## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 15 - Exercises - Page 715: 44

#### Answer

$K_c = 0.018$

#### Work Step by Step

1. At equilibrium, these are the concentrations of each compound: $[ SO_2Cl_2 ] = 0.020 \space M - x$ $[ SO_2 ] = 0 \space M + x$ $[ Cl_2 ] = 0 \space M + x$ 2. Using the concentration of $Cl_2$ at equilibrium, find x: $0 + x = 1.2 \times 10^{-2}$ $x = 0.012$ $[ SO_2Cl_2 ] = 0.020 \space M - 0.012 =8.0 \times 10^{-3}$ $[ SO_2 ] = 0 \space M + 0.012 =0.012$ $[ Cl_2 ] = 0 \space M + 0.012 =0.012$ - The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ SO_2 ][ Cl_2 ]}{[ SO_2Cl_2 ]}$$ 3. Substitute the values and calculate the constant value: $$K_C = \frac{( 0.012 )( 0.012 )}{( 8.0 \times 10^{-3} )} = 0.018$$

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