Burdge/Overby, Chemistry: Atoms First, 2e Ch14 | Page 15

584

CHAPTE R 14? Entropy and Free Energy

Student Annotation: The concept of
equilibrium will be examined in detail in
Chapters 15, 16, and 17.

and describes a specific type of process known as an equilibrium process. An equilibrium process
is one that does not occur spontaneously in either the net forward or net reverse direction but can
be made to occur by the addition or removal of energy to a system at equilibrium. An example of
an equilibrium process is the melting of ice at 0°C. (Remember that at 0°C, ice and liquid water
are in equilibrium with each other [9 Section 12.5].)
With Equations 14.6 and 14.7, we can calculate the entropy changes for both the system and
surroundings in a process. We can then use the second law of thermodynamics (Equation 14.8)
to determine if the process is spontaneous or nonspontaneous as written or if it is an equilibrium
process.
Consider the synthesis of ammonia at 25°C:
N2(g) + 3H2(g)

2NH3(g)?????H° = –92.6 kJ/mol
rxn

From Worked Example 14.2(b), we have ?S ° = –199 J/K?????mol, and substituting ?H °
sys
sys
(–92.6?kJ/mol) into Equation 14.7, we get
–(–92.6 × 1000)?J/mol
___________________
?Ssurr = ?? ?? = 311 J/K ? mol
??
??
298 K
The entropy change for the universe is
?S° = ?S° + ?S°
univ
sys
surr
= –199 J/K ? mol + 311 J/K ? mol
= 112 J/K ? mol
Student Annotation: The spontaneity
that we have seen as favored by a process
being exothermic is due to the spreading
out of energy from the system to the
surroundings; thus, the negative ?Hsys
corresponds to a positive ?Ssurr. It is this
positive contribution to the overall ?Suniv
that actually favors spontaneity.

Because ?S ° is positive, the reaction will be spontaneous at 25°C. Keep in mind, though, that
univ
just because a reaction is spontaneous does not mean that it will occur at an observable rate. The
synthesis of ammonia is, in fact, extremely slow at room temperature. Thermodynamics can tell
us whether or not a reaction will occur spontaneously under specific conditions, but it does not
tell us how fast it will occur.
Worked Example 14.4 lets you practice identifying spontaneous, nonspontaneous, and equilibrium processes.

Worked Example 14.4
Determine if each of the following is a spontaneous process, a nonspontaneous process,
or an equilibrium process at the specified temperature: (a) H2(g) + I2(g)
2HI(g) at 0°C,
(b) CaCO3(s)
CaO(s) + CO2(g) at 200°C, (c) CaCO3(s)
CaO(s) + CO2(g) at 1000°C,
(d) Na(s)
Na(l) at 98°C. (Assume that the thermodynamic data in Appendix 2 do not vary
with temperature.)

Strategy For each process, use Equation 14.6 to determine ?S° and Equations 10.18 and 14.7 to
sys
determine ?H° and ?S° . At the specified temperature, the process is spontaneous if ?Ssys and ?Ssurr
sys
surr
sum to a positive number, nonspontaneous if they sum to a negative number, and an equilibrium
process if they sum to zero. Note that because the reaction is the system, ?Srxn and ?Ssys are used
interchangeably.
Setup From Appendix 2,
(a) S°[H2(g)] = 131.0 J/K ? mol, S°[I2(g)] = 260.57 J/K ? mol, S°[HI(g)] = 206.3 J/K ? mol;
?H°[H2(g)] = 0 kJ/mol, ?H°[I2(g)] = 62.25 kJ/mol, ?H°[HI(g)] = 25.9 kJ/mol.
f
f
f
(b), (c) In Worked Example 14.2(a), we determined that for this reaction, ?S° = 160.5 J/K ? mol,
rxn
?H°[CaCO3(s)] = –1206.9 kJ/mol, ?H°[CaO(s)] = –635.6 kJ/mol, ?H°[CO2(g)] = –393.5 kJ/mol.
f
f
f
(d) S°[Na(s)] = 51.05 J/K ? mol, S°[Na(l)] = 57.56 J/K ? mol; ?H°[Na(s)] = 0 kJ/mol,
f
?H°[Na(l)] = 2.41 kJ/mol.
f

Solution?
(a) ?S° = [2S°(HI)] ? [S°(H2) + S°(I2)]
rxn
= (2)(206.3 J/K ? mol) ? [131.0 J/K ? mol + 260.57 J/K ? mol] = 21.03 J/K ? mol

bur11184_ch14_570-603.indd 584

9/10/13 12:01 PM