Why does the amount of an inert gas affect the equilibrium conversion? I had a problem in which sulfur dioxide reacts with oxygen gas to form sulfur trioxide. However, air was involved to supply the oxygen. Hence, about 79% of the air was nitrogen.
I had to do an expression for the equilibrium constant:
SO2 + 0.5 O2 = SO3
K = ySO3*P/[(ySO2*P) (yO2*P)1/2]
K = ySO3/[(ySO2) (yO2*P)1/2]
However, in order to calculate the partial pressures and/or percent compositions (the y values), I had to account for the amount of amout of material coming out, I had to do a calculation for amount coming in and amount coming out with the conversion.
However, I had to find percent composition. I could find the expression for the amount that comes out for each compound. However, I had to divide that by the total amount of moles to get percent composition. The problem was that I also had to include the nitrogen gas (which does not take part in the reaction) in the total moles.
This seems incredibly against what I have always believed in what I learned in chemistry. In a constant volume, additon of an inert (inert substance that takes no part in reaction) should have no affect on the equilibrium states. Can anyone answer this, or should I specify with the actual example I had to do???
Thanks in advance... I hope....