Ppt04 (PS8), Thermodynamics
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energy changes (heat flows)
Very global (“the universe”)
Applies to all fields of science
Led to creation of heat
engines/refrigerators/air conditioners
• In chemistry, focus is on physical and
chemical changes
• How can we use “spontaneous processes” to
“do work”?
• We can use tabulated data to make
predictions (and even calculate K’s for
possible processes!)
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“Spontaneous”  “fast” !!
• Spontaneity refers to “directionality”
– Does it occur (on its own) in the forward [or stated]
direction, or does it not?
• Thermodynamics will answer / address this question.
• Does not depend on how the process is carried out.
• Fast refers to rate at which it occurs (in the
spontaneous direction)
– How fast will it occur?
• Kinetics will answer / address this question
• Does depend on how the process occurs (“pathway”;
transition state, activation energy, etc.)
• Recall that we should be spontaneously combusting right now!!
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Figure 18.4
Thermodynamics domain vs. Kinetics domain
THERMODYNAMICS
Initial and final states,
spontaneity
“Path independent”
“Pathway
dependent”
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Reminder: 1st Law of Thermodynamics
• The total amount of energy in the universe
never changes (it’s constant)
• Euniv= 0  Esys = -Esurr
• “If energy leaves the system, it must go to
the surroundings, and vice versa”
• Has nothing to do with “spontaneity”
– 1st Law is consistent with a casserole dish
coming out of the oven colder than when it
was put in, as long as the oven would get
hotter!
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Spontaneity and the 2nd Law
• A spontaneous process is one that occurs
(at a specified set of conditions) “without
external intervention”.
– Gases expand into the volume of their
containers
– Ice will melt at 10°C
– Water will freeze at -10°C
– A chemical reaction will occur in the forward
direction if Q < K
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Spontaneity and the 2nd Law (continued)
• At the same conditions, the reverse
process of a spontaneous one is not
spontaneous
– Gases won’t spontaneously “contract” into a
smaller volume (perfume molecules going
back into the bottle after opened?)
– Water will not freeze at 10°C
– Ice will not melt at -10°C
– A chemical reaction will not occur in the
reverse direction if Q < K
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Spontaneity and the 2nd Law (continued)
• What determines whether a given process
is spontaneous or not?
– This question is not addressed by the 1st Law;
it is addressed by the 2nd Law.
• 2nd Law:
For a spontaneous process to occur,
the entropy (S) of the universe must
increase.
Suniv > 0 for any spontaneous process
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**Suniv  Ssys!** Be careful!
• The universe is made up of two “parts”—
system and surroundings:
ΔSuniv = ΔSsys + ΔSsurr
• It doesn’t ultimately matter if ΔSsys is
positive, or if ΔSsurr is positive. The key is
whether ΔSuniv is positive! (next slide).
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Interplay of ΔSsys and ΔSsurr in
Determining the Sign of ΔSuniv
(Table 16.3, Zumdahl)
ΔSsys + ΔSsurr = ΔSuniv
This turns out to be very temperature
dependent (See Section 18.5 in Tro). We
will come back to this issue later.
plus
equals
See next slide
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Bar plots to visualize idea on prior slide
(system and surroundings both contribute to Suniv!)
ΔSsys + ΔSsurr = ΔSuniv
|ΔSsurr| > |ΔSsys|
|ΔSsys| > |ΔSsurr|
NOTE: These two examples both have Ssys < 0 and Ssurr > 0,
but all possible variations are possible!
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What is entropy?
• Hard thing to define conceptually!
– Something like “energy dispersal”--not necessarily “spatial”
dispersal, but energy dispersed over the various motions of
atoms & molecules. Reflects quality or usefulness of energy.
• Let’s start with individual substances
– Look at what affects the amount of entropy in a sample of a
single substance (without necessarily understanding “what” it is”).
• Then we’ll look at mixtures of substances (in a reactive
system, e.g.)
– When a change occurs in the system, Ssys changes because
there are new substances present (or new physical states)
• We’ll consider the surroundings last
– Just a “bunch of substances not doing anything” (Prof. Mines’
view)
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Patterns for Entropies of Substances
(see handout/outline for details; Tro does this later; Section 18.7)
• Entropy of a sample of a substance increases with:
– Energy added (T increase or phase change)
• Solid < Liquid << Gases
– Volume (gas or solution species)
– Complexity of basic unit of substance
– Number of moles of the substance
• Entropy, like energy, is a “per mole” kind of quantity
• Recall that we’ll typically only specify substances
that undergo some change to be part of the
system.
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Differences in entropy of physical states
(assume a given T)
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Fig. 18.5 Entropy of a substance increases with T,
and depends (significantly) on state (s, l, g)
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Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse?) E
System A (4 J)
System B (4 J)
E=5J
E=5J
One “way” to
arrange the energy
(one [micro]state)
E=5J
Two “ways” to
arrange the energy
(two [micro]states)
Both systems have 4 J of energy, but the entropy of System B is greater
because it has two ways to “arrange” the energy. A greater # of “accessible”
(i.e., “low”) energy levels leads to greater entropy (see Slide 17)!
