Scott L.

Consider the reaction below: 4NH3(g) + 3O2(g) -> 2N2 + 6H20, K=10^80 @ certain temp initially, all the reactants and products have concentrations equal to 12M. At equilibrium, what is the approximate concentration of oxygen?

At a particular temp, K = 1.6 x 10^-5 for the reaction: 2SO3(g) <-> 2SO2(g) + O2 (g) If 4 mol of SO2 and 2 mol of O2 are placed into a 2.0L flask, calculate the equilibrium concentrations of all species. I set up an ICE table with SO3 to start with 0M and...

i set up an ICE table then got: 10^80 = [(12+2x)^2 (12+6x)^6]/[(12-4x)^4 (12-3x)^3] But then it looks waaay to complicated. Am I doing it wrong? Please help. Thanks

Oh ok I understand now. Thanks!

I don't understand where you got the 4 on the left side of your equation from? May you please explain?