# Arrhenius Equation Colliding molecules or atoms must have a minimum amount of energy for a chemical reaction to occur. Temperature is a measure of the energy of reactants. The Arrhenius equation provides a relationship between rate constant and temperature.

An important factor affecting reaction rates is temperature. At a higher temperature, the kinetic energy of the molecules is greater, which means they move faster. According to collision theory, reaction rate is higher at higher temperatures. More molecules are able to overcome the activation energy, and the number of collisions increases.

The relationship between temperature and rate is nonlinear. A broad rule of thumb is that reaction rate doubles with every 10°C increase in temperature.

Svante Arrhenius, a Swedish chemist, experimentally determined the relationship between temperature and rate in 1889. This was before a publication of collision theory. The Arrhenius equation relates temperature to the rate constant:
$k=Ae^{-\frac{E_a}{RT}}$
• Ea is the activation energy, a constant that is dependent on the nature of the reaction. It has energy units which are most often joules.
• T is absolute temperature in kelvins.
• R is the gas constant, 8.314 J/K·mol.
• A is a constant that is dependent on the nature of the reaction. This constant represents how often molecules collide and what percentage of the collisions are in the correct orientation. A is independent of temperature. Its units vary, depending on the reaction.
• k is the rate constant.
If the natural logarithm of both sides is taken, the Arrhenius equation can be written as
$\ln{k}=-\frac{E_a}R\frac1T+\ln{A}$
This equation is in the form $y=mx+b$, which is the equation of a straight line. Plotting $\ln{k}$ versus $\frac{1}{T}$ produces a straight-line graph. Furthermore, the slope of the line is $\frac{-E_{\text{a}}}{R}$, and the $\ln{k}$ term is the y-intercept of the graph. This allows scientists to determine Ea and A from measured rate constants and temperature.

#### Graph of Arrhenius Equation A graph is useful for helping solve for quantities in the Arrhenius equation. Using measured values of the rate constant k and the temperature T, a graph of ln⁡k \ln{k} lnk versus 1T\frac{1}{T}T1​ allows for the calculation of the activation energy EaE_{\rm a}Ea​ and the coefficient A. The slope of the graph is equal to −EaR\frac{-E_{\rm{a}}}{R}R−Ea​​, where R is the gas constant.
With two measured values of k, the rate constant, and T, temperature, for the same reaction, it is possible to eliminate the $\ln{A}$ term and calculate Ea mathematically. Two measurements are taken, k1 and T1, and k2 and T2.
\begin{aligned}{\ln{k}_1=-\frac{E_{\rm a}}R\frac1{T_1}+\ln{A}}\\{\ln{k}_2=-\frac{E_{\rm a}}R\frac1{T_2}+\ln{A}}\end{aligned}
The slope of the Arrhenius equation can be calculated from the change in k and T by subtracting the first equation from the second equation.
$\ln{k}_2-\ln{k}_1=\left(\!-\frac{E_{\rm a}}R\frac1{T_2}+\ln{A}\right)-\left(\!-\frac{E_{\rm a}}R\frac1{T_1}+\ln{A}\right)$
To simplify the equation, rearrange the terms on the right.
$\ln{k}_2-\ln{k}_1=-\frac{E_{\rm a}}R\left(\frac1{T_2}-\frac1{T_1}\right)+\ln{A}-\ln{A}$
The $\ln{A}$ term cancels, and the equation can be rearranged to calculate Ea from the measured k and T values, as well as R, the ideal gas constant.
$\ln\frac{k_2}{k_1}=-\frac{E_{\rm a}}R\left(\frac1{T_2}-\frac1{T_1}\right)$
Step-By-Step Example
Calculating Activation Energy for the Decomposition of Dinitrogen Pentoxide
Dinitrogen pentoxide (N2O5) breaks down into nitrogen dioxide (NO2) and oxygen gas (O2) with the following reaction:
$2{\rm N}_2{\rm O}_5(\mathit g)\rightarrow4{\rm{NO}}_2(\mathit g)+{\rm O}_2(\mathit g)$
For this reaction, at 25.0°C the rate constant is measured to be $3.46\times10^{-5}\,\rm{s}^{-1}$. At 55.0°C the rate constant is measured to be $1.50\times10^{-3}\,\rm s^{-1}$. Calculate Ea for this reaction.
Step 1
Use the modified version of the Arrhenius equation, relating k1 and T1, and k2 and T2.
\begin{aligned}\ln\frac{k_2}{k_1}&=-\frac{E_{\rm{a}}}R\left(\frac1{T_2}-\frac1{T_1}\right)\\-\ln\frac{1.50\times10^{-3}\;{\rm{s}}^{-1}}{3.46\times10^{-5}\;{\rm{s}}^{-1}}&=\frac{E_{\rm{a}}}{8.314\;\rm{J/K}{\cdot}{\rm{mol}}}\left(\frac1{328.15\;\rm{{K}}}-\frac1{298.15\;{\rm{K}}}\right)\end{aligned}
Note the temperatures are in K, not °C.
Solution
This equation can be solved for $E_{\rm a}$.
\begin{aligned}-3.769&=\frac{E_{\rm{a}}}{8.314\;\rm{J/K}{\cdot}{\rm{mol}}}(-3.0663\times10^{-4}\text{ K}^{-1})\\E_{\rm{a}}&=102{,}193\;\rm{J/mol}\\&=102\;\rm{kJ/mol}\end{aligned}
Note that by using any of the data pairs of k and T and Ea, it is possible to calculate the value of A.