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Arrhenius Equation

The Arrhenius equation is one of the most important equations in physical chemistry. It describes the relationship between the reaction rate and the temperature for many chemical reactions. It is a simple and accurate formula used to study the temperature dependence of the rate constant [1-4].

Swedish scientist Svante Arrhenius proposed the equation in 1889 by combining the concept of activation energy and the Boltzmann distribution law.

Form of Arrhenius Equation

The Arrhenius equation is rooted in experimental observations and cannot be derived logically. The common form of the Arrhenius equation is given by an exponential equation, as shown below [1-5].

k = A exp(-E/RT)

Where

k : rate constant

A : pre-exponential factor

E : activation energy

T : temperature

R : universal gas constant ( its value is 8.314 kJ M-1 K-1)

The Arrhenius equation has a massive and significant application in determining chemical reaction rates and calculating the activation energy. The equation tells us that the rate constant is decaying, not with time but with E/RT. The term RT represents the average kinetic energy. So, the rate constant decays with the ratio of the activation energy over average kinetic energy. The larger the ratio, the smaller the rate constant. High temperature and low activation energy result in a large rate constant and thus accelerate the chemical reaction.

Units

The unit of the various variables and constants of Arrhenius equations are as follows:

Activation energy: J/mol

Temperature: K or Kelvin

Universal gas constant (R): J/mol·K

Pre-exponential factor and rate constant: s-1 for a first-order reaction

Arrhenius Plot

Another way to write the Arrhenius equation is by taking the logarithm of both sides [1-5].

ln (k) = ln (A) – E/RT

Rearranging the equation gives

ln (k) = (– E/R) (1/T) + ln (A)

The above equation is more convenient and easier to interpret graphically than the exponential equation. It is in the form of a straight line: y = mx + c. Let us look at the various terms in this equation.

Dependent variable: y = ln (k)

Independent variable: x = (1/T)

Slope: m = (-E/R)

Intercept: c = ln (A)

A plot of ln (k) vs. 1/T gives a straight line with slope = -E/R and intercept = ln (A). This plot is known as the Arrhenius plot and is shown in the image below.

Arrhenius Equation

Arrhenius plot is conveniently used to determine the activation energy E and the pre-exponential factor A.

Multiplying the slope m by the negative of the universal gas constant R gives E.

E = m x (-R)

Taking the exponential of the intercept c gives A.

A = exp (c)

Exponential Term

The exponential term in the Arrhenius equation implies that the rate constant, hence the reaction rate, of a chemical reaction increases exponentially when the activation energy decreases. A chemical reaction with small activation energy does not require much energy to reach the transition state. It proceeds faster than a reaction with large activation energy. As the activation energy of a catalyzed reaction is lower than that of an uncatalyzed reaction, the former occurs faster than the latter. The reaction rate is impacted more by temperature changes than the activation energy for the uncatalyzed reaction [1].

Pre-exponential Factor

The pre-exponential factor A is also known as the frequency factor since it includes the frequency of collisions and their orientations. It is not directly involved in the determination of the activation energy. However, when multiplied by the exponential term, it contributes to the rate constant and thus the reaction rate [1].

Combining the Arrhenius equation with Maxwell-Boltzmann law, we find that the term exp (-E/RT) represents the fraction of reactant molecules that possess enough kinetic energy to react. This fraction can be anywhere between 0 and 1, depending on E and T.

Suppose E = 0. Then,

k = A

It means that A is the fraction of molecules that would react if the activation energy were zero.

Using Arrhenius Equation to Determine Unknown Variable

The Arrhenius equation can determine an unknown rate constant or temperature without using the pre-exponential factor. For this purpose, the rate constant is measured for two different temperatures T1 and T2. The two-point form of the Arrhenius equation is written as follows [1,2]:

ln (k1) = (– E/R) (1/T1) + ln (A)

ln (k2) = (– E/R) (1/T2) + ln (A)

Rewriting the 2nd equation

ln (A) = ln (k2) – (E/R) (1/T2)

Substituting ln (A) in the 1st equation

ln (k1) = – (E/R) (1/T1) (1/T1) + ln (k2) – (E/R) (1/T2)

Or, ln (k1) – ln (k2) = – (E/R) (1/T1 – 1/T2)

Or, ln (k1/k2) = – (E/R) (1/T1 – 1/T2)

Hence, one can determine the unknown variable by knowing any three variables, k1, k2, T1, and T2.

Example Problems and Solutions

Problem 1: The rate constant of a reaction at 25.0 °C was measured to be 7.4 × 10-2 s-1. If the frequency factor is 1.8 × 1014 s-1, what is the activation energy? (R = 8.314 JK-1M-1)

Solution:

Given

k = 7.4 × 10-2 s-1

A = 1.8 × 1014 s-1

T = 25.0 °C = 25 + 273 = 298 K

From Arrhenius equation

ln (k) = (– E/R) (1/T) + ln (A)

Or, (E/R) (1/T) = ln (A) – ln (k)

Or, E = RT (ln (A) – ln (k))

Or, E = 8.314 JK-1M-1 x 298 K x (ln (1.8 × 1014 s-1) – ln (7.4 × 10-2 s-1))

Or, E = 74873 JM-1 or 74.9 kJM-1

Problem 2: The activation energy for a reaction is 33.7 kJ/mol. The rate constant for the reaction is 4.5 x 10-3 s-1 at 47 °C. Calculate the rate constant at 125 °C. (R = 8.314 JK-1M-1)

Solution:

Given

E = 33.7 kJ/mol

k1 = 4.5 x 10-3 s-1

T1 = 47 °C = 273 + 47 = 320 K

T2 = 125 °C = 273 + 125 = 398 K

From Arrhenius equation,

ln (k1/k2) = – (E/R) (1/T1 – 1/T2)

Or, ln (4.5 x 10-3 s-1/k2) = – (33.7 x 103 J M-1 / 8.314 JK-1M-1) (1/320 – 1/398)

Or, ln (4.5 x 10-3 s-1/k2) = 2.48 JM-1

Or, 4.5 x 10-3 s-1/k2 = 11.9

Or, k2 = 4.5 x 10-3 s-1/11.9

Or, k2 = 3.78 x 10-4 s-1

References

  1. Chem.libretexts.org
  2. Courses.lumenlearning.com
  3. Chemguide.co.uk
  4. Chemistrytalk.org
  5. Thoughtco.com

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