# Why does an increase in Kinetic Energy increase the likelihood of successful collisions?

As far as I understand bond formation, bonds must be broken between reactant particles to allow reactant particles to form new bonds and create product particles. This involves increasing the potential energy of the reactant particles, such that the attractive forces between say, reactant A and reactant B > the attractive forces between reactant A particles.

My textbook states that reactant particles must have sufficient kinetic energy so the collisions are energetic enough to break reactant bonds. What I don't understand is how the KE of the particles is related to breaking bonds. KE describes the energy needed to shift a body from rest to a given velocity, not the magnitude of a distance between particles (potential energy). So my question is: how is KE related to collisions, potential energy and product formation?

• You buy 2000 iPhones and start throwing them against a brick wall at different speed, just out of curiosity. You notice that at low speed most of them are but slightly cracked or visibly unharmed at all. At higher speed (and consequently, higher kinetic energy) the picture is different: they start smashing into pieces. Well, molecules are like that. Mar 18 '17 at 12:07
• So has it got something to do with Newton's third law of motion? Mar 18 '17 at 12:11
• @TomBrooks an intuitive answer: (1) particles are not billiard balls -- they don't "collide" physically, they repel each other electrostatically when they are close. (2) molecules are held together by electrostatic attraction, (3) this electrostatic attraction can be cancelled by electrostatic repulsion of a "colliding" particle nearby, (4) high-speed particles will tend to get really close to our molecule. Feb 10 '18 at 7:19

The kinetic energy of a particle is an alternative way of measuring how fast it moves (the mass of the particle also matters but for a given type of particle of a given mass the only thing that matters is how fast it is moving).

Potential energy has nothing to do with how far particles are apart (unless they are very close) and is pretty irrelevant in this situation.

In a gas or liquid, molecules are moving around a lot and banging into each other a lot. The speeds of the molecules are distributed statistically but their average speed is related to temperature. Whether a reaction happens when they collide depends on whether the molecules hit each other with enough energy to cause a reaction to happen (the orientation of the molecules might also matter but there are usually enough collisions in many different orientations that we don't think about this explicitly other than as part of the rate of the reaction).

In general, more energy in the collisions will lead to more reactions as more molecules will have enough energy to make the reaction happen (rather than just bounce off each other). More energy means more molecules have faster speeds. So, in general, more kinetic energy leads to a higher probability that any given collision will lead to a reaction.

Consider you have only two molecules, A and B, enclosed in a box and that the temperature is low. Furthermore, we suppose that A and B will react forming AB, only if (1): A and B collide and if (2): A and B have the right kinetic energy. A and B are moving randomly in the box. If the temperature is low, the kinetic energy of A and B is low. It will then take a long time before A and B will collide. This is because A and B are moving randomly quite short distances per unit time, i.e. the likelihood they will collide is low. Now, we increase the temperature => the kinetic energy of the molecules will increase. A and B are now moving at a much higher velocity, i.e. they move much longer distances per unit time. The probability that they will collide has increased per unit time. A and B will now collide once and a while, but no AB is formed. We therefore increase the temperature even more. By that, A and B will collide frequently, resulting in the forming of AB. A and B has now the kinetic energy required for colliding within a reasonable time, as well as, the required energy to react with each other.

Consider a simple reaction of the following type:

$$\ce{A-A(g)->2A(g)}$$

In a concentrated enough sample, the molecules are colliding all the time.

We can then invoke the equipartition theorem to say that the kinetic energy in the system is equally distributed between all of the vibrational, rotational, and translational modes of the system. Now, if you vibrate the central single bond in $\ce{A2}$ hard enough, the molecule will fragment and the reaction will occur.

This idea also applies to how reactions work in general. Reactants must reach a high energy transition state before transforming into product. Having the energy to deform to this high energy state allows the reaction to proceed, but with enough molecular collisions, higher kinetic energy in the system translates to stronger deformations and the ability to reach the transition state.