[PFScience]

Evening all.. I was wondering whether anyone could help me with something that has been bugging me for a long while and I haven't yet managed to get a decent answer...

Exactly why is it that some endothermic reactions get colder??

Now Ive seen all the usual explanations like "because it draws heat from the surroundings"... etc. But that for me isnt an answer, all it is, is a rough definition of an endothermic reaction, not an explanation of the mechanics.

I know enthalpy is a combination of kinetic and potential energy and that as one of them increases the other decreases. I also understand that temperature is a function of kinetic energy.

So here are my queries...

• If the system is getting colder then the amount of kinetic energy the reactants have must be decreasing
• Endothermic reactions typically have a significant E_A and thus energy absorbed from the surroundings must go to increasing the potential energy of the reactants in order for the reaction to proceed
• What is the mechanism for the reduction of the K.E of the reactants? Is it caused by a flow of heat from the kinetic aspects of the compound to being used to break bonds and therefore via conversion, increasing the potential energy of the compound?
• Do reactants in endothermic reactions, therefore, have low levels of P.E but high levels of K.E, and thus similarly do reactants in exothermic reactions have high levels of P.E but low levels of K.E?

What exactly are the changes going on in PE and KE during an endothermic reaction?

For an exothermic reaction (burning a piece of paper) I have written my explanation (please do give me feedback if Im wrong!) If anyone could give me an equivalent for an endothermic reaction I would be hugely grateful (and kudos for solving my 2 year itch!!)

Energy in the form of a flame is applied to a piece of paper. This provides the initial input of additional energy (thermal energy converting to PE) needed to overcome the small E_A so that the reaction may then proceed spontaneously. The initial energy breaks bonds in the paper, this decreases the PE and increases the KE as the separated atoms are free to move more.
The atoms combine with oxygen and since the newly formed products have a lower enthalpy than the reactants, excess heat energy is released (KE converting to thermal energy) and the temperature of the surroundings increase (thermal energy converting to KE). Part of that released excess heat (thermal energy) is now used to beak further bonds in the paper  the cycle repeats itself over until the paper is consumed. 

[mjc123]

I'm not sure this is the most useful way of looking at things. For example:

energy absorbed from the surroundings must go to increasing the potential energy of the reactants in order for the reaction to proceed

This is true for all reactions, whether endothermic or exothermic.

Do reactants in endothermic reactions, therefore, have low levels of P.E but high levels of K.E, and thus similarly do reactants in exothermic reactions have high levels of P.E but low levels of K.E?

All reactants at the same temperature have the same average KE - that's what temperature is. Besides, how do they know whether they're about to participate in an endothermic or exothermic reaction? But in an endothermic reaction, the reactants have lower PE than the products (not "than reactants in an exothermic reaction" - that's a meaningless comparison, unless you mean the reverse of the first reaction).
Consider the potential energy diagram below, for an exothermic forward reaction (or an endothermic back reaction). The products have lower PE than the reactants, hence ΔH is negative. The reduction in PE means an increase in KE, i.e. temperature, or heat released to the surroundings. The reverse is clearly true for the back (endothermic) reaction. The activation energy for the forward reaction is E_af, and for the backward reaction E_ab; obviously E_ab > E_af, and ΔH = E_af - E_ab.
What happens during a reaction? The average KE, as said, depends on temperature, but there is actually a spread of KE values (Boltzmann distribution), and a small fraction of molecules have enough energy to get over the activation barrier (which may involve e.g. kinetic energy from collision being converted to potential energy of breraking a bond), and go through the transition state to the products. Random collisions cause the Boltzmann equilibrium to be re-established, so reaction continues until completion (or equilibrium). The process is the same for the reverse reaction (this is called the principle of microscopic reversibility), except that the activation barrier is higher, so the fraction of molecules with sufficient energy is smaller and the reaction is slower. Increasing the temperature increases the fraction of energetic molecules and speeds up the reaction (in either direction). There is nothing special, in that sense, about an endothermic reaction.

[swintarka]

Try to imagine on of the simplest cases - evaporation. Molecules with high kinetic energy leave out the surface of a liquid. If the pressure surrounding the liquid is low, the chance for a molecule to come back to the bulk of the liquid is pretty low.  So in effect, molecules with lower kinetic energy stay in the liquid, molecules with high kinetic energy stay in the gas phase.
Similar argument can be used for reactions that evolve gases.
I guess the argument can be put even further, to more complicated cases, but that's just a hunch.

[PFScience]

Thanks for the reply!

All reactants at the same temperature have the same average KE - that's what temperature is.

I know, I meant from a relative point of view ie) in an exothermic reaction, the reactants have higher level of PE relative to KE

What happens during a reaction? The average KE, as said, depends on temperature, but there is actually a spread of KE values (Boltzmann distribution), and a small fraction of molecules have enough energy to get over the activation barrier (which may involve e.g. kinetic energy from collision being converted to potential energy of breraking a bond), and go through the transition state to the products. Random collisions cause the Boltzmann equilibrium to be re-established, so reaction continues until completion (or equilibrium). The process is the same for the reverse reaction (this is called the principle of microscopic reversibility), except that the activation barrier is higher, so the fraction of molecules with sufficient energy is smaller and the reaction is slower. Increasing the temperature increases the fraction of energetic molecules and speeds up the reaction (in either direction). There is nothing special, in that sense, about an endothermic reaction.

