I am trying to understand the atmospheric ppm data, specifically around how CH4 decays to CO2.

This data says 2017 CH4 is about 1850ppm, and 2017 CO2 is 402ppm increasing about 3ppm/year. (https://www.eea.europa.eu/data-and-maps/daviz/atmospheric-concentration-of-carbon-dioxide-5)

I've also read that CH4 has a half-life in the atmosphere of about 9 years, decaying with the chemical equation CH4 + O2 → CO2 + H4

So if you'd expect methane's 1850ppm to decay to 925ppm over 9 years, that means you'd also expect to see an increase of 925ppm of CO2, or about 100ppm/yr just from methane decay. That'd be insanely high. But the data I linked earlier shows an increase of 3ppm.

So what am I missing in this analysis?

  • 7
    $\begingroup$ Reread your source. The CH4 (methane) numbers are in parts per billion while CO2 is in parts per million. $\endgroup$ Feb 8, 2022 at 8:33
  • $\begingroup$ Methane emissions are far from as quantitative as CO2 emissions. Annual CO2 emissions were around 34 Billion tons. Annual methane emissions are 570 million tons, or 1.6% as much decomposition reaction: CH4 + O2 → CO2 + H4 $\endgroup$
    – LazyReader
    Feb 9, 2022 at 3:44

1 Answer 1


So what am I missing in this analysis?

The linked data says 2017 CH4 was about 1850 ppb, not 1850 ppm. Dividing parts per billion (ppb) by 1000 results in parts per million (ppm). In other words, the atmospheric methane concentration was about 1.850 ppm in 2017. Assuming an exponential decay with a half life of 9 years means that about 0.14 ppm $\left(1850*(1-2^{-1/9}\,)\right)$ of the 2.4 ppm increase in the 2017 to 2018 CO2 concentration was due to methane decomposition. That's a small but not insignificant portion of the increase.

  • $\begingroup$ That makes so much more sense. It's not 100% insignificant but like 99% insignificant then. If methane decays relatively quickly, and it's tiny compared to the amount of CO2, it seems to me it shouldn't be much of a concern. $\endgroup$
    – Some Guy
    Feb 8, 2022 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.