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If all chemical processes are equilibria at least to some extent, then where does the energy to overcome the activation energy come from?

Is a process with a very high activation energy (i.e. the ionization of sodium vapor into sodium ion) still considered an equilibrium process at least to some extent?

If so, where might the energy to furnish the forward reaction come from naturally - i.e. with no human input of energy?

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3 Answers 3

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I don't think the statement that all processes are in equilibrium to some extent is true. By definition, dynamic equilibrium means that the rate of change in one direction is equal to the rate of change in the opposite direction.

This is certainly not true for all processes, otherwise there would not be a process - nothing would appear to change.

On the other hand, for any given reaction equation that is known to occur, we can say:

for a reaction at equilibrium, neither the concentration of reactants nor the concentration of products will be zero

I think that may be what you meant in your question?

If that is the case, then let's look at your example:

Is a process with a very high activation energy (i.e. the ionization of sodium vapor into sodium ion) still considered an equilibrium process at least to some extent?

It's not an equilibrium process. However, the reaction:

$\ce{Na(g)->Na(g)+ + e-}$

could be at equilibrium. In other words, the ratio of reactants to products could be such that the Gibbs free energy is at a minimum for this temperature and pressure. A more mathematical way to say that is:

$Q = K_{eq}$

where $Q=\frac{[Na^+]}{[Na]}$ and $K_{eq}$ is the equilibrium constant under these conditions. (Brackets mean "partial pressure" - not concentration here.)

Under normal lab conditions (1 atm, 298 K), this is not thermodynamically favorable. The reaction quotient heavily favors the reactants, or another way to say it is $K_{eq}$ is really close to zero. So close you may not be able to measure it. The activation energy is extremely high (equal to the ionization energy), and so any electrons that are removed would have to get that energy from somewhere - which was your next question:

If so, where might the energy to furnish the forward reaction come from naturally - i.e. with no human input of energy?

In an idealized, isolated system, you only have one source of energy for things like that: kinetic. Thermal motion of the atoms could conceivably put enough energy into a single atom that an electron could be "kicked off" and result in an ion. However - the odds of that happening are extremely low, given that the distribution of speeds would have a vanishingly small, although still non-zero, probability of occurring at the magnitude needed to overcome the activation energy barrier. Since the reaction is thermodynamically unfavorable, the probability that the electron would jump right back on is very high - much higher than it would be for leaving in the first place.

Now when you have $6.022 \cdot 10^{23}$ opportunities each femtosecond, very low odds can still work out. But the total amount that would be ionized at any one time is still very, very low. Possibly so low that you would have to wait for an essentially infinite amount of time to see a measurable amount of it.

In a real system, a much larger fraction would be ionized (and this may or may not be reflected in $K_{eq}$, depending on how it is measured). The reason is that there is always ionizing radiation present on earth, and there will be more or less depending on where you are located. Statistical thermodynamics doesn't take this into account, so any prediction of the relative amounts of reactants and products would not be correct. This is kind of a unique example in that respect - most equilibrium reactions, even with very slow kinetics, deal with systems where the thermal energy of the system has a much bigger effect than radiation.

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@ thomij "I don't think it would be a Boltzmann distribution" Why? –  ron Jun 19 at 23:23
    
What's a Boltzmann distribution? –  Dissenter Jun 19 at 23:52
    
@ron - because it's a solid, Maxwell-Boltzmann distributions are derived for the gas phase and apply reasonably well to liquids. I could be wrong though. –  thomij Jun 20 at 0:25
    
@Dissenter - A Maxwell-Boltzmann distribution. Often times people will refer to one or the other when in a hurry. –  thomij Jun 20 at 0:27
    
Actually, I just read the question more carefully and saw that he was talking about gas phase sodium - I will edit my answer. –  thomij Jun 20 at 0:38

Maybe this is another case where semantics enters in. It seems that if a reaction coordinate connects two species, then over time an equilibrium will be achieved between them. "Over time" could mean microseconds or millennia. The kinetic energy distribution of a collection of molecules is described by a Boltzmann distribution. Therefor even at room temperature, there will be some molecules with enough energy to surmount a high barrier along the reaction coordinate. If there is a high barrier connecting two molecules, then it will just take longer for equilibrium to be achieved.

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I think there are a couple conceptual problems here you should consider:

1) It is not true that all reactions are equilibrium in the normal sense of equilibrium: just burn any complex organic compound, and now try to re-make the material from carbon-dioxide and water. In other words: entropy does mater. Technically any elementary reaction can go both direction, but this is a statement about microscopic processes, and you should careful about it when you try to conclusions about ensembles.

2) You are talking about reactions and equilibrium in a macroscopic sense, while you are asking about microscopic mechanism (transfer of activation energy). In microscopic scale there is a fluctuation of the energy what a molecule can get from its environment. The kinetic energy, vibration energies etc of molecules are not a single value, rather a distribution of values, so there is a non-zero probability that two molecules collide with enough energy or enough energy is stored in a vibration mode so the reaction can go.

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