Sulfur dioxide gas reacts with oxygen gas for form sulfur trioxide gas. the equilibrium constant for the reaction at 6000c is 4.34. an equilibrium mixture is found to have [so3] = 0.391 m and [o2] = 0.125 m at 6000c. what is the concentration of sulfur dioxide?

When the balanced equation for this reaction is:

            2SO3 → 2SO2 + O2 so, by using ICE table:

initial    0.391              0          0.125m

change  -X               +2X           +X

Equ     (0.391-X)         2X           0.125+X

when Keq = concentration of products / concentration of reactants

so, by substitution

    4.34  = [2X]^2 * [0.125+X] / [0.391-X]^2  by solving for X

∴ X = 0.357 m 

∴ the concentration of SO2 = 2x = 2 * 0.357 

                                               = 0.714 m 

Rethink and reinvent