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: a


$K^{\circ}=4.9\times10^{-9}$ $K^{\circ}$ is close to $K_{c}$.

Work Step by Step

$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta_{f}G^{\circ}(reactants)$ $=[\Delta_{f}G^{\circ}(Ca^{2+},aq)+\Delta _{f}G^{\circ}(CO_{3}^{2-},aq)]-[\Delta_{f}G^{\circ}(CaCO_{3},s)]$ $=[(-553.58\,kJ/mol)+(-527.81\,kJ/mol)]-(-1128.79\,kJ/mol)$ $=47.4\,kJ/mol$ $\Delta _{r}G^{\circ}=-RT\ln K^{\circ}$ $\implies K^{\circ}=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{47.4\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}}$ $=4.9\times10^{-9}$ From table 12.1, $K_{c}=2.8\times10^{-9}$ $K^{\circ}$ is close to $K_{c}$.
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