Answer
The $I$ concentration on this equilibrium is equal to 0.015M
Work Step by Step
1. Write the equilibrium constant expression:
$K_{eq} = \frac{[Products]}{[Reactants]} = \frac{[I]^2}{[I_2]}$
2. Find the $I$ concentration:
$1.1 \times 10^{-2} = \frac{[I]^2}{(0.0205)}$
$1.1 \times 10^{-2} \times 0.0205 = [I]^2$
$2.255 \times 10^{-4} = [I]^2$
$[I] = \sqrt {2.255 \times 10^{-4}} = 0.015M$
