Answer
The $SO_2$ concentration in equilibrium is equal to 0.531M.
Work Step by Step
1. Write the $K_{eq}$ expression:
$K_{eq} = \frac{[Products]}{[Reactants]} = \frac{[SO_3]^2}{[SO_2]^2][O_2]}$
2. Use the given values to find the $SO_2$ concentration:
$4.34 = \frac{(0.391)^2}{[SO_2]^2 \times 0.125}$
$[SO_2]^2 = \frac{(0.391)^2}{4.34 \times 0.125}$
$[SO_2] = \sqrt {\frac{(0.391)^2}{4.34 \times 0.125}} = 0.531M$
