Chemistry · General chemistry II · Concept

Activation energy from Arrhenius plots

Rate constants rise with temperature by an amount set by the activation energy, $E_{a}$ , the energy barrier a reaction must cross. Plot $ln k$ against $1 / T$ and the slope is $- E_{a} / R$ , or use the two-point Arrhenius equation when only two rate constants are known.

A concrete case

Twenty degrees, four times as fast

A reaction has $k = 2.5 \times 10^{- 3} s^{- 1}$ at $298 K$ , about room temperature. At $318 K$ , only $20 K$ warmer, $k = 1.1 \times 10^{- 2} s^{- 1}$ . The reaction runs $4.4$ times as fast.

The molecules are not colliding $4.4$ times as often. Their average speed grows with the square root of the kelvin temperature, and $\sqrt{318 / 298} = 1.033$ , so collisions become only about $3$ percent more frequent.

What changes a lot is the share of collisions with enough energy to react. With a barrier of $58.4 kJ / mol$ , about $5.8$ in every $100$ billion collisions clear it at $298 K$ , and about $25.5$ in every $100$ billion at $318 K$ . That share grew $4.4$ times, the whole speed-up. A small rise in temperature makes the rare, energetic collisions much less rare.

The same reaction at two temperatures
$T$ $(K)$	$k$ $(s^{- 1})$	Share of collisions over a $58.4 kJ / mol$ barrier
$298$	$2.5 \times 10^{- 3}$	$5.8 \times 10^{- 11}$
$318$	$1.1 \times 10^{- 2}$	$2.55 \times 10^{- 10}$
Ratio	$4.4$	$4.4$

The rule

The Arrhenius equation

The Arrhenius equation writes a rate constant as two factors. $E_{a}$ is the activation energy, the height of the energy barrier, in $J / mol$ . $R$ is the gas constant, $8.314 J / (mol \cdot K)$ , and $T$ is the temperature in kelvin.

The exponential factor, $e^{- E_{a} / R T}$ , is the share of collisions with at least the energy $E_{a}$ . The frequency factor, $A$ , counts how often molecules meet in a way that could react. It has the same units as $k$ .

k = A e^{- E_{a} / R T}

A straight line, and two points

Taking the natural logarithm of both sides turns the equation into a straight line. Plot $y = ln k$ against $x = 1 / T$ . The slope is $- E_{a} / R$ and the intercept is $ln A$ .

With only two rate constants, write the line at each temperature and subtract. $ln A$ cancels, and what is left links the two rate constants to $E_{a}$ . This is the two-point form.

ln k = - \frac{E_{a}}{R} (\frac{1}{T}) + ln A, ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

\begin{array}{c} ln k = - \frac{E_{a}}{R} (\frac{1}{T}) + ln A \\ ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \end{array}

Reading an Arrhenius plot
Plot feature	Equals	So
x-axis	$1 / T$ $(K^{- 1})$	Temperatures must be in kelvin
y-axis	$ln k$	Keep one set of $k$ units
Slope	$- E_{a} / R$	$E_{a}$ $= - slope \times 8.314 J / (mol \cdot K)$
Intercept	$ln A$	$A = e raised to the intercept$ , in the units of $k$

Why it works

Why the share of fast collisions grows so steeply

Molecules in a gas or a solution have a spread of energies. The typical energy is of the order of $R T$ , $2.5 kJ / mol$ at room temperature. Only a thin tail of the spread reaches a barrier of $50$ or $100 kJ / mol$ .

The share of collisions with more than an energy $E$ falls off as $e^{- E / R T}$ . When $E_{a}$ is twenty or more times $R T$ , that share is tiny, and a small rise in $T$ moves the whole tail outward. The share beyond $E_{a}$ can then double or quadruple, while the average energy rises by only a few percent.

