Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 15 - Exercises - Page 715: 43

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: $$K_C = \frac{( 1.7 \times 10^{-4} )}{( 8.3 \times 10^{-4} )( 6.3 \times 10^{-4} )} = 3.3 \times 10^2$$
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