Arrhenius Equation

The Arrhenius equation demonstrates the affect of various properties on the rate constant (and consequently, the rate) of a reaction.
This equation is grounded in collision theory. The equation is given as

$k = Ae^{\frac{-E_{\mathrm{a}}}{RT}}$

Variable Description
k rate constant
A pre-exponential factor (constant; reaction dependent)
Ea activation energy (J mol–1)
R gas constant (8.315 J mol–1 K–1)
T temperature (in K)

It is important to note the following qualitative relationships:

• As Ea gets larger, k gets smaller, rate gets slower
• As T gets larger, k gets larger, rate gets faster
• As A gets larger, k gets larger, rate gets faster

Therefore, as the activation energy of a reaction increases, the rate constant (and subsequently the rate of the reaction) decreases. As the temperature or pre-exponential factor increases, the rate constant and rate of reaction increases.

The pre-exponential factor (or frequency factor), A, represents the frequency of effective collisions between molecules that can undergo a chemical change. This factor is determined via experiment and extrapolation.

Arrhenius Equation Two-point Form

The Arrhenius equation can be given in a two-point form (similar to the Clausius-Claperyon equation).

$\ln\dfrac{k_2}{k_1} = \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )$

Below are the algebraic steps to solve for any variable in the Clausius-Clapeyron two-point form equation.

Ea

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\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

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\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

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\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

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\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

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\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 Problems

Problem 1

For the following reaction, the rate constant at 701 K is 2.57 M–1 s–1 and at 895 K is 567 M–1 s–1. What is the activation energy (in kJ mol–1)?

NO2 (g) + CO (g) ⟶ NO (g) + CO2 (g)

Solution

\begin{align*} \ln \dfrac{k_2}{k_1} &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) \longrightarrow \\ E_{\mathrm{a}} &= R \left ( \dfrac{\ln\left ( \dfrac{k_2}{k_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \\ &= 8.315~\mathrm{J~mol^{-1}~K^{-1}} \left (\dfrac{\ln\left (\dfrac{567~M^{-1}~\mathrm{s}^{-1}}{2.57~M^{-1}~\mathrm{s}^{-1}}\right )} {\dfrac{1}{701~\mathrm{K}} - \dfrac{1}{895~\mathrm{K}}} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\ &= 1.5\times 10^{2}~\mathrm{kJ~mol^{-1}} \end{align*}

Timed Assessments

VIDEO

1. 66.0 °C
2. 31.0 °C
3. 38.0 °C

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1. 52.0 °C
1. °C
2. 112 °C

Solve for k1

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1. 0.1324 M s–1
2. 5.218 M–1 s–1
3. 0.001430 s–1

Solve for k2

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1. 0.8341 M s–1
2. 3.418 M–1 s–1
3. 0.003460 s–1

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