Answer
402 J/mol
Work Step by Step
We find:
$\Delta S^{\circ}_{tr}=\frac{\Delta H^{\circ}_{tr}}{T_{tr}}$
$\implies \Delta H^{\circ}_{tr}=\Delta S^{\circ}_{tr}\times T_{tr}$
$=(1.09\,J\,mol^{-1}K^{-1})(95.5+273.15)K=402\,J/mol$
General Chemistry: Principles and Modern Applications (10th Edition)
by Petrucci, Ralph H.; Herring, F. Geoffrey; Madura, Jeffry D.; Bissonnette, Carey
Published by
Pearson Prentice Hal
ISBN 10:
0132064529
ISBN 13:
978-0-13206-452-1
Chapter 19 - Spontaenous Change: Entropy and Gibbs Energy - Example 19-2 - Determining the Entropy Change for a Phase Transition - Page 829: Practice Example B
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