Methane undergoes combustion with 100 theoretical pure oxyge

Methane undergoes combustion with 100% theoretical pure oxygen. In the product mixture, carbon dioxide undergoes dissociation into 2 kmols of carbon monoxide and 1 kmol of oxygen. If the total number of kmols in the product mixture is 3.65 kmols, what is the mole fraction of carbon monoxide at equilibrium?

Solution

The combustion of methane (full combustion) follows

CH4+ 2O2 => 2H2O + CO2

Further dissociation of CO2 follows 2CO2=> 2CO +O2

Overall eqn CH4+ 2O2 => 2H2O + CO +0.5O2

Hence in 3.5 kmoles of product formed CO is 1 kmole

For 3.65 kmoles , CO comprises (1/3.5) * 3.65 kmoles = 1.043 Kmoles

Molefraction of CO is 1/3.5 =.286