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 Effect of Temperature

Whether or not a reaction is product-favored can be affected by raising or lowering the temperature. Since:

DSsurr = -DHsyst/T

this term can be made smaller by raising the temperature or larger by lowering the temperature. If it is made small or large enough by raising or lowering the temperature, a reaction can be made product-favored if it is reactant-favored, or reactant-favored if it is product-favored.

On the previous page it was determined that the decomposition of calcium carbonate to carbon dioxide and calcium oxide is reactant-favored at 25 ºC. What if the temperature is raised to 6000.0 ºC?

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

DSuniv =
Correct! By raising the temperature, the reactant-favored reaction has been made product-favored. See the solution below.
You have correctly selected product-favored, but your value of Duniv is incorrect. See the solution below.
Your value of Duniv is correct, but you have incorectly chosen reactant-favored. The positive value of Duniv means the reaction is now product-favored.
That is incorrect. See the solution below.
How can you decide if the reaction is product- or reactant-favored if you haven't calculated Duniv?
Reactant-favored Product-favored

DHº = 178.321 kJ
DSº = 160.59 J/K (from previous page)
DSunivº = DSº - DHº/T
DSunivº = 160.59 J/K - 178.321 kJ/(6000+273.15)K • 1000 J/kJ = 132.16 J/K > 0
Therefore, the reaction is product-favored.

The Effect of Temperature