Relative atmospheric lifetimes of carbon monoxide and methane 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  

$$\ce{CO + OH -&gt; CO2 + H*}$$ 
$$\ce{H* + O2 -&gt; HO2}$$

I've also come across notes from a pure chemistry class that mentions a different reaction with an intermediate:

$$\ce{CO + OH -&gt; OCOH*}$$
$$\ce{OCOH* + O2 -&gt; CO2 + HO2}$$

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 $\ce{CH4}$.
The dominant oxidation (via hydroxyl radical) pathways for CO don't have any intermediate species as unstable as the methyl radical, which is step one for methane oxidation via the hydroxyl radical.