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:$k=Ae^{-\frac{E_a}{RT}}$
- E_{a} 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.
$\ln{k}=-\frac{E_a}R\frac1T+\ln{A}$
Graph of Arrhenius Equation
$\begin{aligned}{\ln{k}_1=-\frac{E_{\rm a}}R\frac1{T_1}+\ln{A}}\\{\ln{k}_2=-\frac{E_{\rm a}}R\frac1{T_2}+\ln{A}}\end{aligned}$
$\ln{k}_2-\ln{k}_1=\left(\!-\frac{E_{\rm a}}R\frac1{T_2}+\ln{A}\right)-\left(\!-\frac{E_{\rm a}}R\frac1{T_1}+\ln{A}\right)$
$\ln{k}_2-\ln{k}_1=-\frac{E_{\rm a}}R\left(\frac1{T_2}-\frac1{T_1}\right)+\ln{A}-\ln{A}$
$\ln\frac{k_2}{k_1}=-\frac{E_{\rm a}}R\left(\frac1{T_2}-\frac1{T_1}\right)$
Step-By-Step Example
Calculating Activation Energy for the Decomposition of Dinitrogen Pentoxide
Dinitrogen pentoxide (N_{2}O_{5}) breaks down into nitrogen dioxide (NO_{2}) and oxygen gas (O_{2}) with the following reaction:
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 E_{a} for this reaction.
$2{\rm N}_2{\rm O}_5(\mathit g)\rightarrow4{\rm{NO}}_2(\mathit g)+{\rm O}_2(\mathit g)$
Step 1
Use the modified version of the Arrhenius equation, relating k_{1} and T_{1}, and k_{2} and T_{2}.
Note the temperatures are in K, not °C.
$\begin{aligned}\ln\frac{k_2}{k_1}&=-\frac{E_{\rm{a}}}R\left(\frac1{T_2}-\frac1{T_1}\right)\\-\ln\frac{1.50\times10^{-3}\;{\rm{s}}^{-1}}{3.46\times10^{-5}\;{\rm{s}}^{-1}}&=\frac{E_{\rm{a}}}{8.314\;\rm{J/K}{\cdot}{\rm{mol}}}\left(\frac1{328.15\;\rm{{K}}}-\frac1{298.15\;{\rm{K}}}\right)\end{aligned}$
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 E_{a}, it is possible to calculate the value of A.