Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions - Study Questions - Page 217i: 108


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Work Step by Step

a) $\begin{smallmatrix} CaBr_2(s)+H_2O(g)&\rightarrow &CaO(s)+2\ HBr(g)\\ HgBr_2(s)+\underline{CaO}(s)&\rightarrow &HgO(s)+\underline{CaBr_2}(s)\\ \underline{HgO}(s)&\rightarrow&{Hg}(l)+1/2\ O_2(g)\\ \underline{Hg}(l)+\underline{2\ HBr}(s)&\rightarrow &\underline{HgBr_2}(s)+H_2(g)\\ \\ \hline\\ H_2O(g)&\rightarrow&H_2(g)+1/2\ O_2(g) \end{smallmatrix}$ b) From the overall stoichiometry: $1000\ kg\div 18.015\ kg/kmol\ H_2O\times 1\ kmol\ H_2/kmol\ H_2O\times2.016\ kg/kmol\ H_2=112\ kg$ c) Enthalpies of formation (kJ/mol): $CaO(s): -635.09, H_2O(g): -241.83, HBr(g): -36.29, HgO(s): -90.83$ First reaction: $-635.09+2\times -36.29--241.83--683.2=+217.36\ kJ/mol$ Endothermic Second reaction: $-169.5-2\times-36.29=-96.92\ kJ/mol$ Exothermic Third reaction: $-683.2-90.83--635.09--169.5=+30.56\ kJ/mol$ Endothermic Fourth reaction: $--90.83=+90.83\ kJ/mol$ Endothermic. d) This method is subject to a series of industrial inefficiencies and logistical problems, which could be worth it in a cost-benefit analysis, but an alternative method, like electrolysis, should be considered.
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