3.6 Arrhenius Equation
We have already seen how temperature affects the rate of reaction via the rate constant. The relationship between temperature, activation energy and rate constant is given by the Arrhenius equation.
Exponential Form
\[k = Ae^{\frac{-E_{\mathrm{a}}}{RT}}\]
As temperature increases, the rate constant increases and therefore the rate of reaction (as seen with the rate law).
The Arrhenius equation has a similar form to the Clausius-Clapeyron equation and can therefore be written in a linear and two-point form.
Linear Form
\[\ln k = \dfrac{-E_{\mathrm{a}}}{R} \left (\dfrac{1}{T}\right) + \ln A\]
Two-point Form
\[\ln\dfrac{k_2}{k_1} = \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\]
The explanations for these equations are similar to the Clausius-Clapeyron equations except here we relate rate constant with temperature.
We can numerically determine a set of rate constants for a range of temperatures and then plot these via the linear form (ln(k) vs 1/T). The resulting slope allows us to determine the energy of activation. The pre-exponential factor, A, is discussed in Collision Theory.
\[\mathrm{slope} = \dfrac{-E_{\mathrm{a}}}{R}\]
Additionally, we can determine the energy of activation by evaluating the two-point Arrhenius equation. The activation energy does not change with temperature.
You should be able to solve for any variable in these equations. The algebra for isolating each term is given below.
Ea
\[\begin{align*} \ln \left ( \dfrac{k_2}{k_1} \right ) &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[4ex] \dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} &= \dfrac{E_{\mathrm{a}}}{R} \\[2ex] R\left (\dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) &= E_{\mathrm{a}} \\[2ex] E_{\mathrm{a}} &= R\left (\dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \end{align*}\]
k1
\[\begin{align*} \ln \left ( \dfrac{k_2}{k_1} \right ) &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[4ex] \ln k_2 - \ln k_1 &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[2ex] - \ln k_1 &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) - \ln k_2\\[2ex] \ln k_1 &= -\dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) + \ln k_2\\[2ex] e^{(\ln k_1)} &= e^{\left(-\frac{E_{\mathrm{a}}}{R} \left ( \frac{1}{T_1} - \frac{1}{T_2} \right ) + \ln k_2\right)}\\[2ex] k_1 &= e^{\left(-\frac{E_{\mathrm{a}}}{R} \left ( \frac{1}{T_1} - \frac{1}{T_2} \right ) + \ln k_2\right)} \end{align*}\]
k2
\[\begin{align*} \ln \left ( \dfrac{k_2}{k_1} \right ) &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[4ex] \ln k_2 - \ln k_1 &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[2ex] \ln k_2 &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) + \ln k_1\\[2ex] e^{(\ln k_2)} &= e^{\left(\frac{E_{\mathrm{a}}}{R} \left ( \frac{1}{T_1} - \frac{1}{T_2} \right ) + \ln k_1\right)}\\[2ex] k_2 &= e^{\left(\frac{E_{\mathrm{a}}}{R} \left ( \frac{1}{T_1} - \frac{1}{T_2} \right ) + \ln k_1\right)} \end{align*}\]
T1
\[\begin{align*} \ln \left ( \dfrac{k_2}{k_1} \right ) &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[4ex] \dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{E_{\mathrm{a}}}{R}} &= \dfrac{1}{T_1} - \dfrac{1}{T_2} \\[2ex] \dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{E_{\mathrm{a}}}{R}} + \dfrac{1}{T_2} &= \dfrac{1}{T_1} \\[2ex] \dfrac{1}{\left (\dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{E_{\mathrm{a}}}{R}} + \dfrac{1}{T_2}\right )} &= T_1 \\[2ex] T_1 &= \dfrac{1}{\left (\dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{E_{\mathrm{a}}}{R}} + \dfrac{1}{T_2}\right ) } \end{align*}\]
T2
\[\begin{align*} \ln \left ( \dfrac{k_2}{k_1} \right ) &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\\[4ex] \dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{E_{\mathrm{a}}}{R}} &= \dfrac{1}{T_1} - \dfrac{1}{T_2} \\[2ex] \dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{E_{\mathrm{a}}}{R}} - \dfrac{1}{T_1} &= -\dfrac{1}{T_2} \\[2ex] -\dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{E_{\mathrm{a}}}{R}} + \dfrac{1}{T_1} &= \dfrac{1}{T_2} \\[2ex] \dfrac{1}{\left ( -\dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{E_{\mathrm{a}}}{R}} + \dfrac{1}{T_1} \right )} &= T_2 \\[2ex] T_2 &= \dfrac{1}{\left (-\dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{E_{\mathrm{a}}}{R}} + \dfrac{1}{T_1}\right )} \end{align*}\]
Practice
As seen previously, the following reaction has a rate constant of 0.3387 M–1 s–1 at 326.85 °C (or 600 K).
