2
$\begingroup$

The intergovernmental panel on climate change's fifth assessment report (IPCC AR 5) lists the radiative forcing (RF) of methane in chapter 8 as being $28\times$ and $84\times$ as potent as $\ce{CO2}$ by weight (or tonne of methane for tonne of $\ce{CO2}$) for the global warming potential integrated over $20$ and $100$ years, $\mathrm{GWP}_{20}$ and $\mathrm{GWP}_{100}$ respectively. In order to calculate the ratios of the gases RF potential molecule for molecule in the 20 and 100 year time spans, I'd need to apply the ratio of atomic weights for these molecules. But do I need to consider the average concentration in the atmosphere of $\ce{CO2}$ when multiplying the RF ratios.

Adding to the complexity of the problem might be the fact that at various levels in the atmosphere methane may have a different lifespan, concentration and RF, i.e. in the troposphere methane has a role as a precursor to tropospheric ozone, which takes the overall RF ratio of atmospheric methane to $105\times$ $\ce{CO2}$ (Shindell et al, IPCC AR5, NASA). Or can I ignore this complexity and just derive an indicative molecule for molecule ratio of RF potency for $\ce{CH4}$ to $\ce{CO2}$ by dividing each side of the equation by the molecular weight?


My attempt at answering:

  • Molecular weight of methane is $\pu{16.04 g/mol}$ and $\ce{CO2}$ is $\pu{44.01 g/mol}$.
  • Assuming it’s safe not to consider atmospheric concentrations in this, $\pu{1 kg}$ of of methane has $44.01/16.04 = 2.744$ fewer molecules than $\ce{1 kg}$ of $\ce{CO2}$
  • If methane is $105\times$ as potent as $\ce{CO2}$ by weight, it is $105 \times 2.744 = 288\times$ as potent as $\ce{CO2}$ molecule for molecule
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.