## General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal

# Chapter 15 - Priciples of Chemical Equilibrium - Exercises - Experimental Determination of Equilibrium Constants - Page 689: 17

#### Answer

$$K_C = 0.106$$

#### Work Step by Step

1. Calculate both concentrations: $$[PCl_5]_{initial} = \frac{1.00 \times 10^{-3} \space mol}{250.0 \space mL} \times \frac{1000 \space mL}{1 \space L} = 4.00 \times 10^{-3} \space M$$ $$[Cl_2]_{eq} = \frac{9.65 \times 10^{-4} \space mol}{250.0 \space mL} \times \frac{1000 \space mL}{1 \space L} = 3.86 \times 10^{-3} \space M$$ 2. At equilibrium, these are the concentrations of each compound: $[ PCl_5 ] = 4.00 \times 10^{-3} \space M - x$ $[ PCl_3 ] = 0 \space M + x$ $[ Cl_2 ] = 0 \space M + x$ 3. Using the concentration of $Cl_2$ at equilibrium, find x: $0 + x = 3.86 \times 10^{-3}$ $x = 3.86 \times 10^{-3} - 0$ $x = 3.86 \times 10^{-3}$ $[ PCl_5 ] = 4.00 \times 10^{-3} \space M - 3.86 \times 10^{-3} =1.40 \times 10^{-4}$ $[ PCl_3 ] = 0 \space M + 3.86 \times 10^{-3} =3.86 \times 10^{-3}$ $[ Cl_2 ] = 0 \space M + 3.86 \times 10^{-3} =3.86 \times 10^{-3}$ - The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ PCl_3 ][ Cl_2 ]}{[ PCl_5 ]}$$ 4. Substitute the values and calculate the constant value: $$K_C = \frac{( 3.86 \times 10^{-3} )( 3.86 \times 10^{-3} )}{( 1.40 \times 10^{-4} )} = 0.106$$

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