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Textbook Notes
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McMaster University
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Chemistry
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CHEM 1E03
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Weisner
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Chapter

Description

Entropy, the 2 ndLaw of Thermodynamics
Gibbs Free Energy & Spontaneity
& the connection to Equilibrium
→ Which rxns go & which do not?
ST
→ 1 Law just keeps track of energy
There is a natural direction of physical & chemical
processes not explained by the 1 Law.
Eg. 1 A gas expands spontaneously into a vacuum ♣
The reverse of this process is never seen though it is allowed
by the 1stlaw.
Eg. 2 Heat flows from a hot body (on the right) to a cold
body (on the left).
⇒ T i1creases & T de2reases until the two systems achieve
the same temperature – thermal equilibrium
T1 T 2 T1< T 2
q
→Heat never flows the other way, though the 1 Lawst
would allow it.
115 Egs. 3. Spontaneous mixing of gases, miscible liquids,
(natural)
dissolution of soluble salts
♣
Methanol & water are miscible - they
spontaneously form a solution - a
more disordered state when added
together.
NaCl(s) dissolves spontaneously
when added to water - the resulting
solution is more disordered than solid
salt + water.
Though these examples include energy
considerations, they illustrate that
increasing disorder is a driving force
causing some processes to proceed,
while prohibiting the reverse
Spontaneous mixing of two ideal gases is the best
illustration. In this case, interactions between molecules
can be neglected (i.e. no energy considerations), yet the
Random motions of individual molecules leads to
irreversible mixing of gases.
116 • an ordered state → a disordered state
• disorder increases
In Eg. 2, kinetic energy spreads from the hot body to the cold
body.
• energy is more spread out
• a more disordered state
In Eq. 3, dissolution of salt in water ...
• the ions have greater freedom in solution - due to
increased volume (in comparison to solid)
• final state has more freedom more disorder
In a natural (i.e. spontaneous) process there is a spontaneous
increase in disorder.
→ Entropy, S, measures this disorder
→ S increases
123
true for S of entire universe
How do we measure "disorder"?
or
How is S defined?
117 Consider a deck of cards - continuously reshuffled.
→ If the card shuffler doesn’t cheat, then any ordering of
the card is possible ⇒ the cards are disordered.
→ However, if the card shuffler cheats - eg. producing an
ordered deck of cards with each shuffle (eg. always
ensuring that the dealer gets a royal flush) ⇒ the cards
are more ordered.
supplementary - you are not responsible for the formulae from here to the
next dividing line ...
Boltzmann proposed the following formula for entropy in
terms of the "number of arrangements of the system which
produce the same state"
S = k ln W
number of arrangements ...
entropy Boltzmann constant
k = R/N A
= 1.3807 × 10-23J/K
This formula is engraved on Boltzmann's headstone.
In the case of the shuffled cards, the "number of arrangements
...", W, is just the number deck states sampled in the
continuous reshuffling.
118 Unshuffled deck:
W = 1 → completely ordered case
⇒ S = O
Properly shuffled deck:
W = n! → disordered deck of n cards
⇒ S = k ln n!
→ proper shuffling produces much larger S
Always-royal-flush-to-dealer shuffled deck:
W' = 4 (n-5)! → There are 4 royal flushes - the remaining
n - 5 cards are arranged randomly
⇒ S' ≅ k [ln 4 + ln (n-5)!]
∆S = S' - S
= k [ln 4 + ln 47! − ln 52!← case of 52 cards
= -2.51 × 10-2J/K
→ negative ∆S ⇒ the system (deck of cards in this case)
has become more ordered
Of course the dealer might get a royal flush without
cheating. However, the probability is
W'/W = exp[∆S /k] -8
= 4/[5251 50•49•48] = 1.3 × 10
i.e. 1 chance in about 100 million
- very unlikely
119 |∆S| for processes we consider in chemistry are much larger -
the corresponding probabilities of these events are mind-
bogglingly small.
eg. What is the probability of all the air, at STP, inside a
(10 m) × (10 m) × (10 m) room, finding itself inside an
open 1 L pop bottle (soda pop having been removed)?
Solution:
3 6
The room has a volume of 1000 m = 1.0 × 10 L.
⇒ There is 1 chance in 1.0 × 10 of finding one specific air
6 -6
molecule in the bottle - prob. = 1/(1.0 × 10 ) = 1.0 × 10
However, there are pV/RT =
6 -1 -1
= (1.0 × 10 L atm)/[(0.0821 L atm K mol )(298 K)]
