Answer
$$K_C =3.3 \times 10^2$$
Work Step by Step
1. The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ FeSCN^{2+} ]}{[ Fe^{3+} ][ SCN^{-} ]}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ Fe^{3+} ] = 1.0 \times 10^{-3} \space M - x$
$ [ SCN^{-} ] = 8.0 \times 10^{-4} \space M - x$
$ [ FeSCN^{2+} ] = 0 \space M + x$
3. Using the concentration of $ FeSCN^{2+} $ at equilibrium, find x:
$ 0 + x = 1.7 \times 10^{-4} $
$x = 1.7 \times 10^{-4} $
$ [ Fe^{3+} ] = 1.0 \times 10^{-3} \space M - 1.7 \times 10^{-4} =8.3 \times 10^{-4} $
$ [ SCN^{-} ] = 8.0 \times 10^{-4} \space M - 1.7 \times 10^{-4} =6.3 \times 10^{-4} $
$ [ FeSCN^{2+} ] = 0 \space M + 1.7 \times 10^{-4} =1.7 \times 10^{-4} $
4. Substitute the values and calculate the constant value:
$$K_C = \frac{( 1.7 \times 10^{-4} )}{( 8.3 \times 10^{-4} )( 6.3 \times 10^{-4} )} = 3.3 \times 10^2$$
