# In what circumstances and why do bonds break without an external energy source? [closed]

Edit: I've flagged my own question because it's too broad instead I'll split it up into more specific questions.

So I'm trying to work out exactly how bonds are able to break when there is no external energy source such as heat or light.

I am already aware of displacement reactions but I am unsure exactly why this happens. So in general it seems that a more reactive element will replace a less reactive one in a molecule. Why does this happen? Is it because the more reactive element is more electronegative and yanks electrons away from the previous bond? Does this happen in every reaction?

It seems as if it is a similar case with free radicals. I've read that this is to do with "yanking" electrons also. Do they always attack the weakest bond?

What other circumstances are there and why? Ideally I would like a mechanistic answer.

## closed as too broad by Klaus-Dieter Warzecha, Ben Norris, Todd Minehardt, getafix, NotEvans.Dec 28 '16 at 18:58

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• This question is too general. Specific, indivudual Bonds form and break under specific circumstances, when the geometry and energy allows for the rearrangement. – Karl Dec 28 '16 at 17:04
• Heat cannot be an "external energy" in the sense you seem to think. A reaction mixture simply has a temperature. – Karl Dec 28 '16 at 17:06

You don't need a mechanistic answer. You need thermodynamics.

You're missing one key insight which is that a well-defined process has a non-zero probability of happening. Universe entropy changes and Gibbs free energy changes only tell you how likely something is to happen, and while the corresponding probabilities from these values may be close to zero, they are not zero. Therefore, bonds have some nonzero probability of cleavage.

Unless you are at zero kelvin, there is always heat energy. Most reactions have an activation energy or energy barrier between reactants and products that has to be overcome if a reactant is to transform into product.

The Boltzmann distribution shows us that at a given temperature T there is always a distribution of energies between molecules. The chance of having a given energy say E decreases exponentially with energy as $\exp(-E/(k_BT)$ where $k_B$ is Boltzmann's constant.

Thus if the activation energy is large then there is only a small chance in any given second that by random motion of molecules, producing by collision in a solvent a short lived fluctuation in energy, that the barrier can be overcome and a product formed. If the barrier is smaller then its height (in energy) may be more easily surmounted and thus molecules react many more times /second. (You may associate 'weaker' bonds with lower barriers towards a particular product).

If the energy of the product is lower than that of reactants then the reaction back to reactants will have a larger barrier that that which led to products, thus the equilibrium between products and reactants lies on the product side. Vice versa for endothermic reactions.

Some reactions have a very low or no barrier between reactant and product, radical reactions can be of this form as can some electron transfer reactions. In the gas phase these species react on contact and the reaction rate is very large, e.g. $\ce{Cl\cdot + H2 }$. In solution the products are also formed as soon as reactants come into contact with one another, however, in these cases the rate of reaction is limited by how fast the reactants can diffuse together in whatever solvent is used.

In reactions that proceed by many steps each step has one of the forms described.