Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 15 - Chemical Equilibrium - Exercises: 15.26a

Answer

Kc = 281.03

Work Step by Step

$ Kp = Kc(RT)^{\Delta n} $ If we have Kp = 3.423 then we need to find the value of $\Delta n$ 2SO$_2$(g) + O$_2$(g) 2SO$_3$(g) Sicne we have 3 moles on the left and 2 moles on the right, it means we have a $\Delta n$ of -1. Therefore, we have: $ 3.423 = Kc(RT)^-1 $ R = 0.0821 and T = 1000K so we have RT = 82.1 and then Kc = 281.03
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