Factors That Affect the Rate of Reactions

Lessons

Senior 6 · Chemistry

Factors That Affect the Rate of Reactions

3 lessons

Lesson 3 of 3

Temperature, Activation Energy and the Arrhenius Equation

25 min

A light stick in hot and cold water

Bend a chemical light stick, so that the glass capsule inside breaks and the two solutions mix, and it glows for hours. Drop one into hot water and it glows much more brightly but fades within an hour; put an identical one in a freezer and it is dim but still glowing the next day. The light comes from a chemical reaction, and the brightness tracks its rate.

In Lesson 2 you saw why temperature matters so much: the fraction of collisions with $E \ge E_\text{a}$ is $e^{-\frac{E_\text{a}}{RT}}$ . In 1889 the Swedish chemist Svante Arrhenius turned this into an equation for the rate constant, which lets you measure $E_\text{a}$ and predict a rate at any temperature.

Definition

Rate constant, $k$ : the constant of proportionality in a rate equation, e.g. $\text{rate} = k[\mathrm{A}]$ . It does not depend on concentration but does depend on temperature (and on a catalyst). Its unit depends on the order of the reaction (Unit 13); for a first-order reaction it is $\mathrm{s^{-1}}$ .

Arrhenius equation: $k = A\,e^{-\frac{E_\text{a}}{RT}}$ , where $A$ is the pre-exponential (frequency) factor, which includes the collision frequency and the orientation factor and has the same unit as $k$ ; $E_\text{a}$ is in $\mathrm{J\,mol^{-1}}$ ; $R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$ ; and $T$ is in kelvin.

Three useful forms

Taking natural logarithms of $k = A\,e^{-\frac{E_\text{a}}{RT}}$ :

\ln k = \ln A - \frac{E_\text{a}}{R}\cdot\frac{1}{T}

This has the form $y = c + m x$ with $y = \ln k$ and $x = \frac{1}{T}$ . A graph of $\ln k$ against $\frac{1}{T}$ , an Arrhenius plot, is a straight line with

\text{gradient} = -\frac{E_\text{a}}{R}, \qquad \text{intercept} = \ln A

Writing the logarithmic form at two temperatures and subtracting removes $A$ :

\ln\frac{k_2}{k_1} = \frac{E_\text{a}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Because a rate measured at fixed concentrations is proportional to $k$ , you may put a ratio of rates (or the inverse ratio of times for a fixed amount of reaction) in place of $\frac{k_2}{k_1}$ .

The steeper the Arrhenius line, the larger $E_\text{a}$ and the more sensitive the rate is to temperature. A catalyst lowers $E_\text{a}$ , so a catalysed reaction is both faster and less temperature-sensitive.

Worked example 1: activation energy from an Arrhenius plot

Problem. The first-order rate constant of a decomposition was measured at five temperatures (illustrative data). Find $E_\text{a}$ and $A$ .

$T$ / K	298	308	318	328	338
$k$ / $\mathrm{s^{-1}}$	$3.46 \times 10^{-5}$	$1.33 \times 10^{-4}$	$4.73 \times 10^{-4}$	$1.55 \times 10^{-3}$	$4.74 \times 10^{-3}$
$\frac{1}{T}$ / $10^{-3}\,\mathrm{K^{-1}}$	3.356	3.247	3.145	3.049	2.959
$\ln k$	−10.27	−8.93	−7.66	−6.47	−5.35

Solution. The points lie on a straight line (Figure 3.1). Using two points on the line, far apart:

\text{gradient} = \frac{-5.35 - (-10.27)}{(2.959 - 3.356) \times 10^{-3}\,\mathrm{K^{-1}}} = \frac{4.92}{-3.97 \times 10^{-4}\,\mathrm{K^{-1}}} = -1.24 \times 10^{4}\,\mathrm{K}

E_\text{a} = -R \times \text{gradient} = 8.314 \times 1.24 \times 10^{4} = 1.03 \times 10^{5}\,\mathrm{J\,mol^{-1}} = 103\,\mathrm{kJ\,mol^{-1}}

For $A$ , substitute one point: $\ln A = \ln k + \frac{E_\text{a}}{RT} = -10.27 + \frac{1.24 \times 10^{4}}{298} = 31.3$ , so $A = e^{31.3} \approx 4 \times 10^{13}\,\mathrm{s^{-1}}$ .

Figure 3.1: Arrhenius plot of the data in worked example 1.

Worked example 2: the "doubles every 10 degrees" rule

Problem. The rate of a reaction doubles when the temperature rises from 25 °C to 35 °C. Calculate $E_\text{a}$ .

Solution. $T_1 = 298\,\mathrm{K}$ , $T_2 = 308\,\mathrm{K}$ , $\frac{k_2}{k_1} = 2$ .

E_\text{a} = \frac{R\ln\frac{k_2}{k_1}}{\frac{1}{T_1} - \frac{1}{T_2}} = \frac{8.314 \times \ln 2}{\frac{1}{298} - \frac{1}{308}} = \frac{5.763}{1.090 \times 10^{-4}} = 5.29 \times 10^{4}\,\mathrm{J\,mol^{-1}} = 52.9\,\mathrm{kJ\,mol^{-1}}

So the popular rule "the rate doubles for every 10 °C" is true only for reactions with $E_\text{a}$ near $53\,\mathrm{kJ\,mol^{-1}}$ close to room temperature. It is a rough guide, not a law.

