Posted by **Scott L.** on Sunday, May 23, 2010 at 9:07pm.

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

- Chemistry -
**bobpursley**, Sunday, May 23, 2010 at 9:22pm
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

- Chemistry -
**Scott L.**, Sunday, May 23, 2010 at 10:41pm
I don't understand where you got the 4 on the left side of your equation from? May you please explain?

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