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 854: 62b

Answer

$10.2\,kJ$ The reaction is non-spontaneous.

Work Step by Step

$\Delta H^{\circ}=\Sigma n_{p}\Delta H_{f}^{\circ}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$ $=[1\,mol\times\Delta H_{f}^{\circ}(CaO,s)+1\,mol\times\Delta H_{f}^{\circ}(CO_{2},g)]-[1\,mol\times\Delta H_{f}^{\circ}(CaCO_{3},s)]$ $=[1\,mol(-634.9\,kJ/mol)+1\,mol(-393.5\,kJ/mol)]-[1\,mol(-1207.6\,kJ/mol)]$ $=179.2\,kJ$ $\Delta S^{\circ}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$ $=[S^{\circ}(CaO,s)+S^{\circ}(CO_{2},g)]-[S^{\circ}(CaCO_{3},s)]$ $=[1\,mol(38.1\,Jmol^{-1}K^{-1})+1\,mol(213.8\,Jmol^{-1}K^{-1})]-[1\,mol(91.7\,Jmol^{-1}K^{-1})]$ $=160.2\,JK^{-1}=0.1602\,kJK^{-1}$ $\Delta G=\Delta H-T\Delta S$ $=(179.2\,kJ)-(1055\,K)(0.1602\,kJK^{-1})$ $=10.2\,kJ$ Because $\Delta G$ is positive, the reaction is non-spontaneous.
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