# Arrhenius Plot

The Arrhenius plot is used to study the effect temperature has on reaction rates.

### Introduction

In 1889 Svante Arrhenius proposed the Arrhenius equation from his direct observations of the plots of rate constants vs temperatures.

$k = Ae^{\frac{-E_a}{RT}}$

The activation energy, Ea, is the minimum energy molecules must possess in order to form a product. As noted above, we can use the slope of the Arrhenius plot to find the activation energy. We can also use the Arrhenius plot by extrapolating the line back to the y-intercept, to obtain the pre-exponential factor, A. A= p x Z where p is a steric factor and Z is the collision frequency. The pre-exponential, or frequency, factor is related to the amount of times molecules will hit in the orientation necessary to cause a reaction. It is important to note that the Arrhenius equation is based on the collision theory. It states that particles must collide with proper orientation and with enough energy. Now that we have obtained the activation energy and pre-exponential factor from the Arrhenius plot, we can solve for the rate constant at any temperature using the Arrhenius equation.

We obtain the Arrhenius plot by plotting the logarithm of the rate constant, K, versus the inverse temperature, 1/T. The resulting negatively sloped line is useful for finding the missing components of the Arrhenius equation. Extrapolation of the line back to the y-intercept yields the value for ln A. The slope of the line is equal to the negative activation energy divided by the gas constant, R. In most biological and chemical reactions, the reaction rate doubles when the temperature increases every 10 degree Celsius.

Looking at the Arrhenius equation, we can see that the denominator of the exponential function contains the gas constant, R, and the temperature, T. This is only the case when we are dealing with moles of a substance since R has the units of J/mol*k. When we are dealing with molecules of a substance, the gas constant in the dominator of the exponential function of the Arrhenius equation is replaced by the constant, K. The constant, K, has the units J/K. At room temperature, KT, is the available energy for a molecule at 25 C or 273K, and is equal to approximately 200 wave numbers.

It is important to note that the decision to use the gas constant or the constant, K, in the Arrhenius equation depends primarily on the cancelling of the units. To take the inverse log of a number it requires the number to be unitless. Therefore all the units in the exponential factor need to cancel out. If the activation energy is in terms of joules per moles, then the gas constant should be used in the dominator. However, if the activation energy is in unit of joules per molecule, then the constant, K, should be used.

• Arrhenius Equation per Mole    $$k = Ae^{\frac{-Ea}{RT}}$$
• Arrenhius Equation per Molecule    $$k = Ae^{\frac{-Ea}{KT}}$$

### Variations to Arrhenius Equation

#### Graphical Form

$lnk = \dfrac{-E_a}{RT}+lnA$

The Arrhenius equation can be rearranged to deal with specific situations. For example, taking the logarithm of both sides yields the equation above in the form y=-mx+b. Then, plot lnk vs 1/T and all variables can be found.

y=lnk

m=-Ea/RT

x=1/T

b=lnA

This form of the Arrhenius equation makes it easy to determine the slope and y-intercept from an Arrhenius plot. It is also convenient to note that the above equation shows the connection between temperature and rate constant. As the temperature increases, the rate constant decreases when the above equation is plotted. The same is true when the temperature decreases, the rate constant increases. From this connection we can infer that the rate constant is inversely proportional to temperature.

#### Integrated Form

$$ln \dfrac{k_2}{k_1} = \dfrac{E_a}{R} \left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$$

We can also use the integrated form of the Arrhenius equation. This variation of the Arrhenius eqation involves the use of two Arrhenius plots constructed on the same graph to determine the activation energy. From the above equation, we can see the temperature's effect on multiple rate constants. This allows us to easily infer the rate constants' sensitivity to activation energy and temperature changes. If the activation energy is high for a given temperature range, then the rate constant will be highly sensitive. This means that changes in temperature will have a big effect on the rate constant. If the activation energy is low for a given temperature range, then the rate constant will not be as sensitive. This means that changes in temperature will have little effect on the rate constant. We can graphically see this phenomenon in the example below:

 Experimental Data 1/Temp 0.0085 0.0075 0.0065 0.0055 0.0045 0.0035 lnk (Big Ea) 3 2.5 2 1.5 1 0.5 lnk (Small Ea) 1.8 1.7 1.6 1.5 1.4 1.3

The graph above shows that the plot with the greater slope has a higher activation energy and the plot with the less steep slope has a smaller activation energy. This means that over the same temperature range, a reaction with a higher activation energy will change more rapidly than a reaction with a lower activation energy.

#### Biological Significance

The Arrhenius plot may become non-linear if steps becoming rate-limiting at different temperatures. Such an example can be found with Fox and co-workers in 1972 with beta-glycoside transport in E. coli. The differences in the transition temperatures is due to fatty acid composition in cell membranes. The transition state difference is a result of the sharp change of fluidity of the membrane. Another example includes a sudden drop at low 1/T (high Temperatures) which is a result of protein denaturation.

### Key Points

• Arrhenius plot shows that reaction rates are inversely proportional to temperature changes
• The negative slope from the Arrhenius plot gives the activation energy, Ea: slope = -Ea/R
• Extrapolation of the Arrhenius plot back to the y-intercept gives lnA
• Arrhenius plot shows activation energy and temperature affect the sensitivity of the reaction rate

### Practice Problems

1. T/F   The Ea calculated from the Arrhenius equation gives an exact value.

2. Describe the relationship between temperature and Ea and give examples.

3. Using the following information:

A= 1x1014sec-1

Ea= 75x103 J/mol

R= 8.314 J mol/K

Calculate k at 27° C with proper units.

4. Using information from problem 3, calculate k at 37° C with proper units.

5. Using the integrated equation solve for Ea using:

k1=7.78x10-7 at T1=273 K

k2=3.46x10-5 at T2=298 K

1. F the Ea calculated is an average or "apparent" value.
2. As the temperature increases, the rate constant decreases when the above equation is plotted. The same is true when the temperature decreases, the rate constant increases. From this connection we can infer that the rate constant is inversely proportional to temperature.
3. k= 8.727 sec-1 (Remember to convert from Joules to Kelvin)
4. k=23.02 sec-1
5. Ea=1.026x105 J/mol

### References

1. Alberty, R. A. and R. J. Silbey (1997). Physical chemistry. New York, Wiley.
2. Petrucci, R. H., W. S. Harwood, et al. (2002). General chemistry: principles and modern applications. Upper Saddle River, N.J., Prentice Hall.
3. Dawber, J. G. and A. T. Moore (1973). Chemistry for the life sciences. London, New York, McGraw-Hill.
4. Atkins, P. W. and J. De Paula (2006). Physical chemistry for the life sciences. New York, Oxford University Press; Freeman.
5. Stiller, W. (1989). Arrhenius equation and non-equilibrium kinetics: 100 years Arrhenius equation. Leipzig, BSB B.G. Teubner.
6. Segel, Irwin H. (1975). Enzyme Kinetics: Behavior and Analysis of Rapid Equilibrium and Steady-State Enzyme Systems. John Wiley and Sons Inc.

### Contributors

• David Johns, Andra Hutton UC Davis Undergraduate

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