Bad Science

There may be some legitimate criticisms to be made on the current science behind global warming, however that does not excuse bad science being used to make incorrect counter arguements. One such arguement has been posted a couple of times in various comments left by readers of Kiwi Blog. The flawed arguement goes like this:

Methane (CH4), rated as a stronger greenhouse gas than Carbon Dioxide (CO2) is not such a major concern because, if one assumes a chemical equilibrium between CO2 and CH4, then reducing CO2 will drive the eqilibrium to consume more CH4, so increasing concentrations of CH4 can be delt with by reducing the CO2 concentration.

One can see this by assuming thermodynamic equilibrium is reached between the concentrations of CO2 and CH4 through the standard combustion reaction:

CH4 + 2O2 -> CO2 + 2H2O

Were this the case, the equlibrium is dictated by an equation every 5th form chemistry student at least used to know:

K=[CO2][H2O]^2 / ([CH4] [O2]^2 )

Where the square brackets represent the concentration of each species and K is a constant.

Proponents of this bad science then point out that if K is a constant, then any measure to reduce CO2 will also cause a drop in CH4, as is seen by a simple rearrangement:
[CH4]=[CO2][H2O]^2 / (K [O2]^2 )

Problem is, CH4 is NOT IN CHEMICAL EQUILIBRIUM with CO2 in the atmosphere, nor is it anywhere near chemical equilibrium. The forward direction of the reaction dominates and the reaction will essentially go to completion. To see this, one needs to work out (at least approximately) what K should be, then plug in the known concentrations of H2O and O2 in the atmosphere to get the equilibrium ratio of [CO2]/[CH4] and compare it with the observed ratio.

Now K can be found by the following equaiton

K=q(CO2)q(H2O)^2 /( q(CH4) q(O2)^2) *exp(-E/(RT) )

where the varous q's are single particle partition functions, E is the reaction energy for the forward reaction, R is the Gas Constant and T is the temperature.

Without going into too much more detail, the various q's tend to cancel each other out to within an order of magnitude in this case but the ominous exponential function sitting on the end gives a huge value, as the reaction energy (using the lower heat value) is -802 340 J/mol.

Being generous, we shall assume the atmosphere is at 300K (lower values are probably more reasonable given CH4's propensity to be upwardly displaced and the general lowering of temperatures as one moves upward)

So we have K~ 1*exp(802340/(8.314*300)=5*10^139

Plugging that into our (rearranged) original equation, we have

[CO2]/[CH4]=5*10^139*[O2]^2/[H2O^2]

Now there is much more O2 in the atmosphere than H2O, but lets be further generous (and not waste time looking up extra stuff) and assume 1/1 ratio of O2 to H2O

We then Have an equilibrium ratio of [CO2]/[CH4]=5*10^139 (given our generosity, a gross underestimation)

The actual ratio is about [CO2]/[CH4]=2*10^2

So we can very strongly conclude CO2 is in no way in equilibrium with CH4.

To those who wish to argue against global warming, or any aspect thereof, thats fine; but please do not undermine yourselves by relying in part or full on bad science.