Chemistry (4th Edition)

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

Chapter 15 - Section 15.3 - Equilibrium Expressions - Checkpoint: 15.3.4

Answer

c) 0.14

Work Step by Step

$K_{c} = 1.1\times10^{-3}$ $R = \frac{0.08206L.atm}{K.mol}$ T = (1280+273) K = 1553 K $ \Delta n = 2-1 =1 $ Using the relation, $K_{p} = K_{c}(RT)^{\Delta n}$ We get, $K_{p} = 1.1\times10^{-3} (\frac{0.08206L.atm}{K.mol} \times 1553 K) \approx 0.14$
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