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 217g: 96


$-27.3\ kJ/mol$

Work Step by Step

Number of moles of lead nitrate: $200.\ mL\times 0.75\ M=150\ mmol$ Number of moles of NaBr: $200.\ mL\times 1.5\ M=150\ mmol$ Both are in stoichiometric proportions. From the definition of heat capacity: $(200+200)\ mL\times 1.0\ g/mL \times 4.2\ J/g.K \times 2.44\ K=4099.2\ J$ With a negative sign since it's exothermic. Per mol: $-4099.2\ J/150\ mmol=-27.3\ kJ/mol$
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