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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Kinetics > Modeling Reaction Kinetics > Temperature Dependence of Reaction Rates > The Arrhenius Law > The Arrhenius Law: Activation Energies

The Arrhenius Law: Activation Energies

All molecules possess a certain minimum amount of energy. The energy can be in the form of Kinetic Energy and/or Potential Energy. When molecules collide, the kinetic energy of the molecules can be used to stretch, bend, and ultimately break bonds, leading to chemical reactions. If molecules are moving too slowly with little kinetic energy, or collided with an improper orientation, they will not react and just simply bounce off each other. However, if the molecules are moving at a fast enough velocity with a proper collision orientation, such that the kinetic energy upon collision is greater than the minimum energy barrier, then a reaction will occur.  The minimum energy barrier that must be met for a chemical reaction to happen is called the activation energy, \(E_a\).

Figure 1

The reaction pathway is similar to what happens in Figure 1. In order to get to the other end of the road, object A needs to roll with a fast enough speed to be able to completely roll over the hill of a certain height, h. In other words, the faster the object move, the more kinetic energy it has. If the object move too slowly, it does not have enough kinetic energy necessary to overcome the barrier, as a result, it will eventually go back down. In the same way, there is a minimum amount of energy needed in order for molecules to break existing bonds during a chemical reaction. If the kinetic energy of the molecules upon collision is greater than this minimum energy, then bond breaking and forming will happen, forming a new product (provided that the molecules collide with the proper orientation).

Reaction diagram.bmp

Figure 2

The activation energy, Ea (on graph) is the energy difference between the reactants and the activated complex, also known as transition state. In a chemical reaction, the transition state is defined as the highest-energy state of the system. If the molecules in the reactants collide with enough kinetic energy and this energy is higher than the transition state energy, then the reaction will happen and products will form. In other words, the higher the activation energy, the harder it is for a reaction to occur and vice versa. However, if a catalyst is added to the reaction, the activation energy is lowered because of the lowered transition state. (Catalyst lowers the transition state energy, therefore reducing the activation energy barrier).  (Figure 3)

Figure 3

As indicated by figure 3 above, a catalyst helps lower the activation energy barrier, increasing the reaction rate and allowing the reaction to occur faster. In the case of a biological reaction, when an enzyme (a form of catalyst) binds to a substrate, the activation energy necessary to overcome the barrier will be lowered, thus increasing the rate of the reaction for both the forward and reverse reaction. See below for the effects of an enzyme on activation energy.

Activation Enthalpy, Entropy and Gibbs Energy

In Thermodynamics, changes in Gibbs free energy, \( \Delta G \), is defined as:

\[ \Delta G = \Delta H  - T \Delta S \]
  • \( \Delta G \) = change in Gibbs free energy of the reaction
  • \( \Delta H \) = change in enthalpy
  • \( \Delta S \) = change in entropy

\( \Delta G^o \) is the change in Gibbs energy when the reaction happens at Standard State (1 atm, 298 K, pH 7). To calculate a reaction's change in Gibbs free energy that did not happen in standard state, the Gibbs free energy equation can be written as:

\[ \Delta G = \Delta G^o + RTlnk \]
  • \( \Delta G \) = change in Gibbs free energy of the reaction
  • \( \Delta G^o \) = standard Gibbs free energy
  • \( R \) = 8.314 J/molK
  • \( k \) = equilibrium constant

When the reaction is at equilibrium, \( \Delta G = 0\). The equation above turns into:

\[ 0 = \Delta G^o + RTlnk \]

Solve for \( \Delta G^o \)

\[ \Delta G^o = -RT lnk \]

Similarly, in transition state theory, Gibbs energy of activation,\( \Delta G ^{\ddagger} \), is defined by:

\[ \Delta G ^{\ddagger} = -RTln K^{\ddagger} \] (1)


\[ \Delta G ^{\ddagger} = \Delta H^{\ddagger} - T\Delta S^{\ddagger} \] (2)
  • \( \Delta G^{\ddagger} \) = Gibbs Energy of Activation
  • \( \Delta H^{\ddagger} \) = enthalpy of activation
  •  \( \Delta S^{\ddagger} \) = entropy of activation
Combining (1) and (2), solve for \(ln K^{\ddagger}\) we have the Eyring equation:
\[ lnK^{\ddagger} = -\dfrac{\Delta H^{\ddagger}}{RT} + \dfrac{\Delta S^{\ddagger}}{R} \]

Transitionstate graph.bmp

As you can see in the figure above, activation enthalpy, \(\Delta{H}^{\ddagger} \), represents the difference in energy between the ground state and the transition state in a chemical reaction. The higher the activation enthalpy, the more energy is required for the products to form. Note that this activation enthalpy quantity, \( \Delta{H}^{\ddagger} \), is similar to the activation energy quanity, \(E_a\), when comparing the Arrehinus equation (described below) with the Eyring equation. Specifically,

\(E_a = \Delta{H}^{\ddagger} + RT\)

In general, a reaction will take place faster if the values of \(E_a\) and \(\Delta{H}^{\ddagger} \) are low. Conversely, if \(E_a\) and \( \Delta{H}^{\ddagger} \) are high, the reaction rate is slower.

Effects of Temperature on Activation Energy, Calculation of Ea using Arrhenius Equation

As temperature increases, gas molecules' velocity also increases (Kinetic Theory of Gas). This is also true for substances in the form of liquid and solid. Kinetic energy of molecules is directly proportional to the velocity of the molecules (KE = 1/2 mv2) . So, when temperature increases, KE also increases. This means that as temperature increases, more molecules will have higher KE, thus the fraction of molecules that have high enough KE to exceed the minimum energy needed for a reaction also increase.

