Thermodynamics. III. Entropy of the System
Questions for Quiz #6.
Note: Questions 5 and 10 are omitted for Chem 122.
1. How is DS(surrounds) measured for a closed system?
We measure DS(surrounds) for a closed system using the relationship with the energy change and the temperature:
We have: DS(surrounds) = -DE / T
2a. Could this method be used to measure DS (that is the entropy change within the system)?
We can use this approach to measure DS only at equilibrium when DS(universe) = 0
in which case DS = DE / T.
2b. Under what special conditions can DS of the system be equated with DS(surrounds) and so (possibly) determined?
As in the answer to 2(a).
2c. If DS(surrounds) = -DS (of the system), what does DS(universe) equal, and what does this mean in terms of spontaneity of the change being studied?
It means that DS(universe) = 0 and the system is at equilibrium.
3a. What is meant by the phrase "thermodynamic reversibility"?
This refers to a process at equilibrium which can move microscopically in either direction without changing that equilibrium.
3b. What changes can occur when a system is at equilibrium?
The changes introduced here are: phase changes and heating a sample.
3c. How can heating or cooling a body be considered to be taking place at equilibrium?
Kinetic energy is transferred during molecular collisions, which is microscopically.
3d. How can a phase change (s to l, l to g, for example) be considered to be taking place at equilibrium?
Because if heat is added or subtracted, some of the sample will change phase but the equilibrium will be maintained.
4. A phase change takes place at a constant temperature, so:
DS(univ) = 0 = DS - DH/T
DS = DH/T.
4a. Calculate the entropy change when 1 mole of water freezes at 0º C. The heat of fusion for water is 6.01 kJ/mol.
Note: watch the units. The usual units for entropy change are J/mol.K, whilst those for enthalpy change are kJ/mol.
For a phase change: DS = DH / T = 6010 J / mol / 273 K = 22.0 J /mol.K.
4b. For the sublimation of iodine, I2(s) Ž I2(g), DSº = 145 J/mol.K, and DHº = 62.4 kJ/mol. Calculate the sublimation temperature under 100 kPa pressure.
For this phase change: DS = DH / T , so T = DH / DS = 62400 J / mol / 145 J / mol.K = 430 K.
5. Cpº for I2(s) = 54.4 J/mol.K. Assuming this value to be constant over the temperature range involved, calculate DS for heating solid I2 from -20º C to +20º C.
For heating a solid when Cp is constant, DS = Cp ln ( T2 / T1 ) = 54.4 J/mol.K * ln ( 293/253) = 7.99 J/mol.K.
6a. What is the "Third Law of Thermodynamics"?
6b. What is meant by the term "perfect crystal"?
A crystal in which there is only one way to order the particles.
6c. What is meant by the term "third law entropy value"?
Entropy values based on the perfect crystal at 0 K having a zero entropy. The entropy value is calculated by graphical integration of the heating curve for the substance from 0 K to 298 K.
7. Tables of thermodynamic parameters list enthalpy values as DHºf and entropy values as Sº298. Why DH and not DS?
Because the S values are third law values obtained by the relationship
that DS (heating) = S (at 298K) - S (at 0K)
or (since S at 0K = 0), DS (heating) = S ( at 298K).
Thus the S values listed are absolute entropies, not differences.
8. From thermodynamic tables, calculate the normal boiling point of mercury, Hg.
At the normal boiling point (which is the boiling point at 100 kPa), DGº = 0, so DSº = DHº / T
which means that T = DHº / DSº, and these latter values can be obtained from the tables for the process:
Hg(l) --> Hg(g)
From tables: DHº = DHºf(Hg(g)) - DHºf(Hg(l)) = 61.30 kJ - 0 kJ = 61.3 kJ
and: DSº = Sº(Hg(g)) - Sº(Hg(l)) = 175 J/K - 76.0 J/K = 99.0 J/K
From which T = DHº / DSº = 61300 J / 99.0 J/K = 619 K.
(Note the literature value is 629.6 K).
9. Now that both DHº and DSº are available from tables for many reactions, DSº(univ) may be obtained. Note that, since the tables refer to standard states of concentration (and pressure) and temperature, the DSº(univ) calculated from them must also have these restrictions. Note also that they apply to both reactants and to products, that is the spontaneity or not of the process is found only for the situation where all chemicals involved are at unit activity (standard concentration and pressure) and at 298 K.
Calculate the spontaneity or not under standard conditions of the following processes.
9a. 2 MnO2(s) Ž 2 MnO(s) + O2(g)
We need: DS(universe) = DS - DH/T
From tables: DHº = (-2)(-520.9) + (+2)(-385.5) + (+1)(0) = 270.8 kJ
and: DSº = (-2)(53.1) + (+2)(59.7) + (+1)(205) = 218.3 J/K.
So DS(universe) = 218.3 J/K - 270800 J / 298 K = -690 J/K
The negative value indicates a non-spontaneous process.
9b. SiH4(g) + 4 Br2(l) Ž SiBr4(l) + 4 HBr(g)
Following the same route:
From tables: DHº = (-1)(34.3) + (-4)(0) + (+1)(-457.6) + (+4)(-36) =-944.9 kJ.
and DSº = (-1)(204.7) + (-4)(152.2) + (+1)(278.0) + (+4)(198.6) = 258.9 J/K.
So DS(universe) = 258.9 J/K - (-944900 J)/298K = 3430 J/K.
The positive value indicates a spontaneous process.
9c. 2 CuCl(s) Ž CuCl2(s) + Cu(s)
Following the same route as part a:
From tables: DHº = (-2)(-137.3) + (+1)(-206.0) + (+1)(0) = 68.6 kJ.
and DSº = (-2)(86.3) + (+1)(108.2) + (+1)(33.2) =--31.2 J/K.
So DS(universe) = -31.2 J/K - 68600J / 298K = -261.4 J/K.
The negative value indicates a non-spontaneous process.
10 Entropy values for aqueous solutions of ions and molecules have no known zero value (solutions are never, of course, pure substances, so the Third Law cannot be applied). To overcome this difficulty, the zero of entropy for an aqueous solution is arbitrarily defined as the entropy of a 1.00 M solution of H+ ions in water.
a. Can the tabulated entropy value for a pure substance be less than zero? Why or why not?
No. Tabulated entropies for pure substances are absolute entropies, calculated by finding the increase in entropy from 0K to 298K: this must always be greater than the value at 0K which is 0 J/K
b. Can the tabulated entropy value for a substance or ion in solution be less than zero? Why or why not? If yes, explain what this actually means.
Yes. In this case the Sº298 value is comparing the entropy of the ion in question with the entropy of H3O+ (1 M) in water, and the system may be more ordered (lower entropy) than the H3O+ solution which is arbitrarily given a 0 J/K value.