Chemistry · General chemistry II · Worked example

Find Ea and A from an Arrhenius plot

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

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

Transform the data

The plot needs $1 / T$ in $K^{- 1}$ and $ln k$ , so compute both for every row before fitting. $ln k$ is negative here because every $k$ is less than $1 s^{- 1}$ .

Rate constants and Arrhenius coordinates
$T$ $(K)$	$k$ $(s^{- 1})$	$1 / T$ $(K^{- 1})$	$ln k$
$300$	$1.92 \times 10^{- 3}$	$3.3333 \times 10^{- 3}$	$- 6.255$
$310$	$4.46 \times 10^{- 3}$	$3.2258 \times 10^{- 3}$	$- 5.413$
$320$	$9.81 \times 10^{- 3}$	$3.1250 \times 10^{- 3}$	$- 4.624$
$330$	$2.06 \times 10^{- 2}$	$3.0303 \times 10^{- 3}$	$- 3.882$

Fit a straight line

The four points lie on a straight line, with $R^{2} = 0.9999998$ , so one activation energy describes this range. The least-squares slope is $- 7.83 \times 10^{3} K$ , and the intercept is $19.844$ . The slope has units of K because $ln k$ has none and $1 / T$ is in $K^{- 1}$ .

slope = - 7.83 \times 10^{3} K, intercept = 19.844

\begin{array}{c} slope = - 7.83 \times 10^{3} K \\ intercept = 19.844 \end{array}

Activation energy from the slope

The slope equals $- E_{a} / R$ , so $E_{a} = - R \times slope$ . Call the slope $m$ . Kelvins cancel, leaving $J / mol$ , and $6.51 \times 10^{4} J / mol$ is $65.1 kJ / mol$ .

E_{a} = - (8.314 J {mol}^{- 1} K^{- 1}) (- 7.83 \times 10^{3} K) = 65.1 kJ/mol

\begin{array}{c} E_{a} = - (8.314) (- 7.83 \times 10^{3}) \\ = 6.51 \times 10^{4} J/mol \\ = 65.1 kJ/mol \end{array}

Frequency factor from the intercept

The intercept is $ln A$ , so $A = e^{19.844}$ . Keep the third decimal of the intercept, because rounding it to $19.84$ would give $4.1 \times 10^{8}$ . $A$ has the units of $k$ , and because it is an extrapolation to $1 / T = 0$ , two significant figures are plenty.

A = e^{19.844} \approx 4.2 \times 10^{8} s^{- 1}

Check one point

Put the results back in at $320 K$ . $(4.15 \times 10^{8} s^{- 1}) e^{- 65, 100 / (8.314 \times 320)}$ $= 9.8 \times 10^{- 3} s^{- 1}$ , the measured $9.81 \times 10^{- 3} s^{- 1}$ .

Result

$E_{a} = 65.1 kJ / mol$ and $A \approx 4.2 \times 10^{8} s^{- 1}$ .

E_{a} = 65.1 kJ/mol

Your turn

1. One like the example

An Arrhenius plot of $ln k$ against $1 / T$ has a slope of $- 6.00 \times 10^{3} K$ . What is $E_{a}$ ?

Show the answer and explanation

$49.9 kJ / mol$ .

The slope is $- E_{a} / R$ , so $E_{a} = - R \times slope$ $= - (8.314 J / (mol \cdot K)) (- 6.00 \times 10^{3} K)$ $= 4.99 \times 10^{4} J / mol$ . Divide by $1000$ to get $49.9 kJ / mol$ . A negative slope gave a positive $E_{a}$ , as it must.

\begin{array}{l} E_{a} = - R m \\ m = - 6.00 \times 10^{3} K \\ R = 8.314 J {mol}^{- 1} K^{- 1} \\ E_{a} = - (8.314 J {mol}^{- 1} K^{- 1}) (- 6.00 \times 10^{3} K) = 4.99 \times 10^{4} J/mol \end{array}

\begin{array}{c} E_{a} = - R m \\ m = - 6.00 \times 10^{3} K \\ R = 8.314 J {mol}^{- 1} K^{- 1} \\ E_{a} = - (8.314 J {mol}^{- 1} K^{- 1}) \\ \times (- 6.00 \times 10^{3} K) \\ = 4.99 \times 10^{4} J/mol \end{array}

2. A variation

Two points on an Arrhenius line are $(1 / T, ln k)$ $= (3.300 \times 10^{- 3} K^{- 1}, - 7.20)$ and $(3.100 \times 10^{- 3} K^{- 1}, - 5.40)$ . Find the slope and $E_{a}$ .

Show the answer and explanation

The slope is $- 9.00 \times 10^{3} K$ , and $E_{a} = 74.8 kJ / mol$ .

