Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 15 - Questions and Problems - Page 708: 15.25


a) $8.2\times10^{-2}$ b) 0.29

Work Step by Step

a) $K_{c} = \frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}} = \frac{(0.25)^{2}}{0.11(1.91)^{3}} \approx 8.2\times10^{-2}$ b) The equation is divided by 2. So, the new equilibrium constant is the square root of the original. i.e. $K_{c}= (8.2\times10^{-2})^{\frac{1}{2}} \approx 0.29$
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