KAR - Collision Theory Revisited: Activation Energy and Orientation Factor (Lesson)
Collision Theory Revisited: Activation Energy and Orientation Factor
Potential Energy Diagrams
This module began with a discussion of collision theory and how an increase in the number of collisions coincides with an increase in reaction rate. This module will explore this idea a bit further as well as expand upon it. In this lesson, the intricacies of collision theory will be explored as well as the relationship that exists between reaction rate and temperature.
Activation Energy
One of the important ideas that originates from collision theory is the fact that in order for chemical reactions to occur molecules must come into physical contact with one another. They must collide. However, not all collisions will lead to a reaction due to the fact that every reaction has what is referred to as an activation energy; a minimum amount of energy that must be overcome in order for a chemical process to proceed. It is this activation energy that controls the kinetics of a chemical reaction. The higher this barrier, the slower the reaction will proceed. Activation energies are often depicted graphically on an energy-level diagram (also often referred to as a reaction profile).
In the diagram above, the activation energy is understood to be the difference between the energy level of the reactants and the top of the energy "hill" or barrier. Every chemical reaction has a unique activation energy that remains constant regardless of temperature or any other factors. We can also see from the energy diagram that after the reactants obtain enough energy to overcome this energy barrier they pass through a transition state or activated complex before completely reacting to form products. If a snapshot of the process were taken at the top of this energy barrier, the transition state would appear as almost a hybrid between reactants and products having partially broken bonds and partially formed bonds.
Temperature and Activation Energy
So, if temperature typically increases the rates of reaction for different chemical processes, then how does this happen? What exactly does this increase in energy do? First, it is important to understand that an increase in temperature does not lower the activation barrier for the reaction. Instead, an increase in temperature increases the fraction of molecules that will actually collide with enough energy to overcome this activation energy.
Using the illustration above it can be clearly seen from the shaded regions that the sample contains more particles with sufficient energy to overcome the activation barrier at high temperature compared to low temperature. However, this model needs further refinement as this would actually predict a faster reaction rate for reactions than are actually observed.
Collision Orientation and the Arrhenius Equation
The final piece to the puzzle comes from the understanding that while collisions between particles are necessary in order for a reaction to occur, those collisions must occur with the proper orientation. Take for example the illustration shown below. In this image can be seen two different collisions between the same two reactant molecules. In the top portion, we see a collision between reactant molecules that did not lead to a reaction. These types of collisions are referred to as ineffective collisions. The bottom portion, on the other hand, represents an effective collision where the reactant molecules not only collide but collide with the proper orientation relative to one another in order to produce product molecules.
The two parameters of effective collisions and activation energy can be further analyzed using the equation shown below referred to as the Arrhenius equation.
k=Ae−Ea/RT
In this equation, k refers to the rate constant and A is often referred to as the Arrhenius constant which incorporates both the frequency and orientation of collisions.
The Integrated Form of the Arrhenius Equation
Like the rate laws discussed in the previous lesson, the Arrhenius equation can also be integrated to obtain an equation in the form of y = mx + b.
ln(k)=−EaRT+ln(A)
A plot of ln (k) vs. 1/t will yield a straight line where the slope of that line equals -Ea/R and can be used to determine the activation energy for a particular chemical process.
It is important not to confuse an Arrhenius plot with the integrated rate laws discussed previously. An Arrhenius plot will always yield a straight line no matter what the order of reaction is.
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