Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 17 - Sections 17.1-17.9 - Exercises - Problems by Topic - Page 855: 70b

Answer

7.83

Work Step by Step

$\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$ $=[2\Delta G_{f}^{\circ}(BrCl,g)]-[\Delta G_{f}^{\circ}(Br_{2},g)+\Delta G_{f}^{\circ}(Cl_{2},g)]$ $=[2(-1.0\,kJ/mol)]-[(3.1\,kJ/mol)+(0\,kJ/mol)]$ $=-5.1\,kJ/mol$ $\Delta G^{\circ}=-RT\ln K\implies \ln K=-\frac{\Delta G^{\circ}}{RT}$ $=-\frac{-5.1\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(25+273)K}=2.058467$ Then, $K=e^{2.058467}=7.83$
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