General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 14 - Chemical Equilibrium - Questions and Problems - Page 629: 14.83


Please see the work below.

Work Step by Step

We know that $K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$ We plug in the known values to obtain: $K_c=\frac{0.015}{(0.096)(0.191)^2}=4.28$
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