More About Entropy
More About Entropy
For the SAT II Chemistry exam, you’ll be expected to be have an understanding of all of the laws of thermodynamics, so to refresh your memory, the first law of thermodynamics says that energy can neither be created nor destroyed. The second law of thermodynamics says that the disorder of the universe, meaning its entropy, or DS, is constantly increasing. The third law of thermodynamics says that the entropy of a perfect crystal at 0K is zero. What does the third law mean to you? It means that we can calculate the entropy of any substance that’s at a temperature higher than 0K. Here are some rules about determining the entropy of a system:
  1. The greater the disorder or randomness in a system, the larger the entropy.
  2. The entropy of a substance always increases as it changes state from solid to liquid to gas.
  3. When a pure solid or liquid dissolves in a solvent, the entropy of the substance increases.
  4. When a gas molecule escapes from a solvent, there is an increase in entropy.
  5. Entropy generally increases with increasing molecular complexity.
  6. Reactions that increase the number of moles of particles often increase the entropy of the system.
Example
3. Which of the following reactions results in the largest increase in entropy?
(A) CO2(s)CO2(g)
(B) H2(g) + Cl2(g)2HCl(g)
(C) KNO3(s)KNO3(l)
(D) C(diamond)C(graphite)
Explanation
All of the reactions result in an increase in disorder, but A, in which CO2 moves from a solid state to a gaseous one, represents the largest change in disorder.
You can calculate entropy using a table of standard values in much the same way that you calculated enthalpy earlier by using the equation below:
The units of entropy are J/K. The higher the S value, the more disordered the system, so a positive (+) S value is more disordered, and a –S value is less disordered. Remember that disorder is the favored condition, according to the second law of thermodynamics. Now try a problem that involves using standard S values.
Example
Calculate the entropy change at 25ºC in J/K for
2SO2(g) + O2(g)2SO3(g)
given the following data:
SO2(g): 248.1 J/mol-K
O2(g): 205.3 J/mol-K
SO3(g): 256.6 J/mol-K
Explanation
[2(256.6)] - [2(248.1) + 1(205.3)] = -188.3 J/K
Don’t forget to multiply by coefficients, just as you did in enthalpy, because the given data are still per 1 mole.
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