Answer
$$K_c = 3.3 \times 10^2$$
Work Step by Step
1. Write the equilibrium constant expression:
- 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$
As we know, at equilibrium: $[ FeSCN^{2+} ] = 1.7 \times 10^{-4} \space M$
Thus, $x = 1.7 \times 10^{-4}$. Substituting:
$ [ Fe^{3+} ] = 8.3 \times 10^{-4} \space M$
$ [ SCN^- ] = 6.3 \times 10^{-4} \space M$
$$K_c = \frac{( 1.7 \times 10^{-4} )}{( 8.3 \times 10^{-4})( 6.3 \times 10^{-4})} = 3.3 \times 10^2$$
