## Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company

# Chapter 15 - Questions and Problems - Page 709: 15.46

#### Answer

(a) $$K_c = 0.50$$ (b) $[ CO_2 ] = 0.48 \space M$ $[ H_2 ] = 0.020 \space M$ $[ CO ] = 0.075 \space M$ $[ H_2O ] = 0.065 \space M$

#### Work Step by Step

(a) 1. Write the equilibrium constant expression: - The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ CO ][ H_2O ]}{[ CO_2 ][ H_2 ]}$$ 2. Substitute the values and calculate the constant value: $$K_c = \frac{( 0.050 )( 0.040 )}{( 0.086 )( 0.045 )} = 0.50$$ ------- (b) 3. At equilibrium, these are the concentrations of each compound: $[ CO_2 ] = 0.50 \space M - x$ $[ H_2 ] = 0.045 \space M - x$ $[ CO ] = 0.050 \space M + x$ $[ H_2O ] = 0.040 \space M + x$ $$0.50 = \frac{(0.050 + x)(0.040 + x)}{(0.50 - x)(0.045 - x)}$$ x = 0.025 $[ CO_2 ] = 0.48 \space M$ $[ H_2 ] = 0.020 \space M$ $[ CO ] = 0.075 \space M$ $[ H_2O ] = 0.065 \space M$

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