Right so what your saying is:

1) As per the Boltzmann constant, for every 1K increase in temp, a certain % will go towards increasing the KE of the molecules and the rest to increasing PE?

What is misleading/confusing I find is that a lot of texts refer to the importance of increasing KE in overcoming the E_A barrier but it seems to me that increasing PE carries a lot more importance, as demonstrated by its y-axis presence on many energy diagrams?

So in terms of the observed temperature drop in the reaction system, you are saying there are 2 interlinked possibilities:

1) In an endothermic reaction, only a small fraction of the molecules have sufficient overall energy (KE + PE) to overcome the E_A. Due to the lower relative levels of PE, there is a flow of heat energy from the kinetic aspect of the molecule to the potential aspect  thus causing the temp drop? Heat absorbed from the surroundings also flows in this manner?

2) The small fraction of molecules that do have sufficient overall energy to overcome the E_A go on to react which releases heat energy  this energy is further absorbed and stored as PE. As these reactions occur the larger fraction that did not have sufficient energy to overcome the E_A effectively now represent an even larger % of the total molecules, and so the temperature drops further due to the absence of the higher energy molecules?

Sorry if Im coming across a bit muddled!

[mjc123]

No, that's not what I'm saying. I'm afraid you do seem rather muddled; you appear to be confused about the relationship between kinetic and potential energy, and between thermodynamics and kinetics (whether a reaction is endo- or exothermic is purely a matter of thermodynamics). I will reply at more length when I have time, but as I said I don't think this is a very useful way of approaching the problem.

[mjc123]

OK, let's see...

As per the Boltzmann constant, for every 1K increase in temp, a certain % will go towards increasing the KE of the molecules and the rest to increasing PE?

No, it will virtually all go to increasing KE. Translational and rotational energies are purely kinetic (to a first approximation, ignoring intermolecular forces and centrifugal distortion). Vibrational energy is a combination of KE and PE, but for most molecules at ambient temperatures the population of vibrational states above the ground state is very small, and changes little with temperature. (So much for gases; it's a bit more complicated for liquids and solids where intermolecular/ionic forces are stronger and translational and rotational motions can be more like vibrations because there are potential energy barriers. But that's the general picture.)

What is misleading/confusing I find is that a lot of texts refer to the importance of increasing KE in overcoming the EA barrier but it seems to me that increasing PE carries a lot more importance, as demonstrated by its y-axis presence on many energy diagrams?

Suppose you are rolling a ball up a slope. It doesn't quite get to the top, so next time you roll it harder. Or you fire a rocket into space, to escape the Earth's gravity. What you are doing each time is putting KE into something in order to overcome a PE barrier. As it rises through the gravitational field, KE is converted to PE. Would it make sense to say "You're doing it wrong; you're increasing the KE, but you should be increasing the PE, because it's a PE barrier you have to cross"? How would you even add PE other than by adding KE? (Well, you could lift the ball up the slope, but not the rocket!) Similarly with a reaction - how would you increase the PE other than by increasing KE? There are ways of injecting PE - e.g. you can tune a laser to add vibrational energy to a specific bond - but these are rather exceptional, and you are talking about ordinary thermal reactions. The energy to overcome the PE barrier comes from KE.

Due to the lower relative levels of PE, there is a flow of heat energy from the kinetic aspect of the molecule to the potential aspect  thus causing the temp drop

I'm not sure what you mean here. Why should energy "flow" from kinetic to potential? Systems naturally tend to go to a minimum of potential energy, converting the PE to KE, either of an object or of its molecules (= thermal energy). A ball will roll down a slope, not up.

The small fraction of molecules that do have sufficient overall energy to overcome the EA go on to react which releases heat energy  this energy is further absorbed and stored as PE.

Reaction, after overcoming the E_a barrier, means coming down the PE curve on the other side, i.e. PE being converted to KE (heat energy), not the other way round. It is not "further absorbed and stored as PE". Note that the overall release or absorption of heat energy in a reaction is the result of the difference in PE between products and reactants, i.e. ΔH, and is not related to E_a. That is because you put in E_af and recover E_ab, and the energy change is the difference between them, ΔH, irrespective of the height of the barrier.

As these reactions occur the larger fraction that did not have sufficient energy to overcome the EA effectively now represent an even larger % of the total molecules, and so the temperature drops further due to the absence of the higher energy molecules?

Yes, if the system is isolated from the surroundings, or heat transfer is slow. If it is in thermal equilibrium with the surroundings, the temperature will stay constant and collisions will result in the Boltzmann equilibrium being maintained, so there will always be a fraction of the molecules with energy over E_a. Even in the former case, it is not strictly correct to say that T drops "further", because depletion of the most energetic reactant molecules creates product molecules more energetic than the thermal average, but collisions will quickly even things out, so that a new equilibrium is established at a lower temperature determined solely by ΔH. (The case of evaporation referred to above is different because the energetic molecules escape from the system.)