\[2\mathrm{NO_2}(g) \longrightarrow \mathrm{2NO}(g) + \mathrm{O_2}(g)\]
At 526.85 °C (800 K), the rate constant is 93.823 M–1 s–1.
What is the energy of activation (in kJ mol–1)?
Solution
We have two rate constants and two temperatures. We can solve the Arrhenius two-point form.
\[\begin{align} \ln \left ( \dfrac{k_2}{k_1} \right ) &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) \longrightarrow\\[4ex] E_{\mathrm{a}} &= R\left (\dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \\[2ex] &= 8.314~\mathrm{J~mol^{-1}~K^{-1}} \left (\dfrac{\ln\left (\dfrac{93.82~M^{-1}~\mathrm{s}^{-1}}{0.3387~M^{-1}~\mathrm{s}^{-1}}\right )} {\dfrac{1}{600~\mathrm{K}} - \dfrac{1}{800~\mathrm{K}}} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\ &= 112.22~\mathrm{kJ~mol^{-1}} \end{align}\]
Math String: 8.314 * ln(93.82/0.3387) / (1/600 - 1/800) / 1000
Rationalize
The energy of activation for this reaction is quite high. Many reactions that take place at around room temperature have activation energies of less than 80 kJ mol–1. This reaction is very fast at high temperatures (300+ °C) and is very slow at low temperatures (such as room temperature).
Extend
Estimate the rate constant for this reaction at room temperature (25 °C). The pre-exponential factor for the reaction is 2 × 109.
Solve
\[\begin{align*} k &= Ae^{\frac{-E_{\mathrm{a}}}{RT}} \\[1.5ex] &= (2\times 10^{9}) e^{\frac{-(112.22~\mathrm{kJ~mol^{-1}})}{\frac{8.314~\mathrm{J}}{\mathrm{mol~K}} \left (\frac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \left ( 298.15~\mathrm{K} \right ) }}\\[1.5ex] &= 4.364\times 10^{-11}~M^{-1}~\mathrm{s^{-1}} \end{align*}\]
Math String: 2e9 * e^( -112.22 / (8.314 / 1000 * 298.15))
The reaction is ten orders of magnitude slower at room temperature!
Plot
Above we’ve solved for two data points on a k vs. T plot. Let’s plot the rate constant vs. T from 600 K to 2000 K (326.85 °C to 1,726.85 °C).
Figure 3.21: Rate constant vs temperature for this second-order reaction.
This reaction is FAST at higher temperatures.
Let’s transform the data to give a straight line as the linear-form of the Arrhenius equation is set up to do. We simply plot ln(k) vs. 1/T.
Figure 3.22: Linear-form of the Arrhenius equation.
The resulting slope is
\[\mathrm{slope} = -13,498\]
The energy of activation (in kJ mol–1) can be determined.
\[\begin{align*} \dfrac{-E_{\mathrm{a}}}{R} &= \mathrm{slope} \\[1.5ex] E_{\mathrm{a}} &= -\mathrm{slope} \times R \\[1.5ex] E_{\mathrm{a}} &= - (-13,498) \left ( \dfrac{8.314~\mathrm{J}}{\mathrm{mol~K}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[1.5ex] &= 112.22~\mathrm{kJ~mol^{-1}} \end{align*}\]
Math String: -1 * -13498 * 8.314 / 1000