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

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 15 - Priciples of Chemical Equilibrium - Example 15-2 - Relating K to the Balanced Chemical Equation - Page 664: Practice Example A


$$K = 83$$

Work Step by Step

1. If we take all the coefficients of the following equation: $$N_2(g) + 3H_2(g) \leftrightharpoons 2 NH_3(g)$$ And divide them by 3, we get the reaction we have. Thus, we just need to find the cube root (**1/3) of K for the equation above. $$K'' = \sqrt [3] {K} = \sqrt [3] {5.8 \times 10^5} = 83$$
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