In atmospheric chemistry, the lifetime of most substances is dictated by their reactions with the hydroxyl ($\ce{OH}$) radical. The rates of these reactions are listed by NIST as about $\mathrm{1.5e^{-13} cm^{3}\ mol^{-1}\ s^{1}}$ for carbon monoxide ($\ce{CO}$) and $\mathrm{6.7e^{-15}\ cm^{3}\ mole^{-1}\ s^{1}}$ for methane ($\ce{CH4}$). With atmospheric $\ce{OH}$ concentrations of about $\mathrm{1.2e^6\ mole\ cm^{-3}}$, this leads to an atmospheric lifetime of about 2 months for $\ce{CO}$ and 4 years for $\ce{CH4}$. This is indeed what's observed, so the relative reaction rates are definitely experimentally true.
My question is: what's the theoretical explanation for the $\ce{CO}$ reaction being faster? I would expect the $\ce{C-O}$ triple bond to be much more stable than any $\ce{C-H}$ bond. In atmospheric chemistry texts I've seen, the reaction goes like
$$\ce{CO + OH -> CO2 + H*}$$ $$\ce{H* + O2 -> HO2}$$
I've also come across notes from a pure chemistry class that mentions a different reaction with an intermediate:
$$\ce{CO + OH -> OCOH*}$$ $$\ce{OCOH* + O2 -> CO2 + HO2}$$
I can't get my head around how the intermediate would/should increase the reaction rate, though, or how something like electronic configuration could lead to $\ce{CO}$ being more susceptible to $\ce{OH}$ than $\ce{CH4}$.
