# Not sure why, but I can't modify/balance this equation

I'm working on modeling a few reactions for my research. For example:

$$\ce{CO2 + H2S + CH4 <=> CH3SH + H2O}$$

was one I grabbed from the literature. However, this reaction is taking place in the aqueous phase around neutral pH, so most of the $\ce{CO2}$ will actually be $\ce{HCO3-}$. So in an attempt to model modified versions, I swap out the $\ce{CO2}$:

$$\ce{HCO3- + H2S + CH4 <=> CH3SH}$$

I've tried a redox-equation balancing approach but this hasn't really gotten me anywhere- sulfur doesn't change oxidation states.

I think my brain might just be fried, but can anyone lend any insight on this?

• You can't just swap out the carbon dioxide. For starters you lost the water to balance all the oxygen atoms... The equation presented should be balanced as is. You're free to balance a different equation, but that's a different equation (try to make sure that one can actually be balanced). – Zhe Aug 10 '18 at 21:02
• If sulfur doesn't change oxidation states, then what does? – Ivan Neretin Aug 11 '18 at 4:48
• There's no way such reaction happens in such conditions! Where did you hear something like that? – Mithoron Aug 11 '18 at 18:52
• Temp. could only have indirect influence - it looks it's a complicated enzymatic reaction proceeding in vivo - saying it happens in aq. in mild pH is like saying you detonated a piece of TNT with a touch of a finger, when you actually used a button on a bomb. – Mithoron Aug 12 '18 at 18:38
• @Mithoron- I see. Yeah, we are looking at the thermodynamics of the reaction- nothing kinetically/enzymatically, so it may not proceed quickly. I was just trying to modify the equation from literature that had used aqueous CO2 to actually use bicarb instead. Just in the modeling phase at this point- haven't carried this out in lab yet. – cdarwin Aug 12 '18 at 18:42

For starters, You have a problem because of a lack of charge balancing. You should also consider that at neutral pH $\ce{H2S}$ is present as $\ce{HS-}$ and $\ce{CH3SH}$ is present as $\ce{CH3S-}$. Modifying the reaction to account for charge you get: $$\ce{HCO3- + HS- + CH4 <=> CH3S-}$$
This will give you a start. The other piece that will help is what the $\ce{HCO3-}$ molecule becomes. For the sake of being able to provide an answer I will assume formate ($\ce{HCO2-})$ is formed. Thus the reaction is now: $$\ce{HCO3- + HS- + CH4 <=> CH3S- + HCO2-}$$
We know that the molecule of $\ce{CH4}$ becomes $\ce{CH3S-}$ and the molecule of $\ce{HCO3-}$ becomes $\ce{HCO2-}$ thus to balance the lost oxygen of $\ce{HCO3-}$ and the lost two hydrogens of $\ce{CH4}$ we add water to the products. $$\ce{HCO3- + HS- + CH4 <=> CH3S- + HCO2- + H2O}$$