In an experiment, 0.0300 moles each of SO3(g), S02(g), and O2(g) were placed in a 10.0L flask at a certain temperature. When the reaction came to equilibrium, the cocentration of SO2(g) in the flask was 3.50x10^-5 molar. What is Kc for the reaction? 2SO2(g) + O2(g) 2SO3(g)

Question

In an experiment, 0.0300 moles each of SO3(g), S02(g), and O2(g) were placed in a 10.0L flask at a certain temperature. When the reaction came to equilibrium, the cocentration of SO2(g) in the flask was 3.50x10^-5 molar. What is Kc for the reaction? 2SO2(g) + O2(g) <--> 2SO3(g)
a. 3.5x10^-5
b. 1.9x10^7
c. 5.2x10^-8
d. 1.2X10^-9
e. 8.2x10^8

The answer is B but i am not sure why

DrBob222 · Answer

I change these to M initially = 0.0300mols/10L = 0.00300 M.
.........2SO2 + O2 ==> 2SO3
I....0.00300.0.00300..0.00300
C........-2x.....-x......+2x
E..0.00300-2x..0.00300-x..0.00300+2x

At equil (SO2) = 3.5E-5M.
Therefore, 0.00300-2x = 3.5E-5.
Solve for x, the calculate concns of the other reactants/products and set up for Kc. b is the correct answer but post your work if you still don't understand how to obtain that answer.