Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 6 - Thermochemistry - Exercises - Page 288: 64


$26\text{ kJ/mol}$

Work Step by Step

$q=mc\Delta T$ $q=75.0*4.18*(23.34-25.00)=-520.41\text{ J}$ $1.60\text{ grams $NH_4NO_3$}*\frac{1\text{ mol $NH_4NO_3$}}{80.043\text{ grams $NH_4NO_3$}}=0.0200$ We switch the signs of the enthalpy change as the system is the $NH_4OH$ which is absorbing energy. $\frac{520.14\text{ J}}{0.200\text{ mols}}=26\text{ kJ/mol}$
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