Posted by Scott L. on .
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 SO2 to start with 2M and O2 with 1M.
At equilibrium, I got SO3 = 2x, SO2 = 22x, and O2 = 1x.
After I got:
1.6 x 10^5 = [(22x)^2 (1x)]/(2x)^2
After that do I just solve for x? If so, is there an faster way to determine x because that looks a bit complex. Thanks

Chemistry 
bobpursley,
I didnt' check the equation, but yes, solve for x.
4*1.6E5 x^2=(22x)^2 (1x)
The left side had to be close to zero, so assume
0=(22x)^2 (1x)
or the solution is x=1M 
Chemistry 
Scott L.,
I don't understand where you got the 4 on the left side of your equation from? May you please explain?