The fraction of molecules that has an energy equal to or greater than \(E_a\) is given by the exponential term \(e^{\frac{-E_a}{RT}}\) in the Arrhenius equation:

\(k = Ae^{\frac{-E_a}{RT}}\) (3) 

  • \(k\)  is the rate constant
  • \(E_a\) is the activation energy
  • \(R\) is the gas constant
  • \(T\) is temperature in Kelvin
  • \(A\) is frequency factor constant or also known as Pre-exponential factor or Arrhenius factor (\(A\)  indicates the rate of collision and the fraction of collisions with the proper orientation for the reaction to occur)

Take the natural log of both side in equation (3):

\(\ln k = \ln A - \frac{E_a}{RT}\) (4) 

Equation (4) has the form of a straight line, y = mx + b. When we graph lnk vs 1/T, the equation represents a straight line with a slope of -Ea/R and a y-intercept of lnA. (figure 4 below)


Figure 4

high EA vs low EA.bmp

Figure 5

As indicated in Figure 5 above, the reaction which has a higher \(E_a\) has a steeper slope, thus the reaction rate is very sensitive to temperature change. In contrast, the reaction that has a lower \(E_a\) is less sensitive to a temperature change. For a radical reaction, since radicals are extremely reactive, its \(E_a\) is 0, therefore radical reaction has no slope and is independent of temperature.

There is also another method to calculate \(E_a\) when there are two known temperatures and a rate constant at each temperature. Using Equation (2), suppose that at two different temperatures T1 and T2, reaction rate constants k1 and k2, we have:

\[\ln\; k_1 = - \frac{E_a}{RT_1} + \ln A \]


\[\ln\; k_2 = - \frac{E_a}{RT_2} + \ln A \]

Subtract \(ln\; k_2\) from \(ln\; k_1\) 

\[ \ln\; k_1 - \ln\;  k_2 = \left (- \dfrac{E_a}{RT_1} + \ln A  \right ) - \left(- \dfrac{E_a}{RT_2} + \ln A \right)\]   

After rearrangement, we have:

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

Effects of Enzyme on Activation Energy

Similar to a catalyst mentioned above, enzymes can be thought of as biological catalysts which lower activation energy. Enzymes are proteins or RNA molecules that are able to lower activation energy by providing an alternate reaction pathway which have a lower activation energy than the original pathway. Enzyme will affect the rate of the reaction in both forward and reverse direction. Therefore, the reaction will go faster because less energy is required for molecules to react when they collide. Thus, the rate constant (k) will also increase. Notice that \(\Delta G\) of transition state and \(\Delta G >\Delta{H}\) of transition state also decreased. However, \(\Delta{G}\) and \(\Delta{G}">\Delta{H}\) of the overall reaction does not change. As shown in the diagram below, the activation energy barrier, \(\Delta{G}\) and \(\Delta{G}>\Delta{H}\) of transition state have decreased. 

New graoh.jpg

Additional information on enzymes: General principles and introduction to enzymes.

Practice Questions

  1. Given that the rate constant is 11M-1s-1 at 345K and the pre-exponential factor is 20M-1s-1 calculate the activation energy.
  2. If a reaction's rate constant at 298K is 33M-1s-1 and 45M-1s-1 at 675K what is the activation energy? 
  3. What is the Gibbs Free Energy at transition state when delta H at transition state is 34 kJ/mol and delta S at transition state is 66 J/mol at 334K?
  4. What is the Activation Energy of a reverse reaction at 679K if the foward reaction have an rate constant of 50M-1s-1, pre-exponential factor of 30M-1s-1and delta H reaction of 23 kJ/mol?    
  5. We know that the enzymes lower activation energy. Thus, increase the rate constant and the speed of the reaction. However, we also know that increasing the temperature can increase the rate of the reaction. Does that mean that at extremely high temperature, enzymes can operate at extreme speed? 

Practice Questions Solutions

1. Use the Arrhenius Equation: lnk = Ae-Ea/RT

  • k is the rate constant, A is the pre-exponential factor, T is temperature and R is gas constant which equals to 8.314 J/molK
  • ln(11) = (20)e-Ea/(8.314)(345)
  • Ea = 6084.1 J/mol

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

  • ln(33/45) = [-Ea/8.314](1/298 - 1/675)
  • Ea = 1375.8 J/mol

3. Use the equation \(\Delta{G} = \Delta{H} - T \Delta{S}\)

  • \(\Delta{G} = (34 \times 1000) - (334)(66)\)
  • \(\Delta{G} = 11956\, J/mol\)

4. Use the equation lnk = Ae-Ea/RT to calculate the activation energy of the foward reaction

  • ln(50) = (30)e-Ea/(8.314)(679)
  • Ea = 11500 J/mol
  • Because the reverse reaction's activation energy is the activation energy of the foward reaction plus the delta H of the reaction:
  • 11500 J/mol + (23 kJ/mol X 1000) = 34500 J/mol

5. No, most of the enzyme will denature at high temperature. At some point, the rate of the reaction and rate constant will derease signigicantly and eventually drop to zero. Once the enzyme is denatured, the alternate pathway is lost, the original pathway will take more time to complete.      


  1. Atkins P., de Paua J.. Physical Chemistry for the Life Sciences. pg 256-259. New York. Oxford Univeristy Press. 2006.
  2. Garrett R., Grisham C. Biochemistry. 3rd Edition. pg 64. California. Thomson Learning, Inc. 2005
  3. Wade L.G. Organic Chemistry. 6th Edition. pg 139-142. New Jersey. Pearson Prentice Hall. 2006.


  • Matthew Bui, Kan, Chin Fung Kelvin, Sinh Le, Eva Tan

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