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Here, ν represents stoichiometric coefficients in the balanced equation representing the process. For example, Δ S ° for the following reaction at room temperature
is computed as the following:
[link] lists some standard entropies at 298.15 K. You can find additional standard entropies in Appendix G .
| Standard Entropies (at 298.15 K, 1 atm) | |
|---|---|
| Substance | (J mol −1 K −1 ) |
| carbon | |
| C( s , graphite) | 5.740 |
| C( s , diamond) | 2.38 |
| CO( g ) | 197.7 |
| CO 2 ( g ) | 213.8 |
| CH 4 ( g ) | 186.3 |
| C 2 H 4 ( g ) | 219.5 |
| C 2 H 6 ( g ) | 229.5 |
| CH 3 OH( l ) | 126.8 |
| C 2 H 5 OH( l ) | 160.7 |
| hydrogen | |
| H 2 ( g ) | 130.57 |
| H( g ) | 114.6 |
| H 2 O( g ) | 188.71 |
| H 2 O( l ) | 69.91 |
| HCI( g ) | 186.8 |
| H 2 S( g ) | 205.7 |
| oxygen | |
| O 2 ( g ) | 205.03 |
The value for is negative, as expected for this phase transition (condensation), which the previous section discussed.
−120.6 J mol −1 K −1
24.7 J/mol·K
The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, S univ >0. If Δ S univ <0, the process is nonspontaneous, and if Δ S univ = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.
What is the difference between Δ S , Δ S °, and for a chemical change?
Calculate for the following changes.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(a) 107 J/K; (b) −86.4 J/K; (c) 133.2 J/K; (d) 118.8 J/K; (e) −326.6 J/K; (f) −171.9 J/K; (g) −7.2 J/K
Determine the entropy change for the combustion of liquid ethanol, C 2 H 5 OH, under standard state conditions to give gaseous carbon dioxide and liquid water.
Determine the entropy change for the combustion of gaseous propane, C 3 H 8 , under standard state conditions to give gaseous carbon dioxide and water.
100.6 J/K
“Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.
(a) −198.1 J/K; (b) −348.9 J/K
From the following information, determine for the following:
By calculating Δ
S
univ at each temperature, determine if the melting of 1 mole of NaCl(
s ) is spontaneous at 500 °C and at 700 °C.
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
As Δ S univ <0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.
Use the standard entropy data in Appendix G to determine the change in entropy for each of the reactions listed in [link] . All are run under standard state conditions and 25 °C.
Use the standard entropy data in Appendix G to determine the change in entropy for each of the reactions listed in [link] . All are run under standard state conditions and 25 °C.
(a) 2.86 J/K; (b) 24.8 J/K; (c) −113.2 J/K; (d) −24.7 J/K; (e) 15.5 J/K; (f) 290.0 J/K
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