## Introductory Chemistry (5th Edition)

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