Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 16 - Thermodynamics: Directionality of Chemical Reactions - Problem Solving Practice 16.7 - Page 718: b



Work Step by Step

$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta_{f}G^{\circ}(reactants)$ $=[\Delta_{f}G^{\circ}(HCO_{3}^{-},aq)+\Delta _{f}G^{\circ}(H^{+},aq)]-[\Delta_{f}G^{\circ}(H_{2}CO_{3},aq)]$ $=[(-586.77\,kJ/mol)+(0\,kJ/mol)]-(-623.08\,kJ/mol)$ $=36.31\,kJ/mol$ $\Delta _{r}G^{\circ}=-RT\ln K^{\circ}$ $\implies K^{\circ}=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{36.31\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}}$ $=4.3\times10^{-7}$ From table 12.1, $K_{c}=4.3\times10^{-7}=K^{\circ}$
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