Answer
$\Delta G=266.77 kJ$
Work Step by Step
We know that
$\Delta G=\Sigma n\Delta G_f(products)-\Sigma n\Delta G_f(reactants)$
We plug in the known values to obtain:
$\Delta G=(-33.33)-(-300.1)$
$\Delta G=266.77 kJ$
General Chemistry 10th Edition
by Ebbing, Darrell; Gammon, Steven D.
Published by
Cengage Learning
ISBN 10:
1-28505-137-8
ISBN 13:
978-1-28505-137-6
Chapter 18 - Thermodynamics and Equilibrium - Questions and Problems - Page 773: 18.79
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