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

The Second Law of Thermodynamics

The situations described in the second and third pages of this tutorial illustrate the fact that product-favored reactions tend to increase disorder simply because they are much more likely to occur. The formal statement of this fact is the Second Law of Thermodynamics: in any product-favored process the entropy of the universe increases.

How can the Second Law of Thermodynamics be used to predict whether a reaction is product- or reactant-favored? The entropy change of the universe can be broken up into two parts, the entropy change of the system and the entropy change of the surroundings:

DSuniv = DSsyst + DSsurr

DSsyst, the entropy change of the system, represents the change in order of the molecules of the system, similar to what was discussed in Entropy 2. It can be calculated using absolute entropies as has been described on the previous page.

It may seem unlikely that the entropy change of the surroundings can be calculated just from what is known about the system. But, the only way the system affects the surroundings is by a transfer of heat. Since the surroundings are so much bigger than the system, its temperature is certain to stay constant. Thus, equation (1) can be used to calculate DSsurr:

DSsurr = qsurr/T = -DHsyst/T

Why does DHsyst have a negative sign? Any heat lost by the system is gained by the surroundings, and conversely, any heat gained by the system is lost by the surroundings. So, qsurr and DHsyst must always have opposite signs, which is why DHsyst is given a negative sign. DHsyst can be calculated in the way described in the Thermochemical Equations module.

Now that DSsyst and DSsurr are known, DSuniv can be determined. If it is greater than zero, the reaction is product-favored. If it is less than zero, the reaction is reactant-favored.

Determine whether the reaction:

CaCO3 (s) CaO (s) + CO2 (g)

is product- or reactant-favored at 25 ºC.

Product-favored Reactant-favored
The complete solution
is shown below.
Correct!
The complete solution is shown below.
Dsyst = 1 mol • DHfº(CaO(s)) + 1 mol • DHfº(CO2(g)) - 1 mol • DHfº(CaCO3(s))
Dsyst = 1 mol • (-635.09 kJ/mol) + 1 mol • (-393.509 kJ/mol) - 1 mol • (-1206.92 kJ/mol)
Dsyst = 178.321 kJ

Dsyst = 1 mol • Sº(CaO(s)) + 1 mol • Sº(CO2(g)) - 1 mol • Sº(CaCO3(s))
Dsyst = 1 mol • 39.75 J/mol•K + 1 mol • 213.74 J/mol•K - 1 mol • 92.9 J/mol•K
Dsyst = 160.59 J/K

Duniv = Dsyst - Dsyst/T
Duniv = 160.59 J/K - 178.321 kJ/(273.15 + 25)K • 1000 J/1 kJ = -437.50 J/K < 0

The Second Law of Thermodynamics