## Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company

# Chapter 15 - Questions and Problems: 15.26

#### Answer

a) $K_{c} = 2.6\times10^{4}$ b) $K_{p} = 3.2\times10^{2}$

#### Work Step by Step

a) $K_{c} = \frac{1}{3.8\times10^{-5}} = 2.6\times10^{4}$ b) T = (727+273) K = 1000 K, $\Delta n$ = 1-2 = -1 , $K_{p} = K_{c}(RT)^{\Delta n} =(2.6\times10^{4})\times (\frac{0.08206L.atm}{K.mol}\times1000K)^{-1}=3.2\times10^{2}$

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