Answer
$+103.61\ kJ/mol$
Work Step by Step
Enthalpies of formation (kJ/mol):
$H_2O(l):-285.83, CO_2(g):393.509$
Enthalpy of combustion:
$-4395.0=4\times-285.83+8\times-393.509-x$
$x=+103.61\ kJ/mol$, which is the enthalpy of formation of styrene.
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
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 217d: 64
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