What if an energy state at 5 J was present? Would the # of microstates
change in either system?
 No, because there isn’t enough energy (in either system) to
access that 5 J state. But…
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Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse?) E
System A (6-7 J)
System B (6-7 J)
E=5J
E=5J
E
Two “ways” to
arrange the energy
(vs. one before);
two[micro]states
Three “ways” to
arrange the energy
(vs two before);
three [micro]states
Both systems have 6-7 J of energy, but the entropy of System B is (still) greater
because it has three ways to “arrange” the energy. A greater # of “accessible”
(i.e., “low”) energy levels leads to greater entropy (see next slide)!
What if 2-3 more units of energy were added (to make a total of 6-7 J)?
In both cases, adding
energy (raising T)
leads to increased S!
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More accessible
energy levels in gas!
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Patterns for Entropies of Substances
(revisit earlier slide, with nanoscopic “explanation”)
• Entropy of a sample of a substance increases with:
– Energy added (T increase or phase change)
Adding units
of energy
• Solid < Liquid << Gases
Adding more
accessible energy
states
– Volume (gas or solution species)
– Complexity of basic unit of substance
– Number of moles of the substance
(at a given T)
• Entropy, like energy, is a “per mole” kind of quantity
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(NOTE: Appendix has lots more data. Use appendix for PS8!)
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(From another text [McMurry])
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Predicting whether processes are “entropy
increasing” or “decreasing” (in the system)
(the most dominant effect in most cases is the number of moles of
gases made and lost [ngas], not the complexity of the substance[s]):
• C(s) + 2 H2(g)  CH4(g)
• N2(g) + 3 H2 (g)  2 NH3(g) [see next slide]
• 2 CO(g) + O2(g)  2 CO2(g)
• CaCO3(s)  CaO(s) + CO2(g)
• 2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)
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Can quantify the change in S (at a given T)
using standard entropies of substances
• C(s) + 2 H2(g)  CH4(g) Srxn° = 186.2 – (2.4 + 2 x 130.6)
2.4
2 x 130.6
186.2
= -77.4 J/K (per mol of C reacted)
(consistent with prediction, S decreases)
– Although CH4 is more complex than H2, it doesn’t have twice the
entropy (per mole) as H2, so the dominant effect here (as usual) is the
change in the number of moles of gases on reaction, ngas)
• N2(g) + 3 H2 (g)  2 NH3(g)
191.5
3 x 130.6
2 x192.3
Srxn° = 2 x 192.3 – (191.5 + 3 x 130.6)
= -198.7 J/K (per mol of N2)
– Again, consistent with prediction (S decreases). Though NH3 more
complex, only 2 moles of it form compared to 4 moles of gas “lost”).
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• Recall (1st semester):
• Now:
S
S°
S°
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The entropy of a perfect crystal at 0 K is zero (3rd Law).
W = “the number of microstates”
(ways to arrange energy)
Tro, Fig. 17.8
Zumdahl, Fig 16.5
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Entropy increases for a substance as energy is
added to it (either T increases, or a phase change occurs)
Curve differs
for different
substances.
This is how
standard
entropies of
substances are
determined (at,
say, 298 K)
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Relating Entropy of the Surroundings to a
Property of the System--What is “free
energy” (G)?
• See Section IV of Handout Outline
• Up through the derivation of the
expression:
G = H –TS
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EXAMPLE 17.3 Computing Gibbs Free Energy
Changes and Predicting Spontaneity from H and S
Consider the reaction for the decomposition of carbon tetrachloride
gas:
(a) Calculate G at 25 C and determine whether the reaction is
spontaneous.
(b) If the reaction is not spontaneous at 25 C, determine at what
temperature (if any) the reaction becomes spontaneous.
Consider only (a)
for now.
SOLUTION
(a) Use Equation 17.9 to calculate G
from the given values of H and S.
The temperature must be in kelvins. Be
sure to express both H and S in the
same units (usually joules).
The reaction is not
spontaneous.
© 2011 Pearson Education, Inc.
Ppt04_Thermo
At the specific conditions /
concentrations of the
reaction system!
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Back to Outline
• Conceptual connections…
• Now, return to discuss T-dependence of
spontaneity (specifically, Suniv and Gsys)
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EXAMPLE 17.3 Computing Gibbs Free Energy
Changes and Predicting Spontaneity from H and S
Consider the reaction for the decomposition of carbon tetrachloride
gas:
(a) Calculate G at 25 C and determine whether the reaction is
spontaneous.
(b) If the reaction is not spontaneous at 25 C, determine at what
temperature (if any) the reaction becomes spontaneous.
Earlier did (a)
Now begin to
consider (b)!
SOLUTION
(a) Use Equation 17.9 to calculate G
from the given values of H and S.
The temperature must be in kelvins. Be
sure to express both H and S in the
same units (usually joules).