Worked example 3: a medicine at body temperature

Problem. A drug decomposes in solution with $k = 2.5 \times 10^{-4}\,\mathrm{s^{-1}}$ at 25 °C and $E_\text{a} = 75\,\mathrm{kJ\,mol^{-1}}$ . Calculate $k$ at 37 °C.

Solution.

\ln\frac{k_2}{k_1} = \frac{75\,000}{8.314}\left(\frac{1}{298} - \frac{1}{310}\right) = 9021 \times 1.299 \times 10^{-4} = 1.172

\frac{k_2}{k_1} = e^{1.172} = 3.23, \qquad k_2 = 3.23 \times 2.5 \times 10^{-4} = 8.1 \times 10^{-4}\,\mathrm{s^{-1}}

A 12 K rise makes the decomposition more than three times faster.

Worked example 4: how much does a catalyst help?

Problem. A catalyst lowers $E_\text{a}$ from 75 to $50\,\mathrm{kJ\,mol^{-1}}$ . Assuming $A$ is unchanged, by what factor does the rate increase at 298 K?

Solution. $A$ cancels in the ratio:

\frac{k_\text{cat}}{k} = \frac{e^{-\frac{50\,000}{RT}}}{e^{-\frac{75\,000}{RT}}} = e^{\frac{25\,000}{8.314 \times 298}} = e^{10.09} = 2.4 \times 10^{4}

Lowering the barrier by only a third makes the reaction about 24 000 times faster. Enzymes, which can lower $E_\text{a}$ much further, speed up reactions by factors of millions or more.

Warning

Common mistakes

Using °C in the Arrhenius equation. Always convert: $T/\mathrm{K} = \theta/^\circ\mathrm{C} + 273$ .
Mixing kJ and J. With $R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$ , $E_\text{a}$ must be in $\mathrm{J\,mol^{-1}}$ ; convert to kJ only at the end.
Reading the gradient from the axis labels without the power of ten: if the axis is $\frac{1}{T}$ in $10^{-3}\,\mathrm{K^{-1}}$ , the run is in thousandths.
Getting the sign wrong in the two-point form. Check that a higher temperature gives a larger $k$ for any positive $E_\text{a}$ .

Tip

Before a two-point calculation, predict: if $T_{2} > T_{1}$ , then $k_{2} > k_{1}$ . If your answer disagrees, you have swapped $T_{1}$ and $T_{2}$ . For a plot, choose two points on the line, as far apart as possible, not two data points that lie off it.

Allery Chemistry: Arrhenius Equation. Watch how the equation is rearranged into the straight-line form and how the gradient of ln k against 1/T gives the activation energy.

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Practice

State the Arrhenius equation and give the meaning and unit of each symbol.
Explain why a graph of $\ln k$ against $\frac{1}{T}$ is a straight line, and how $E_\text{a}$ is found from it.
A rate constant is $1.0 \times 10^{-3}\,\mathrm{s^{-1}}$ at 300 K and $5.0 \times 10^{-3}\,\mathrm{s^{-1}}$ at 330 K. Calculate $E_\text{a}$ .
Milk souring has an effective $E_\text{a}$ of about $50\,\mathrm{kJ\,mol^{-1}}$ (illustrative value). How many times faster does it go at 30 °C than in a cooling tank at 4 °C?
A reaction has $E_\text{a} = 120\,\mathrm{kJ\,mol^{-1}}$ . At what temperature is its rate constant ten times its value at 300 K?
Use $A = 4.0 \times 10^{13}\,\mathrm{s^{-1}}$ and $E_\text{a} = 103\,\mathrm{kJ\,mol^{-1}}$ from worked example 1 to predict $k$ at 350 K.
Two reactions have $E_\text{a} = 40$ and $120\,\mathrm{kJ\,mol^{-1}}$ . Calculate the factor by which each rate increases from 300 K to 310 K, and explain what this tells you about the slopes of their Arrhenius plots.

Show answersHide answers

$k = A\,e^{-\frac{E_\text{a}}{RT}}$ : $k$ rate constant (unit depends on order); $A$ pre-exponential factor (same unit as $k$ ); $E_\text{a}$ activation energy ( $\mathrm{J\,mol^{-1}}$ ); $R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$ ; $T$ temperature (K).
$\ln k = \ln A - \frac{E_\text{a}}{R}\cdot\frac{1}{T}$ is linear in $\frac{1}{T}$ ; gradient $= -\frac{E_\text{a}}{R}$ , so $E_\text{a} = -R \times \text{gradient}$ .
$E_\text{a} = \frac{8.314 \ln 5}{\frac{1}{300} - \frac{1}{330}} = 4.42 \times 10^{4}\,\mathrm{J\,mol^{-1}} = 44.2\,\mathrm{kJ\,mol^{-1}}$ .
$e^{\frac{50\,000}{8.314}\left(\frac{1}{277} - \frac{1}{303}\right)} = 6.4$ times faster.
$\frac{1}{T_2} = \frac{1}{300} - \frac{8.314 \ln 10}{120\,000}$ , so $T_2 = 315\,\mathrm{K}$ (42 °C).
$k = 4.0 \times 10^{13}\,e^{-\frac{103\,000}{8.314 \times 350}} = 1.7 \times 10^{-2}\,\mathrm{s^{-1}}$ .
Factors 1.68 and 4.72. The reaction with the larger $E_\text{a}$ is more temperature-sensitive, so its Arrhenius plot is steeper (gradient $-\frac{E_\text{a}}{R}$ three times as large).

1A light stick in hot and cold water

2Three useful forms

A light stick in hot and cold water

Three useful forms