General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

Published by Pearson
ISBN 10: 0321967461
ISBN 13: 978-0-32196-746-6

Chapter 10 - Reaction Rates and Chemical Equilibrium - 10.3 Equilibrium Constants - Questions and Problems - Page 383: 10.20

Answer

$$K_c = 170$$

Work Step by Step

1. Write the equilibrium constant expression: $$K_c = \frac{[Products]}{[Reactants]} = \frac{[ NH_3 ]^{ 2 }}{[ N_2 ][ H_2 ]^{ 3 }}$$ 2. Substitute the values and calculate the constant value: $$K_c = \frac{( 2.2 )^{ 2 }}{( 0.44 )( 0.40 )^{ 3 }} = 170$$
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