Here we will discuss collision theory to aid in the explanation of how bonds are broken and new bonds are formed in a chemical reaction. In a previous General Chemistry 1 study guide, it was stated a chemical reaction is a rearrangement of atoms. In other words, bonds break and new bonds are formed. In order for this to happen, reactant molecules must collide with one another. The collisions must be forceful enough and the molecules must be in the proper orientation. Not all collisions are effective collisions.
In order for a collision to be an effective collision:
2. The collisions must have sufficient energy for the chemical bonds to break. This energy is called the activation energy, Ea
3. Molecules must collide with the proper orientation.
The reason why different reactions occur at different rates is explained by collision theory. Collision theory gives us information about how to change a reaction rate.
Consider the following bimolecular step.
A + B → P with Rate = k[A][B]
The molecules A and B must collide in order to disrupt existing bonds and to allow for the formation of new bonds. The frequency of collisions between the reactant molecules is proportional to the concentration of [A] and [B]. If the concentration of [A] is doubled, the reaction rate will also double. If the concentration of [B] is doubled, the reaction rate will also double. The higher concentration will result in a higher collision frequency between the reactants.
Next, temperature is important. Just a 10 degree increase in temperature can double or triple the reaction rate. The rate constant, k, increases exponentially with the temperature as shown in the plot below.
Increasing the temperature will increase the rate constant, k, and result in an increased rate of reaction.
Below is a plot of the fraction of molecules
All molecues possess either kinetic energy or potential energy. During collisions, the kinetic energy is used to bend, stretch and possibly break bonds. Once a bond is broken, this can lead to a chemical reaction. If a molecule has a low kinetic energy or if they collide with an improper orientation, they will not react. If the molecules collide with the proper orientation and have enough kinetic energy to overcome the activation energy, then a reaction will occur.
The Arrhenius equation below can be used to determine the activation energy, Ea for a process or reaction. The activation energy is the minimum amount of energy required for the reactants to undergo a chemical reaction.
where k is the rate constant, A is the frequency factor, and includes the frequency of collisions as well as the orientation of the molecules during the collision, Ea is the activation energy. The quantity, \(\displaystyle e^{-\frac{E_a}{RT}}\), is the fraction of collisions with enough energy for a reaction to occur and is symbolized as f. The temperature, T, is in Kelvin, and the gas constant R = 8.314 \(\displaystyle \frac{\mathrm J}{\mathrm mol⋅K}\).
Here we can calculate f. If Ea = 45 kJ/mol and the temperature is 345 K we calculate f as:
This means, if we have one billion, 109, collisions only
\(\displaystyle \frac{x}{10^9}\;=\;1.54\times 10^{-7}\)
109 x (1.54 x 10-7) = 154 collisions out of one billion would lead to a reaction.
Next, let’s try this with a lower activation energy of 5 kJ/mol and the same temperature of 345 K.
In this case, \(\displaystyle \frac{x}{10^9}\;=\;0.175\)
Out of a billion collisions 1.75 x 108 collisions would have enough energy to react.
Now, we will change the temperature, to 425 K and keep the activation energy at 5 kJ/mol.
In this case, 2.43 x 108 molecules out of a billion would have enough energy to lead to a reaction. We have seen that lowering the activation energy or increasing the temperature will increase the fraction of molecules with enough energy to react.
Below, is a potential energy diagram. The y-axis is potential energy while the x-axis is the progress of the reaction going from reactants to products.
The potential energy of the reactants up to the maximum of the curve is the activation energy, Ea. The reaction is exothermic because the energy of the products is lower than the energy of the reactants — Hproducts < Hreactants, and ΔH < 0.
At the maximum energy is the transition state. The transition state, also called the activated complex, is a configuration that has partial bonds. There are bonds that are breaking and other bonds that are forming. The transition state cannot be isolated because it is very unstable. If the kinetic energy from collisions of the reactant particles is less than the activation energy, the reactant molecules will just bounce apart–in other words they cannot surmount the barrier.
Consider the following reaction:
When methyl bromide reacts with hydroxide ion, we see the bond between carbon and bromine is broken and hydroxide ion forms a bond with the carbon to form methanol and bromide ion. The transition state has two partial bonds — 1) the carbon-bromine bond is breaking, and 2) the hydroxide ion is forming a bond with carbon.
The carbon in the transition state has 3 bonds and two partial bonds which is very unstable since carbon only forms 4 bonds.
In the energy diagrams below, we see the first diagram corresponds to an exothermic reaction while the second diagram corresponds to an endothermic reaction. An endothermic reaction will have the product energy higher than the reactant energy, Hproducts > Hreactants, and ΔH > 0.
The two energy diagrams above show the activation energy in the first diagram as larger than the activation energy in the endothermic diagram (second diagram). In this case, the endothermic reaction is faster than the exothermic reaction. A larger activation energy results in a slower rate of reaction while a smaller activation energy results in a faster rate of reaction.
In the figure below, orientations of the molecules in the reaction are shown. There is only one orientation, out of many orientations, that will allow the products of the reaction to be formed. The molecules in the reaction are small so you can imagine that as molecules get more complex, there will be many more ineffective orientations of the molecules.
Recall the Arrhenius equation below:
The frequency factor, A, takes into account the collision frequency, Z, and the orientation probability factor, p also called the steric factor.
The orientation probability factor, p, is the ratio of effective collisions to all other possible collisions. Generally, p will have values between 0 and 1.
Below is a plot of the rate constant, k, vs temperature. The Arrhenius equation can be used to calculate the activation energy, Ea.
To calculate the activation energy from the Arrhenius equation, take the natural log of both sides of the equation.
In the plot below, ln(k) is plotted against 1/T. The temperature is in Kelvin. The slope is equal to the negative of the activation energy divided by R, ΔHvap/R; R = 8.314 J/(mol⋅K). The two point form of the equation is derived below.
In the equation, ln A is constant. We can set two equations with different vapor pressures and temperatures equal to one another. We then rearrange to get the final two-point equation.
If we know the vapor pressure at two different temperatures, we can calculate the activation energy from the equation. If we know the activation energy and the vapor pressure at some temperature, we can determine the vapor pressure at a different temperature.
Exercises
Exercise 1. Sketch a potential energy diagram, properly labeled, for the reaction between nitrogen monoxide and ozone.
NO (g) + O3 (g) → NO2 (g) + O2 (g)
At a given temperature, Ea is 25.0 kJ, and ΔH is -250 kJ. What is Ea for the reverse reaction?
Check Solution/Answer to Exercise 1
Exercise 2. Plot the data below to determine the activation energy, Ea, and the frequency factor, A.
Check Solution/Answer to Exercise 2
Exercise 3. A first order reaction has a rate constant of 2.3 x 10,-3 s-1 at 35oC. The temperature is increased to 45oC and the reaction rate doubles. What is the activation energy for this reaction?
Check Solution/Answer to Exercise 3
Exercise 4. Consider the following reaction.
2 HI (g) → H2 (g) + I2 (g)
The rate constant at 226.9oC is 9.50 x 10-9 M-1 s-1 and 1.11 x 10-5 M-1 s-1 at 326.9oC. What is the activation energy?
Check Solution/Answer to Exercise 4
Exercise 5. Using the following energy diagram, determine Ea(forward), Ea(reverse), the transition state, and ΔHrxn. Draw the transition state.
Check Solution/Answer to Exercise 5
Exercise 6. Calculate the number of effective collisions out of 10 billion collisions for the following:
b) the same reaction with Ea = 24 kJ at a temperature of 98oC.
Check Solution/Answer to Exercise 6
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