The logarithm makes this a straight line because $ln (e^{- E_{a} / R T})$ is just $- E_{a} / R T$ , which is proportional to $1 / T$ . The steeper the line, the larger $E_{a}$ , and the more strongly the rate depends on temperature.

ln (e^{- E_{a} / R T}) = - \frac{E_{a}}{R} \cdot \frac{1}{T}

The method

From an Arrhenius plot

1. Convert every temperature to kelvin, and keep every $k$ in the same units.

2. Make two columns, $x = 1 / T$ and $y = ln k$ .

3. Fit a straight line, or read its slope from two points far apart on it.

4. Multiply the slope by $- R$ . With $R = 8.314 J / (mol \cdot K)$ , $E_{a}$ comes out in $J / mol$ . Divide by $1000$ for $kJ / mol$ .

5. If you need $A$ , raise $e$ to the intercept and give $A$ the units of $k$ , to two significant figures at most.

From two rate constants

1. Label the pairs, $k_{1}$ at $T_{1}$ and $k_{2}$ at $T_{2}$ , with both temperatures in kelvin.

2. Write the two-point form and put in the values.

3. Solve for $E_{a}$ . $E_{a} = R ln (k_{2} / k_{1})$ $\div (1 / T_{1} - 1 / T_{2})$ .

4. To predict $k$ at a third temperature, use the same equation with the $E_{a}$ you found, and solve for the new $k$ .

E_{a} = \frac{R ln (k_{2} / k_{1})}{\frac{1}{T_{1}} - \frac{1}{T_{2}}}

Where it breaks

Temperatures must be in kelvin. In $° C$ , $1 / T$ has no meaning, and the temperature of $0 ° C$ would put a division by zero in the plot.

The plot must be straight. A curved Arrhenius plot means no single $E_{a}$ fits, often because a different step limits the rate at different temperatures. Two points always lie on a line, so the two-point form cannot show that curve. Use it only when one $E_{a}$ is known to fit.

$A$ is an extrapolation. The intercept sits at $1 / T = 0$ , an infinite temperature far outside any data, so $A$ is much less certain than $E_{a}$ . Report it to one or two significant figures.

The rule of thumb that $k$ doubles for every $10 K$ holds only for an $E_{a}$ near $50 kJ / mol$ close to room temperature. For $E_{a} = 100 kJ / mol$ the same $10 K$ nearly quadruples $k$ .

Enzyme reactions break the pattern at high temperature. Above an optimum the protein unfolds and the rate falls, a change in the catalyst rather than a failure of the Arrhenius equation.

Check your answer

Check the sign and the size. $E_{a}$ must be positive, so the slope of $ln k$ against $1 / T$ must be negative. Most reactions studied in a course have $E_{a}$ from about $20$ to $200 kJ / mol$ . An $E_{a}$ of $0.0651$ or of $65, 100 kJ / mol$ means a slip with the factor of $1000$ .

Check the direction. A higher temperature must give a larger $k$ . If your predicted $k$ at a higher temperature came out smaller, the two temperatures are swapped in $1 / T_{1} - 1 / T_{2}$ .

Recompute one point. Put $E_{a}$ , $A$ and one of the measured temperatures back into $k = A e^{- E_{a} / R T}$ , and you should get back the measured $k$ to within its precision.

In a Chemistry box, write the two-point form, then each $k$ , $T$ and $R$ on its own row, then $E_{a}$ with the numbers put in. The Checker reads that row as “Substitution matches row 6 with the values defined above.” in the worked example’s box.

$E_{a}$ can go into the exponent in $J / mol$ or in $kJ / mol$ . Next to $R$ in $J / (mol \cdot K)$ , Chalk converts the $kJ$ to $J$ , and the row reads ✓ either way.

A rearranged two-point form, $E_{a} = R ln (k_{2} / k_{1}) / (1 / T_{1} - 1 / T_{2})$ , reads ✓ under the two-point form, “Rearranges row 1 correctly (treating every quantity as positive).” With the temperatures or the rate constants swapped it reads ✗. The numbers can also go straight under the two-point form. For the slope method, write $E_{a} = - R m$ with the slope as $m$ on its own row.

Barrier height and the transition state

$E_{a}$ from a plot is an empirical number, defined by the slope of $ln k$ against $1 / T$ . IUPAC keeps it apart from the threshold energy, $E_{0}$ , the actual height of the barrier on a potential energy diagram. For most reactions the two are close but not equal.

