How To Calculate The Frequency Factor In Chemical Kinetics

If you've ever wondered how engineers calculate the strength of concrete they create for their projects or how chemists and physicists measure the electrical conductivity of materials, much of it comes down to how fast chemical reactions occur.

Figuring out how fast a reaction happens means looking at the reaction kinematics. The Arrhenius equation lets you do such a thing. The equation involves the natural logarithm function and accounts for the rate of collision between particles in the reaction.

In one version of the Arrhenius equation, you can calculate the rate of a first-order chemical reaction. First-order chemical reactions are ones in which the rate of reactions depends only on the concentration of one reactant. The equation is:

\(K=Ae^{-E_a/RT}\)

Where ​K​ is the reaction rate constant, the energy of activation is ​E​​__a​ (in joules), ​R​ is the reaction constant (8.314 J/mol K), ​T​ is the temperature in Kelvin and ​A​ is the frequency factor. To calculate the frequency factor ​A​ (which is sometimes called ​Z​), you need to know the other variables ​K​, ​E_a​, and ​T_​.

The activation energy is the energy that the reactant molecules of a reaction must possess in order for a reaction to occur, and it's independent of temperature and other factors. This means that, for a specific reaction, you should have a specific activation energy, typically given in joules per mole.

The activation energy is often used with catalysts, which are enzymes that speed up the process of reactions. The ​R​ in the Arrhenius equation is the same gas constant used in the ideal gas law ​PV = nRT​ for pressure ​P​, volume ​V​, number of moles ​n​, and temperature ​T​.

The Arrhenius equations describes many reactions in chemistry such as forms of radioactive decay and biological enzyme-based reactions. You can determine the half-life (the time required for the reactant's concentration to drop by half) of these first-order reactions as ln (2) / ​K​ for the reaction constant ​K​. Alternatively, you can take the natural logarithm of both sides to change the Arrhenius equation into ln (​K​) ​=​ ln (​A​) ​_− E_a/RT​​._​ This lets you calculate the activation energy and temperature more easily.

The frequency factor is used to describe the rate of molecular collisions that occur in the chemical reaction. You can use it to measure the frequency of the molecular collisions that have the proper orientation between particles and appropriate temperature so that the reaction can occur.

The frequency factor is generally obtained experimentally to make sure the quantities of a chemical reaction (temperature, activation energy and rate constant) fit the form of the Arrhenius equation.

The frequency factor is temperature-dependent, and, because the natural logarithm of the rate constant ​K​ is only linear over a short range in temperature changes, it's difficult to extrapolate the frequency factor over a broad range of temperatures.

As an example, consider the following reaction with rate constant ​K​ as 5.4 × 10 ⁻⁴ M ⁻¹s ⁻¹ at 326 °C and, at 410 °C, the rate constant was found to be 2.8 × 10 ⁻² M ⁻¹s ⁻¹. Calculate the activation energy ​_E_a​ and frequency factor ​A_​.

H₂(g) + I₂(g) → 2HI(g)

You can use the following equation for two different temperatures ​T​ and rate constants ​K​ to solve for activation energy ​_E_a_​.

\(\ln\bigg(\frac{K_2}{K_1}\bigg) = -\frac{E_a}{R}\bigg(\frac{1}{T_2} – \frac{1}{T_1}\bigg)\)

Then, you can plug the numbers in and solve for ​_E_a_​. Make sure to convert the temperature from Celsius to Kelvin by adding 273 to it.

\(\ln\bigg(\frac{5.4 ×10^{-4} \;\text{M}^{-1}\text{s}^{-1}}{2.8 ×10^{-2}\; \text{M}^{-1}\text{s}^{-1}}\bigg) = -\frac{E_a}{R}\bigg(\frac{1}{599 \;\text{K}} – \frac{1}{683 \;\text{K}}\bigg)\)

\(\begin{aligned}
E_a&= 1.92 × 10^4 \;\text{K} × 8.314 \;\text{J/K mol} \
&= 1.60× 10^5 \;\text{J/mol}
\end{aligned}\)

You can use either temperature's rate constant to determine the frequency factor ​A​. Plugging in the values, you can calculate ​A​.

\(k = Ae^{-E_a/RT}\)

\(5.4 × 10^{-4} \;\text{M}^{-1}\text{s}^{-1} =A e^{-\frac{1.60 × 10^5 \;\text{J/mol}}{8.314 \;\text{J/K mol} ×599 \;\text{K}}}\)
\(A = 4.73 × 10^{10} \;\text{M}^{-1}\text{s}^{-1}\)