Question

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?

Answer 1

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