Transition-state theory writes $k$ with an enthalpy and an entropy of activation, in the Eyring equation. Its plot uses $ln (k / T)$ against $1 / T$ , and its slope gives the enthalpy of activation. That enthalpy differs from $E_{a}$ by about $R T$ , $2.5 kJ / mol$ at room temperature.

Collision theory reads $A$ as a collision frequency times a steric factor, $p$ , the fraction of collisions with a workable orientation. Large, awkward molecules can have $p$ far below $1$ , which is why two reactions with the same $E_{a}$ can run at very different rates.

Common mistakes

Plotting $k$ against $T$ instead of $ln k$ against $1 / T$ . Raw $k$ against $T$ curves upward and has no slope that means anything. The straight line, with slope $- E_{a} / R$ , is $ln k$ against $1 / T$ .
Using temperatures in $° C$ . $1 / 25$ and $1 / 45$ are not $1 / 298$ and $1 / 318$ , and the $E_{a}$ they give is far off. The Arrhenius equation measures temperature from absolute zero.
Reporting the slope itself as $E_{a}$ . A slope of $- 7.83 \times 10^{3} K$ is not an energy. $E_{a} = - R \times slope$ $= 6.51 \times 10^{4} J / mol$ , which is $65.1 kJ / mol$ .
Mixing $kJ$ and $J$ . With $R = 8.314 J / (mol \cdot K)$ , $E_{a} = 58.4$ must go in as $58, 400 J / mol$ . Left as $58.4$ , the exponential factor comes out a thousand times too close to $1$ .
Quoting $A$ to as many figures as $E_{a}$ . $A$ comes from an intercept at $1 / T = 0$ , far from every data point. $A \approx 4.2 \times 10^{8} s^{- 1}$ is as far as the data go.
Assuming a $10 K$ rise always doubles $k$ . The factor is $e^{(E_{a} / R) (1 / T_{1} - 1 / T_{2})}$ , which from $20 ° C$ to $30 ° C$ is $1.97$ at $50.0 kJ / mol$ but $3.87$ at $100 kJ / mol$ .

Key terms

Arrhenius equation: $k = A \cdot e^{- E_{a} / R T}$ : how a rate constant $k$ grows with temperature, where $A$ is the frequency factor, $E_{a}$ the activation energy, $R$ the gas constant and $T$ the kelvin temperature. A plot of $ln k$ against $1 / T$ is a line with slope $- E_{a} / R$ , which gives the activation energy.
Frequency factor: The factor $A$ in the Arrhenius equation $k = A \cdot e^{- E_{a} / R T}$ , where $k$ is the rate constant, $E_{a}$ the activation energy, $R$ the gas constant and $T$ the kelvin temperature. It measures how often reactant molecules collide with the right orientation.
Exponential factor: The term $e^{- E_{a} / R T}$ in the Arrhenius equation, where $E_{a}$ is the activation energy, $R$ the gas constant and $T$ the kelvin temperature. It is the fraction of collisions with enough energy to get over the activation barrier. It grows as the temperature rises.
Collision model: The model in which a reaction occurs only when particles collide with enough energy and a suitable orientation. It explains the Arrhenius equation $k = A \cdot e^{- E_{a} / R T}$ , where $k$ is the rate constant, $E_{a}$ the activation energy, $R$ the gas constant and $T$ the kelvin temperature. The exponential factor is the fraction of collisions with energy of at least $E_{a}$ , and $A = p z$ combines the orientation factor $p$ with the collision frequency $z$ .
Collision frequency: In the collision model, the number of collisions between reactant particles per unit time ( $z$ ). The frequency factor is the orientation factor times the collision frequency, $A = p z$ .
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.

Work through an example

A first-order reaction has these rate constants at four temperatures. Find the activation energy and the frequency factor.

Find

E_{a}

and

A

from an Arrhenius plot →

Find $E_{a}$ from two rate constants →

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Study the 6 key terms Check Ea from the slope in a Chemistry box Fit the Arrhenius plot in Kinetics & Arrhenius See what Ea does to energetic collisions Open worked example on a board Chemistry formulas: experimental analysis

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