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Wize University Chemistry Textbook > Kinetics

Arrhenius Equation

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Arrhenius Equation

  • Rates of reaction are highly temperature dependent! Imagine the speed of ice melting at 0oC vs 40oC
  • Svante Arrhenius discovered a simple equation that relates the rate of a chemical reaction to the temperature which it is performed at. The Arrhenius equation is shown below.
ln⁡k=ln⁡A−EaRT\ln k=\ln A-\frac{Ea}{RT}

k=Ae−Ea/RTk=Ae^{-E_a/RT}
  • k is still the rate constant
  • A is called the pre-exponential factor, it is unique for all reactions and is constant across all temperatures. It is related to the probability of a successful collision.
  • T is temperature and R is the Gas constant
  • Ea is the activation energy, which is shown below graphically

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  • Activation energy, Ea, is the amount of energy it takes to get to the products
  • The Arrhenius equation on its own is not very useful because of the pre-exponential factor A is difficult to calculate. However by dividing one rate by another and rearranging we end up with the following equation. Notice that A has been cancelled out and is no longer necessary.
ln⁡(k1k2)=−EaR(1T1−1T2)\ln \bigg(\cfrac{k_1}{k_2}\bigg)=-\cfrac{E_a}{R}\bigg(\cfrac{1}{T_1}-\cfrac{1}{T_2}\bigg)
  • We can use this equation to compare the rate of a reaction at two different temperatures.
a. A chemical reaction increases its rate by a factor of 15 when it is warmed from room temperature to 100oC. What is the activation energy of this process? R = 8.314 J/Kmol

ln⁡(k1k2)=−EaR(1T1−1T2)\ln \bigg(\cfrac{k_1}{k_2}\bigg)=\cfrac{-E_a}{R}\bigg(\cfrac{1}{T_1}-\cfrac{1}{T_2}\bigg)
Ea=−ln⁡(k1k2)R(1T1−1T2)=ln⁡(115)(8.314JK−1mol−1)(1298K−1373K)=33.4kJmol−1E_a=\cfrac{-\ln\bigg(\cfrac{k_1}{k_2}\bigg)R}{\bigg(\cfrac{1}{T_1}-\cfrac{1}{T_2}\bigg)}=\cfrac{\ln \bigg(\cfrac{1}{15}\bigg)(8.314 JK^{-1}mol^{-1})}{\bigg(\cfrac{1}{298K}-\cfrac{1}{373K}\bigg)}=33.4kJmol^{-1}
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If the activation energy of a reaction is measured to be 59.6 kJ/mol, by what factor will the rate of this reaction be increased when the temperature is raised from 305 K to 405 K?

We start with the adapted Arrhenius equation,
ln⁡(k1k2)=−EaR(1T1−1T2)\ln \bigg(\cfrac{k_1}{k_2}\bigg)=-\cfrac{E_a}{R}\bigg(\cfrac{1}{T_1}-\cfrac{1}{T_2}\bigg)
the question is asking for the ratio of k2 to k1
k1k2=e−5.8=3.02×10−3\frac{k_1}{k_2}=e^{-5.8}=3.02\times 10^{-3}

k2k1=13.02×10−3=331\frac{k_2}{k_1}=\frac{1}{3.02\times10^{-3}}=331
The reaction will be 331 times faster at 405 K than it will be at 305 K. That's quite a bit!
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A Schiff base reaction between 3-propanone and aniline proceeds with rate constants of 0.0065 M-1 s -1 and 0.0084 M-1 s -1 at 30oC and 80oC respectively. Determine the activation energy and preexponential factor for this reaction.




Extra Practice
Kinetics: Arrhenius Equation
What is the activation energy of a reaction if the rate constant at 40oC is double the value of the rate constant at 25oC?
A Schiff base reaction between 3-propanone and aniline proceeds with rate constants of 0.0065 M-1 s -1 and 0.0084 M-1 s -1 at 30oC and 80oC respectively. Determine the activation energy for this reaction.
Report your answer in kJ/mol to two decimal places.
In the gas phase, carbon can react with oxygen as shown below.
C+O2→k1CO2Ea=24.6 kJ/mol\rm C\quad+\quad O_2\quad\overset{k_1}{\overrightarrow{\hspace{1.5cm}}}\quad CO_2\qquad E_a=24.6\ kJ/mol
a) If the reaction is run at room temperature the rate constant is found to be 0.04 atm s−10.04\ atm\ s^{-1}. Determine the pre-exponential factor AA.
A first order gas phase reaction N2O4 → 2 NO2 was carried out isothermally and the activation barrier was found to be 23.5 kJ/mol. Initially p(N2O4) = 2.6 atm and p(NO2) = 0. After 1 hour p(NO2) = 2.0 atm and p(N2O4) =1.6 atm. Assume that both NO2 and N2O4 are ideal gases.
Determine the first order rate constant k, and the half life t1/2.
The above reaction was performed at 220oC. Calculate the pre-exponential factor.
In the gas phase, carbon can react with oxygen in two different ways as shown below.
C+O2→k1CO2Ea=24.6kJ/molC+O_2 \xrightarrow{k_1} CO_2 \hspace{20pt} E_a=24.6kJ/mol
C+0.5O2→k2COEa=19.6kJ/molC+0.5O_2 \xrightarrow{k_2} CO \hspace{20pt} E_a=19.6kJ/mol
Kinetics: Arrhenius Equation
What is the activation energy of a reaction if the rate constant at 40oC is double the value of the rate constant at 25oC?