= 4.1 × 10 moles of air
28
= 2.5 × 10 molecules of air
Probability of all air molecules inside bottle =
29
-6 2.5 × 1028 -1.5 × 10
(1.0 × 10 ) = 1.0 × 10
→ this is such a ridiculously small number, it is hard to
take it seriously
→ Gases simply do not voluntarily constrain themselves
in this way - evidenced by the extraordinary unlikelihood of
such an event ...
2ndLaw of Thermodynamics:
Entropy increases for all spontaneous processes - i.e. those
actually occuring in nature - natural processes
120 But wait a minute ... Don't we ever see processes with
negative ∆S ?
Yes, but the entropy of one system decreases only at the
expense of an even greater entropy increase in another system
(the surroundings).
Change in the total entropy of the universe,
∆ S = ∆ S + ∆ S > O
total system surroundings
⇒ ∆ S system> ∆ S surroundings
♣
A. A process for which ∆ system 0 A
is always accompanied by q < 0.
Heat leaving the system enters the
surroundings & heat increases
entropy (hotter ⇒ more disorder).
In fact, the entropy increase of the B
surroundings always exceeds the
decrease in entropy of the system.
B. If the system increases its entropy,
then the surroundings can actual
decrease in entropy, but not by more
than the system increased its entropy. C
C. There are restrictions, if both
system & surroundings increase in
entropy.
121 Properties of entropy:
→ S is proportional to amount
sofstance (like E & H - also n & V)
eg.
SSS total2 S
→ S depends on temperature & on phase
- increases with T, including discrete (positive)
jumps when solid → liquid & liquid → gas
♣
5
4
3
2
1
0 K
122 1. 3 Law of Thermodynamics: S = O for a pure
crystalline substance at T = 0 K.
→ a perfectly ordered state
2. Heating increases disorder of solid.
→ atoms vibrate about mean positions
3. When the melting point is reached further heating
produces a liquid wherein molecules are much freer to
move about (diffusive motion).
4. Entropy of liquid (like the solid) increases with T until
the b.p. is reached.
5. Vaporization increases S significantly - molecules are
free to explore the entire containment vessel.
∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗
Standard molar entropy , S °
°
→ entropy of 1 mol of substance at 1 atm & 25 C
→ units of S = J K −1 mol −1
note - Joules not kJ
egs. S °[H2O(l)] = 70.0 J K −1 mol −1
S °[H2O(g)] = 188.7 J K −1 mol−1
123 ° ° °
NOTE ∆S vap = S [H O2g)] − S [H O2l)]
= 118.7 J K −1 > 0
more egs.
° −1 −1
S [H2(g)] = 130.6 J K mol
S ° [O2(g)] = 205.0 J K −1 mol−1
→ both gases - however, S [O 2g)] > S [H 2g)]
→ heavier molecules have more entropy
→ more complex molecules also have more entropy
eg.
♣
Molecules with more bonds have more ways of spreading out
vibrational energy. Thermal energy takes the form of
translational, rotational & vibrational energy - at the
microscopic level.
124 Standard reaction entropy
° ° °
∆S = Σ nprodS prod − Σ nreacS reac
prod reac
eg. H 2(g) + ½ O2(g) → H 2(l)
∆S = 1 × S [H O(2)] − (1 × S [H2(g)] + ½ × S [O 2g)])
= 70.0 -(130.6 + 205.0/2) J K−1
−1
= - 163.1 J K
→ S is a state function like E & H
→ ∆S add for sequential processes - i.e. Hess's Law
→ Gases have the most entropy
⇒ a rxn increasing moles gas generally has a positive ∆S
while decreasing moles of gas means negative ∆S
eg. H 2(g) + ½ O2(g) → H 2(l)
123 123
1½ moles of gas → 0 moles of gas
∆n gas= -1½
° −1
Recall ∆S = −163.1 J K < 0
125 another eg.
2 NH 4O (3) → 2 N (g2 + O (g2 + 4 H O(2)
°
∆n gas= +7 & ∆S > 0
Let's return to the 2w of Thermodynamics.
It's utilization requires knowledge surroundings
Entropy change in surroundings is due to heat transferred
to/from system - hence from/to surroundings
It can be shown that
∆Ssurr= q surr = −q /sys
= − ∆H /T (under constant pressure conditions)
sys
→ exothermic ⇒ ∆S surr> 0
→ system heats surroundings
→ endothermic ⇒ ∆S surr< 0
→ system cools surroundings
(system entropy must increase in this case)
NOTE: effect of transferred heat is greater at low T
126 Consider H 2g) + ½ O (2) → H O(l)2
∆H ° = −285.8 kJ ( heat of formation of H O(l) )
sys 2
⇒ ∆S = + 285.8 × 1000 J K −1
surr
298.15
= 958.6 J K −1
° ° °
∆S tot ∆S sys ∆S surr
= −163.1 + 958.6 J K −1
−1
= 795.5 J K
> 0 ⇒ a spontaneous process
Consider 4 Fe(s) + 3 O 2g) → 2Fe O (2) 3
°
∆H = −1649 kJ ← −ve ⇒ exothermic
° −1
∆S = −549 JK ← −ve ( note - ∆n gas = -3 )
° ° ° −1
∆S tot= ∆S − ∆H = 4980 JK > 0 at 298 K
T
⇒ spontaneous
→ energetics (i.e. ∆H) drives this rxn
127 Gibbs free energy
Criterion for spontaneity i∆S tot> 0
or ∆S −

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