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 = 2-2x, and O2 = 1-x.
After I got:
1.6 x 10^-5 = [(2-2x)^2 (1-x)]/(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

I didnt' check the equation, but yes, solve for x.

4*1.6E-5 x^2=(2-2x)^2 (1-x)
The left side had to be close to zero, so assume
0=(2-2x)^2 (1-x)
or the solution is x=1M

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

To determine the equilibrium concentrations of all species, you correctly set up an ICE table and expressed the equilibrium concentrations as variables (SO3 = 2x, SO2 = 2-2x, and O2 = 1-x).

Next, you substituted these expressions into the equilibrium constant expression:
1.6 x 10^-5 = [(2-2x)^2 (1-x)] / (2x)^2

Solving this equation for x will give you the value of x, which represents the amount of SO3 that reacts to form SO2 and O2 at equilibrium.

This equation might appear complex, but there are a few simplifications you can make to simplify the math.

First, you can expand the squared terms in the equation to get rid of the parentheses:
1.6 x 10^-5 = [(4 - 8x + 4x^2) (1-x)] / 4x^2

Next, you can multiply both sides of the equation by 4x^2 to eliminate the denominator:
(4x^2)(1.6 x 10^-5) = (4 - 8x + 4x^2) (1-x)

After multiplying out the terms, you will end up with a quadratic equation. You can solve this equation using the quadratic formula or by factoring.

Once you find the value of x, you can substitute it back into the equilibrium expressions for SO3, SO2, and O2 to find their equilibrium concentrations.