General Chemistry 10th Edition

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: 18.80


$159.10 kJ/mol$

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: $\Delta G=(-16.40+124.7)-(-50.80)$ $\Delta G=159.10 kJ/mol$
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