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By the end of this section, you will be able to:
  • Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates
  • Define the concepts of activation energy and transition state
  • Use the Arrhenius equation in calculations relating rate constants to temperature

We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates.

Collision theory is based on the following postulates:

  1. The rate of a reaction is proportional to the rate of reactant collisions:

    reaction rate # collisions time
  2. The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.

  3. The collision must occur with adequate energy to permit mutual penetration of the reacting species’ valence shells so that the electrons can rearrange and form new bonds (and new chemical species).

We can see the importance of the two physical factors noted in postulates 2 and 3, the orientation and energy of collisions, when we consider the reaction of carbon monoxide with oxygen:

2 CO( g ) + O 2 ( g ) 2 CO 2 ( g )

Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient quantity, the reaction is spontaneous at high temperature and pressure.

The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules:

CO( g ) + O 2 ( g ) CO 2 ( g ) + O( g )

Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in [link] . In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms ( O = C = O ) . This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction.

A diagram is shown that illustrates two possible collisions between C O and O subscript 2. In the diagram, oxygen atoms are represented as red spheres and carbon atoms are represented as black spheres. The diagram is divided into upper and lower halves by a horizontal dashed line. At the top left, a C O molecule is shown striking an O subscript 2 molecule such that the O atom from the C O molecule is at the point of collision. Surrounding this collision are a mix of molecules of C O, and O subscript 2 of varying sizes. At the top middle region of the figure, two separated O atoms are shown as red spheres with the label, “Oxygen to oxygen,” beneath them. To the upper right, “No reaction” is written. Similarly in the lower left of the diagram, a C O molecule is shown striking an O subscript 2 molecule such that the C atom from the C O molecule is at the point of collision. Surrounding this collision are a mix of molecules of C O, and O subscript 2 of varying sizes. At the lower middle region of the figure, a black sphere and a red spheres are shown with the label, “Carbon to oxygen,” beneath them. To the lower right, “More C O subscript 2 formation” is written and three models of C O subscript 2 composed each of a single central black sphere and two red spheres in a linear arrangement are shown.
Illustrated are two collisions that might take place between carbon monoxide and oxygen molecules. The orientation of the colliding molecules partially determines whether a reaction between the two molecules will occur.

If the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. Every reaction requires a certain amount of activation energy for it to proceed in the forward direction, yielding an appropriate activated complex along the way. As [link] demonstrates, even a collision with the correct orientation can fail to form the reaction product. In the study of reaction mechanisms, each of these three arrangements of atoms is called a proposed activated complex    or transition state .

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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Adjanou
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Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Magreth
progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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