Chemistry 9th Edition

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

Chapter 17 - Spontaneity, Entropy, and Free Energy - Exercises: 31

Answer

A)$\ H_{2}$ at 100°C and 0.5 atm B)$\ N_{2}$ at STP C)$\ H_{2}O (l)$ at 20°C

Work Step by Step

A)$\ H_{2}$ at 100°C and 0.5 atm is considered to be the right answer here because the water molecule is much more active at 100°C versus the 0°C the hydrogen is at STP. Positional Probability depends on the number of configurations in space that yield a particular state. Because the temperature is higher at 100°C as opposed to 0°C, the hydrogen molecules are much more active and have a greater likelihood of being in multiple positions or states. B)$\ N_{2}$ at STP can be explained to have a larger positional probability than 1 mol of$\ H_{2}$ at 100k and 2.0 atm through entropy: the measure of randomness or disorder (the number of variation in the arrangments/positions that are available to a system). As the pressure increases in a system, the entropy and thus positional probability decreases. Moreover, because the temperature is greater at STP, the positional probability increases as the nitrogen molecule is excited and more active. C)$\ H_{2}O (l)$ at 20°C has a greater positional probability than$\ H_{2}O (s)$ at 0°C. This is because of the temperature and state of matter of the two molecules. Frozen ice, at 0°C exemplifies a lower positional probability because a solid molecule has a lower positional probability: a solid is not as excited as liquid or gaseous molecules. Moreover,$\ H_{2}O (l)$ at 20°C is at a higher temperature and can thus contain more positions than the ice.
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