This time the slope is not given. Rise over run is $(- 5.40 - (- 7.20))$ $\div (3.100 \times 10^{- 3} - 3.300 \times 10^{- 3})$ $= 1.80 \div (- 2.00 \times 10^{- 4})$ $= - 9.00 \times 10^{3} K$ . Then $E_{a} = - R \times slope$ $= 8.314 \times 9.00 \times 10^{3}$ $= 7.48 \times 10^{4} J / mol$ , or $74.8 kJ / mol$ .

\begin{array}{l} \frac{- 5.40 - (- 7.20)}{3.100 \times 10^{- 3} K^{- 1} - 3.300 \times 10^{- 3} K^{- 1}} = - 9.00 \times 10^{3} K \\ E_{a} = - R m \\ m = - 9.00 \times 10^{3} K \\ R = 8.314 J {mol}^{- 1} K^{- 1} \\ E_{a} = - (8.314 J {mol}^{- 1} K^{- 1}) (- 9.00 \times 10^{3} K) = 7.48 \times 10^{4} J/mol \end{array}

\begin{array}{c} \frac{- 5.40 - (- 7.20)}{(3.100 - 3.300) \times 10^{- 3} K^{- 1}} \\ = - 9.00 \times 10^{3} K \\ E_{a} = - R m \\ m = - 9.00 \times 10^{3} K \\ R = 8.314 J {mol}^{- 1} K^{- 1} \\ E_{a} = - (8.314 J {mol}^{- 1} K^{- 1}) \\ \times (- 9.00 \times 10^{3} K) \\ = 7.48 \times 10^{4} J/mol \end{array}

3. Somewhere new

Use the worked results, $E_{a} = 6.51 \times 10^{4} J / mol$ and $A = 4.2 \times 10^{8} s^{- 1}$ , to estimate $k$ at $340 K$ .

Show the answer and explanation

$k \approx 4.2 \times 10^{- 2} s^{- 1}$ .

Now the equation runs forward, from $E_{a}$ and $A$ to $k$ . $E_{a} / R T$ $= 65, 100 \div (8.314 \times 340)$ $= 23.03$ , and $e^{- 23.03} = 1.0 \times 10^{- 10}$ . Then $k = 4.2 \times 10^{8} \times 1.0 \times 10^{- 10}$ $= 4.2 \times 10^{- 2} s^{- 1}$ . $340 K$ is just past the measured $330 K$ , so this is a short extrapolation, about twice the $k$ at $330 K$ .

\begin{array}{l} k = A e^{- \frac{E_{a}}{R T}} \\ A = 4.2 \times 10^{8} s^{- 1} \\ E_{a} = 6.51 \times 10^{4} J/mol \\ R = 8.314 J {mol}^{- 1} K^{- 1} \\ T = 340 K \\ k = (4.2 \times 10^{8} s^{- 1}) e^{- \frac{6.51 \times 10^{4} J/mol}{(8.314 J {mol}^{- 1} K^{- 1}) (340 K)}} = 4.2 \times 10^{- 2} s^{- 1} \end{array}

\begin{array}{c} k = A e^{- \frac{E_{a}}{R T}} \\ A = 4.2 \times 10^{8} s^{- 1} \\ E_{a} = 6.51 \times 10^{4} J/mol \\ R = 8.314 J {mol}^{- 1} K^{- 1} \\ T = 340 K \\ k = (4.2 \times 10^{8} s^{- 1}) \\ \times e^{- \frac{6.51 \times 10^{4} J/mol}{(8.314 J {mol}^{- 1} K^{- 1}) (340 K)}} \\ = 4.2 \times 10^{- 2} s^{- 1} \end{array}

Keep exploring

Fit the same four points in Kinetics & Arrhenius, then add $340 K$ as the prediction temperature. It predicts $k = 0.0414 s^{- 1}$ . The rounded $E_{a}$ and $A$ give $4.2 \times 10^{- 2} s^{- 1}$ , and both are a short extrapolation past the data.

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Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

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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Transform the data

Fit a straight line

Activation energy from the slope

Frequency factor from the intercept

Check one point

Result

Your turn

1. One like the example

49.9 kJ⁢/⁢mol49.9 kJ/mol.

2. A variation

The slope is −9.00×103 K−9.00×103 K, and Ea=74.8 kJ⁢/⁢molEa​=74.8 kJ/mol.

3. Somewhere new

k≈4.2×10−2 s−1k≈4.2×10−2 s−1.

Keep exploring

Study it, then try it

$49.9 kJ / mol$ .

The slope is $- 9.00 \times 10^{3} K$ , and $E_{a} = 74.8 kJ / mol$ .

$k \approx 4.2 \times 10^{- 2} s^{- 1}$ .