# Arrhenius Equation

Whether it is through the Collision Theory, Transition State Theory, or just common sense, chemical reactions are typically expected to proceed faster at higher temperatures and slower at lower temperatures.

### Introduction

The "Arrhenius Equation" was physical justification and interpretation in 1889 by Svante Arrhenius, a Swedish chemist. Arrhenius performed experiments that correlated chemical reaction rate constants with temperature. After observing that many chemical reaction rates depended on the temperature, Arrhenius developed this equation to characterize the temperature-dependent reactions.

$\large k=Ae^{^{\frac{-E_{a}}{k_{B}T}}}$

or

$\large \ln k=\ln A - \frac{E_{a}}{k_{B}T}$

With the following terms:

• In unit of s-1(for 1st order rate constant) or M-1s-1(for 2nd order rate constant)

A: The pre-exponential factor or frequency factor

• Specifically relates to molecular collision
• Deals with the frequency of molecules that collide in the correct orientation and with enough energy to initiate a reaction.
• It is a factor that is determined experimentally, as it varies with different reactions.
• In unit of L mol-1s-1 or M-1s-1(for 2nd order rate constant) and s-1(for 1st order rate constant)
• Because frequency factor A is related to molecular collision, it is temperature dependent
• Hard to extrapolate pre-exponential factor because lnk is only linear over a narrow range of temperature

Ea: The activation energy is the threshold energy that the reactant(s) must acquire before reaching the transition state.

• Once in the transition state, the reaction can go in the forward direction towards product(s), or in the opposite direction towards reactant(s).
• A reaction with a large activation energy requires much more energy to reach the transition state.
• Likewise, a reaction with a small activation energy doesn't require as much energy to reach the transition state.
• In unit of kJ/mol.
• -Ea/RT resembles the Boltzmann distribution law.

R: The gas constant.

• Its value is 8.314 J/mol K.

T:The absolute temperature at which the reaction takes place.

• In units of Kelvin (K).

### Implications

The exponential term in the Arrhenius Equation implies that the rate constant of a reaction increases exponentially when the activation energy decreases. Because the rate of a reaction is directly proportional to the rate constant of a reaction, the rate increases exponentially as well. Because a reaction with a small activation energy does not require much energy to reach the transition state, it should proceed faster than a reaction with a larger activation energy.

In addition, the Arrhenius Equation implies that the rate of an uncatalyzed reaction is more affected by temperature than the rate of a catalyzed reaction. This is because the activation energy of an uncatalyzed reaction is greater than the activation energy of the corresponding catalyzed reaction. Since the exponential term includes the activation energy as the numerator and the temperature as the denominator, a smaller activation energy will have less of an impact on the rate constant compared to a larger activation energy. Hence, the rate of an uncatalyzed reaction is more affected by temperature changes than a catalyzed reaction.

### The Math in Eliminating the Constant A

To eliminate the constant A, there must be two known temperatures and/or rate constants. With this knowledge we write:

$\ln k_{1}=\ln A - \frac{E_{a}}{k_{B}T_1}$

at T1 and

$\ln k_{2}=\ln A - \frac{E_{a}}{k_{B}T_2}$

at T2 . By rewriting the second equation:

$\ln A = \ln k_{2} + \frac{E_{a}}{k_{B}T_2}$

and substitute for ln A into the first equation:

$\ln k_{1}= \ln k_{2} + \frac{E_{a}}{k_{B}T_2} - \frac{E_{a}}{k_{B}T_1}$

This simplifies to:

$\ln k_{1} - \ln k_{2} = -\frac{E_{a}}{k_{B}T_1} + \frac{E_{a}}{k_{B}T_2}$

$\ln \frac{k_{1}}{k_{2}} = -\frac{E_{a}}{k_{B}} \left (\frac{1}{T_1}-\frac{1}{T_2} \right )$

### Graphically determining the Activation Energy of a Reaction

If we look at the Arrhenius equation more carefully, one notices that the natural logarithm form of the Arrhenius equation is in the form of y = mx + b. In other words, it is similar to the equation of a straight line.

$\ln k=\ln A - \frac{E_{a}}{k_{B}T}$

where temperature is the independent variable and the rate constant is the dependent variable.

So if one were given a data set of various values of k, the rate constant of a certain chemical reaction at varying temperature T, one could graph ln(k) versus 1/T. From the graph, one can then determine the slope of the line and realize that this value is equal to -Ea/R. One can then solve for the activation energy by multiplying through by -R, where R is the gas constant.

### Practice Problems

1. Find the activation energy (in kJ/mol) of the reaction if the rate constant at 600K is 3.4 M-1s-1 and 31.0 at 750K.
2. Find the rate constant if the temperature is 289K, Activation Energy is 200kJ/mol and pre-exponential factor is 9 M-1s-1
3. Find the new rate constant at 310K if the rate constant is 7 M-1s-1 at 370K, Activation Energy is 900kJ/mol
4. Calculate the activation energy if the pre-exponential factor is 15 M-1s-1, rate constant is 12M-1s-1 and it is at 22K
5. Find the new temperature if the rate constant at that temperature is 15M-1s-1 while at temperature 389K the rate constant is 7M-1s1,  the Activation Energy is 600kJ/mol

For more practice go through problems 9.26-9.37 in Physical Chemistry for the Biosciences, as well as more help on understanding the Collision Theory and the Transition State Theory.

### Solutions to Practice Problems

1.     Ea is the factor the question asks to be solved. Therefore it is much simpler to use

To find Ea, subtract ln A from both sides and multiply by -RT.

This will give us:

Ea = (ln A - ln k)RT

2.     Substitute the numbers into the equation:

lnk = -(200 X 1000) / (8.314)(289) + ln9

k = 6.37X10-36 M-1s-1

3.     Use the equatioin ln(k1/k2)=-Ea/R(1/T1-1/T2)

ln(7/k2)=-[(900 X 1000)/8.314](1/370-1/310)

k2=1.788X10-24 M-1s-1

4.     Use the equation k = Ae-Ea/RT

12 = 15e-Ea/(8.314)(22)

Ea = 40.82J/mol

5.     Use the equatioin ln(k1/k2)=-Ea/R(1/T1-1/T2)

ln(15/7)=-[(600 X 1000)/8.314](1/T1 - 1/389)

T1 = 390.6K

### References

1. Chang, Raymond. 2005. Physical Chemistry for the Biosciences. Sausalito (CA): University Science Books. p. 311-347.
2. Segal, Irwin. 1975. Enzyme Kinetics. John Wiley & Sons, Inc. p.931-933.
3. Ames, James. 2010. Lecture 7 Chem 107B. University of California, Davis.
4. Laidler, Keith. "The Development of the Arrhenius Equation." J. Chem. Educ., 1984, 61 (6), p 494
5. Logan, S. R. "The orgin and status of the Arrhenius Equation." J. Chem. Educ., 1982, 59 (4), p 279

### Contributors

•  Guenevieve Del Mundo, Kareem Moussa, Pamela Chacha, Florence-Damilola Odufalu, Galaxy Mudda, Kan, Chin Fung Kelvin

09:29, 2 Oct 2013

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