In my next-to-last post on this subject, I left the question hanging in the air: how does a reaction proceed when the Gibb's Free Energy is positive? I am now going to try to answer this.
The reaction in question is the decomposition of methane in the presence of CO2:
CH4 + CO2 ===> 2CO + 2H2
As I recall I determined that at STP conditions (25 C and 1 atm) the Free Energy was definitely unfavorable. I think it worked out to something on the order of 90 kiloJoules. There was a complication however: the reaction takes place in a nuclear reactor at a temperature of 300 C. The effect of this is to bring the Free Energy down quite a bit (because the entropy term is favored by the smaller mole fragments on the product side of the equation) but not nearly enough to make it go negative. Let's say it came to around 65 kJ.
The funny thing is that these numbers lead to all kinds of consequences in terms of measuring the actual gas concentrations. To see what happens we have to convert the Gibbs Free Energy to the corresponding Equilibrium Constant. The equilibrium constant k is given by the formula:
k = exp(-(Gibbs Free Energy) /RT)
where RT is the gas constant product from the formula PV=nRT. It's easy to remember that RT comes to exactly 2.2 kJ because one litre of air at STP occupies just 22 litres.
Now it gets interesting. The equilibrium constant in the first case (room temperature) works out, on the ratio of 90/2.2, to around e^-40, or dividing by 2.3 you get 10^-17. At the higher temperature, however, the numbers come out a little less extreme, at around 10^-12. These are all still very tiny numbers. How could they make much difference?
It turns out that the difference is critical in the case of interest. In a previous blog I showed how to use the equilibrium constant to calculate the concentrations of a reaction. Now lets apply it to the trace gasses in the nuclear reactor. We assume that reactor coolant is leaking into the CO2 purge gas stream through a tiny crack in a pressure tube. The coolant, a complex hydrocarbon, will typically break down into methane-like fragments. Let us suppose the initial concentration of methane in CH4 is 20 ppm, or 2 * 10^-5.
How can we balance the equilibrium equation in this case? We will write the equation as
k = [CO]^2 * [H2]^2 / [CH4] * [CO2]
For a reactor temperature of 300 degrees, we have k = 10^-13 and we need to make the concentrations in the above equation work out to that value. By toying with numbers, we can see that if the initial concentration of methane is 90% dissociated, so that only 2 ppm remains, that the CO and H2 concentrations must each be close to 40 ppm: we then get for the numerator a value of 256 *10^-20, which requires the denominator to be 256 * 10^-8, or just over 2 ppm. That's where the equation becomes balanced.
The situation is quite different, however, when the gas returns to room temperature. Now the equilibrium constant is 10^-17. The math is a little hard to write out in ASCII, but it turns out if only 10% of the methane dissociates, you get a concentration of 4 ppm for both the CO and the H2. Multiply out the terms and the equation balances at these concentrations. You have 18 ppm methane.
These numbers describe exactly what was seen to be happening in the leak detection system! When gas was initially sampled at the detector, it showed definite methane readings in the tens of parts per million. But after drawing for several minutes on a sample line, the reading "settled down" to zero, reassuring everyone that there was no cause for concern. What was actually happening was that the initial reading came from the room temperature gasses that had been sitting in the tubes: the methane that had initially decomposed at high temperatures had recombined into its original form. But as you kept drawing on the sample line, you then got to the fresh gasses which had just decomposed at reactor temperatures. Even if the sample point was at room temperature, the gasses had not yet had enough time to recombine, so the methane reading was basically zero.
I started off telling this story because almost thirty years later, it would have critical consequences in my development of the theory of Quantum Siphoning. I was involved in an online discussion of the quantum nature of the photographic process, and my opponents had thrown in my face the fact that the photo-reduction of silver bromide to silver was thermodynamically non-spontaneous. It was only because of my previous experience in analyzing the decomposition of trace methane in CO2 that I was able to see the possibility of doing the same thing for trace concentrations of silver in a silver bromide crystal. This insight was to have critical consequences for my ability to see the way to Quantum siphoning.
There was one other physics problem from my younger days that gave me an unusual perspective on processes involve waves and the interchange of energy. That was the problem of the crystal radio. But that's a story for another day.
There is a final postscript to the story of the nuclear reactor that needs to be told. Having figured out what was the problem with the leak monitoring system, I naturally wanted to tell people all about it; in no small measure, to show people how smart I was. I called a meeting of the interested parties. The department manager was the chair of the meeting. I began my presentation but almost immediately the manager interrupted me. "Before we go too far, I want to ask one question". He then turned to the reactor superintendant and said: "Even if what Marty says is right, would it make any difference to how you operate the reactor?".
"Absolutely not," replied the superintendant. The manager then turned to me and said, "Well, I guess there's no need to continue with this meeting". And that was it.
I didn't object because at the time I was actually impressed at what a good job the manager had done: he had gotten right down to the essentials without wasting a lot of time over theoretical issues. I was young and I wanted to be a team player, so I accepted the result.
It wasn't until I started writing this series of articles several weeks ago that it struck me: I had been set up! The manager and the superintendant had gotten together and figured out how to shut me up. It's dirty trick that people use when they want to dismiss what you're saying without letting you make your point: they start off with "even if what you say is true...". Of course, if what I was saying were true, it should have made a difference. They just didn't want to admit it.