## Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company

# Chapter 15 - Questions and Problems: 15.22

#### Answer

$1.76\times10^{20}$

#### Work Step by Step

$K_{c} = 2.24\times10^{22}$ R = $\frac{0.0821 L.atm}{K.mol}$ (Universal gas constant) T = (1273+273) K = 1546 K $\Delta n$ = 2 - 2 - 1 = -1 Substituting the values in the relation, $K_{p} = K_{c}(RT)^{\Delta n} = (2.24\times10^{22})(\frac{0.0821L.atm}{K.mol}\times 1546 K)^{-1} = 1.76\times10^{20}$

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