Chemistry (4th Edition)

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

Chapter 15 - Questions and Problems: 15.23

Answer

$3.5\times10^{-7}$

Work Step by Step

$K_{p} = 1.8\times10^{-5}$ T = (350+273) K = 623 K $\Delta n$ = (2+1)-2 = 1 $K_{p} = K_{c} (RT)^{\Delta n}$ Rearranging the equation, $K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{1.8\times10^{-5}}{0.08206L.atm.K^{-1}mol^{-1}\times 623 K} = 3.5\times10^{-7}$
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