The reaction is not
spontaneous.
© 2011 Pearson Education, Inc.
Ppt04_Thermo
At the specific conditions /
concentrations of the
reaction system!
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RECALL: Interplay of ΔSsys and ΔSsurr in
Determining the Sign of ΔSuniv
(Table 16.3, Zumdahl)
ΔSsys + ΔSsurr = ΔSuniv
This turns out to be very temperature
dependent (See Section 18.5 in Tro). We
will come back to this issue later.
plus
equals
See next slide
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Revisiting Earlier Example
Hint: Ssurr 
Hsys
T
If both of these plots represent the same exact process under the
same exact conditions EXCEPT THAT ONE IS AT A HIGHER
TEMPERATURE, then which plot represents the higher T and
which the lower T? How do you know?
What would the bar plot look like if the T went to infinity? To zero?
(see next slides →)
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As T → , |Ssurr| → 0!
(so Suniv → Ssys)
Ssurr 
Hsys
T
T~ 
T increasing
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As T → 0, |Ssurr| →  !
(so Suniv → Ssurr)
Ssurr 
Hsys
T
(Y-axis scale much bigger (zoomed
out); Ssys same as on right.)
T very small
T decreasing
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Summary, T-dependence of Suniv
• At very high T, Suniv approaches Ssys
– Because Ssurr becomes tiny (and Ssys
remains essentially unchanged)
• At very low T, Suniv approaches Ssurr
– Because Ssurr becomes huge, while Ssys
remains essentially unchanged
**A more detailed summary table of this will be
discussed a bit later.
NOTE: Many people (most, actually) end up taking a different
approach to this “T-dependence” issue: They look at
Gsys rather than Suniv. We’ll explore this next!
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G = H –TS
Thus: As T goes to…
…infinity, G → -TS
…zero, G → H
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(Ssys dominates)
(Ssurr dominates)
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Tro’s “version” of this
G = H –TS
Thus: As T goes to…
…infinity, G → -TS
…zero, G → H
Gsys “view”
H
-TS
opposites
NOTE: You could also look at each of these from a Suniv “view”!
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“Global” (entropy) view (Review of earlier slide)
Suniv = Ssys + Ssurr; Ssurr 
Hsys
T
Thus: As T goes to…
…infinity, Suniv → Ssys
…zero, Suniv → Ssurr
Suniv “view”
Ssys
Ssurr
Ssurr
Ssys
(Ssurr > 0)
(Ssys > 0)
+
+


(Ssurr < 0)
(Ssys < 0)

+

(Ssurr > 0)
(Ssurr < 0)
(Ssys < 0)
+
(Ssys > 0)
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T-Dependence Explored Further
Table 16.4 (Zumdahl) Results of the Calculation of ΔSuniv and
ΔG° for the Process H2O(s)  H2O(l) at
-10°C, 0°C, and 10°C
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EXAMPLE 17.3 Computing Gibbs Free Energy Changes and
Predicting Spontaneity from H and S
Consider the reaction for the decomposition of carbon tetrachloride gas:
Now consider
part (b)
(a) Calculate G at 25 C and determine whether the reaction is spontaneous.
(b) If the reaction is not spontaneous at 25 C, determine at what temperature (if any)
the reaction becomes spontaneous.
SOLUTION
(a) Use Equation 17.9 to calculate G from the given values
of H and S. The temperature must be in kelvins. Be sure
to express both H and S in the same units (usually
joules).
At the specific conditions /
The reaction is not spontaneous. concentrations of the
reaction system!
(b) Since S is positive, G will become more negative with
increasing temperature. To determine the temperature at
which the reaction becomes spontaneous, use Equation
17.9 to find the temperature at which G changes from
positive to negative (set G = 0 and solve for T). The
reaction is spontaneous above this temperature.
FOR PRACTICE 17.3
Consider the reaction:
C2H4(g) + H2(g)  C2H6(g) ∆H = –137.5 kJ; ∆S = –120.5 J/K
Calculate G at 25 C and determine whether the reaction is spontaneous. Does G
become more negative or more positive as the temperature increases?
© 2011 Pearson Education, Inc.
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Relationship of G to Q (and G)
• See Outline (and board)
• ΔG = ΔG° + RT ln Q
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Relationship of G to K
• See Outline—Derive from prior slide’s
relationship!
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Table 16.6(Zumdahl) Qualitative Relationship Between
the Change in Standard Free Energy and the
Equilibrium Constant for a Given Reaction
G = -RT lnK
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From another text
These data provide a more direct way to calculate the G
for a chemical reaction equation (i.e., you don’t need to
calculate H and S if you know Gf values!
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Fig. 17.10 (Tro). Why “free” energy?
H = -74.6 kJ
S = -80.8 J/K
G = -50.5 kJ
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Table 16.1 The Microstates That Give
a Particular Arrangement (State)
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Table 16.2 Probability of Finding All the
Molecules in the Left Bulb as a Function of
the Total Number of Molecules
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