Thermodynamics:
Entropy

Thermodynamics Gateway Page
In this module:
Introduction
Disorder in Atoms
Disorder in Energy
Measuring Entropy
Entropy of Phase Changes
Patterns in the Entropies of Substances
Entropy in Thermochemical Equations
The Second Law of Thermodynamics
The Effect of Temperature
Predicting How Reactions will Go
Two Examples

Two Examples

In the introduction to this tutorial you encountered two reactions that were endothermic yet still product-favored. The second reaction was between barium hydroxide and ammonium chloride:

Ba(OH)2• 8H2O (s) + 2 NH4Cl (s) BaCl2• 2H2O (s) + 2 NH3 (aq) + 8 H2O (l)

Three moles of two solids react to form one mole of a solid, eight moles of liquid, and two moles of an aqueous solution. Using Patterns in the Entropies of Substances, would you expect the entropy change of this reaction to be positive or negative?

Calculate DSuniv for the reaction and verify that it is, indeed, positive, making the reaction product-favored.

DSuniv =
DHº = 1 mol • DHfº(BaCl2• 2H2O(s)) + 2 mol • DHfº(NH3(aq)) + 8 mol • DHfº(H2O(l)) - 1 mol • DHfº(Ba(OH)2• 8H2O(s)) - 2 mol • DHfº(NH4Cl)(s)
DHº = 1 mol • -1460.1 kJ/mol + 2 mol • -80.29 kJ/mol + 8 mol • -285.83 kJ/mol - 1 mol • -3342 kJ/mol - 2 mol • -314.4 kJ/mol
DHº = 63 kJ

DSº = 1 mol • Sº(BaCl2• 2H2O(s)) + 2 mol • Sº(NH3(aq)) + 8 mol • Sº(H2O(l)) - 1 mol • Sº(Ba(OH)2• 8H2O(s)) - 2 mol • Sº(NH4Cl)(s)
DSº = 1 mol • 203 J/mol•K + 2 mol • 111 J/mol•K + 8 mol • 69.91 J/mol•K - 1 mol • 427 J/mol•K - 2 mol • 94.6 J/mol•K
DSº = 368 J/K

Duniv = DSº - DHº/T
Duniv = 368 J/K - 63 kJ/(273.15 + 25)K • 1000 J/1 kJ = 157 J/K > 0

When ice melts to liquid water, there is a similar increase in entropy in the system that offsets its increase in enthalpy:

H2O (s) H2O (l)

Thus, this reaction is also product-favored at 25 ºC.


Two Examples