I ended my last post rather abruptly when I got confused about the equilibrium equation for the reaction we were discussing. We had:

CH4 + CO2 ===> 2CO + 2 H2

The Gibbs Free Energy for this reaction is positive, so conventionally it "shouldn't go". I had considered what would actually happen in the case of a reaction whose Free Energy was exactly zero, and I said it should be in equilibrium just as it was written. That's where things got messed up.

For a Free Energy of zero, you get k=1 for the equilibrium constant. I plugged in the concentrations as written above into the equilibrium formula

k = [CO]^2 * [H2]^2 / [CH4] * [CO2]

and it comes to k = 16, so something is wrong.

Here's what I've figured out so far. Free energy is calculated for gasses at STP conditions: Standard Temperature and Pressure, or 1 atmosphere at 25 degrees Celsius. If you think about it, you'll see that it is physically impossible to combine these four gasses, in their stoichiometric ratios, in a single container at STP conditions. Just think about it. There is twice as much H2 as there is CH4, so the partial pressure of hydrogen must be twice the partial pressure of methane. They can't both be at STP.

By confining the gasses in a cylinder with a piston, we can vary the total pressure and it is not surprising that the equilibrium concentrations will change. This is just an application of Le Chatelier's principle. The value of k remains constant but the equilibrium point moves to the left or right: as the piston is compressed, the heavier molecules are favored because they take up less space, and vice versa. At some arbitray position, the gasses will inevitably be present in their stoichiometric proportions: but it is physically impossible for them to be present in the precise conditions specified in the balanced chemical equation: namely, all of them simultaneoulsy at STP.

By fiddling with the numbers, you can in fact verify that the equilibrium equation is satisfied for the following concentrations:

1/4 CH4 + 1/4 CO2 ===> 1/2 CO + 1/2 H2

It's just the balanced equation divided through by 4. So whatever enthalpies and entropies we had for the balanced equation, they are all altered in proportion for this modified equation. The free energy change is clearly zero.

What is still bothering the hell out of me is the fact that you still cannot put these four components into a single container at the exact conditions represented in the equation. The equation is written for STP conditions, and when you combine these gasses the partial pressures of the product side are still goind to be double the pressures of the reactant side. So it still needs to be explained: why does the equilibrium hold?

Let's look closely at what it means to combine these gasses.You have in the reaction as written 1/4 litre of methane, 1/4 litre of CO2, 1/2 litre of CO, etc; all of them at one atmosphere. You put them into a 1 litre container, and they fill the container: now, the partial pressures are respectively 1/4 atmosphere, 1/4 atmosphere, 1/2 atmosphere, etc. (You may notice that the total pressure inside the container is now 1.5 atmospheres but that is neither here nor there.) The point is that none of the gasses are at STP any more, so their enthalpies and entropies are all different. Why then does the equilibrium hold?

The problem is not so much with the enthalpies as with the entropies. In fact, the enthalpy of an ideal gas at a given temperature does not depend on the pressure or size of the containment vessel. (This fact in itself ought to be surprising but that is a story I'm not going to open up at this point.) But the entropies certainly change with expansion. And the gasses on the left hand of the equation expand by a factor of four, while those on the right hand side (the products) expand only by a factor of two. So the entropy changes ought to be different.

Most troubling of all is the fact that the entropy change tends to be logarithmic with the change in volume. So if we are looking to balance out the competing energy changes, it seems odd to me that the physical solution should come out in terms of nice fractional ratios of the original chemical equation. Clearly I haven't fully understood the situation yet. Maybe someone else will have a better explanation. But for now I'm going to put this little side issue to rest and continue with my story about Atomic Energy of Canada and the hydrocarbon detection system.

## Tuesday, June 22, 2010

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