

8.7 Problem: The Shelf Life of a Medicine

Problem 8.1

Weekend problem — an accelerated stability test: first-order degradation at 60 °C, three temperatures, the Arrhenius plot, and the shelf life at room temperature

To predict the shelf life $t_{90}$ of a tablet (the time after which 10 % of the active ingredient has degraded), a laboratory stores samples at high temperature and measures the percentage of ingredient left. At $60\,{}^{\circ}\mathrm{C}$ :

time (days)	0	10	20	30	40
ingredient left (%)	100.0	91.4	83.5	76.3	69.8

At 40 and $50\,{}^{\circ}\mathrm{C}$ the rate constants found are $1.27 \times 10^{-3}\,\mathrm{d}^{-1}$ and $3.48 \times 10^{-3}\,\mathrm{d}^{-1}$ . Take $R = 8.314\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$ and one month $= 30.4\,\mathrm{d}$ .

Part I — The order at 60 °C.

Compute the losses over each 10-day interval. Is the reaction of order 0?
Compute $\ln(\%\text{ left})$ at each time. What do you conclude?
Deduce the rate constant $k_{60}$ .
Compute the half-life at $60\,{}^{\circ}\mathrm{C}$ .
What percentage is left after 60 days at $60\,{}^{\circ}\mathrm{C}$ ?
Compute $t_{90}$ at $60\,{}^{\circ}\mathrm{C}$ .

Part II — Shelf life and order.

Show that for a first-order degradation $t_{90} = \ln(10/9)/k$ .
Express $t_{90}$ as a fraction of the half-life.
Why does $t_{90}$ not depend on the dose of the tablet?
Why do laboratories measure degradation at high temperature rather than at $25\,{}^{\circ}\mathrm{C}$ ?
If the degradation were of order 2 in the ingredient, would a tablet twice as strong keep longer or less long? Explain.

Part III — The Arrhenius plot.

Tabulate $1/T$ and $\ln k$ for the three temperatures.
Compute the slope of the line through the 40 and $60\,{}^{\circ}\mathrm{C}$ points.
Deduce the activation energy.
Check that the $50\,{}^{\circ}\mathrm{C}$ point lies on the line.
Compute the pre-exponential factor $A$ .
By what factor does the rate increase between 25 and $35\,{}^{\circ}\mathrm{C}$ ? Compare with the rule of thumb.

Part IV — Room temperature.

Extrapolate the rate constant to $30\,{}^{\circ}\mathrm{C}$ .
Compute $t_{90}$ at $30\,{}^{\circ}\mathrm{C}$ , in months.
A box is forgotten three days in a car at $60\,{}^{\circ}\mathrm{C}$ . What fraction of the ingredient is lost?
Explain the instruction “store below $25\,{}^{\circ}\mathrm{C}$ ”.
Extrapolate the rate constant to $25\,{}^{\circ}\mathrm{C}$ .
Compute the shelf life $t_{90}$ at $25\,{}^{\circ}\mathrm{C}$ , in days and in months.

Solution

Solution of Problem 8.1.

1. Losses 8.6, 7.9, 7.2 and 6.5 points: not constant, so not order 0. 2. $\ln$ : 4.605, 4.515, 4.425, 4.335, 4.245; it drops by 0.090 every 10 days: a straight line, order 1. 3. $k_{60} = 0.090/10 = 9.0 \times 10^{-3}\,\mathrm{d}^{-1}$ . 4. $t_{1/2} = \ln2/k_{60} = 77\,\mathrm{d}$ . 5. $100\,\eu^{-0.54} = 58.3\,\%$ . 6. $t_{90} = \ln(10/9)/k_{60} = 0.1054/0.0090 = 11.7\,\mathrm{d}$ . 7. 90 % left: $\eu^{-kt_{90}} = 0.9$ , so $t_{90} = \ln(10/9)/k$ . 8. $t_{90}/t_{1/2} = \ln(10/9)/\ln2 = 0.152$ . 9. For order 1 the fraction left depends on $kt$ only, not on the initial amount. 10. At $25\,{}^{\circ}\mathrm{C}$ the degradation takes years; heating accelerates it (Arrhenius) so that it can be measured in weeks. 11. For order 2 the half-life is $1/(ak\,a_0)$ : a tablet twice as strong would degrade twice as fast in relative terms and keep less long. 12. $1/T$ ( $\mathrm{K}^{-1}$ ): $3.1934 \times 10^{-3}$ , $3.0945 \times 10^{-3}$ , $3.0017 \times 10^{-3}$ ; $\ln k$ : $-6.669$ , $-5.661$ , $-4.711$ . 13. Slope $= (-4.711 + 6.669)/(3.0017 \times 10^{-3} - 3.1934 \times 10^{-3}) = -1.021 \times 10^{4}\,\mathrm{K}$ . 14. $E_a = 8.314 \times 1.021 \times 10^4 = 85\,\mathrm{kJ}/\mathrm{mol}$ . 15. Predicted $\ln k_{50} = -4.711 - 1.021 \times 10^4 \times (3.0945 - 3.0017) \times 10^{-3} = -5.658$ , measured $-5.661$ : on the line. 16. $A = k_{60}\,\eu^{E_a/RT} = 0.0090\,\eu^{10215/333.15} = 1.9 \times 10^{11}\,\mathrm{d}^{-1}$ . 17. $\exp\big(1.021 \times 10^4(\frac{1}{298.15} - \frac{1}{308.15})\big) = 3.0$ : more than the factor 2 of the rule, because $E_a$ is larger than $53\,\mathrm{kJ}/\mathrm{mol}$ . 18. $k_{30} = k_{60}\exp\big(-1.021 \times 10^4(\frac{1}{303.15} - \frac{1}{333.15})\big) = 4.3 \times 10^{-4}\,\mathrm{d}^{-1}$ . 19. $t_{90} = 0.1054/4.33 \times 10^{-4} = 243\,\mathrm{d}$ , 8.0 months. 20. $1 - \eu^{-3 \times 0.0090} = 2.7\,\%$ . 21. The shelf life falls quickly with temperature (8 months at $30\,{}^{\circ}\mathrm{C}$ against 14 at $25\,{}^{\circ}\mathrm{C}$ ): the expiry date is valid only below $25\,{}^{\circ}\mathrm{C}$ . 22. $k_{25} = k_{60}\exp\big(-1.021 \times 10^4(\frac{1}{298.15} - \frac{1}{333.15})\big) = 2.46 \times 10^{-4}\,\mathrm{d}^{-1}$ . 23. $t_{90} = 0.1054/2.46 \times 10^{-4} = 428\,\mathrm{d}$ , that is 14 months: the date printed on the box, obtained from six weeks of measurements.

Terms defined in this chapter

See all 852 terms in the glossary