My book gives the equation for synthesis of producer gas as follows:

$$ \ce{2C{(s)} + O2{(g)} + N2{(g)} -> 2CO{(g)} + 4N2{(g)}} $$

And it says that the stuff on the right-hand side is called producer gas.

My question is why does $\ce{N2}$ even appear in the equation? It isn't reacting is it? It's just a mixture.

  • $\begingroup$ en.wikipedia.org/wiki/Producer_gas $\endgroup$ – Mithoron May 12 '17 at 18:20
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    $\begingroup$ In the wikipedia page, the nitrogen has a (seemingly) empirically derived stoichiometric coefficient, so I would also be interested in knowing why that level of detail is presented (since I believe it is not facilitating/participating in the reaction). I would hazard a guess and say that if you had a process that made producer gas, you would track $\ce{N2}$ per unit volume because that would effect the caloric value of the mixture to be used as fuel. $\endgroup$ – J. Ari May 12 '17 at 20:17
  • $\begingroup$ I can tell you why the coefficient is four, or rather four times that of oxygen: precisely because in air, they're found in roughly a 4:1 ratio. Yes, the reaction could definitely be written without the presence of N2. But as it stands, the N2 has been included to emphasise the presence of N2 in the reaction mixture (which, technically, is open air) and to indicate that our focus is on the collective mixture of carbon monoxide and nitrogen. $\endgroup$ – Parth Kohli May 13 '17 at 14:27

As the comments allude, the coefficient is 4 to represent the approximate 4:1 ratio of oxygen to nitrogen in air (more precisely 21:78 or 7:26). This added nitrogen must be heated along with the oxygen and carbon. From a stoichiometric perspective, nitrogen is just a spectator and could be excluded, but from a thermal/total system perspective as nitrogen increases the specific heat of the system requiring more heat to be inputted for heating the mixture which subsequently requires more heat dumping to cool the system, reducing the efficiency of the system compared to a system of pure reactants.


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