Chemistry (4th Edition)

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

Chapter 15 - Questions and Problems: 15.27

Answer

$K_{p} = 0.105$ $K_{c} = 2.05\times10^{-3}$

Work Step by Step

Given, $K_{p}$ = 0.105 T = (350+273) K = 623 K $\Delta n = 1+1-1 = 1$ We know, $K_{p} = K_{c}(RT)^{\Delta n}$ Or $K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{0.105}{\frac{0.08206L.atm}{K.mol}\times623 K} = 2.05\times10^{-3}$
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